["19.11. V\u00f2i III \u1edf 1 b\u1ec3 n\u01a3\u1edbc (t\u1eeb \u0111\u00e1y) n\u00ean l\u00fac \u0111\u1ea7u hai v\u00f2i I v\u00e0 II c\u00f9ng ch\u1ea3y \u0111\u1ec3 \u0111\u1ea7y 1 b\u1ec3 n\u01a3\u1edbc. Sau \u0111\u00f3 3 v\u00f2i c\u00f9ng ch\u1ea3y 33 \u0111\u1ea7y 2 b\u1ec3 c\u00f2n l\u1ea1i v\u00e0 l\u01a3\u1ee3ng n\u01a3\u1edbc trong b\u1ec3 \u0111\u01a3\u1ee3c th\u00eam s\u1ebd b\u1eb1ng t\u1ed5ng l\u01a3\u1ee3ng n\u01a3\u1edbc hai v\u00f2i ch\u1ea3y v\u00e0o tr\u1eeb \u0111i l\u01a3\u1ee3ng n\u01a3\u1edbc 3 ch\u1ea3y ra. Th\u1eddi gian hai v\u00f2i ch\u1ea3y \u0111\u1ea7y 1 b\u1ec3 n\u01a3\u1edbc v\u00e0 th\u1eddi gian ba v\u00f2i ch\u1ea3y \u0111\u1ea7y 1 b\u1ec3 n\u01a3\u1edbc ch\u00ednh l\u00e0 2 gi\u1edd 48 ph\u00fat = 14 3 35 gi\u1edd Ta c\u00f3 c\u00e1ch gi\u1ea3i sau . G\u1ecdi th\u1eddi gian v\u00f2i th\u1ee9 hai ch\u1ea3y v\u00e0o m\u1ed9t m\u00ecnh \u0111\u1ea7y b\u1ec3 l\u00e0 x gi\u1edd (x >0) Suy ra 1 gi\u1edd v\u00f2i th\u1ee9 hai ch\u1ea3y 1 b\u1ec3 . M\u1ed9t gi\u1edd v\u00f2i I ch\u1ea3y m\u1ed9t m\u00ecnh \u0111\u01a3\u1ee3c 1 b\u1ec3 . M\u1ed9t gi\u1edd hai v\u00f2i c\u00f9ng ch\u1ea3y \u0111\u01a3\u1ee3c x4 1 \uf02b 1 b\u1ec3. 4x Ba v\u00f2i c\u00f9ng ch\u1ea3y m\u1ed9t gi\u1edd l\u01a3\u1ee3ng n\u01a3\u1edbc trong b\u1ec3 c\u00f2n 1 \uf02b 1 \uf02d 1 \uf03d 1 \uf02b 1 . 4 x 12 6 x Ta c\u00f3 ph\u01a3\u01a1ng tr\u00ecnh 1 : \uf0e6 1 \uf02b 1 \uf0f6 \uf02b 2 : \uf0e6 1 \uf02b 1 \uf0f6 \uf03d 14 . 3 \uf0e8\uf0e7 4 x \uf0f7\uf0f8 3 \uf0e7\uf0e8 6 x \uf0f8\uf0f7 5 Gi\u1ea3i ph\u01a3\u01a1ng tr\u00ecnh ta s\u1ebd c\u00f3 : 19x2 \u2013 30x \u2013 504 = 0 \uf0db (x \u2013 6)(19x + 84) = 0 T\u00ecm \u0111\u01a3\u1ee3c x = 6 th\u1ecfa m\u00e3n \u0111i\u1ec1u ki\u1ec7n. V\u1eady th\u1eddi gian ch\u1ea3y m\u1ed9t m\u00ecnh \u0111\u1ea7y b\u1ec3 c\u1ee7a v\u00f2i II l\u00e0 6 gi\u1edd. 19.12. B\u00e0i to\u00e1n li\u00ean quan \u0111\u1ebfn c\u1ea5u t\u1ea1o s\u1ed1. S\u1ed1 c\u00f3 hai ch\u1eef s\u1ed1 l\u00e0 ab \uf03d 10a \uf02b b ; \u0110\u1ed5i ch\u1ed7 \u0111\u01a3\u1ee3c s\u1ed1 ba \uf03d 10b \uf02b a v\u1edbi a,b\uf0ce N; 0 \uf03c a \uf0a3 9; 0 \uf0a3 b \uf0a3 9 ). Ta c\u00f3 c\u00e1ch gi\u1ea3i: G\u1ecdi ch\u1eef s\u1ed1 h\u00e0ng ch\u1ee5c l\u00e0 x (x \uf0ce N ; 3 < x \uf0a3 9) th\u00ec ch\u1eef s\u1ed1 h\u00e0ng \u0111\u01a1n v\u1ecb l\u00e0 (x \u2013 3). S\u1ed1 \u0111\u00e3 cho : x(x \uf02d 3) \uf03d10x \uf02b (x \uf02d 3) ; \u0110\u1ed5i ch\u1ed7 c\u00e1c ch\u1eef s\u1ed1: (x \uf02d3) x \uf03d10(x \uf02d3) \uf02b x Ta c\u00f3 ph\u01a3\u01a1ng tr\u00ecnh 10\uf028x \uf02d 3\uf029 \uf02b x \uf02d 10x \uf02b (x \uf02d 3) \uf03d 37 3 Gi\u1ea3i ph\u01a3\u01a1ng tr\u00ecnh \u0111\u01a3\u1ee3c x = 9 ph\u00f9 h\u1ee3p \u0111i\u1ec1u ki\u1ec7n c\u1ee7a \u1ea9n.S\u1ed1 c\u1ea7n t\u00ecm l\u00e0 96. 19.13. B\u00e0i to\u00e1n li\u00ean quan \u0111\u1ebfn c\u1ea5u t\u1ea1o s\u1ed1. S\u1ed1 c\u00f3 b\u1ed1n ch\u1eef s\u1ed1 m\u00e0 ch\u1eef s\u1ed1 h\u00e0ng \u0111\u01a1n v\u1ecb l\u00e0 6 l\u00e0 abc6 \uf03d10.abc \uf02b 6 ; Chuy\u1ec3n 6 l\u00ean \u0111\u1ea7u \u0111\u01a3\u1ee3c s\u1ed1 6abc \uf03d 6000 \uf02b abc v\u1edbi a,b,c\uf0ce N; 0 \uf03c a \uf0a3 9; 0 \uf0a3 b,c \uf0a3 9 ) . T\u1eeb \u0111\u00f3 c\u00f3 c\u00e1ch gi\u1ea3i : G\u1ecdi s\u1ed1 c\u00f3 ba ch\u1eef s\u1ed1 \u0111\u1ee9ng tr\u01a3\u1edbc s\u1ed1 6 l\u00e0 x (x \uf0ce N ; 99 < x < 1000) th\u00ec s\u1ed1 \u0111\u00e3 cho l\u00e0 x6 \uf03d10x \uf02b 6 ; Chuy\u1ec3n 6 l\u00ean \u0111\u1ea7u \u0111\u01a3\u1ee3c s\u1ed1 6x \uf03d 6000 \uf02b x Ta c\u00f3 ph\u01a3\u01a1ng tr\u00ecnh 10x + 6 + 6000 + x = 8217 Gi\u1ea3i ph\u01a3\u01a1ng tr\u00ecnh \u0111\u01a3\u1ee3c x = 201 ph\u00f9 h\u1ee3p \u0111i\u1ec1u ki\u1ec7n c\u1ee7a \u1ea9n S\u1ed1 c\u1ea7n t\u00ecm l\u00e0 2016. 19.14. G\u1ecdi k\u1ebft qu\u1ea3 sau khi bi\u1ebfn \u0111\u1ed5i c\u1ee7a b\u1ed1n s\u1ed1 l\u00e0 x (x \uf0ce R) th\u00ec : S\u1ed1 th\u1ee9 nh\u1ea5t l\u00e0 x \u2013 5. S\u1ed1 th\u1ee9 hai l\u00e0 x + 5. S\u1ed1 th\u1ee9 ba l\u00e0 x : 5. S\u1ed1 th\u1ee9 t\u01a3 l\u00e0 x . 5 Ta c\u00f3 ph\u01a3\u01a1ng tr\u00ecnh (x \u2013 5) + (x + 5) + x : 5 + x . 5 = 720 Gi\u1ea3i ph\u01a3\u01a1ng tr\u00ecnh \u0111\u01a3\u1ee3c x = 100 th\u1ecfa m\u00e3n \u0111i\u1ec1u ki\u1ec7n c\u1ee7a \u1ea9n. 300","Website: tailieumontoan.com V\u1eady s\u1ed1 th\u1ee9 nh\u1ea5t l\u00e0 95; s\u1ed1 th\u1ee9 hai l\u00e0 105; s\u1ed1 th\u1ee9 ba l\u00e0 20; s\u1ed1 th\u1ee9 t\u01a3 l\u00e0 500. 19.15. S\u1ed1 em \u0111\u01a3\u1ee3c chia \u1edf c\u00e1ch chia th\u1ee9 hai \u00edt h\u01a1n s\u1ed1 em \u0111\u01a3\u1ee3c chia \u1edf c\u00e1ch chia th\u1ee9 nh\u1ea5t l\u00e0 1 em. T\u1eeb \u0111\u00f3 c\u00f3 c\u00e1ch gi\u1ea3i: G\u1ecdi x l\u00e0 s\u1ed1 qu\u1ea3 b\u00f2ng \u0111em chia. S\u1ed1 em \u0111\u01a3\u1ee3c chia \u1edf c\u00e1ch th\u1ee9 nh\u1ea5t l\u00e0 x \uf02d 5 em. S\u1ed1 em \u0111\u01a3\u1ee3c chia \u1edf c\u00e1ch th\u1ee9 hai 5 l\u00e0 x . Ta c\u00f3 ph\u01a3\u01a1ng tr\u00ecnh x \uf02d 5 \uf02d x \uf03d 1 6 56 Gi\u1ea3i ph\u01a3\u01a1ng tr\u00ecnh \u0111\u01a3\u1ee3c x = 60 (qu\u1ea3 b\u00f2ng) v\u00e0 s\u1ed1 tr\u1ebb l\u00e0 11 em. 19.16. V\u1edbi h\u00ecnh ch\u1eef nh\u1eadt : Chu vi = (d\u00e0i + r\u1ed9ng)\uf0b4 2; Di\u1ec7n t\u00edch = d\u00e0i \uf0b4 r\u1ed9ng . Di\u1ec7n t\u00edch c\u0169 \u2013 di\u1ec7n t\u00edch m\u1edbi = 200 m2. Ta c\u00f3 c\u00e1ch gi\u1ea3i : N\u1eeda chu vi l\u00e0 100m. G\u1ecdi chi\u1ec1u d\u00e0i th\u1eeda ru\u1ed9ng l\u00e0 x (m);( 0 < x < 100) th\u00ec chi\u1ec1u r\u1ed9ng l\u00e0 (100 \u2013 x) (m). Chi\u1ec1u d\u00e0i sau khi gi\u1ea3m l\u00e0 (x \u2013 10) (m), chi\u1ec1u r\u1ed9ng sau khi t\u0103ng l\u00e0 100 \u2013 x + 4 = 104 \u2013 x (m). Ta c\u00f3 ph\u01a3\u01a1ng tr\u00ecnh : x . (100 \u2013 x) \u2013 (x \u2013 10).(104 \u2013 x) = 200 Gi\u1ea3i ph\u01a3\u01a1ng tr\u00ecnh \u0111\u01a3\u1ee3c x = 60 th\u1ecfa m\u00e3n \u0111i\u1ec1u ki\u1ec7n c\u1ee7a \u1ea9n. V\u1eady th\u1eeda ru\u1ed9ng c\u00f3 chi\u1ec1u d\u00e0i l\u00e0 60m, chi\u1ec1u r\u1ed9ng l\u00e0 40m. 19.17. V\u1edbi \u0111\u01a3\u1eddng tr\u00f2n : Chu vi = \u0110\u01a3\u1eddng k\u00ednh \uf0b4 \uf070 ; G\u1ecdi b\u00e1n k\u00ednh \u0111\u01a3\u1eddng tr\u00f2n ban \u0111\u1ea7u l\u00e0 x cm. ( x > 0), th\u00ec b\u00e1n k\u00ednh sau khi k\u00e9o d\u00e0i l\u00e0 (x + 5) (cm). Chu vi \u0111\u01a3\u1eddng tr\u00f2n ban \u0111\u1ea7u l\u00e0 \uf070 .2x (m); Chu vi \u0111\u01a3\u1eddng tr\u00f2n sau l\u00e0 \uf070 .2(x + 5) (m); Ta c\u00f3 ph\u01a3\u01a1ng tr\u00ecnh : \uf070 .2x + \uf070 .2(x + 5) = \uf070 . 90 \uf0db 2x + 2(x + 5) = 90 Gi\u1ea3i ph\u01a3\u01a1ng tr\u00ecnh t\u00ecm \u0111\u01a3\u1ee3c x = 20 th\u1ecfa m\u00e3n \u0111i\u1ec1u ki\u1ec7n c\u1ee7a \u1ea9n. V\u1eady b\u00e1n k\u00ednh \u0111\u01a3\u1eddng tr\u00f2n ban \u0111\u1ea7u l\u00e0 20 cm. 19.18. N\u1ed3ng \u0111\u1ed9 ph\u1ea7n tr\u0103m (C%) c\u1ee7a m\u1ed9t dung d\u1ecbch l\u00e0 s\u1ed1 gam ch\u1ea5t tan ch\u1ee9a trong 100 gam dung d\u1ecbch C% \uf03d mct .100% . mdd Kh\u1ed1i l\u01a3\u1ee3ng NaCl trong dung d\u1ecbch lo\u1ea1i I + Kh\u1ed1i l\u01a3\u1ee3ng NaCl trong dung d\u1ecbch lo\u1ea1i II = Kh\u1ed1i l\u01a3\u1ee3ng NaCl trong 1000 gam dung d\u1ecbch n\u1ed3ng \u0111\u1ed9 27%. Ta c\u00f3 c\u00e1ch gi\u1ea3i : G\u1ecdi kh\u1ed1i l\u01a3\u1ee3ng dung d\u1ecbch NaCl lo\u1ea1i I l\u00e0 x gam (0 < x < 1000) th\u00ec G\u1ecdi kh\u1ed1i l\u01a3\u1ee3ng dung d\u1ecbch NaCl lo\u1ea1i II l\u00e0 (1000 \u2013 x) (gam). Ta c\u00f3 ph\u01a3\u01a1ng tr\u00ecnh : 30% x + 25% (1000 \u2013 x) = 27% . 1000 Gi\u1ea3i ph\u01a3\u01a1ng tr\u00ecnh t\u00ecm \u0111\u01a3\u1ee3c x = 400 th\u1ecfa m\u00e3n \u0111i\u1ec1u ki\u1ec7n c\u1ee7a \u1ea9n. V\u1eady kh\u1ed1i l\u01a3\u1ee3ng dung d\u1ecbch NaCl lo\u1ea1i I l\u00e0 400g; lo\u1ea1i II l\u00e0 600g. 19.19. B\u00e0i to\u00e1n li\u00ean quan \u0111\u1ebfn vi\u1ec7c t\u00ecm nhi\u1ec7t l\u01a3\u1ee3ng t\u1ecfa ra, thu v\u00e0o c\u1ee7a n\u01a3\u1edbc theo c\u00f4ng th\u1ee9c Qt\u1ecfa = C.m (t2 \u2013 t1) v\u00e0 Qthu = C.m (t1 \u2013 t2) v\u1edbi C l\u00e0 nhi\u1ec7t dung ri\u00eang c\u1ee7a n\u01a3\u1edbc, m l\u00e0 kh\u1ed1i l\u01a3\u1ee3ng c\u1ee7a n\u01a3\u1edbc.. Nhi\u1ec7t l\u01a3\u1ee3ng t\u1ecfa ra c\u1ee7a 10kg n\u01a3\u1edbc \u1edf 900C b\u1eb1ng nhi\u1ec7t l\u01a3\u1ee3ng thu v\u00e0o c\u1ee7a 5kg n\u01a3\u1edbc \u1edf 240C. Ta c\u00f3 c\u00e1ch gi\u1ea3i: G\u1ecdi t (\u0111\u1ed9 C)l\u00e0 nhi\u1ec7t \u0111\u1ed9 cu\u1ed1i c\u00f9ng c\u1ee7a n\u01a3\u1edbc sau khi pha 24 < t < 90). Nhi\u1ec7t l\u01a3\u1ee3ng t\u1ecfa ra c\u1ee7a 10kg n\u01a3\u1edbc \u1edf 900C l\u00e0 C. 10 (90 \u2013 t) (J) v\u00e0 nhi\u1ec7t l\u01a3\u1ee3ng thu v\u00e0o c\u1ee7a 5kg n\u01a3\u1edbc \u1edf 240C l\u00e0 C. 5 (t \u2013 24) (J) Ta c\u00f3 ph\u01a3\u01a1ng tr\u00ecnh C. 10 (90 \u2013 t) = C. 5 (t \u2013 24) \uf0db 10 (90 \u2013 t) = 5 (t \u2013 24) Gi\u1ea3i ph\u01a3\u01a1ng tr\u00ecnh \u0111\u01a3\u1ee3c t = 68 th\u1ecfa m\u00e3n \u0111i\u1ec1u ki\u1ec7n c\u1ee7a \u1ea9n. V\u1eady nhi\u1ec7t \u0111\u1ed9 cu\u1ed1i c\u00f9ng sau khi h\u00f2a c\u1ee7a n\u01a3\u1edbc l\u00e0 680C. 301","Website: tailieumontoan.com 19.20. M\u1ed9t \u0111\u1ea1i l\u01a3\u1ee3ng b\u00e0i to\u00e1n kh\u00f4ng cho nh\u01a3ng coi nh\u01a3 \u0111\u00e3 bi\u1ebft, \u0111\u00f3 l\u00e0 g\u00e0 v\u00e0 v\u1ecbt \u0111\u1ec1u c\u00f3 2 ch\u00e2n; ch\u00f3 v\u00e0 th\u1ecf \u0111\u1ec1u c\u00f3 4 ch\u00e2n. S\u1ed1 v\u1ecbt + s\u1ed1 g\u00e0 + s\u1ed1 th\u1ecf + s\u1ed1 ch\u00f3 = 290 S\u1ed1 ch\u00e2n v\u1ecbt + s\u1ed1 ch\u00e2n g\u00e0 + s\u1ed1 ch\u00e2n th\u1ecf + s\u1ed1 ch\u00e2n ch\u00f3 = 290. Ta c\u00f3 c\u00e1ch gi\u1ea3i: G\u1ecdi s\u1ed1 v\u1ecbt l\u00e0 x con ( 0 < x < 100) th\u00ec s\u1ed1 th\u1ecf l\u00e0 x con , s\u1ed1 ch\u00f3 l\u00e0 2x con , s\u1ed1 g\u00e0 l\u00e0 100 \u2013 4x (con). Ta c\u00f3 ph\u01a3\u01a1ng tr\u00ecnh : 2x + 4x + 8x + 2(100 \u2013 4x) = 290 Gi\u1ea3i ph\u01a3\u01a1ng tr\u00ecnh \u0111\u01a3\u1ee3c x = 15 th\u1ecfa m\u00e3n \u0111i\u1ec1u ki\u1ec7n c\u1ee7a \u1ea9n. V\u1eady s\u1ed1 v\u1ecbt l\u00e0 15 con; s\u1ed1 g\u00e0 l\u00e0 40 con; s\u1ed1 th\u1ecf l\u00e0 15 con; s\u1ed1 ch\u00f3 l\u00e0 30 con. 19.21. Trong b\u00e0i to\u00e1n t\u00ednh tu\u1ed5i, khi cha th\u00eam 1 tu\u1ed5i th\u00ec con c\u0169ng th\u00eam m\u1ed9t tu\u1ed5i n\u00ean hi\u1ec7u gi\u1eefa tu\u1ed5i cha v\u00e0 con lu\u00f4n kh\u00f4ng \u0111\u1ed5i. Ta c\u00f3 c\u00e1ch gi\u1ea3i : a) G\u1ecdi tu\u1ed5i con hi\u1ec7n nay l\u00e0 x (tu\u1ed5i; x > 0) th\u00ec tu\u1ed5i cha hi\u1ec7n nay l\u00e0 x + 30. Tr\u01a3\u1edbc \u0111\u00e2y 4 n\u0103m tu\u1ed5i con l\u00e0 x \u2013 4 v\u00e0 tu\u1ed5i cha l\u00e0 x + 30 \u2013 4 = x + 26. Ta c\u00f3 ph\u01a3\u01a1ng tr\u00ecnh : x + 26 = 4(x \u2013 4) Gi\u1ea3i ph\u01a3\u01a1ng tr\u00ecnh \u0111\u01a3\u1ee3c x = 14 th\u1ecfa m\u00e3n \u0111i\u1ec1u ki\u1ec7n c\u1ee7a \u1ea9n. V\u1eady tu\u1ed5i con hi\u1ec7n nay l\u00e0 14 v\u00e0 tu\u1ed5i cha l\u00e0 44. b) G\u1ecdi y l\u00e0 tu\u1ed5i con l\u00fac tu\u1ed5i cha g\u1ea5p 2,5 tu\u1ed5i con ( y > 0), do cha lu\u00f4n h\u01a1n con 30 tu\u1ed5i n\u00ean tu\u1ed5i cha l\u00fac \u1ea5y l\u00e0 y + 30. Ta c\u00f3 ph\u01a3\u01a1ng tr\u00ecnh y + 30 = 2,5y Gi\u1ea3i ph\u01a3\u01a1ng tr\u00ecnh t\u00ecm \u0111\u01a3\u1ee3c y = 20 th\u1ecfa m\u00e3n \u0111i\u1ec1u ki\u1ec7n c\u1ee7a \u1ea9n. V\u1eady sau \u0111\u00e2y 20 \u2013 14 = 6 n\u0103m n\u1eefa th\u00ec tu\u1ed5i cha g\u1ea5p 2,5 l\u1ea7n tu\u1ed5i con. * Ghi ch\u00fa : a) C\u00f3 th\u1ec3 ch\u1ecdn \u1ea9n gi\u00e1n ti\u1ebfp l\u00e0 tu\u1ed5i cha (ho\u1eb7c con) khi tu\u1ed5i cha g\u1ea5p 4 l\u1ea7n tu\u1ed5i con. (b\u1ea1n \u0111\u1ecdc t\u1ef1 gi\u1ea3i). b) N\u1ebfu ch\u1ecdn z l\u00e0 s\u1ed1 n\u0103m t\u1eeb nay \u0111\u1ebfn khi tu\u1ed5i cha g\u1ea5p 2,5 l\u1ea7n tu\u1ed5i con ( z > 0 l\u00e0 sau \u0111\u00e2y z n\u0103m , c\u00f2n z < 0 l\u00e0 tr\u01a3\u1edbc \u0111\u00e2y z n\u0103m). Ta c\u00f3 ph\u01a3\u01a1ng tr\u00ecnh 2,5(14 + z) = 44 + z ta t\u00ecm \u0111\u01a3\u1ee3c z = 6. 19.22. T\u01a3\u01a1ng t\u1ef1 v\u00ed d\u1ee5 10. \u0110\u00e1p s\u1ed1 : S\u1ed1 qu\u00fdt \u0111em b\u00e1n : 150 qu\u1ea3, s\u1ed1 l\u1ea7n b\u00e1n l\u00e0 5 l\u1ea7n. S\u1ed1 qu\u00fdt thu ho\u1ea1ch : 160 qu\u1ea3. 19.23. N\u1eeda chu vi t\u1ea5m t\u00f4n l\u00e0 57cm. G\u1ecdi k\u00edch th\u01a3\u1edbc th\u1ee9 nh\u1ea5t c\u1ee7a t\u1ea5m t\u00f4n l\u00e0 x (cm); (10 < x < 57). Th\u00ec k\u00edch th\u01a3\u1edbc th\u1ee9 hai l\u00e0 57 \u2013 x (cm). Sau khi g\u1ea5p th\u00e0nh h\u00ecnh h\u1ed9p ch\u1eef nh\u1eadt, ba k\u00edch th\u01a3\u1edbc c\u1ee7a n\u00f3 l\u00e0 x \u2013 10 (cm); 47 \u2013 x (cm); 5cm.Ta c\u00f3 ph\u01a3\u01a1ng tr\u00ecnh (x \u2013 10)(47 \u2013 x).5 = 1500 \uf0db x2 \u2013 57x +770 = 0 \uf0db (x \u2013 35)(x \u2013 22) = 0 \uf0db x = 35 v\u00e0 x = 22. C\u1ea3 hai gi\u00e1 tr\u1ecb \u0111\u1ec1u th\u1ecfa m\u00e3n. V\u1eady k\u00edch th\u01a3\u1edbc c\u1ee7a t\u1ea5m t\u00f4n l\u00e0 35cm v\u00e0 22 cm. 19.24. N\u1eeda chu vi l\u00e0 61m. G\u1ecdi m\u1ed9t chi\u1ec1u l\u00e0 x (m) ( 0 < x < 61) th\u00ec chi\u1ec1u kia l\u00e0 61 \u2013 x (m). Ta c\u00f3 ph\u01a3\u01a1ng tr\u00ecnh x(61 \u2013 x) = 900 \uf0db x2 \u2013 61x + 900 = 0 \uf0db (x \u2013 25)(x \u2013 36) = 0 \uf0db x = 25 ho\u1eb7c x = 36. C\u1ea3 hai gi\u00e1 tr\u1ecb \u0111\u1ec1u th\u1ecfa m\u00e3n. V\u1eady chi\u1ec1u d\u00e0i v\u00e0 chi\u1ec1u r\u1ed9ng c\u1ee7a khu v\u01a3\u1eddn l\u00e0 36m v\u00e0 25m. 302","Website: tailieumontoan.com 19.25. G\u1ecdi s\u1ed1 chi ti\u1ebft m\u00e1y th\u00e1ng th\u1ee9 nh\u1ea5t t\u1ed5 I s\u1ea3n xu\u1ea5t l\u00e0 x (chi ti\u1ebft m\u00e1y, 0<x< 900) th\u00ec t\u1ed5 II s\u1ea3n xu\u1ea5t l\u00e0 900 \u2013 x (chi ti\u1ebft m\u00e1y). Ta c\u00f3 ph\u01a3\u01a1ng tr\u00ecnh : 115%x + 110%(900 \u2013 x) = 1010 hay 115x \uf02b 110(900 \uf02d x) \uf03d 1010 100 100 Gi\u1ea3i ph\u01a3\u01a1ng tr\u00ecnh t\u00ecm \u0111\u01a3\u1ee3c x = 400. V\u1eady th\u00e1ng th\u1ee9 nh\u1ea5t t\u1ed5 I s\u1ea3n xu\u1ea5t l\u00e0 400 chi ti\u1ebft m\u00e1y v\u00e0 th\u00ec t\u1ed5 II s\u1ea3n xu\u1ea5t l\u00e0 500 chi ti\u1ebft m\u00e1y. 19.26. G\u1ecdi v\u1eadn t\u1ed1c c\u1ee7a gi\u00f3 l\u00e0 x (km\/h), 0< x < 280. Th\u1eddi gian bay t\u1eeb A \u0111\u1ebfn B l\u00e0 960 (h). Th\u1eddi gian bay t\u1eeb 280 \uf02d x B \u0111\u1ebfn A l\u00e0 960 (h). Ta c\u00f3 ph\u01a3\u01a1ng tr\u00ecnh : 960 = 960 + 1 bi\u1ebfn \u0111\u1ed5i th\u00e0nh x2 + 1920x \u2013 78400 =0 280 \uf02b x 280 \uf02d x 280 \uf02b x \uf0db (x \u2013 40)(x + 1960) = 0 . Nghi\u1ec7m x = 40 th\u1ecfa m\u00e3n \u0111i\u1ec1u ki\u1ec7n \u0111\u1ea7u b\u00e0i. V\u1eady v\u1eadn t\u1ed1c c\u1ee7a gi\u00f3 l\u00e0 40km\/h. 19.27. G\u1ecdi th\u1eddi gian l\u00e0m m\u1ed9t m\u00ecnh xong c\u00f4ng vi\u1ec7c c\u1ee7a ng\u01a3\u1eddi th\u1ee9 nh\u1ea5t l\u00e0 x gi\u1edd ( x > 0) th\u00ec m\u1ed9t gi\u1edd ng\u01a3\u1eddi \u0111\u00f3 l\u00e0m \u0111\u01a3\u1ee3c 1 c\u00f4ng vi\u1ec7c. M\u1ed9t gi\u1edd ng\u01a3\u1eddi th\u1ee9 hai l\u00e0m \u0111\u01a3\u1ee3c \uf0e6 1 \uf02d 1 \uf0f6 c\u00f4ng vi\u1ec7c. x \uf0e7\uf0e8 18 x \uf0f7\uf0f8 Theo b\u00e0i ra ta c\u00f3 ph\u01a3\u01a1ng tr\u00ecnh : 6 \uf02b12 \uf0e6 1 \uf02d 1 \uf0f6 \uf03d 1 \uf0db x = 36 . x \uf0e8\uf0e7 18 x \uf0f7\uf0f8 2 Ng\u01a3\u1eddi th\u1ee9 nh\u1ea5t l\u00e0m m\u1ed9t m\u00ecnh trong 36 gi\u1edd xong c\u00f4ng vi\u1ec7c. Ng\u01a3\u1eddi th\u1ee9 hai l\u00e0m m\u1ed9t m\u00ecnh trong 1: \uf0e6 1 \uf02d 1 \uf0f6 \uf0e7\uf0e8 18 36 \uf0f7\uf0f8 = 36 (gi\u1edd) xong c\u00f4ng vi\u1ec7c. 19.28. G\u1ecdi n\u0103ng su\u1ea5t d\u1ef1 ki\u1ebfn l\u00e0 x (s\u1ea3n ph\u1ea9m \/ ng\u00e0y) (x \uf0ce N*). Th\u1eddi gian ho\u00e0n th\u00e0nh theo k\u1ebf ho\u1ea1ch l\u00e0 200 x (ng\u00e0y). B\u1ed1n ng\u00e0y \u0111\u1ea7u h\u1ecd l\u00e0m \u0111\u01a3\u1ee3c 4x s\u1ea3n ph\u1ea9m. Nh\u1eefng ng\u00e0y sau n\u0103ng su\u1ea5t l\u00e0 (x + 10) s\u1ea3n ph\u1ea9m \/ ng\u00e0y. S\u1ed1 ng\u00e0y ho\u00e0n th\u00e0nh s\u1ed1 s\u1ea3n ph\u1ea9m c\u00f2n l\u1ea1i l\u00e0 200 \uf02d 4x . Theo b\u00e0i ra ta c\u00f3 ph\u01a3\u01a1ng tr\u00ecnh : x \uf02b10 200 \u2013 2 = 200 \uf02d 4x + 4. Bi\u1ebfn \u0111\u1ed5i ph\u01a3\u01a1ng tr\u00ecnh th\u00e0nh x2 + 30x \u2013 1000 = 0 \uf0db (x \u2013 20)(x + 50) = 0 \uf0db x = 20 x x \uf02b10 do x \uf0ce N*. V\u1eady n\u0103ng su\u1ea5t d\u1ef1 ki\u1ebfn l\u00e0 20 s\u1ea3n ph\u1ea9m \/ ng\u00e0y . 19.29. G\u1ecdi v\u1eadn t\u1ed1c ng\u01a3\u1eddi \u0111i xe \u0111\u1ea1p khi \u0111i t\u1eeb A \u0111\u1ebfn B l\u00e0 x (km\/h), x > 0. Ta c\u00f3 ph\u01a3\u01a1ng tr\u00ecnh 36 \uf02d 36 \uf03d 36 . x x \uf02b 3 60 Bi\u1ebfn \u0111\u1ed5i th\u00e0nh x2 + 3x \u2013 180 = 0 \uf0db (x \u2013 12)(x + 15) = 0. Nghi\u1ec7m x = 12 th\u1ecfa m\u00e3n \u0111i\u1ec1u ki\u1ec7n. V\u1eady v\u1eadn t\u1ed1c ng\u01a3\u1eddi \u0111i xe \u0111\u1ea1p khi \u0111i t\u1eeb A \u0111\u1ebfn B l\u00e0 12 km\/h. 19.30. N\u1ebfu C l\u00e0 v\u1ecb tr\u00ed xe m\u00e1y b\u1ecb h\u1ecfng th\u00ec AC = 90km ; CB = 30km. G\u1ecdi v\u1eadn t\u1ed1c (km\/h) c\u1ee7a xe m\u00e1y khi \u0111i t\u1eeb A \u0111\u1ebfn C l\u00e0 x , x > 10 th\u00ec v\u1eadn t\u1ed1c c\u1ee7a xe m\u00e1y khi \u0111i t\u1eeb C \u0111\u1ebfn B l\u00e0 (x \u2013 10) (km\/h). Xe m\u00e1y \u0111i qu\u00e3ng \u0111\u01a3\u1eddng AC h\u1ebft 303","Website: tailieumontoan.com 90 (h) v\u00e0 CB h\u1ebft 30 (h). Th\u1eddi gian s\u1eeda xe m\u00e1y 10 ph\u00fat = 1 h. Th\u1eddi gian xe \u0111i h\u1ebft qu\u00e3ng \u0111\u01a3\u1eddng AB (k\u1ec3 c\u1ea3 s\u1eeda x x \uf02d10 6 xe) l\u00e0 4 gi\u1edd 40 ph\u00fat = 14 h. 3 Ta c\u00f3 ph\u01a3\u01a1ng tr\u00ecnh 90 \uf02b 30 \uf02b 1 \uf03d 14 . Bi\u1ebfn \u0111\u1ed5i th\u00e0nh 3x2 \u2013 110x + 600 = 0 x x \uf02d10 6 3 \uf0db (x \u2013 30)(3x \u2013 20) = 0. Nghi\u1ec7m x = 30 th\u1ecfa m\u00e3n \u0111i\u1ec1u ki\u1ec7n. Th\u1eddi gian \u0111i t\u1eeb A \u0111\u1ebfn C l\u00e0 90 : 30 =3(h). Th\u1eddi \u0111i\u1ec3m b\u1ecb h\u1ecfng xe l\u00fac 10 gi\u1edd s\u00e1ng c\u00f9ng ng\u00e0y. 19.31. G\u1ecdi qu\u00e3ng \u0111\u01a3\u1eddng AB d\u00e0i l\u00e0 x km, x > 0. Th\u1eddi gian xe t\u1ea3i \u0111i h\u1ebft qu\u00e3ng \u0111\u01a3\u1eddng AB l\u00e0 x (h ). Th\u1eddi gian 40 d\u1ef1 ki\u1ebfn c\u1ee7a xe kh\u00e1ch t\u1eeb A \u0111\u1ebfn B l\u00e0 x (h). Th\u1eddi gian xu\u1ea5t ph\u00e1t sau c\u1ee7a xe kh\u00e1ch so v\u1edbi xe t\u1ea3i l\u00e0 x \u2013 x . \u0111i 50 40 50 Th\u1eddi gian xe kh\u00e1ch th\u1ef1c t\u1ebf \u0111i l\u00e0 1 . x \uf02b 1 . x (h) ; 16 ph\u00fat = 4 h 2 50 2 60 15 Ta c\u00f3 ph\u01a3\u01a1ng tr\u00ecnh x \uf03d x \uf02b x \uf02b 4 \uf02b \uf0e6 x \uf02d x \uf0f6 \uf0db x = 160 th\u1ecfa m\u00e3n \u0111i\u1ec1u ki\u1ec7n. V\u1eady qu\u00e3ng \u0111\u01a3\u1eddng AB 40 100 120 15 \uf0e8\uf0e7 40 50 \uf0f7\uf0f8 d\u00e0i 160 km. Chuy\u00ean \u0111\u1ec1 20. PH\u01a2\u01a0NG TR\u00ccNH NGHI\u1ec6M NGUY\u00caN 20.1. a) H\u1ec7 s\u1ed1 c\u1ee7a \u1ea9n y l\u00e0 \u20131. \u0110\u00e1p s\u1ed1 : \uf0ecx \uf03d t ; t\uf0ceZ \uf0ee\uf0edy \uf03d 8t \uf02d15 b) V\u1ec1 gi\u00e1 tr\u1ecb tuy\u1ec7t \u0111\u1ed1i th\u00ec h\u1ec7 s\u1ed1 c\u1ee7a x nh\u1ecf h\u01a1n h\u1ec7 s\u1ed1 c\u1ee7a y . Do \u0111\u00f3 ta t\u00ednh x theo y. sau \u0111\u00f3 t\u00e1ch ph\u1ea7n nguy\u00ean . \u0110\u00e1p s\u1ed1 : \uf0ecx \uf03d 12u \uf02b 9 ;u\uf0ceZ \uf0ed\uf0eey \uf03d 5u \uf02b1 c) Ta c\u00f3 (14; 21) = 7 n\u00ean y 7. \u0110\u1eb7t y = 7t ta c\u00f3 ph\u01a3\u01a1ng tr\u00ecnh 2x \u2013 9t = 3.Ti\u1ebfp t\u1ee5c l\u00e0m nh\u01a3 b) ta t\u00ecm \u0111\u01a3\u1ee3c \uf0ecx \uf03d 9u \uf02d 3 ; u\uf0ceZ. \uf0ed\uf0eey \uf03d 14u \uf02d 7 d) C\u00e1ch 1 : Ph\u01a3\u01a1ng tr\u00ecnh \u0111\u00e3 cho (vi\u1ebft t\u1eaft l\u00e0 PT) : 29x + 15y = 20 \uf0db y = 20 \uf02d 29x \uf03d 15 \uf02d 30x+5 \uf02b x \uf03d 1\uf02d 2x \uf02b 5 \uf02b x 15 15 15 \u0110\u1eb7t 5 \uf02b x \uf03d u , (u\uf0ce Z) ta c\u00f3 x = 15u \u2013 5 15 Nghi\u1ec7m nguy\u00ean t\u1ed5ng qu\u00e1t c\u1ee7a PT l\u00e0 \uf0ecx \uf03d 15u \uf02d 5 \uf0ed\uf0eey \uf03d 11\uf02d 29u C\u00e1ch 2 : Ta c\u00f3 (15; 20) = 5 n\u00ean x 5. \u0110\u1eb7t x = 5t ta c\u00f3 ph\u01a3\u01a1ng tr\u00ecnh 29t + 3y = 4. Ti\u1ebfp t\u1ee5c l\u00e0m nh\u01a3 b) ta t\u00ecm \u0111\u01a3\u1ee3c \uf0ecx \uf03d 15u \uf02d 5 ; u\uf0ceZ. \uf0ed\uf0eey \uf03d 11\uf02d 29u 20.2. Ta ch\u1ee9ng minh b\u1eb1ng ph\u1ea3n ch\u1ee9ng : 304","Website: tailieumontoan.com Gi\u1ea3 s\u1eed ph\u01a3\u01a1ng tr\u00ecnh ax + by = c (a; b \uf0b9 0) c\u00f3 nghi\u1ec7m nguy\u00ean l\u00e0 (x0; y0) t\u1ee9c l\u00e0 ax0 + by0 = c . G\u1ecdi (a ; b) = d th\u00ec a = dm ; b = dn (m, n \uf0ce Z) . Ta c\u00f3 dmx0 + dny0 = c \uf0de mx0 + ny0 = c d Do c \uf02f d n\u00ean c \uf0cfZ \uf0de mx0 + ny0 \uf0cf Z \u0111i\u1ec1u n\u00e0y v\u00f4 l\u00fd v\u00ec m ; n; x0 ; y0\uf0ce Z \uf0de \u0111pcm. d 20.3. a) Nghi\u1ec7m nguy\u00ean t\u1ed5ng qu\u00e1t l\u00e0 \uf0ecx \uf03d 5t \uf02b1 ; t\uf0ceZ . Nghi\u1ec7m nguy\u00ean d\u01a3\u01a1ng l\u00e0 (x ; y) = (1 ; 3). \uf0ed\uf0eey \uf03d 3 \uf02d 4t b) Bi\u1ebfn \u0111\u1ed5i PT th\u00e0nh 3x + 7y = 45. Do (3; 45) = 3 n\u00ean y 3. Ph\u01a3\u01a1ng tr\u00ecnh c\u00f3 nghi\u1ec7m nguy\u00ean t\u1ed5ng qu\u00e1t l\u00e0 \uf0ecx \uf03d 15 \uf02d 7t (t \uf0ce Z) . \uf0ee\uf0edy \uf03d 3t \u0110\u1ec3 x > 0 v\u00e0 y > 0 ta ph\u1ea3i c\u00f3 0 \uf03c t \uf03c 15 v\u1eady t = 1 ; 2. Ph\u01a3\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m nguy\u00ean d\u01a3\u01a1ng l\u00e0 (x ; y) = 7 (8 ; 3) ; (1 ; 6). c) Bi\u1ebfn \u0111\u1ed5i PT th\u00e0nh 8x \u2013 9y = 30. Nghi\u1ec7m t\u1ed5ng qu\u00e1t l\u00e0 \uf0ecx \uf03d 9t \uf02b 6 ; t\uf0ceZ v\u00e0 t = 0 ; 1; 2; 3; ...Ph\u01a3\u01a1ng tr\u00ecnh c\u00f3 v\u00f4 s\u1ed1 nghi\u1ec7m nguy\u00ean d\u01a3\u01a1ng (x ; \uf0ed\uf0eey \uf03d 8t \uf02b 2 y) = (6 ; 2) ; (15 ; 10) ; (24 ; 18) ; ... d) Bi\u1ebfn \u0111\u1ed5i PT th\u00e0nh x2 + 2xy + y2 = 25 \uf0db (x + y)2 = 52 Hay x \uf02b y \uf03d 5 . Do x > 0 ; y > 0 n\u00ean x + y = 5 v\u00e0 ph\u01a3\u01a1ng tr\u00ecnh c\u00f3 4 nghi\u1ec7m : (x ; y) = (1 ; 4) ; (2 ; 3) ; (3 ; 2) ; (4 ; 1). 20.4. Ta c\u00f3 \u2013 4x + 3y + 8z = 9 \uf0db x + 8(y + z) \u2013 5(x + y) = 9 . \u0110\u1eb7t u = y + z ; v = x + y \uf0ecx \uf03d 9 \uf02d 8u \uf02b 5v (u \uf0ce Z; v \uf0ce Z) Nghi\u1ec7m nguy\u00ean t\u1ed5ng qu\u00e1t c\u1ee7a PT l\u00e0 \uf0ef\uf0edy \uf03d \uf02d9 \uf02b 8u \uf02d 4v \uf0ee\uf0efz \uf03d 9 \uf02d 7u \uf02b 4v 20.5. Ta chuy\u1ec3n v\u1ebf \u0111\u01a3a v\u1ec1 d\u1ea1ng A(x) = 0 sau \u0111\u00f3 ph\u00e2n t\u00edch A(x) th\u00e0nh nh\u00e2n t\u1eed b\u1eb1ng t\u00e1ch v\u00e0 th\u00eam b\u1edbt c\u00e1c h\u1ea1ng t\u1eed. a) 3x2 \u2013 14x = 5 \uf0db 3x2 + x \u2013 15x \u2013 5 = 0 \uf0db (x \u2013 5)(3x + 1) = 0 Nghi\u1ec7m nguy\u00ean c\u1ee7a ph\u01a3\u01a1ng tr\u00ecnh l\u00e0 x = 5. a) x(2x2 + 9x + 7) = 6 \uf0db 2x3 + 9x2 + 7x \u2013 6 = 0 \uf0db (x + 3)(x + 2)(2x \u2013 1) = 0. T\u1eadp nghi\u1ec7m l\u00e0 S = \uf07b\uf02d3; \uf02d2\uf07d c) x4 + 2x3 \u2013 19x2 + 8x + 60 = 0 \uf0db (x \u2013 2)(x + 2)(x + 3)(x \u2013 5) = 0. T\u1eadp nghi\u1ec7m S = \uf07b\uf02d3 ; \uf02d2 ; 2 ; 5\uf07d . b) (x4 \u2013 13x2 + 36)(x2 + 2x) = 65x2 \u2013 5x4\u2013 180 \uf0db (x4 \u2013 13x2 + 36)(x2 + 2x + 5) = 0 \uf0db (x2 \u2013 4)(x2 \u2013 9)(x2 + 2x + 5) = 0 \uf0db (x \u2013 2)(x + 2)(x + 3)(x \u2013 3)(x2 + 2x + 5) = 0 305","Website: tailieumontoan.com Do x2 + 2x + 5 = (x + 1)2 + 4 > 0 , \uf022 x n\u00ean nghi\u1ec7m nguy\u00ean c\u1ee7a ph\u01a3\u01a1ng tr\u00ecnh PT l\u00e0 x = \uf0b12 ; x = \uf0b13 . 20.6. a) \u0110KX\u0110 : x \uf0b9 \u2013 1 ; x \uf0b9 \u2013 2 ; x \uf0b9 \u2013 3 . Ph\u01a3\u01a1ng tr\u00ecnh bi\u1ebfn \u0111\u1ed5i th\u00e0nh 1 \uf02d 1 \uf03d1 . x2 \uf02b 4x \uf02b 3 x2 \uf02b 4x \uf02b 4 12 \u0110\u1eb7t x2 + 4x + 3 = y ph\u01a3\u01a1ng tr\u00ecnh tr\u1edf th\u00e0nh 1 \uf02d 1 \uf03d 1 suy ra y y \uf02b1 12 y2 + y \u2013 12 = 0 \uf0db (y \u2013 3)(y + 4) = 0. * x2 + 4x + 3 \u2013 3 = 0 \uf0db x(x + 4) = 0 \uf0db x = 0 ho\u1eb7c x = 4. * x2 + 4x + 3 + 4 = 0 \uf0db (x + 2)2 + 3 = 0 v\u00f4 nghi\u1ec7m v\u00ec v\u1ebf tr\u00e1i > 0 \uf022 x . V\u1eady ph\u01a3\u01a1ng tr\u00ecnh c\u00f3 2 nghi\u1ec7m l\u00e0 0 v\u00e0 \u2013 4. b) C\u00e1c m\u1eabu \u0111\u1ec1u d\u01a3\u01a1ng n\u00ean \u0110KX\u0110 l\u00e0 x \uf0ce R . Bi\u1ebfn \u0111\u1ed5i ph\u01a3\u01a1ng tr\u00ecnh th\u00e0nh x2 \uf02b 4x \uf02b 4 \uf02b x2 \uf02b 4x \uf02b 5 \uf03d 1 . x2 \uf02b 4x \uf02b 5 x2 \uf02b 4x \uf02b 6 2 \u0110\u1eb7t x2 + 4x + 5 = y ta c\u00f3 y \uf02d1 \uf02b y \uf03d 1 suy ra 3y2 \u2013 y \u2013 2 = 0 \uf0db (y \u2013 1)(3y + 2) = 0. T\u1eeb \u0111\u00e2y t\u00ecm \u0111\u01a3\u1ee3c y y\uf02b1 2 nghi\u1ec7m c\u1ee7a ph\u01a3\u01a1ng tr\u00ecnh l\u00e0 x = \u2013 2. c) \u00c1p d\u1ee5ng h\u1eb1ng \u0111\u1eb3ng th\u1ee9c (a + b)3 = a3 + 3a2b + 3ab2 + b3 Ta c\u00f3 (x + 1)3 + (x + 2)3 + (x + 3)3 + (x + 4)3 \u2013 (x + 5)3 = 0 \uf0db 3x3 + 15x2 + 15x = 25 V\u1ebf tr\u00e1i chia h\u1ebft cho 3; v\u1ebf ph\u1ea3i kh\u00f4ng chia h\u1ebft cho 3. Ph\u01a3\u01a1ng tr\u00ecnh kh\u00f4ng c\u00f3 nghi\u1ec7m nguy\u00ean. Ch\u00fa \u00fd : C\u00e2u c) c\u00f3 th\u1ec3 \u0111\u1eb7t x + 3 = y (\uf0ce Z). Ph\u01a3\u01a1ng tr\u00ecnh tr\u1edf th\u00e0nh (y \u2013 2)3 + (y \u2013 1)3 + y3 + (y + 1)3 \u2013 (y + 2)3 = 0 , r\u00fat g\u1ecdn th\u00e0nh 3y3 \u2013 12y2 + 6y = 16 20.7. a) 6(x + y) = xy + 33 \uf0db (x \u2013 6)(y \u2013 6) = 3. V\u00ec 3 = 1.3 = 3.1 = (\u20131)(\u2013 3) = (\u2013 3)(\u2013 1) Gi\u1ea3i c\u00e1c c\u1eb7p ta t\u00ecm \u0111\u01a3\u1ee3c c\u00e1c nghi\u1ec7m nguy\u00ean sau : (x ; y) = (7 ; 9) ; (9 ; 7) ; (5; 3) ; (3 ; 5). b) 3(x + y) = 2xy \uf0db 4xy \u2013 6x \u2013 6y = 0 \uf0db (2x \u2013 3)(2y \u2013 3) = 9 Ta bi\u1ebft 9 = 1.9 = 9.1 = (\u20131)(\u2013 9) = (\u2013 9)(\u2013 1) = 3.3 = (\u2013 3)(\u2013 3) Gi\u1ea3i t\u1eebng c\u1eb7p ta c\u00f3 nghi\u1ec7m t\u1ef1 nhi\u00ean c\u1ee7a ph\u01a3\u01a1ng tr\u00ecnh tr\u00ean l\u00e0 : (x; y) = (2; 6) ; (6; 2) ; (3 ; 3); (0 ; 0) 20.8. a) G\u1ecdi ba s\u1ed1 nguy\u00ean d\u01a3\u01a1ng l\u00e0 x ; y; z. Theo \u0111\u1ea7u b\u00e0i x + y + z = xyz. Do vai tr\u00f2 x, y, z nh\u01a3 nhau n\u00ean gi\u1ea3 s\u1eed 1 \uf0a3 x \uf0a3 y \uf0a3 z . Chia hai v\u1ebf c\u1ee7a ph\u01a3\u01a1ng tr\u00ecnh cho xyz ta c\u00f3 1 = 1\uf02b1\uf02b1 \uf0a3 3 hay x2 \uf0a3 3 \uf0de x = 1. yz zx xy x2 Thay x = 1 v\u00e0o ph\u01a3\u01a1ng tr\u00ecnh ta c\u00f3 1 + y + z = yz \uf0db yz \u2013 y \u2013 z = 1 \uf0db (y \u2013 1)(z \u2013 1) = 2 = 1. 2 . T\u1eeb \u0111\u00f3 t\u00ecm \u0111\u01a3\u1ee3c y = 2 ; z = 3. Nghi\u1ec7m nguy\u00ean d\u01a3\u01a1ng c\u1ee7a ph\u01a3\u01a1ng tr\u00ecnh l\u00e0 : (x; y; z) = (1 ; 2; 3) ; (1 ; 3; 2) ; (2 ; 3; 1) ; (2 ; 1; 3) ; (3 ; 2; 1) ; (3 ;1 ; 2). 306","Website: tailieumontoan.com b) Gi\u1ea3 s\u1eed x \uf0b3 y \uf0b3 z \uf0b3 t \uf0b31, chia hai v\u1ebf c\u1ee7a ph\u01a3\u01a1ng tr\u00ecnh cho xyzt ta c\u00f3 6 = 5 \uf02b 5 \uf02b 5 \uf02b 5 \uf02b 4 \uf0a3 24 \uf0de t3 \uf0a3 4 \uf0de t = 1 . xyz xzt xyt yzt xyzt t3 * Thay t = 1 v\u00e0o ph\u01a3\u01a1ng tr\u00ecnh ta c\u00f3 : 5(x + y + z) + 9 = 6xyz. Ta c\u0169ng c\u00f3 6 = 5 \uf02b 5 \uf02b 5 \uf02b 9 \uf0a3 24 \uf0de z2 \uf0a3 4 \uf0de z = 1 ; 2 . yz xz xy xyz z2 + V\u1edbi z = 1 th\u00ec 5(x + y) + 14 = 6xy \uf0db 6xy \u2013 5x \u2013 5y = 14 \uf0db 36xy \u2013 30x \u2013 30y + 25 = 109 \uf0db (6x \u2013 5)(6y \u2013 5) = 109 .1 \uf0db \uf0e96x \uf02d5 \uf03d 109 \uf0db \uf0e9x \uf03d 19 \uf0ea\uf0eb6x \uf02d5 \uf03d1 \uf0ea\uf0eby \uf03d1 + V\u1edbi z = 2 gi\u1ea3i t\u01a3\u01a1ng t\u1ef1 , kh\u00f4ng c\u00f3 nghi\u1ec7m nguy\u00ean d\u01a3\u01a1ng. V\u1eady nghi\u1ec7m nguy\u00ean d\u01a3\u01a1ng c\u1ee7a ph\u01a3\u01a1ng tr\u00ecnh l\u00e0 (x; y; z; t) = (19 ; 1; 1; 1) v\u00e0 c\u00e1c ho\u00e1n v\u1ecb (1; 1; 1; 19) ; (1; 1; 19; 1) ; (1; 19; 1; 1) . 20.9. a) PT \uf0db x2(y2 \u2013 3) \u2013 2y2x + 2y2 + 6x = 7 \uf0db x2(y2 \u2013 3) \u2013 2x(y2 \u2013 3) + 2(y2 \u2013 3) = 1 \uf0db (y2 \u2013 3)(x2\u2013 2x + 2) = 1 Do x , y\uf0ceZ n\u00ean y2 \u2013 3 \uf0ceZ ; x2\u2013 2x + 2 \uf0ceZ. v\u00ec th\u1ebf \uf0ef\uf0ecy2 \uf02d3\uf03d1 \uf0db \uf0ef\uf0ecy2 \uf03d4 \uf0db \uf0ec\uf0e9y \uf03d 2 . \uf0ed \uf02d 2x \uf02b 2 \uf03d1 \uf0ed \uf02d1)2 \uf0ef\uf0ed\uf0eb\uf0eay \uf03d \uf02d2 \uf0ee\uf0efx 2 \uf0ee\uf0ef(x \uf03d 0 \uf0ee\uf0efx \uf03d 1 v\u00e0 \uf0ef\uf0ecy2 \uf02d 3 \uf03d \uf02d1 \uf0db \uf0ef\uf0ecy2 \uf03d2 (kh\u00f4ng c\u00f3 nghi\u1ec7m nguy\u00ean). \uf0ed \uf02d 2x \uf02b 2 \uf0ed \uf02d1)2 \uf0ef\uf0eex 2 \uf03d \uf02d1 \uf0ef\uf0ee(x \uf02b 2 \uf03d 0 V\u1eady nghi\u1ec7m c\u1ee7a ph\u01a3\u01a1ng tr\u00ecnh l\u00e0 (x ; y) = (1 ; \u2013 2) ; (1 ; 2). b) xy2 +2xy \u2013 27y + x = 0 \uf0db x(y2 +2y + 1) = 27y \uf0db x(y + 1)2 = 27y \uf0de x \uf03d 27y (y \uf02b1)2 Ta bi\u1ebft hai s\u1ed1 nguy\u00ean d\u01a3\u01a1ng y v\u00e0 y + 1 nguy\u00ean t\u1ed1 c\u00f9ng nhau n\u00ean y \uf02f (y + 1)2 v\u00ec th\u1ebf 27 (y + 1)2 \uf0de (y + 1)2 = 1 ho\u1eb7c (y + 1)2 = 9 . T\u1eeb \u0111\u00f3 t\u00ecm \u0111\u01a3\u1ee3c nghi\u1ec7m c\u1ee7a ph\u01a3\u01a1ng tr\u00ecnh l\u00e0 (x; y) = (6 ; 2). 20.10. Ch\u00fa \u00fd n\u1ebfu A2 + B2 + C2 = 0 th\u00ec A = 0 ; B = 0 v\u00e0 C = 0 a) Bi\u1ebfn \u0111\u1ed5i PT th\u00e0nh (x \u2013 2y)2 + (y + 1)2 = 0. Nghi\u1ec7m (x ; y) = (\u2013 2 ; \u2013 1) b) Bi\u1ebfn \u0111\u1ed5i PT th\u00e0nh (x + y + 1)2 + (y \u2013 2)2 = 0. Nghi\u1ec7m(x ; y) = (\u2013 3 ; 2) c) Bi\u1ebfn \u0111\u1ed5i PT th\u00e0nh (x \u2013 y)2 + (y \u2013 z)2 + (x \u2013 3)2 = 0 Nghi\u1ec7m: (x ; y ; z) = (3 ; 3 ; 3). d) Bi\u1ebfn \u0111\u1ed5i PT th\u00e0nh (x + y)2 + (y \u2013 2z)2 + (y \u2013 6)2 = 0 307","Website: tailieumontoan.com Nghi\u1ec7m (x ; y ; z) = (\u2013 6 ; 6 ; 3). 20.11. Bi\u1ebfn \u0111\u1ed5i v\u1ec1 d\u1ea1ng 1 + x + x2 + x3 = y3 . Ta x\u00e9t c\u00e1c tr\u01a3\u1eddng h\u1ee3p: 1) x = 0 th\u00ec y = 1. 2) x = \u20131 th\u00ec y = 0 . 3) x = 1 th\u00ec y \uf0cf Z. 4) V\u1edbi x > 0 (1 + x)3 = 1 + 3 x + 3x2 + x3 > 1 + x + x2 + x3 = y3 > x3 V\u1eady (1 + x)3 > y3 > x3 hay 1 + x > y > x \u0111i\u1ec1u n\u00e0y kh\u00f4ng th\u1ec3 x\u1ea3y ra \u0111\u1ed1i v\u1edbi s\u1ed1 nguy\u00ean d\u01a3\u01a1ng. 5) V\u1edbi x < \u2013 1 . \u0110\u1eb7t t = \u2013 1\u2013 x th\u00ec t > 0 v\u00e0 x = \u2013 1\u2013 t. Thay v\u00e0o ph\u01a3\u01a1ng tr\u00ecnh ta c\u00f3 1 + (\u2013 1\u2013 t) + (t2 + 2t + 1) \u2013 (t3 + 3t2 + 3t + 1) = y3 Hay \u2013 (t3 + 2t2 + 2t ) = y3 \uf0de y < 0 hay t3 + 2t2 + 2t = (\u2013y)3 \u0110\u1eb7t \u2013 y = z ta c\u00f3 t3 + 2t2 + 2t = z3 v\u1edbi z > 0. Ta c\u00f3 (t + 1)3 = t3 + 3t2 + 3t + 1 > t3 + 2t2 + 2t = z3 > t3 Hay (t + 1)3 > z3 > t3 \uf0de t + 1 > z > t \u0111i\u1ec1u n\u00e0y v\u00f4 l\u00fd. V\u1eady ph\u01a3\u01a1ng tr\u00ecnh ch\u1ec9 c\u00f3 hai nghi\u1ec7m (x ; y) = (0 ; 1) ; (\u20131 ; 0) 20.12. * V\u1edbi y = 0 th\u00ec x = \u2013 1 ; \u2013 2 ; \u2013 8 ; \u2013 9 . * V\u1edbi y \uf0b9 0 ta c\u00f3 (x2 + 10x + 9)( x2 + 10x + 16) = y2 \u0110\u1eb7t x2 + 10x + 9 = z \uf0ceZ do x \uf0ceZ . Ta c\u00f3 z(z + 7) = y2 hay z2 + 7z = y2 . N\u1ebfu z > 9 th\u00ec z2 + 6z + 9 < z2 + 7z = y2 < z2 + 8z + 16 Hay (z + 3)2 < z2 + 7z = y2 < (z + 4)2 v\u00f4 l\u00fd. V\u1eady z \uf0a3 9 x2 + 10x + 9 \uf0a3 9 \uf0db x(x + 10) \uf0a3 0 \uf0db \uf02d10 \uf0a3 x \uf0a3 0. L\u1ea7n l\u01a3\u1ee3t thay c\u00e1c gi\u00e1 tr\u1ecb c\u1ee7a x ta c\u00f3 nghi\u1ec7m : (x ; y) = (\u2013 1 ; 0) ; (\u2013 2 ; 0) ; (\u2013 8 ; 0) ; (\u2013 9 ; 0) ; (\u2013 5 ; \u2013 12) ; (\u2013 5 ; 12) ; (\u2013 10 ; \u2013 12) ; (\u2013 10 ; 12). 20.13. Ta s\u1eed d\u1ee5ng t\u00ednh ch\u1ea5t chia h\u1ebft v\u00e0 ph\u01a3\u01a1ng ph\u00e1p xu\u1ed1ng thang \u0111\u1ec3 gi\u1ea3i. Gi\u1ea3 s\u1eed (x0 ; y0 ; z0) l\u00e0 nghi\u1ec7m nguy\u00ean c\u1ee7a ph\u01a3\u01a1ng tr\u00ecnh t\u1ee9c l\u00e0 x30 \uf02d 2y30 \uf02d 4z30 \uf03d 0 (2) , khi \u0111\u00f3 x0 2 . \u0110\u1eb7t x0 = 2x1. Thay v\u00e0o ph\u01a3\u01a1ng tr\u00ecnh (2) ta \u0111\u01a3\u1ee3c 4x13 \uf02d y30 \uf02d 2z30 \uf03d 0 ; \u0110\u1eb7t y0 = 2y1 ta l\u1ea1i c\u00f3 2x13 \uf02d 4y13 \uf02d z30 \uf03d 0 \u0110\u1eb7t z0 = 2z1 ta l\u1ea1i c\u00f3 x13 \uf02d 2y13 \uf02d 4z13 \uf03d 0 . Nh\u01a3 v\u1eady (x1 , y1 , z1) = \uf0e6 x0 , y0 , z0 \uf0f6 c\u0169ng l\u00e0 nghi\u1ec7m c\u1ee7a ph\u01a3\u01a1ng tr\u00ecnh. C\u1ee9 ti\u1ebfp t\u1ee5c m\u00e3i ta c\u00f3 \uf0e6 x0 , y0 , z0 \uf0f6 \uf0ce Z, \uf022 k \uf0ceZ. Do \u0111\u00f3 x0 = \uf0e8\uf0e7 2 2 2 \uf0f8\uf0f7 \uf0e7\uf0e8 2k 2k 2k \uf0f8\uf0f7 y0 = z0 = 0. V\u1eady ph\u01a3\u01a1ng tr\u00ecnh (1) c\u00f3 nghi\u1ec7m nguy\u00ean duy nh\u1ea5t l\u00e0 (0; 0; 0). 20.14. G\u1ecdi s\u1ed1 c\u00f3 hai ch\u1eef s\u1ed1 \u0111\u00f3 l\u00e0 xy ( x, y \uf0ceN ; 0 \uf03c x, y \uf0a3 9 ). Ta c\u00f3 xy = 10x + y = kxy (k \uf0ceN*) \uf0de y = x(ky \u2013 10) x do \u0111\u00f3 y = mx \uf0de mx = x(kmx \u2013 10) \uf0de m = kmx \u2013 10 \uf0de 10 = m(kx \u2013 1) \uf0de m = 1 ; 2 ; 5. V\u1edbi m = 1 th\u00ec kx = 11 \uf0de x = y = 1. V\u1edbi m = 2 th\u00ec kx = 6 \uf0de x = 1 ; 2 ; 3 t\u01a3\u01a1ng \u1ee9ng c\u00f3 y = 2 ; 4 ; 6. V\u1edbi m = 5 th\u00ec kx = 3 \uf0de x = 1 t\u01a3\u01a1ng \u1ee9ng c\u00f3 y = 5. S\u1ed1 c\u1ea7n t\u00ecm l\u00e0 xy = 11 ; 12 ; 24 ; 36 ; 15. 308","Website: tailieumontoan.com 20.15. G\u1ecdi chi\u1ec1u d\u00e0i, chi\u1ec1u r\u1ed9ng h\u00ecnh ch\u1eef nh\u1eadt l\u1ea7n l\u01a3\u1ee3t l\u00e0 x v\u00e0 y . C\u1ea1nh h\u00ecnh vu\u00f4ng c\u1ea7n c\u1eaft ra l\u00e0 z . Ta c\u00f3 x; y; z \uf0ce Z+ ; x \uf0b3 y ; z \uf0a3 y v\u00e0 z \uf0a3 3. Ta c\u00f3 xy = 11z2 (1) . T\u1eeb (1) \uf0de x ho\u1eb7c y chia h\u1ebft cho 11. Vai tr\u00f2 c\u1ee7a x v\u00e0 y trong ph\u01a3\u01a1ng tr\u00ecnh nh\u01a3 nhau n\u00ean ta gi\u1ea3 s\u1eed x 11 t\u1ee9c l\u00e0 x = 11d \uf0de 11dy = 11z2 \uf0de dy = z2 . Ta x\u00e9t c\u00e1c tr\u01a3\u1eddng h\u1ee3p c\u00f3 th\u1ec3 c\u1ee7a z : V\u1edbi z = 1, ch\u1ec9 c\u00f3 th\u1ec3 d = 1 ; y = 1 \uf0de x = 11 V\u1edbi z = 2, ch\u1ec9 c\u00f3 th\u1ec3 d = 1 ; y = 4 \uf0de x = 11 d = 2 ; y = 2 \uf0de x = 22 d = 4 ; y = 1 \uf0de x = 44 V\u1edbi z = 3, ch\u1ec9 c\u00f3 th\u1ec3 d = 1 ; y = 9 \uf0de x = 11 d = 3 ; y = 3 \uf0de x = 33 d = 9 ; y = 1 \uf0de x = 99 Trong 7 nghi\u1ec7m c\u1ee7a ph\u01a3\u01a1ng tr\u00ecnh v\u1eeba t\u00ecm ch\u1ec9 c\u00f3 3 nghi\u1ec7m th\u1ecfa m\u00e3n b\u00e0i to\u00e1n \u0111\u00f3 l\u00e0 (x; y) = (11 ; 1) ; (22 ; 2) ; (33 ; 3). 20.16. a) Vai tr\u00f2 x ; y; z nh\u01a3 nhau . Ta gi\u1ea3 s\u1eed x \uf0b3 y \uf0b3 z \uf0b31 Ta c\u00f3 xy \uf02b yz \uf02b zx \uf03d 1 \uf0db xy \uf02b z \uf0e6 y \uf02b x \uf0f6 \uf03d 1 . 3z 3x 3y 3z 3 \uf0e7 x y \uf0f7 \uf0e8 \uf0f8 V\u1edbi x, y > 0 th\u00ec (x \u2013 y)2 \uf0b3 0 \uf0db x2 + y2 \uf0b3 2xy \uf0db x \uf02b y \uf0b3 2 . yx Do \u0111\u00f3 1\uf03d xy \uf02b z \uf0e6 y \uf02b x \uf0f6 \uf0b3 z \uf02b 2z \uf03dz \uf0de z=1 \uf0de y = x = 1. 3z 3 \uf0e7 x y \uf0f7 33 \uf0e8 \uf0f8 V\u1eady (x ; y ; z) = (1 ; 1; 1). b) G\u1ecdi ba s\u1ed1 nguy\u00ean d\u01a3\u01a1ng c\u1ea7n t\u00ecm l\u00e0 x ; y; z. Ta c\u00f3 1 \uf02b 1 \uf02b 1 \uf03d 11 x y z 12 Vai tr\u00f2 x ; y; z nh\u01a3 nhau . Ta gi\u1ea3 s\u1eed x \uf0b3 y \uf0b3 z \uf03e1 ta c\u00f3 3 \uf0b3 1 \uf02b 1 \uf02b 1 \uf03d 11 z x y z 12 \uf0de z \uf0a3 36 \uf0de z = 2 ; 3. V\u1edbi z = 2 thay v\u00e0o v\u00e0 l\u00fd lu\u1eadn t\u01a3\u01a1ng t\u1ef1 ta t\u00ecm \u0111\u01a3\u1ee3c (y = 3 ; x = 12) ; (y = 4; x = 6). 11 V\u1edbi z = 3 ta t\u00ecm \u0111\u01a3\u1ee3c (y = 3 ; x = 4). Nghi\u1ec7m th\u1ecfa m\u00e3n b\u00e0i to\u00e1n l\u00e0 (x ; y ; z) = (12 ; 3 ; 2) ; (6 ; 4 ; 2) ; (4 ; 3 ; 3) c\u00f9ng c\u00e1c ho\u00e1n v\u1ecb c\u1ee7a ch\u00fang. 20.17. Gi\u1ea3 s\u1eed ph\u01a3\u01a1ng tr\u00ecnh c\u00f3 nghi\u1ec7m nguy\u00ean l\u00e0 (x0 ; y0) ta c\u00f3 : 2x 2 \uf02d 9y02 \uf03d11 \uf0de y0 l\u1ebb t\u1ee9c l\u00e0 y0 = 2k + 1 (k\uf0ceN*). Ta c\u00f3 2x02 \uf02d 9(2k \uf02b1)2 \uf03d11 0 \uf0de x 2 \uf02d18k(k \uf02b1) \uf03d 10 \uf0de x0 ch\u1eb5n t\u1ee9c l\u00e0 x0 = 2m (m\uf0ceN*) . T\u1ee9c l\u00e0 0 4m2 \uf02d18k(k \uf02b1) \uf03d10 \uf0de 2m2 \u2013 9k(k + 1) = 5 v\u00f4 l\u00fd v\u00ec k(k + 1) l\u00e0 t\u00edch hai s\u1ed1 nguy\u00ean li\u00ean ti\u1ebfp n\u00ean ch\u1eb5n. V\u1ebf tr\u00e1i ch\u1eb5n, v\u1ebf ph\u1ea3i l\u1ebb. Do \u0111\u00f3 ph\u01a3\u01a1ng tr\u00ecnh 2x2 \u2013 9y2 = 11 kh\u00f4ng c\u00f3 nghi\u1ec7m nguy\u00ean. 20.18. Vai tr\u00f2 x ; y; z nh\u01a3 nhau ta gi\u1ea3 s\u1eed 0 \uf03c x \uf0a3 y \uf0a3 z . Ta c\u00f3 1 \uf03c 1 x 2016 309","Website: tailieumontoan.com \uf0de x > 2016. Ta c\u00f3 1 \uf03d 1 \uf02b 1 \uf02b 1 \uf0a3 3 \uf0de 2016 \uf03c x \uf0a3 3.2016 2016 x y z x V\u1eady c\u00f3 h\u1eefu h\u1ea1n s\u1ed1 nguy\u00ean d\u01a3\u01a1ng x. \u1ee8ng v\u1edbi m\u1ed7i gi\u00e1 tr\u1ecb c\u1ee7a x ta c\u00f3 1 \uf02d 1 \uf03d 1 \uf02b 1 \uf0a3 2 \uf0de y \uf0a3 2.2016x \uf0a3 2.2016x \uf0a3 2.20162. 2016 x y z y x \uf02d 2016 1 V\u1eady y h\u1eefu h\u1ea1n \uf0de z h\u1eefu h\u1ea1n. Do \u0111\u00f3 ph\u01a3\u01a1ng tr\u00ecnh c\u00f3 m\u1ed9t s\u1ed1 h\u1eefu h\u1ea1n nghi\u1ec7m nguy\u00ean d\u01a3\u01a1ng. 20.19. Ta c\u00f3 8x2y2 + x2 + y2 = 10xy \uf0db 8xy(xy \u2013 1) + (x \u2013 y)2 = 0 (*). Do (x \u2013 y)2 \uf0b3 0 n\u00ean n\u1ebfu x , y l\u00e0 nghi\u1ec7m nguy\u00ean c\u1ee7a ph\u01a3\u01a1ng tr\u00ecnh th\u00ec xy(xy \u2013 1) \uf0a3 0 \uf0de 0 \uf0a3 xy \uf0a3 1 . Do x, y nguy\u00ean n\u00ean ch\u1ec9 c\u00f3 hai kh\u1ea3 n\u0103ng : - N\u1ebfu xy = 0 th\u00ec t\u1eeb (*) ta c\u00f3 x = y = 0 - N\u1ebfu xy = 1 th\u00ec t\u1eeb (*) ta c\u00f3 x = y = \uf0b1 1. Ph\u01a3\u01a1ng tr\u00ecnh c\u00f3 3 nghi\u1ec7m nguy\u00ean (x, y) l\u00e0 (0; 0) ; (1 ; 1) ; (\u2013 1; \u2013 1). 20.20. G\u1ecdi s\u1ed1 t\u1ef1 nhi\u00ean c\u1ea7n t\u00ecm l\u00e0 n. Ta c\u00f3: n = 2005x +23 = 2007y + 32 = 2005y + 2y +32 (x; y \uf0ce N) \uf0de 2y + 9 = 2005(x \u2013 y) = 2005k (k \uf0ce N*) \uf0de y = 2005k \uf02d 9 2 n nh\u1ecf nh\u1ea5t khi y nh\u1ecf nh\u1ea5t, y nh\u1ecf nh\u1ea5t l\u00e0 998 khi k = 1. V\u1eady s\u1ed1 t\u1ef1 nhi\u00ean nh\u1ecf nh\u1ea5t c\u1ea7n t\u00ecm l\u00e0 n = 2007.998 + 32 = 2003018. 20.21.G\u1ecdi s\u1ed1 \u00f4 t\u00f4 l\u00fac \u0111\u1ea7u l\u00e0 x (x \uf0ceN v\u00e0 x \uf0b3 2), s\u1ed1 h\u1ecdc sinh \u0111i c\u1eafm tr\u1ea1i s\u1ebd l\u00e0 22x + 1. Theo gi\u1ea3 thi\u1ebft n\u1ebfu s\u1ed1 xe l\u00e0 x \u2013 1 th\u00ec s\u1ed1 h\u1ecdc sinh ph\u00e2n ph\u1ed1i \u0111\u1ec1u cho t\u1ea5t c\u1ea3 c\u00e1c xe . Khi \u0111\u00f3 m\u1ed7i xe ch\u1edf y h\u1ecdc sinh (y \uf0ceN v\u00e0 30 \uf0b3 y > 0). Ta c\u00f3 (x \u2013 1)y = 22x + 1 \uf0db y = 22x \uf02b1 \uf03d 22 \uf02b 23 x \uf02d1 x \uf02d1 V\u00ec x, y \uf0ceN n\u00ean x \u2013 1 ph\u1ea3i l\u00e0 \u01a3\u1edbc s\u1ed1 c\u1ee7a 23, 23 nguy\u00ean t\u1ed1 n\u00ean : * x \u2013 1 = 1 \uf0db x = 2 suy ra y = 22 + 23 = 45 (tr\u00e1i gi\u1ea3 thi\u1ebft) * x \u2013 1 = 23 \uf0db x = 24 suy ra y = 22 + 1 = 23 < 30. V\u1eady \u00f4 t\u00f4 l\u00e0 24 v\u00e0 s\u1ed1 h\u1ecdc sinh l\u00e0 22 . 24 + 1 = 529. 20.22. Do m, n \uf0ce N n\u00ean m2 + n2 = m + n + 8 \uf0db 4(m2 + n2) = 4(m + n + 8) \uf0db (4m2 \u2013 4m + 1) + (4n2 \u2013 4n + 1) = 34 \uf0db (2m \u2013 1)2 + (2n \u2013 1)2 = 32 + 52 (*) T\u1eeb (*) \uf0de \uf0ec2m \uf02d1 \uf03d 3 \uf0db \uf0ecm \uf03d 2 \uf0ec2n \uf02d1 \uf03d 3 \uf0db \uf0ecn \uf03d 2 \uf0ed\uf0ee2n \uf02d1 \uf03d 5 \uf0ed\uf0een \uf03d 3 v\u00e0 \uf0ee\uf0ed2m \uf02d1 \uf03d 5 \uf0ee\uf0edm \uf03d 3 C\u00f3 hai c\u1eb7p (m ; n) th\u1ecfa m\u00e3n l\u00e0 (2 ; 3) v\u00e0 (3 ; 2). 20.23. Nh\u1eadn x\u00e9t : V\u1edbi a, b l\u00e0 c\u00e1c s\u1ed1 nguy\u00ean th\u1ecfa m\u00e3n a2 + b2 3 th\u00ec a 3 v\u00e0 b 3 . Ta c\u00f3 5x2 + 8y2 = 20412 \uf0db (6x2 + 9y2) \u2013 (x2 + y2) = 28. 93 Suy ra x2 + y2 3 \uf0db x 3 v\u00e0 y 3 . \u0110\u1eb7t x = 3x1 ; y = 3y1 (x1, y1 \uf0ceZ) 310","Website: tailieumontoan.com Thay v\u00e0o ph\u01a3\u01a1ng tr\u00ecnh ta \u0111\u01a3\u1ee3c 5x12 \uf02b 8y12 \uf03d 28.92 . T\u01a3\u01a1ng t\u1ef1 ta c\u00f3 x1 = 3x2 ; y1 = 3y2 (x2, y2 \uf0ceZ) ta \u0111\u01a3\u1ee3c 5x22 \uf02b 8y22 \uf03d 28.9 T\u01a3\u01a1ng t\u1ef1 ta c\u00f3 x2 = 3x3 ; y2 = 3y3 (x3, y3 \uf0ceZ) ta \u0111\u01a3\u1ee3c 5x32 \uf02b 8y32 \uf03d 28 Suy ra y32 \uf0a3 28 \uf03c 22 n\u00ean y32 \uf03d 0 ho\u1eb7c y32 \uf03d 1. 8 * V\u1edbi y32 \uf03d0 th\u00ec x 2 \uf03d 28 (lo\u1ea1i). 3 5 * V\u1edbi y32 \uf03d 1 th\u00ec x32 \uf03d 22 \uf0de x22 \uf03d 9.22; y22 \uf03d 9 \uf0de x12 \uf03d 92.22; y12 \uf03d 92 \uf0de x2 = 93. 22 ; y2 = 93. V\u1eady c\u00f3 4 c\u1eb7p s\u1ed1 nguy\u00ean (x; y) th\u1ecfa m\u00e3n l\u00e0 (54; 27) ; (54; \u201327) ; (\u201354; 27) ; (\u201354; \u201327). 20.24. T\u1eeb \u0111i\u1ec1u ki\u1ec7n \u0111\u00e3 cho 6x2 + 5y2 = 74 \uf0de y ch\u1eb5n v\u00e0 x \uf0b9 0 ; y \uf0b9 0. N\u1ebfu c\u1eb7p s\u1ed1 (x0; y0) l\u00e0 m\u1ed9t c\u1eb7p s\u1ed1 nguy\u00ean th\u1ecfa m\u00e3n \u0111i\u1ec1u ki\u1ec7n th\u00ec c\u00e1c c\u1eb7p s\u1ed1 (x0; \u2013y0); (\u2013x0; y0); (\u2013x0; \u2013 y0) c\u0169ng th\u1ecfa m\u00e3n \u0111i\u1ec1u ki\u1ec7n, do \u0111\u00f3 ch\u1ec9 c\u1ea7n x\u00e9t x > 0 , y > 0. T\u1eeb \u0111i\u1ec1u ki\u1ec7n suy ra 5y2 < 74 \uf0de y2 < 15 \uf0de 0 < y < 4 \uf0de y = 2 ( v\u00ec y ch\u1eb5n) \uf0de x = 3. V\u1eady c\u00e1c c\u1eb7p s\u1ed1 nguy\u00ean (x; y) th\u1ecfa m\u00e3n \u0111i\u1ec1u ki\u1ec7n l\u00e0 (3; 2) ; (3; \u20132) ; (\u20133; 2) ; (\u20133; \u20132). 20.25. Do x , y \uf0ceN* n\u00ean ta c\u00f3 (x + y)5 = 120y + 3 < 120(x + y) \uf0de (x + y)4 < 120 < 44 \uf0de x + y < 4. C\u0169ng do x , y \uf0ceN* n\u00ean 2 \uf0a3 x + y < 4 ; m\u00e0 120y + 3 l\u00e0 s\u1ed1 l\u1ebb \uf0de x + y l\u00e0 s\u1ed1 l\u1ebb . Do \u0111\u00f3 x + y = 3. V\u00ec v\u1eady 35 = 120y + 3 \uf0db y = 2 \uf0de x = 1. Nghi\u1ec7m nguy\u00ean d\u01a3\u01a1ng c\u1ee7a ph\u01a3\u01a1ng tr\u00ecnh l\u00e0 (x , y) = (1 ; 2). 20.26. Ta c\u00f3 3x \u2013 2y = 1 \uf0db 3x \u2013 1 = 2y (1) . *N\u1ebfu x ch\u1eb5n t\u1ee9c l\u00e0 x = 2k (k \uf0ce N*) . T\u1eeb (1) ta c\u00f3 (3k + 1)(3k \u2013 1) = 2y Do \u0111\u00f3 \uf0ef\uf0ec3k \uf02b1\uf03d 2a trong \u0111\u00f3 a, b \uf0ce N v\u00e0 a > b. \uf0ee\uf0ef\uf0ed3k \uf02d1 \uf03d 2b X\u00e9t 2a \u2013 2b = (3k + 1) \u2013 (3k \u2013 1) = 2 \uf0de 2b(2a \u2013 b \u2013 1) = 2 n\u00ean \uf0ec\uf0ef2a\uf02db \uf02d1 \uf03d 1 \uf0db \uf0ec2a\uf02db \uf03d 2 \uf0db \uf0eca \uf02d1 \uf03d 1 \uf0db \uf0eca \uf03d 2 \uf0ed\uf0ef\uf0ee2b \uf03d 2 \uf0ed \uf0ed\uf0eeb \uf03d1 \uf0ed\uf0eeb \uf03d 1 \uf0eeb \uf03d 1 Do \u0111\u00f3 \uf0ec\uf0ef3k \uf02b1 \uf03d 22 \uf0db 2.3k \uf03d 6 \uf0db 3k \uf03d 31 \uf0db k \uf03d1 khi \u0111\u00f3 x = 2 \uf0ee\uf0ef\uf0ed3k \uf02d1 \uf03d 21 T\u1eeb (1) c\u00f3 : 2y = 32 \u2013 1 = 8 = 23 \uf0de y = 3. *N\u1ebfu x l\u1ebb t\u1ee9c l\u00e0 x = 2k + 1 (k \uf0ce N). X\u00e9t 32k+1 \u2013 1 = 3(32k \u2013 1) + 2 = 3(9k \u2013 1) + 2 chia cho 8 d\u01a3 2 v\u00ec (9k \u2013 1k) (9 \u2013 1) \uf0de 2y chia cho 8 d\u01a3 2 \uf0de 2y = 2 \uf0de y = 1 311","Website: tailieumontoan.com Ta c\u00f3 3x \u2013 1 = 21 \uf0de x = 1. V\u1eady t\u1ea5t c\u1ea3 c\u00e1c c\u1eb7p s\u1ed1 nguy\u00ean d\u01a3\u01a1ng (x, y) l\u00e0 (2; 3) v\u00e0 (1 ; 1). 20.27. 4 = x(2x + 3 \u2013 y \u2013 xy) \uf0de 4 x \uf0de x \uf0ce\uf07b\uf02d4; \uf02d2; \uf02d1;1; 2; 4\uf07d (*) T\u1eeb ph\u01a3\u01a1ng tr\u00ecnh \uf0de xy(x + 1) = 2x2 + 3x \u2013 4 = (x + 1)(2x + 1) \u2013 5 \uf0de (x + 1)( \u2013 xy + 2x + 1) = 5 \uf0de 5 (x + 1) \uf0de x \uf0ce\uf07b\uf02d6; \uf02d2; 0; 4\uf07d (**) T\u1eeb (*) v\u00e0 (**) \uf0de x \uf0ce\uf07b\uf02d2; 4\uf07d . V\u1edbi x = \u2013 2 th\u00ec y = \u2013 1. V\u1edbi x = 4 th\u00ec y = 2 V\u1eady c\u00f3 hai c\u1eb7p (x; y) th\u1ecfa m\u00e3n l\u00e0 (\u2013 2 ; \u2013 1) v\u00e0 (4 ; 2). 20.28. 3x2 \u2013 2y2 \u2013 5xy + x \u2013 2y \u2013 7 = 0 \uf0db 3x2 \u2013 6xy + xy \u2013 2y2 + x \u2013 2y = 7 \uf0db 3x(x \u2013 2y) + y(x \u2013 2y) + (x \u2013 2y) = 7 \uf0db (x \u2013 2y)(3x + y + 1) = 7. Ta c\u00f3 7 = 7 . 1 = 1 . 7 = (\u2013 7). (\u2013 1) = (\u2013 1). (\u2013 7) Do \u0111\u00f3 ta x\u00e9t tr\u01a3\u1eddng h\u1ee3p sau : x \u2013 2y = 7 (*) v\u00e0 3x + y + 1 = 1 (**) T\u1eeb x \u2013 2y = 7 \uf0db x = 7 + 2y thay v\u00e0o (**) ta c\u00f3 3(7 + 2y) + y + 1 = 1 \uf0db 21 + 7y = 0 \uf0db y = \u2013 3 thay y = \u2013 3 v\u00e0o (*) ta c\u00f3 x + 6 = 7 \uf0db x = 1. T\u01a3\u01a1ng t\u1ef1 v\u1edbi c\u00e1c tr\u01a3\u1eddng h\u1ee3p kh\u00e1c ta kh\u00f4ng t\u00ecm \u0111\u01a3\u1ee3c x ; y nguy\u00ean. V\u1eady nghi\u1ec7m nguy\u00ean c\u1ee7a ph\u01a3\u01a1ng tr\u00ecnh \u0111\u00e3 cho l\u00e0 (x ; y) = (1 ; \u2013 3). Ch\u01a3\u01a1ng IV. B\u1ea4T PH\u01a2\u01a0NG TR\u00ccNH B\u1eacC NH\u1ea4T M\u1ed8T \u1ea8N Chuy\u00ean \u0111\u1ec1 21. B\u1ea4T \u0110\u1eb2NG TH\u1ee8C 21.1. a) V\u1edbi ab > 0 . Ta c\u00f3 (a \u2013 b)2 \uf0b3 0 \uf0db a2 + b2 \uf0b3 2ab . Chia hai v\u1ebf c\u1ee7a b\u1eaft \u0111\u1eb3ng th\u1ee9c cho ab > 0 ta c\u00f3 a2 \uf02b b2 \uf0b3 2ab \uf0db a \uf02b b \uf0b3 2 . D\u1ea5u \u201c =\u201d x\u1ea3y ra \uf0db a ab ab ba =b * V\u1edbi ab < 0 . Ta c\u00f3 (a + b)2 \uf0b3 0 \uf0db a2 + b2 \uf0b3 \u2013 2ab . Chia hai v\u1ebf c\u1ee7a b\u1ea5t \u0111\u1eb3ng th\u1ee9c cho ab < 0 ta c\u00f3 a2 \uf02b b2 \uf0a3 \uf02d2ab \uf0db a \uf02b b \uf0a3 \uf02d2 . D\u1ea5u \u201c =\u201d x\u1ea3y ra \uf0db a = \u2013 b ab ab ba b) Ch\u1ee9ng minh : T\u1eeb (a \u2013 b)2 + (b \u2013 c)2 + (c \u2013 a)2 \uf0b3 0 \uf0de 2(a2 + b2 + c2) \uf0b3 2ab + 2ac + 2bc \uf0de 3(a2 + b2 + c2) \uf0b3 a2 + b2 + c2 + 2ab + 2ac + 2bc . \uf0de 3(a2 + b2 + c2) \uf0b3 (a + b + c)2 Chia 2 v\u1ebf c\u1ee7a b\u1ea5t \u0111\u1eb3ng th\u1ee9c n\u00e0y cho 9 ta c\u00f3 \u0111pcm. D\u1ea5u \u201c = \u201dx\u1ea3y ra \uf0db a = b = c. c) X\u00e9t hi\u1ec7u \uf0e6 a \uf02b b \uf0f63 \uf02d a3 \uf02b b3 \uf03d a3 \uf02b 3a2 b\uf02b 3ab2 \uf02b b3 \uf02d 4a3 \uf02d 4b3 \uf0e8\uf0e7 2 \uf0f7\uf0f8 2 8 312","Website: tailieumontoan.com \uf03d \uf02d3a2(a \uf02d b) \uf02b 3b2(a \uf02d b) \uf03d \uf02d3(a \uf02d b)(a2 \uf02d b2) \uf03d \uf02d3(a \uf02d b)2(a \uf02b b) \uf0a3 0 v\u1edbi a, b \uf0b3 0 8 88 D\u1ea5u \u201c = \u201dx\u1ea3y ra \uf0db a = b. 21. 2. a) Ta c\u00f3 n\u1ebfu nh\u00e2n, chuy\u1ec3n v\u1ebf, t\u00e1ch 3 = 1 + 1 + 1 th\u00ec xu\u1ea5t hi\u1ec7n a2 \u2013 2a + 1 = (a \u2013 1)2 .... Do \u0111\u00f3 : ta c\u00f3 a2 + b2 + c2 + 3 \uf0b3 2(a + b + c) \uf0db a2 \u2013 2a + 1 + b2 \u2013 2b + 1 + c2 \u2013 2c + 1 \uf0b3 0 \uf0db (a \u2013 1)2 + (b \u2013 1)2 + (c \u2013 1)2 \uf0b3 0 \u0111\u00fang. D\u1ea5u \u201c = \u201dx\u1ea3y ra \uf0db a = b = c = 1 b) V\u1ebf ph\u1ea3i c\u00f3 ab + ac + ad. N\u1ebfu nh\u00e2n 4 v\u00e0o hai v\u1ebf, chuy\u1ec3n v\u1ebf v\u00e0 t\u00e1ch 4a2 = a2 + a2 + a2 + a2 k\u1ebft h\u1ee3p v\u1edbi c\u00e1c h\u1ea1ng t\u1eed kh\u00e1c s\u1ebd xu\u1ea5t hi\u1ec7n h\u1eb1ng \u0111\u1eb3ng th\u1ee9c. Do \u0111\u00f3 Nh\u00e2n hai v\u1ebf v\u1edbi 4 ta \u0111\u01a3\u1ee3c 4a2 + 4b2 + 4c2 + 4d2 \uf0b3 4ab +4ac + 4ad \uf0db (a \u2013 2b)2 + (a \u2013 2c)2 + (a \u2013 2d)2 + a2 \uf0b3 0 \u0111\u00fang. D\u1ea5u \u201c = \u201dx\u1ea3y ra \uf0db a = b = c = d = 0 c) Nh\u1eadn x\u00e9t: ab = 2. a .b ; ac = 2. a .c; ... do \u0111\u00f3 ta ngh\u0129 t\u1edbi vi\u1ec7c t\u00e1ch a2 th\u00e0nh a2 \uf02b a2 \uf02b a2 \uf02b a2 \u0111\u1ec3 gh\u00e9p 22 4444 v\u1edbi b2, c2 , d2 , e2. Ta c\u00f3 a2 + b2 + c2 + d2 + e2 \uf0b3 a(b + c + d + e) \uf0db \uf0e6 a2 \uf02d ab \uf02b b2 \uf0f6 \uf02b \uf0e6 a2 \uf02d ac \uf02b c2 \uf0f6 \uf02b \uf0e6 a2 \uf02d ad \uf02b d2 \uf0f6 \uf02b \uf0e6 a2 \uf02d ae \uf02b e2 \uf0f6 \uf0b3 0 \uf0e7 4 \uf0f7 \uf0e7 4 \uf0f7 \uf0e7 4 \uf0f7 \uf0e7 4 \uf0f7 \uf0e8 \uf0f8\uf0e8 \uf0f8\uf0e8 \uf0f8\uf0e8 \uf0f8 \uf0db \uf0e6 a \uf02d b \uf0f62 \uf02b \uf0e6 a \uf02d c \uf0f62 \uf02b \uf0e6 a \uf02d d \uf0f62 \uf02b \uf0e6 a \uf02d e \uf0f62 \uf0b3 0 \u0111\u00fang. \uf0e8\uf0e7 2 \uf0f8\uf0f7 \uf0e7\uf0e8 2 \uf0f7\uf0f8 \uf0e7\uf0e8 2 \uf0f7\uf0f8 \uf0e8\uf0e7 2 \uf0f8\uf0f7 *Ch\u00fa \u00fd: C\u00e1ch kh\u00e1c : N\u1ebfu nh\u00e2n hai v\u1ebf v\u1edbi 4 ta bi\u1ebfn \u0111\u1ed5i t\u01a3\u01a1ng \u0111\u01a3\u01a1ng th\u00e0nh (a \u2013 2b)2 + (a \u2013 2c)2 + (a \u2013 2d)2 + (a \u2013 2e)2 \uf0b3 0 \u0111\u00fang. d) V\u1edbi a, b, c, d > 0 , \u00e1p d\u1ee5ng b\u1ea5t \u0111\u1eb3ng th\u1ee9c Cauchy : a2 + b2 \uf0b3 2ab ; c2 + d2 \uf0b3 2cd do \u0111\u00f3 a2 + b2 + c2 + d2 + ab + cd \uf0b3 3(ab + cd) do abcd = 1 \uf0de ab + cd = ab + 1 \uf0b3 2 . Ta c\u00f3 \u0111pcm. ab 21.3 a) Nh\u1eadn x\u00e9t a3 \uf02d b3 \uf03d a2 \uf02db ... Do \u0111\u00f3 b\u1ea5t \u0111\u1eb3ng th\u1ee9c bi\u1ebfn \u0111\u1ed5i th\u00e0nh a2 \uf02b b2 \uf02b c2 \uf0b3 b \uf02b c \uf02b a \uf0db ab2 b2 a b2 c2 a2 a b c 2 \uf0e6 a \uf0f62 \uf02b 2 \uf0e6 b \uf0f62 \uf02b 2 \uf0e6 c \uf0f62 \uf0b3 2. b \uf02b 2. c \uf02b 2. a \uf0e7\uf0e8 b \uf0f8\uf0f7 \uf0e8\uf0e7 c \uf0f8\uf0f7 \uf0e7\uf0e8 a \uf0f8\uf0f7 a b c \u00c1p d\u1ee5ng b\u1ea5t \u0111\u1eb3ng th\u1ee9c x2 + y2 \uf0b3 2xy c\u00f3 \uf0e6 a \uf0f62 \uf02b \uf0e6 b \uf0f62 \uf0b3 2. a .b \uf03d 2. a . X\u00e9t t\u01a3\u01a1ng t\u1ef1 r\u1ed3i c\u1ed9ng v\u1ebf v\u1edbi v\u1ebf c\u00e1c \uf0e7\uf0e8 b \uf0f7\uf0f8 \uf0e8\uf0e7 c \uf0f8\uf0f7 b c c b\u1ea5t \u0111\u1eb3ng th\u1ee9c c\u00f9ng chi\u1ec1u ta \u0111\u01a3\u1ee3c \u0111pcm. b) V\u00ec a, b, c > 0 n\u00ean a + b + c > a + b > 0. D\u00f9ng ph\u01a3\u01a1ng ph\u00e1p l\u00e0m tr\u1ed9i 313","Website: tailieumontoan.com \uf0de a \uf03c a . T\u01a3\u01a1ng t\u1ef1 b \uf03c b v\u00e0 c \uf03c c . C\u1ed9ng v\u1ebf v\u1edbi v\u1ebf ba b\u1ea5t \u0111\u1eb3ng th\u1ee9c c\u00f9ng a\uf02bb\uf02bc a\uf02bb a\uf02bb\uf02bc b\uf02bc a\uf02bb\uf02bc c\uf02ba chi\u1ec1u ta \u0111\u01a3\u1ee3c a \uf02b b \uf02b c \uf03e a\uf02bb\uf02bc \uf03d1. a\uf02bb b\uf02bc c\uf02ba a\uf02bb\uf02bc 21. 4. a) Bi\u1ebfn \u0111\u1ed5i t\u01a3\u01a1ng \u0111\u01a3\u01a1ng : \uf022 x, y > 0 1 \uf0a31\uf02b1 \uf0db 1 \uf0a3 x\uf02by \uf0db (x + y)2 \uf0b3 4xy x \uf02b y 4x 4y x \uf02b y 4xy \uf0db x2 + 2xy + y2 \uf0b3 4xy \uf0db x2 \u2013 2xy + y2 \uf0b3 0 \uf0db (x \u2013 y)2 \uf0b3 0 \u0111\u00fang. D\u1ea5u \u201c = \u201dx\u1ea3y ra \uf0db x = y . b) \u00c1p d\u1ee5ng b\u1ea5t \u0111\u1eb3ng th\u1ee9c v\u1eeba ch\u1ee9ng minh ta c\u00f3 : 1 \uf0a3 1 \uf02b 1 \uf03d 1 \uf02b 1 \uf0a3 1 \uf02b 1 \uf02b 1 . (1) 2a \uf02b b \uf02b c 8a 4(b \uf02b c) 8a 4b \uf02b 4c 8a 16b 16c T\u01a3\u01a1ng t\u1ef1 1 \uf0a3 1 \uf02b 1 \uf02b 1 . (2) 2b \uf02b c \uf02b a 8b 16c 16a 1 \uf0a31\uf02b 1 \uf02b 1 . (3) 2c \uf02b a \uf02b b 8c 16a 16b C\u1ed9ng v\u1ebf v\u1edbi v\u1ebf c\u1ee7a ba b\u1ea5t \u0111\u1eb3ng th\u1ee9c c\u00f9ng chi\u1ec1u (1) ; (2) ; (3) ta \u0111\u01a3\u1ee3c : 1 \uf02b 1 \uf02b 1 \uf0a3 1 \uf02b 1 \uf02b 1 hay 2a \uf02b b \uf02b c 2b \uf02b c \uf02b a 2c \uf02b a \uf02b b 4a 4b 4c 4 \uf02b 4 \uf02b 4 \uf0a31\uf02b1\uf02b1 2a \uf02b b \uf02b c 2b \uf02b c \uf02b a 2c \uf02b a \uf02b b a b c D\u1ea5u \u201c = \u201dx\u1ea3y ra \uf0db a = b = c 21.5. a) X\u00e9t hi\u1ec7u a3 + b3 + abc \u2013 ab(a + b + c) = (a + b)(a \u2013 b)2 \uf0b3 0 b) X\u00e9t hi\u1ec7u a3 + b3 + c3 \u2013 3abc = (a + b + c)(a2 + b2 + c2 \u2013 ab \u2013 ac \u2013 bc) = (a + b + c). (a \uf02d b)2 \uf02b (b \uf02d c)2 \uf02b (c \uf02d a)2 \uf0b3 0 2 c) Bi\u1ebfn \u0111\u1ed5i th\u00e0nh 4(a3 + b3) \u2013 (a + b)3 + 4(b3 + c3) \u2013 (b + c)3 + 4(c3 + a3) \u2013 (c + a)3 \uf0b3 0 X\u00e9t 4(a3 + b3) \u2013 (a + b)3 = (a + b)[4(a2 \u2013 ab + b2) \u2013 (a + b)2 ] = 3(a + b)(a \u2013 b)2 \uf0b3 0. T\u01a3\u01a1ng t\u1ef1 v\u1edbi 4(b3 + c3) \u2013 (b + c)3 v\u00e0 4(c3 + a3) \u2013 (c + a)3 ta suy ra \u0111pcm. 21.6. a)Vai tr\u00f2 a, b, c nh\u01a3 nhau, kh\u00f4ng m\u1ea5t t\u1ed5ng qu\u00e1t gi\u1ea3 s\u1eed a \uf0b3 b \uf0b3 c > 0. Bi\u1ebfn \u0111\u1ed5i b\u1ea5t \u0111\u1eb3ng th\u1ee9c \u0111\u00e3 cho \u0111\u01a3\u1ee3c b\u1ea5t \u0111\u1eb3ng th\u1ee9c t\u01a3\u01a1ng \u0111\u01a3\u01a1ng : a3 + b3 + c3+ 3abc \u2013 a2(b + c) \u2013 b2(c + a) \u2013 c2(a + b) \uf0b3 0 314","Website: tailieumontoan.com \uf0db a3 + b3 + c3+ 3abc \u2013 a2b \u2013 a2c \u2013 b2c \u2013 b2a \u2013 c2a \u2013 c2 b) \uf0b3 0 \uf0db a2(a \u2013 b) + b2(b \u2013 a) + c(2ab \u2013 a2 \u2013 b2) + c(c2 \u2013 bc + ab \u2013 ac) \uf0b3 0 \uf0db (a \u2013 b)(a2 \u2013 b2) \u2013 c(a \u2013 b)2 + c(c \u2013 a)(c \u2013 b) \uf0b3 0 \uf0db (a \u2013 b)2(a + b \u2013 c) + c( a \u2013 c)(b \u2013 c) \uf0b3 0 Hi\u1ec3n nhi\u00ean \u0111\u00fang v\u00ec a \uf0b3 b ; a + b > c ; a \uf0b3 c b \uf0b3 c >0 b) Tr\u01a3\u1edbc h\u1ebft ta ch\u1ee9ng minh v\u1edbi x, y, k l\u00e0 c\u00e1c s\u1ed1 d\u01a3\u01a1ng v\u00e0 x \uf03c 1 th\u00ec y x \uf03c x \uf02b k .Th\u1eadt v\u1eady x\u00e9t hi\u1ec7u x \uf02d x \uf02b k \uf03d k(x \uf02d y) \uf03c 0 do y(y + k) > 0 v\u00e0 y y\uf02bk y y \uf02b k y(y \uf02b k) x \u2013 y < 0 (do gi\u1ea3 thi\u1ebft x < y). Do a < b + c ; b < c + a ; c < a + b n\u00ean ta c\u00f3 : a \uf03c a\uf02ba ; b \uf03c b\uf02bb ; c \uf03c c\uf02bc b\uf02bc b\uf02bc\uf02ba c\uf02ba c\uf02ba\uf02bb a\uf02bb a\uf02bb\uf02bc C\u1ed9ng v\u1ebf v\u1edbi v\u1ebf ba b\u1ea5t \u0111\u1eb3ng th\u1ee9c c\u00f9ng chi\u1ec1u ta \u0111\u01a3\u1ee3c : a \uf02b b \uf02b c \uf03c 2(a \uf02b b \uf02b c) \uf03d 2. T\u1eeb (1) v\u00e0 (2) \uf0de b\uf02bc c\uf02ba a\uf02bb a\uf02bb\uf02bc 21.7. a) Nh\u1eadn x\u00e9t : n\u1ebfu nh\u00e2n (2x \u2013 3) v\u1edbi (2x \u2013 10) v\u00e0 (2x \u2013 6) v\u1edbi (2x \u2013 7) s\u1ebd c\u00f9ng xu\u1ea5t hi\u1ec7n 4x2 \u2013 26x . Do \u0111\u00f3 c\u00f3 th\u1ec3 \u0111\u1eb7t bi\u1ebfn ph\u1ee5 . Bi\u1ebfn \u0111\u1ed5i v\u1ebf tr\u00e1i (2x \u2013 3)(2x \u2013 6)(2x \u2013 7)(2x \u2013 10) + 36 = (4x2 \u2013 26x + 30)(4x2 \u2013 26x + 42) + 36 \u0110\u1eb7t 4x2 \u2013 26x + 36 = y ta c\u00f3 (y \u2013 6)(y + 6) + 36 = y2 \u2013 36 + 36 = y2 \uf0b3 0 . D\u1ea5u \u201c=\u201d x\u1ea3y ra \uf0db y = 0 \uf0db 4x2 \u2013 26x + 36 = 0 \uf0db 2x2 \u2013 13x + 18 = 0 (x \u2013 2)(2x \u2013 9) = 0 \uf0db x = 2 ; x = 4,5 . b) Ta c\u00f3 M = x2013(x3 \u2013 1) + x(x3 \u2013 1) + 1 V\u1edbi x \uf0b3 1 n\u00ean x3 \uf0b3 1 \uf0de x3\u2013 1 \uf0b3 0 ; x2013 > 1 do \u0111\u00f3 M > 0 . (1) * V\u1edbi x < 1 ta c\u00f3 M = x2016 + x4(1 \u2013 x2009) + (1 \u2013 x) Do 1 > x n\u00ean 1 > x2009 hay 1 \u2013 x2009 > 0 ; 1 \u2013 x > 0 ; x2016 > 0; x4 > 0 n\u00ean M > 0. (2). \u0111pcm. 21.8. Ta c\u00f3 : a\uf02bb\uf02b b\uf02b c\uf02b c\uf02b a \uf03d \uf0e6 a \uf02b b\uf0f6 \uf02b \uf0e6 a \uf02b c\uf0f6 \uf02b\uf0e8\uf0e7\uf0e6 b\uf02b c\uf0f6 \uf0b32 \uf02b2 \uf02b2 \uf03d6. (1) c a b \uf0e8\uf0e7 b a \uf0f7\uf0f8 \uf0e8\uf0e7 c a \uf0f8\uf0f7 c b\uf0f7\uf0f8 Theo ch\u1ee9ng minh \u1edf v\u00ed d\u1ee5 8 th\u00ec : a \uf02b b \uf02b c \uf0b3 3. (2) b\uf02b c c\uf02b a a\uf02b b 2 T\u1eeb (1) v\u00e0 (2) suy ra \u0111pcm . D\u1ea5u \u201c=\u201d x\u1ea3y ra khi a = b = c. C\u00e1ch gi\u1ea3i kh\u00e1c : 315","Website: tailieumontoan.com \u0110\u1eb7t A = A \uf03d a \uf02b b \uf02b c \uf02b a \uf02b b \uf02b b \uf02b c \uf02b c \uf02b a b\uf02bc c\uf02ba a\uf02bb c a b Ta c\u00f3 2A = 2a \uf02b 2b \uf02b 2c \uf02b 2 \uf0e6 a \uf02b b \uf02b b\uf02b c \uf02b c\uf02ba \uf0f6 b\uf02bc c\uf02ba a\uf02bb \uf0e8\uf0e7 c a b \uf0f7\uf0f8 = 2a \uf02b 2b \uf02b 2c \uf02b \uf0e6 1\uf02b 3\uf0f6 \uf0e6 a\uf02b b\uf02b b\uf02b c\uf02b c\uf02b a\uf0f6 b\uf02bc c\uf02b a a\uf02b b \uf0e8\uf0e7 2 2\uf0f7\uf0f8 \uf0e7\uf0e8 c a b \uf0f8\uf0f7 = \uf0e6 2a \uf02b b\uf02b c \uf0f6 \uf02b \uf0e6 2b \uf02b c\uf02b a \uf0f6 \uf02b \uf0e6 2c \uf02b a\uf02b b\uf0f6 \uf02b 3\uf0e6 a\uf02b b\uf02b a\uf02b c\uf02b b\uf02b c\uf0f6 \uf0e7\uf0e8 b\uf02bc 2a \uf0f7\uf0f8 \uf0e8\uf0e7 c\uf02b a 2b \uf0f8\uf0f7 \uf0e8\uf0e7 a\uf02b b 2c \uf0f8\uf0f7 2\uf0e7\uf0e8 b a c a c b\uf0f8\uf0f7 \u00c1p d\u1ee5ng b\u00e0i to\u00e1n : v\u1edbi x > 0 th\u00ec x + 1 \uf0b3 2 ta c\u00f3 : x 2A \uf0b3 2 + 2 + 2 + 3 (2 + 2 + 2) = 6 +9 = 15 \uf0de A \uf0b3 15 . 22 D\u1ea5u \u201c = \u201d x\u1ea3y ra \uf0db a = b = c. * C\u1ea7n tr\u00e1nh sai l\u1ea7m sau \u0111\u00e2y khi gi\u1ea3i b\u00e0i to\u00e1n n\u00e0y : A = a \uf02b b \uf02b c \uf02b a\uf02bb \uf02b b\uf02bc \uf02bc\uf02ba = b\uf02bc c\uf02ba a\uf02bb c a b \uf0e6 a \uf02b b \uf02b c \uf0f6 \uf02b \uf0e6 b \uf02b c\uf02b a \uf0f6 \uf02b \uf0e6 c \uf02b b\uf02b c\uf0f6 \uf0e8\uf0e7 \uf02b a \uf0f8\uf0f7 \uf0e7\uf0e8 c\uf02b b \uf0f8\uf0f7 \uf0e7\uf0e8 a\uf02b b a \uf0f7\uf0f8 b c a Do x + 1 \uf0b3 2 v\u1edbi x > 0 n\u00ean A \uf0b3 2 + 2 + 2 = 6 k\u1ebft qu\u1ea3 sai. Sai l\u1ea7m \u1edf ch\u1ed7 n\u1ebfu x\u00e9t ri\u00eang t\u1eebng c\u1eb7p th\u00ec \u0111\u00fang x nh\u01a3ng x\u00e9t \u0111\u1ed3ng th\u1eddi c\u1ea3 ba c\u1eb7p s\u1ed1 th\u00ec d\u1ea5u \u0111\u1eb3ng th\u1ee9c kh\u00f4ng th\u1ec3 x\u1ea3y ra v\u00ec khi \u1ea5y a = b + c ; b = c + a; c = a + b \uf0de a + b + c = 2( a + b + c) v\u00f4 l\u00fd. 21.9. Bi\u1ebfn \u0111\u1ed5i th\u00e0nh \uf0e9\uf0eb\uf028 x \uf02b y\uf029 \uf02b \uf028 y \uf02b z \uf029 \uf02b \uf028z \uf02b x \uf029\uf0f9\uf0fb \uf0e6 x 1 y \uf02b y 1 z \uf02b z 1 x \uf0f6 \uf0b3 9 \uf0e7 \uf02b \uf02b \uf02b \uf0f7 \uf0e8 \uf0f8 \u0110\u1eb7t x+y=a;y+z=b;z+x=c ta \u0111\u01a3\u1ee3c (a \uf02b b \uf02b c) \uf0e6 1 \uf02b 1 \uf02b 1 \uf0f6 \uf0b3 9 . (B\u1ea1n \u0111\u1ecdc t\u1ef1 l\u00e0m \uf0e7\uf0e8 a b c \uf0f8\uf0f7 t\u01a3\u01a1ng t\u1ef1 v\u00ed d\u1ee5 7). 21.10. a) \u0110\u1eb7t B = 1 \uf02b 1 \uf02b 1 \uf02b ... \uf02b 1 1.3 3.5 5.7 2015.2017 Th\u00ec 2B \uf03d 1\uf02d 1 \uf02b 1 \uf02d 1 \uf02b ... \uf02b 1 \uf02d 1 \uf03d 1\uf02d 1 \uf03c 1 \uf0de \u0111pcm 335 2017 2017 2017 b) Nh\u1eadn x\u00e9t v\u1edbi k \uf0ce N* ; k >1 ta c\u00f3 : k \uf02d1 \uf03d k \uf02d1 \uf03d k \uf02d 1 \uf03d 1 \uf02d 1 k! (k \uf02d1)!k (k \uf02d1)!k (k \uf02d1)!k (k \uf02d1)! k! Do \u0111\u00f3 G = 1 \uf02d 1 \uf02b 1 \uf02d 1 \uf02b ... \uf02b 1 \uf02d 1 = 1\uf02d 1 \uf03c 1 1! 2! 2! 3! 2015! 2016! 2016! 316","Website: tailieumontoan.com c) Ta l\u00e0m tr\u1ed9i b\u1eb1ng c\u00e1ch t\u1eeb h\u1ea1ng t\u1eed th\u1ee9 hai c\u1ee7a H ta b\u1edbt m\u1ed7i m\u1eabu s\u1ed1 1 \u0111\u01a1n v\u1ecb . Ta c\u00f3 H < 1 \uf02b 1 \uf02b 1 \uf02b ...\uf02b 1 12 32 \uf02d1 52 \uf02d1 (2n \uf02d1)2 \uf02d1 = 1\uf02b 1 \uf02b 1 \uf02b ... \uf02b 1 2.4 4.6 (2n \uf02d 2).2n \uf03d1\uf02b 1 \uf0e6 1 \uf02d 1 \uf02b 1 \uf02d 1 \uf02b ... \uf02b 1 2 \uf02d 1 \uf0f6 \uf03d 1\uf02b 1 \uf0e6 1 \uf02d 1 \uf0f6 \uf03c 5 2 \uf0e7\uf0e8 2 4 4 6 2n \uf02d 2n \uf0f7\uf0f8 2 \uf0e7\uf0e8 2 2n \uf0f7\uf0f8 4 21.11. Nh\u1eadn x\u00e9t : 20152+ 2015 = 2015(2015 + 1) = 2015.2016 Ta c\u00f3 : 1 \uf03c1 \uf03c1; 2015.2016 20152 \uf02b1 20152 1 \uf03c1 \uf03c1; 2015.2016 20152 \uf02b 2 20152 1 \uf03c1 \uf03c1 ; 2015.2016 20152 \uf02b 3 20152 ... ... ... 1 \uf03d 1 \uf03c1 . 2015.2016 20152 \uf02b 2015 20152 C\u1ed9ng v\u1ebf v\u1edbi v\u1ebf c\u00e1c b\u1ea5t \u0111\u1eb3ng th\u1ee9c tr\u00ean ta c\u00f3 : 2015 \uf03c S \uf03c 2015 Hay 1 \uf03cS\uf03c 1 . 2015.2016 20152 2016 2015 V\u1edbi S\uf03d 1 \uf02b 1 \uf02b 1 \uf02b ... \uf02b 1 . 20152 \uf02b1 20152 \uf02b2 20152 \uf02b3 20152 \uf02b 2015 21.12. Do x, y, z, t \uf0ce Z n\u00ean ta c\u00f3 : x2 + y2 + z2 + t2 + 13 \u2013 xy \u2013 3y \u2013 2z \u2013 6t \uf0a3 0 \uf0db (x2 \u2013 xy + 0,25y2 )+( 0,75y2\u2013 3y + 3 ) + (z2 \u2013 2z + 1) + (t2 \u2013 6t + 9) \uf0a3 0 \uf0db (x \u2013 0,5y)2 + 3(0,5y \u20131)2 + (z \u20131)2 + (t \u2013 3)2 \uf0a3 0 \uf0de (x, y, z, t) = (1; 2; 1; 3) . 21.13. Ta ch\u1ee9ng minh b\u1eb1ng ph\u01a3\u01a1ng ph\u00e1p quy n\u1ea1p to\u00e1n h\u1ecdc : - V\u1edbi n = 2 th\u00ec S2 = 1\uf02b 1 \uf02b 1 \uf03d 19 \uf03e 37 \u0111\u00fang. 3 4 12 24 - Gi\u1ea3 s\u1eed b\u1ea5t \u0111\u1eb3ng th\u1ee9c \u0111\u00fang v\u1edbi n=k ( k \uf0ceN, k \uf0b3 2) t\u1ee9c l\u00e0 Sk > 37 . 24 Ta ch\u1ee9ng minh b\u1ea5t \u0111\u1eb3ng th\u1ee9c \u0111\u00fang v\u1edbi n = k + 1, t\u1ee9c l\u00e0 Sk+1 > 37 . Th\u1eadt v\u1eady : Sk = 24 1\uf02b 1 \uf02b 1 \uf02b 1 \uf02b ... \uf02b 1 \uf03e 37 k \uf02b1 k \uf02b 2 k \uf02b 3 2k 24 317","Website: tailieumontoan.com Sk+1 = 1\uf02b 1\uf02b 1 \uf02b 1 \uf02b ... \uf02b 1 k\uf02b2 k\uf02b3 k\uf02b4 2k \uf02b 2 Do \u0111\u00f3 Sk+1 \u2013 Sk = 1 \uf02b 1 \uf02d 1 \uf03d 1 \uf03e0 2k \uf02b1 2k \uf02b 2 k \uf02b1 2(k \uf02b1)(2k \uf02b1) 37 V\u1eady b\u1ea5t \u0111\u1eb3ng th\u1ee9c \u0111\u00fang \uf022 n \uf0b3 2 . Suy ra Sk+1 > Sk > 24 . 21.14. Do x \uf03c 2 ; y \uf03c 2 n\u00ean x2 < 4 ; y2 < 4 ngh\u0129a l\u00e0 (4 \u2013 x2) > 0 v\u00e0 (4 \u2013 y2) > 0 . . Do \u0111\u00f3 (4 + xy)2 >[2 (x + y)]2 Hay Ta c\u00f3 (4 \u2013 x2)(4 \u2013 y2) > 0 . M\u00e0 (4 \u2013 x2)(4 \u2013 y2) = 16 + x2y2 \u2013 4( x2 + y2) = (16 + 8xy + x2y2) \u2013 4(x2 + 2xy + y2) = (4 + xy)2 \u2013 [2(x + y)]2 > 0 2(x \uf02b y) \uf03c 4 \uf02b xy . 21.15. * Ta s\u1eed d\u1ee5ng ph\u01a3\u01a1ng ph\u00e1p ph\u1ea3n ch\u1ee9ng : Gi\u1ea3 s\u1eed tr\u00e1i l\u1ea1i, trong ba s\u1ed1 a, b, c c\u00f3 \u00edt nh\u1ea5t m\u1ed9t s\u1ed1 kh\u00f4ng d\u01a3\u01a1ng. Do vai tr\u00f2 c\u1ee7a a, b, c nh\u01a3 nhau n\u00ean kh\u00f4ng m\u1ea5t t\u1ed5ng qu\u00e1t ta coi a \uf0a3 0. Nh\u01a3ng theo (1) a ph\u1ea3i kh\u00e1c 0 v\u1eady a < 0 v\u00e0 ta c\u00f3 bc < 0. Theo (3) ab + bc + ca = a(b + c) + bc > 0 n\u00ean a(b + c) > \u2013 bc > 0 M\u00e0 a < 0 n\u00ean b + c < 0 suy ra a + b + c < 0 tr\u00e1i v\u1edbi (2) V\u1eady a, b, c ph\u1ea3i l\u00e0 ba s\u1ed1 d\u01a3\u01a1ng. 21.16. B\u00e0i to\u00e1n c\u00f3 th\u1ec3 gi\u1ea3i b\u1eb1ng ph\u01a3\u01a1ng ph\u00e1p quy n\u1ea1p to\u00e1n h\u1ecdc (b\u1ea1n \u0111\u1ecdc t\u1ef1 ch\u1ee9ng minh). C\u00e1ch kh\u00e1c l\u00e0 ta s\u1eed d\u1ee5ng t\u00ednh ch\u1ea5t b\u1eafc c\u1ea7u, l\u00e0m tr\u1ed9i bi\u1ec3u th\u1ee9c ho\u1eb7c t\u1eebng nh\u00f3m c\u1ee7a bi\u1ec3u th\u1ee9c : \u0110\u1eb7t A \uf03d1\uf02b 1 \uf02b 1 \uf02b ... \uf02b 1 . 2 3 2n \uf02d1 a) Ch\u1ee9ng minh A < n Ta c\u00f3 A \uf03d 1 \uf02b \uf0e6 1 \uf02b 1 \uf0f6 \uf02b \uf0e6 1 \uf02b ...\uf02b 1 \uf0f6 \uf02b \uf0e6 1 \uf02b ... \uf02b 1 \uf0f6 \uf02b ... \uf02b \uf0e6 1 \uf02b ...\uf02b 1\uf0f6 \uf0e8\uf0e7 2 3 \uf0f8\uf0f7 \uf0e8\uf0e7 22 7 \uf0f7\uf0f8 \uf0e8\uf0e7 23 15 \uf0f7\uf0f8 \uf0e7\uf0e8 2n\uf02d1 2n \uf02d1\uf0f8\uf0f7 Ta l\u00e0m tr\u1ed9i \u1edf t\u1eebng nh\u00f3m b\u1eb1ng c\u00e1ch thay c\u00e1c ph\u00e2n s\u1ed1 trong nh\u00f3m b\u1eb1ng ph\u00e2n s\u1ed1 l\u1edbn nh\u1ea5t c\u1ee7a nh\u00f3m ta c\u00f3 : A \uf03c1\uf02b 1 .2 \uf02b 1 .4 \uf02b 1 .8 \uf02b ... \uf02b 1 .2n\uf02d1 \uf03d1\uf02b1\uf02b1\uf02b ... \uf02b1 \uf03d n . 2 22 23 2n\uf02d1 b) Ch\u1ee9ng minh A > n : 2. Ta c\u00f3 A \uf03d1\uf02b 1 \uf02b \uf0e6 1 \uf02b 1 \uf0f6 \uf02b \uf0e6 1 \uf02b ... \uf02b 1 \uf0f6 \uf02b \uf0e6 1 \uf02b ... \uf02b 1 \uf0f6 \uf02b ... \uf02b \uf0e6 1 \uf02b ... \uf02b 1 \uf0f6 \uf02d 1 2 \uf0e8\uf0e7 3 22 \uf0f8\uf0f7 \uf0e7\uf0e8 5 23 \uf0f8\uf0f7 \uf0e7\uf0e8 9 24 \uf0f8\uf0f7 \uf0e7\uf0e8 2n\uf02d1 \uf02b1 2n \uf0f7\uf0f8 2n Thay m\u1ed7i ph\u00e2n s\u1ed1 trong t\u1eebng nh\u00f3m b\u1eb1ng ph\u00e2n s\u1ed1 nh\u1ecf nh\u1ea5t trong nh\u00f3m ta c\u00f3 : A \uf03e 1 \uf02b 1 \uf02b 1 .2 \uf02b 1 .22 \uf02b ... \uf02b 1 .2n\uf02d1 \uf02d 1 \uf03d1\uf02b n\uf02d 1 \uf03e n 2 22 23 2n 2n 2 2n 2 V\u1eady n \uf03c A \uf03c n . 2 318","Website: tailieumontoan.com 21.17. (1 + ab)2 + (1 + cd)2 + (ac)2 + (bd)2 = = 1 + 2ab + a2b2 + 1 + 2cd + c2d2 + 2(ac).(bd) + (ac)2\u2013 2abcd + (bd)2 1 + (1 + ab + cd)2 + (ac \u2013 bd)2 \uf0b3 1. \uf028 \uf029 \uf028 \uf02921.18. a) Do xy = 1 n\u00ean A \uf03d x3 \uf02b y3 \uf03d x4 \uf02b y4 \uf02b x3 \uf02b y3 \uf03d x2 \uf02b y2 2 \uf02d 2x2y2 \uf02b (x \uf02b y) x2 \uf02d xy \uf02b y2 1\uf02b y 1\uf02b x (1\uf02b x)(1\uf02b y) 1\uf02b xy\uf02b x\uf02b y) \uf028 \uf029 \uf028 \uf029= x2 \uf02b y2 2 \uf02d 2 \uf02b (x \uf02b y) x2 \uf02d xy \uf02b y2 . Ta c\u00f3 x2 + y2 \uf0b3 2xy =2 1\uf02b1\uf02b x \uf02b y) Do \u0111\u00f3 A \uf0b3 4 \uf02d 2 \uf02b (x \uf02b y)\uf0282 \uf02d1\uf029 \uf03d 2 \uf02b x \uf02b y \uf03d 1 (\u0111pcm). 2\uf02b x\uf02b y 2\uf02bx\uf02by b) \u00c1p d\u1ee5ng b\u1ea5t \u0111\u1eb3ng th\u1ee9c 4xy \uf0a3 (x + y)2 ta c\u00f3 : 4(3a + b)(2c + a + b) \uf0a3 (3a + b + 2c + a + b)2 = 4(2a + b + c)2 \uf0de (3a + b)(2c + a + b) \uf0a3 (2a + b + c)2 (\u0111pcm). 21.19. \u00c1p d\u1ee5ng b\u1ea5t \u0111\u1eb3ng th\u1ee9c ab \uf0a3 a2 \uf02b b2 ta c\u00f3 2 (x + y)(1 + xy) \uf0a3 (x \uf02b y)2 \uf02b (1\uf02b xy)2 \uf03d (1\uf02b x2 )(1\uf02b y2 ) \uf02b 2x.2y 22 \uf0a3 (1\uf02b x2 )(1\uf02b y2 ) \uf02b (1\uf02b x2 )(1\uf02b y2 ) \uf03d (1 + x2)(1 + y2) \uf0de (\u0111pcm). 2 21.20. X\u00e9t hi\u1ec7u a5 + b5 \u2013 a3b2 \u2013 a2b3 = a3(a2 \u2013 b2) \u2013 b3(a2 \u2013 b2) = (a2 \u2013 b2)(a3\u2013 b3) = (a \u2013 b)2 (a + b)(a2 \u2013 ab + b2) \uf0b3 0 do a + b \uf0b30 ; (a \u2013 b)2 \uf0b30 ; v\u00e0 a2 \u2013 ab + b2 = \uf0e6 a \uf02d b \uf0f62 \uf02b 3b2 \uf0b30 . \uf0e7\uf0e8 2 \uf0f7\uf0f8 4 21.21. Ta c\u00f3 a2 + 2b2 + 3 = (a2 + b2) + (b2 + 1) + 2 \uf0b3 2ab + 2b + 2. T\u01a3\u01a1ng t\u1ef1 b2 + 2c2 + 3 \uf0b3 2bc + 2c + 2. c2 + 2a2 + 3 \uf0b3 2ca + 2a + 2. Do \u0111\u00f3 a2 \uf02b 1 \uf02b 3 \uf02b b2 \uf02b 1 \uf02b 3 \uf02b c2 \uf02b 1 \uf02b 3 \uf0a3 1 \uf0e6 ab 1 \uf02b1 \uf02b bc 1 \uf02b1 \uf02b ca 1 \uf02b1 \uf0f6 2b2 2c2 2a 2 2 \uf0e8\uf0e7 \uf02bb \uf02bc \uf02ba \uf0f7\uf0f8 V\u1edbi abc = 1 th\u00ec 1 \uf02b 1 \uf02b 1 \uf03d 1 \uf02b ab \uf02b b \uf03d 1 \uf0de \u0111pcm. ab \uf02b b \uf02b1 bc \uf02b c \uf02b1 ca \uf02b a \uf02b1 ab \uf02b b \uf02b1 b \uf02b1\uf02b ab 1\uf02b ab \uf02b b 21.22. T\u1eeb gi\u1ea3 thi\u1ebft \uf0de x \u2013 y > 0 \uf0de x \u2013 y = x3 + y3> x3 \u2013 y3= (x \u2013 y)(x2 + xy + y2). V\u1eady x \u2013 y > (x \u2013 y)(x2 + xy + y2) \uf0de x2 + xy + y2 < 1 \uf0de x2 + y2 < 1 (\u0111pcm). 21.23. T\u1eeb 4ab \uf0a3 (a + b)2 \uf0db 1 \uf0a3 1 \uf0e6 1 \uf02b 1\uf0f6 v\u1edbi a >0;b > 0 a\uf02bb 4 \uf0e7\uf0e8 a b \uf0f7\uf0f8 319","Website: tailieumontoan.com Ta c\u00f3 x \uf03d x\uf0a3 x \uf0e6 x 1 y \uf02b x 1 z \uf0f6 2x \uf02b y \uf02b z (x \uf02b y) \uf02b (x \uf02b z) 4 \uf0e7 \uf02b \uf02b \uf0f7 \uf0e8 \uf0f8 T\u01a3\u01a1ng t\u1ef1 v\u1edbi y v\u00e0 z x \uf02b 2y \uf02b z x \uf02b y \uf02b 2z 2x x \uf02b z \uf02b x \uf02b y \uf02b z \uf02b x \uf02b z 2z \uf0a3 x \uf0e6 x 1 y \uf02b x 1 z \uf0f6 \uf02b y \uf0e6 y 1 x \uf02b y 1 z \uf0f6 \uf02b z \uf0e6 z 1 y \uf02b z 1 x \uf0f6 \uf02by 2y y\uf02b 4 \uf0e7 \uf02b \uf02b \uf0f7 4 \uf0e7 \uf02b \uf02b \uf0f7 4 \uf0e7 \uf02b \uf02b \uf0f7 \uf0e8 \uf0f8 \uf0e8 \uf0f8 \uf0e8 \uf0f8 = 1 \uf0e6 x \uf02b y \uf02b x \uf02b z \uf02b y \uf02b z \uf0f6 \uf03d 3 \uf0de (\u0111pcm). 4 \uf0e7 x \uf02b y x \uf02b z y \uf02b z \uf0f7 4 \uf0e8 \uf0f8 21.24. a) x \uf02b 1 \uf0b3 y \uf02b 1 \uf0db x \uf02d y \uf02b 1 \uf02d 1 \uf0b3 0 \uf0db (x \uf02d y)(xy \uf02d1) \uf0b3 0 \u0111\u00fang v\u00ec x \uf0b3 y \uf0b31. xy xy xy b) Do vai tr\u00f2 a, b, c nh\u01a3 nhau, gi\u1ea3 s\u1eed 1\uf0a3 a \uf0a3 b \uf0a3 c \uf0a3 2 ; \u0110\u1eb7t x = b ; y = c v\u1edbi 1 \uf0a3 x; y \uf0a3 2 ; xy \uf0a3 2 \uf0de y \uf0a3 2 . ab x X\u00e9t hi\u1ec7u hai v\u1ebf v\u00e0 \u00e1p d\u1ee5ng k\u1ebft qu\u1ea3 c\u00e2u a) ta c\u00f3 : (a + b + c) \uf0e6 1 \uf02b 1 \uf02b 1 \uf0f6 \uf02d10 \uf03d \uf0e6 x \uf02b 1 \uf0f6 \uf02b \uf0e6 y \uf02b 1 \uf0f6 \uf02b \uf0e6 xy \uf02b 1 \uf0f6 \uf02d 7 \uf0e8\uf0e7 a b c \uf0f8\uf0f7 \uf0e8\uf0e7 x \uf0f7\uf0f8 \uf0e7 y \uf0f7 \uf0e7 xy \uf0f7 \uf0e8 \uf0f8 \uf0e8 \uf0f8 \uf0a3 \uf0e6 x \uf02b 1 \uf0f6 \uf02b \uf0e6 2 \uf02b x \uf0f6 \uf02b \uf0e6 2 \uf02b 1 \uf0f6 \uf02d 7 \uf03d 3x \uf02b 3 \uf02d 9 \uf03d 3(x \uf02d1)(x \uf02d 2) \uf0a3 0 \uf0e7\uf0e8 x \uf0f7\uf0f8 \uf0e8\uf0e7 x 2 \uf0f7\uf0f8 \uf0e8\uf0e7 2 \uf0f8\uf0f7 2 x 2 2x D\u1ea5u \u201c=\u201d x\u1ea3y ra \uf0db x = 1 ho\u1eb7c x = 2 \u0111\u1ed3ng th\u1eddi xy = 2 \uf0db (a, b, c) = (1; 1; 2) ; (1; 2; 2) v\u00e0 c\u00e1c ho\u00e1n v\u1ecb. 21.25. Ta c\u00f3 (a + b)2 \uf0a3 (a + b)2 + (a \u2013 b)2 = 2(a2 + b2) T\u01a3\u01a1ng t\u1ef1 (a2 + b2)2 \uf0a3 2(a4 + b4) \uf0de (a + b)2 (a2 + b2)2 \uf0a3 4(a2 + b2)(a4 + b4) M\u00e0 (a + b)2 = 4 \uf0de (a2 + b2)2 \uf0a3 (a2 + b2)(a4 + b4) \uf0de a2 + b2 \uf0a3 a4 + b4. 21.26. \u00c1p d\u1ee5ng b\u1ea5t \u0111\u1eb3ng th\u1ee9c 1 \uf0a3 1\uf0e6 1 \uf02b 1 \uf0f6 v\u1edbi x > 0 ; y > 0 v\u00e0 s\u1eed d\u1ee5ng gi\u1ea3 thi\u1ebft abc = 1 ta c\u00f3 : \uf02b \uf0e7 x y \uf0f7 x y 4 \uf0e8 \uf0f8 ab 1 \uf02b 2 \uf03d (ab 1 (a\uf02b1) \uf0a3 1 \uf0e6 1 \uf02b a 1\uf0f6 \uf03d 1 \uf0e6 abc \uf02b a 1\uf0f6 hay \uf02ba \uf02b1) \uf02b 4 \uf0e8\uf0e7 ab \uf02b1 \uf02b 1 \uf0f8\uf0f7 4 \uf0e7\uf0e8 ab \uf02b abc \uf02b 1 \uf0f7\uf0f8 ab 1 \uf02b 2 \uf0a3 1 \uf0e6 c c \uf02b a 1\uf0f6 (*); T\u01a3\u01a1ng t\u1ef1 bc 1 \uf02b 2 \uf0a3 1 \uf0e6 a a \uf02b 1\uf0f6 (**) \uf02ba 4 \uf0e7\uf0e8 \uf02b1 \uf02b 1 \uf0f8\uf0f7 \uf02bb 4 \uf0e7\uf0e8 \uf02b1 b \uf02b1\uf0f7\uf0f8 ca 1 \uf02b 2 \uf0a3 1 \uf0e6 b \uf02b c 1 \uf0f6 (***). C\u1ed9ng v\u1ebf v\u1edbi v\u1ebf c\u1ee7a ba b\u1ea5t \u0111\u1eb3ng th\u1ee9c c\u00f9ng chi\u1ec1u (*), (**) v\u00e0 (***) ta c\u00f3 \uf02bc 4 \uf0e7\uf0e8 b \uf02b1 \uf02b1 \uf0f7\uf0f8 1 \uf02b 1 \uf02b 1 \uf0a3 3. ab \uf02b a \uf02b 2 bc \uf02b b \uf02b 2 ca \uf02b c \uf02b 2 4 Chuy\u00ean \u0111\u1ec1 22. B\u1ea4T PH\u01a2\u01a0NG TR\u00ccNH B\u1eacC NH\u1ea4T M\u1ed8T \u1ea8N 22.1. C\u00e1ch 1: Ta gi\u1ea3i b\u1ea5t ph\u01a3\u01a1ng tr\u00ecnh k\u00e9p 320","Website: tailieumontoan.com \uf0ecx \uf02d1 \uf02d x \uf02d 2 \uf03e 4 \uf0ef\uf0ef \uf02d x 3 2 \uf03c 5 4 < x \uf02d1\uf02d x \uf02d 2 < 5 \uf0db \uf0ed x 2 \uf02d \uf0db \uf0ecx \uf03e 23 23 \uf0ef \uf02d1 3 \uf0ed\uf0eex \uf03c 29 \uf0ee\uf0ef 2 C\u00e1c gi\u00e1 tr\u1ecb nguy\u00ean c\u1ee7a x th\u1ecfa m\u00e3n 23 < x < 29 l\u00e0 x \uf0ce\uf07b24;25;26;27;28\uf07d C\u00e1ch 2: 4 < x \uf02d1 \uf02d x \uf02d 2 < 5 \uf0db 24 < 3(x \u2013 1) \u2013 2(x \u2013 2) < 30 23 \uf0db 24 < x + 1 < 30 \uf0db 23 < x < 29 v\u00e0 c\u0169ng c\u00f3 k\u1ebft qu\u1ea3 tr\u00ean. 22.2. S\u1eed d\u1ee5ng c\u00e1c quy t\u1eafc bi\u1ebfn \u0111\u1ed5i b\u1ea5t ph\u01a3\u01a1ng tr\u00ecnh \u0111\u01a3a c\u00e1c b\u1ea5t ph\u01a3\u01a1ng tr\u00ecnh v\u1ec1 d\u1ea1ng ax + b > 0 a) 3x \u2013 2 > 5(x \u2013 2) + 2(3 \u2013 x) \uf0db 0x > \u2013 2 nghi\u1ec7m \u0111\u00fang \uf022 x. Nghi\u1ec7m c\u1ee7a b\u1ea5t ph\u01a3\u01a1ng tr\u00ecnh l\u00e0 x \uf0ce R b) 5(x + 2)2 < (2x + 3)(2x \u2013 3) + (x \u2013 5)2 + 30x \uf0db 0x < \u2013 4 B\u1ea5t ph\u01a3\u01a1ng tr\u00ecnh v\u00f4 nghi\u1ec7m. c) 4(2,5x2 + 1) \uf0b3 9(x + 3)(x \u2013 3) + ( 2 \u2013 x)2 + 1 \uf0db 4x \uf0b3 \u2013 80 \uf0db x \uf0b3 \u2013 20 d) Th\u00eam v\u00e0o hai v\u1ebf \u2013 64 l\u00e0m xu\u1ea5t hi\u1ec7n d\u1ea1ng x3 \u2013 43 \u1edf v\u1ebf tr\u00e1i v\u00e0 2(x \u2013 4) \u1edf v\u1ebf ph\u1ea3i. Ta c\u00f3 x3 \uf0a3 2x + 56 \uf0db x3 \u2013 64 \uf0a3 2x + 56 \u2013 64 \uf0db ( x \u2013 4)(x2 + 4x + 16) \u2013 2(x \u2013 4) \uf0a3 0 \uf0db (x \u2013 4)( x2 + 4x + 14) \uf0a3 0 Do x2 + 4x + 14 = (x + 2)2 + 10 > 0 , \uf022 x n\u00ean ta c\u00f3 x \u2013 4 \uf0a3 0 hay x \uf0a3 4. 22.3. x \uf02b1 \uf02b x \uf02b 2 \uf03e x \uf02b 3 \uf02b x \uf02b 4 23 45 \uf0db 30(x + 1) + 20(x + 2) > 15(x + 3) + 12 (x + 4) \uf0db 23x > 23 \uf0db x > 1. * Ch\u00fa \u00fd : d) Nh\u1eadn x\u00e9t: n\u1ebfu th\u00eam (\u2013 1) v\u00e0o m\u1ed7i h\u1ea1ng t\u1eed \u1edf hai v\u1ebf r\u1ed1i quy \u0111\u1ed3ng t\u1eebng c\u1eb7p ta th\u1ea5y xu\u1ea5t hi\u1ec7n nh\u00e2n t\u1eed chung l\u00e0 (x \u2013 1). Do \u0111\u00f3 c\u00f2n c\u00e1ch sau : x \uf02b1 \uf02b x \uf02b 2 \uf03e x \uf02b3 \uf02b x \uf02b 4 \uf0db \uf0e6 x \uf02b 1 \uf02d1\uf0f7\uf0f8\uf0f6 \uf02b \uf0e6 x \uf02b 2 \uf02d1\uf0f6\uf0f7\uf0f8 \uf03e \uf0e6 x \uf02b3 \uf02d1\uf0f6\uf0f8\uf0f7 \uf02b \uf0e6 x \uf02b 4 \uf02d1\uf0f7\uf0f8\uf0f6 2 3 4 5 \uf0e8\uf0e7 2 \uf0e8\uf0e7 3 \uf0e8\uf0e7 4 \uf0e8\uf0e7 5 \uf0db \uf028 x \uf02d1\uf029 \uf0e6 1 \uf02b 1 \uf02d 1 \uf02d 1 \uf0f6 \uf03e 0 \uf0db x>1 do \uf0e6 1 \uf02b 1 \uf02d 1 \uf02d 1 \uf0f6 \uf03e 0 \uf0e7\uf0e8 2 3 4 5 \uf0f7\uf0f8 \uf0e8\uf0e7 2 3 4 5 \uf0f7\uf0f8 22.4. a) Gi\u1ea3i b\u1ea5t ph\u01a3\u01a1ng tr\u00ecnh (1) ta c\u00f3 x > 4,6. Gi\u1ea3i b\u1ea5t ph\u01a3\u01a1ng tr\u00ecnh (2) ta c\u00f3 x > 5 . Gi\u00e1 tr\u1ecb x > 4,6 th\u1ecfa 12 m\u00e3n c\u1ea3 hai b\u1ea5t ph\u01a3\u01a1ng tr\u00ecnh. b) Gi\u1ea3i b\u1ea5t ph\u01a3\u01a1ng tr\u00ecnh (3) ta c\u00f3 x \uf0b3 \uf02d1. Gi\u1ea3i b\u1ea5t ph\u01a3\u01a1ng tr\u00ecnh (4) ta c\u00f3 x < 5 . Gi\u00e1 tr\u1ecb \uf02d1\uf0a3 x \uf03c 5 th\u1ecfa m\u00e3n c\u1ea3 hai b\u1ea5t ph\u01a3\u01a1ng tr\u00ecnh. 321","Website: tailieumontoan.com 22.5. a) Gi\u1ea3i b\u1ea5t ph\u01a3\u01a1ng tr\u00ecnh (1) ta c\u00f3 x > 27 . Gi\u1ea3i b\u1ea5t ph\u01a3\u01a1ng tr\u00ecnh (2) ta c\u00f3 x \uf0a3 2 . Gi\u00e1 tr\u1ecb x = 2 th\u1ecfa 20 m\u00e3n c\u1ea3 hai b\u1ea5t ph\u01a3\u01a1ng tr\u00ecnh. b) Gi\u1ea3i b\u1ea5t ph\u01a3\u01a1ng tr\u00ecnh (3) ta c\u00f3 x \uf0b3 \uf02d3,5 Gi\u1ea3i b\u1ea5t ph\u01a3\u01a1ng tr\u00ecnh (4) ta c\u00f3 x \uf0a3 4 . V\u1eady x \uf0ce\uf07b\uf02d3;\uf02d2;\uf02d1;0;1;2;3;4\uf07d . 22.6. Gi\u1ea3i t\u1eebng b\u1ea5t ph\u01a3\u01a1ng tr\u00ecnh ta c\u00f3 : 3(x \uf02d1) \uf02d 4 \uf03c 3x \uf02d11 \uf0db 15x \u2013 15 \u2013 20 < 3x \u2013 11 \uf0db 12x < 24 \uf0db x < 2 5 3x \uf02d11 \uf03c 2 \uf02b 5x \uf02d 2 \uf0db 12x \u2013 44 <10 + 25x \u2013 40 \uf0db \u2013 13x < 14 \uf0db x \uf03e \uf02d 14 54 13 Do \u0111\u00f3 \uf02d14 \uf03c x \uf03c 2 . C\u00e1c gi\u00e1 tr\u1ecb nguy\u00ean c\u1ee7a x th\u1ecfa m\u00e3n l\u00e0 x \uf0ce\uf07b\uf02d1; 0 ;1\uf07d 13 22.7. Sau khi r\u00fat g\u1ecdn bi\u1ec3u th\u1ee9c A ta gi\u1ea3i b\u1ea5t ph\u01a3\u01a1ng tr\u00ecnh A \uf0a3 \u2013 2 v\u00e0 ph\u01a3\u01a1ng tr\u00ecnh ch\u1ee9a tham s\u1ed1 A > ax . Ta \u0111\u1eb7c bi\u1ec7t l\u01a3u \u00fd \u0110KX\u0110 c\u1ee7a A v\u00e0 bi\u1ec7n lu\u1eadn khi gi\u1ea3i b\u1ea5t ph\u01a3\u01a1ng tr\u00ecnh ch\u1ee9a tham s\u1ed1. a) \u0110KX\u0110 : x \uf0b9 \uf0b1 2,5 A \uf03d 4x2 \uf02d10x \uf02b 25 . 5(2x \uf02b 5)(5 \uf02d 2 x) \uf03d \uf02d(2x \uf02b 5) 5(2x \uf02d 5) 4x2 \uf02d10x \uf02b 25 b) \u0110\u1ec3 A \uf0a3 \u2013 2 ta c\u00f3 \u2013 (2x + 5) \uf0a3 \u2013 2 \uf0db 2x + 5 \uf0b3 2 \uf0db 2x \uf0b3 \u2013 3 \uf0db x \uf0b3 \u2013 1,5. c) A > ax t\u1ee9c l\u00e0 \u2013 2x \u2013 5 > ax \uf0db ax + 2x < \u2013 5 \uf0db (a + 2)x < \u2013 5 N\u1ebfu a > \u2013 2 th\u00ec x < \uf02d5 ; N\u1ebfu a < \u2013 2 th\u00ec x > \uf02d5 ; a\uf02b2 a\uf02b2 N\u1ebfu a = \u2013 2 ta c\u00f3 0x < \u2013 5 v\u00f4 l\u00fd. 22.8. \u0110KX\u0110 x \uf0b9 2,5 ta c\u00f3 a2 \uf02d 4 \uf03d 2 \uf02b a 2x \uf02d 5 \uf0db (a \u2013 2)(a + 2) \u2013 (2 + a)(2x \u2013 5) = 0 ; \uf0db (a + 2)(a \u2013 2x + 3 ) = 0 N\u1ebfu a \uf0b9 \u2013 2 th\u00ec a \u2013 2x + 3 = 0 \uf0db x = a \uf02b 3 2 x > 0 khi a + 3 > 0 \uf0db a > \u2013 3 x < 2 khi a\uf02b3 \uf0db a+3<4 \uf0db a < 1. <2 2 V\u1eady \u0111\u1ec3 nghi\u1ec7m c\u1ee7a ph\u01a3\u01a1ng tr\u00ecnh sau l\u00e0 s\u1ed1 d\u01a3\u01a1ng nh\u01a3ng nh\u1ecf h\u01a1n 2 : \u2013 3 < a < 1 v\u00e0 a \uf0b9 \u2013 2 (N\u1ebfu a = \u2013 2 th\u00ec ta c\u00f3 0x = 0 ph\u01a3\u01a1ng tr\u00ecnh c\u00f3 v\u00f4 s\u1ed1 nghi\u1ec7m do \u0111\u00f3 c\u00f3 v\u00f4 s\u1ed1 nghi\u1ec7m d\u01a3\u01a1ng tr\u1eeb x = 2,5) . 22.9. a) a(x \u2013 a) > 5(x \u2013 5) \uf0db ax \u2013 a2 > 5x \u2013 25 \uf0db (a \u2013 5)x > (a \u2013 5)(a + 5) N\u1ebfu a > 5 th\u00ec nghi\u1ec7m c\u1ee7a ph\u01a3\u01a1ng tr\u00ecnh l\u00e0 x > a + 5 N\u1ebfu a < 5 th\u00ec nghi\u1ec7m c\u1ee7a ph\u01a3\u01a1ng tr\u00ecnh l\u00e0 x < a + 5 322","N\u1ebfu a = 5 th\u00ec b\u1ea5t ph\u01a3\u01a1ng tr\u00ecnh tr\u1edf th\u00e0nh 0x > 0, v\u00f4 nghi\u1ec7m. Website: tailieumontoan.com b) Bi\u1ebfn \u0111\u1ed5i b\u1ea5t ph\u01a3\u01a1ng tr\u00ecnh ta c\u00f3 : x \uf02b (a \uf02b b \uf02b1)x \uf03e b \uf02b 2b \uf0db (a + b + 2)x > 3b aa a * N\u1ebfu a + b + 2 > 0 ; th\u00ec x > 3b a(a \uf02b b \uf02b 2) * N\u1ebfu a + b + 2 < 0; th\u00ec x < 3b a(a \uf02b b \uf02b 2) * N\u1ebfu a + b + 2 = 0 ; th\u00ec 0x > 3b khi \u1ea5y : a N\u1ebfu ab \uf0b3 0 : V\u00f4 nghi\u1ec7m. N\u1ebfu ab < 0 : V\u00f4 s\u1ed1 nghi\u1ec7m. 22.10. Th\u00eam 1 v\u00e0o m\u1ed7i h\u1ea1ng t\u1eed \u1edf hai v\u1ebf r\u1ed3i quy \u0111\u1ed3ng m\u1eabu t\u1eebng c\u1eb7p ta th\u1ea5y xu\u1ea5t hi\u1ec7n nh\u00e2n t\u1eed chung 2x + 2015. Ta c\u00f3 c\u00e1ch gi\u1ea3i : 5x+1015 \uf02b 5x+1000 \uf02b 5x+1 \uf03e 5x \uf02d1 \uf02b 5x \uf02d 2 \uf02b 5x \uf02d10 1000 1015 2014 2016 2017 2025 \uf0db 5x+1015 \uf02b1\uf02b 5x+1000 \uf02b1\uf02b 5x+1 \uf02b1 \uf03e 5x \uf02d1 \uf02b1\uf02b 5x \uf02d 2 \uf02b1\uf02b 5x \uf02d10 \uf02b1 1000 1015 2014 2016 2017 2025 \uf0db \uf0285x+2015\uf029 \uf0e6 1 \uf02b 1 \uf02b 1 \uf02d 1 \uf02d 1 \uf02d 1 \uf0f6 \uf03e 0 \uf0e7\uf0e8 1000 1015 2014 2016 2017 2025 \uf0f7\uf0f8 Do 1 \uf02b 1 \uf02b 1 \uf02d 1 \uf02d 1 \uf02d 1 \uf03e 0 1000 1015 2014 2016 2017 2025 n\u00ean 5x + 2015 > 0 \uf0db 5x > \u2013 2015 \uf0db x > \u2013 403. 22.11. 2A = 2 \uf02b 2 \uf02b 2 \uf02b ...\uf02b 2 \uf03d 1\uf02d 1 \uf02b 1 \uf02d 1 \uf02b 1 \uf02d 1 \uf02b ...\uf02b 1 \uf02d 1 \uf03d 10 1.3 3.5 5.7 9.11 3 3 5 5 7 9 11 11 B \uf03d \uf0e7\uf0e6\uf0e81\uf02b 1 \uf0f7\uf0f6\uf0f8 \uf0e7\uf0e8\uf0e61 \uf02b 1 \uf0f8\uf0f6\uf0f7...\uf0e7\uf0e6\uf0e81 \uf02b 1 \uf0f6 \uf0e7\uf0e8\uf0e61 \uf02b 1 \uf0f6 \uf03d 22 . 32 ..... 92 102 \uf03d 20 1.3 2.4 8.10 \uf0f8\uf0f7 9.11 \uf0f8\uf0f7 1.3 2.4 8.10 9.11 11 2A < 2x < B t\u1ee9c l\u00e0 10 \uf03c 2x \uf03c 20 \uf0db 10 \uf03c 2x \uf03c 20 \uf0db 5 \uf03c x \uf03c 10 11 11 11 11 Do \u0111\u00f3 x \uf0ce\uf07b6 ;7 ; 8 ; 9\uf07d 22.12. Ta l\u1eadp c\u00e1c ph\u01a3\u01a1ng tr\u00ecnh bi\u1ec3u th\u1ecb t\u1ed5ng s\u1ed1 tr\u1eadn v\u00e0 t\u1ed5ng s\u1ed1 \u0111i\u1ec3m, x\u00e9t xem x b\u1ecb ch\u1eb7n b\u1edfi hai gi\u00e1 tr\u1ecb n\u00e0o. T\u1eeb \u0111\u00f3 t\u00ecm ra c\u00e1c gi\u00e1 tr\u1ecb c\u1ee7a x v\u00e0 y, z. * G\u1ecdi s\u1ed1 tr\u1eadn th\u1eafng c\u1ee7a \u0111\u1ed9i \u0111\u00f3 l\u00e0 x, s\u1ed1 tr\u1eadn h\u00f2a l\u00e0 y v\u00e0 s\u1ed1 tr\u1eadn thua l\u00e0 z (x, y, z \uf0ce N ). Ta c\u00f3 x + y + z = 20. (1); \u0111\u1ed3ng th\u1eddi 3.x + 1.y + 0.z = 41. (2) T\u1eeb (2) ta c\u00f3 3x + y = 41 suy ra 3x \uf0a3 41 \uf0db x \uf0a3 41 \uf03d 13 2 33 323","Website: tailieumontoan.com T\u1eeb (1) v\u00e0 (2) \uf0de 2x \u2013 z = 21 \uf0de 2x \uf0b3 21 \uf0db x \uf0b3 21 \uf03d 10 1 22 Nh\u01a3 v\u1eady 10 1 \uf0a3 x \uf0a3 13 2 . Do x \uf0ce N \uf0de x = 11; 12 ; 13. 23 Do x l\u00e0 s\u1ed1 ch\u1eb5n n\u00ean x = 12. T\u1eeb \u0111\u00f3 c\u00f3 3. 12 + y = 41 \uf0de y = 5 v\u00e0 z = 3 22.13. Do [a] l\u00e0 s\u1ed1 nguy\u00ean l\u1edbn nh\u1ea5t kh\u00f4ng v\u01a3\u1ee3t qu\u00e1 a n\u00ean n\u1ebfu [a]= n th\u00ec n l\u00e0 s\u1ed1 nguy\u00ean v\u00e0 0 \uf0a3 a \uf02d n \uf03c1. V\u00ec th\u1ebf \uf0e9 8x \uf02d 3\uf0f9 \uf03d 2x+1 \uf0db \uf0ef\uf0ec0 \uf0a3 8x \uf02d 3 \uf02d (2x \uf02b1) \uf03c 1 \uf0eb\uf0ea 5 \uf0fb\uf0fa \uf0ed5 \uf0ee\uf0ef(2x \uf02b1) \uf0ce Z X\u00e9t 0 \uf0a3 8x \uf02d 3 \uf02d (2x \uf02b1) \uf03c 1 \uf0db 0 \uf0a3 8x \u2013 3 \u2013 10x \u2013 5 < 5 5 \uf0db 0 \uf0a3 \u20132x \u2013 8 < 5 \uf0db 8 \uf0a3 \u20132x < 13 \uf0db \u2013 8 \uf0b3 2x > \u2013 13 \uf0db \u2013 7 \uf0b3 2x + 1 > \u2013 12. Do 2x + 1 \uf0ce Z v\u00e0 2x + 1 l\u00e0 s\u1ed1 l\u1ebb n\u00ean 2x + 1 = \u2013 7 \uf0db x = \u2013 4. 2x + 1 = \u2013 9 \uf0db x = \u2013 5 ; 2x + 1 = \u2013 11 \uf0db x = \u2013 6. V\u1eady x \uf0ce\uf07b\uf02d4;\uf02d5;\uf02d6\uf07d 22.14. x \uf02b1 \uf02d 2 \uf03e x \uf02b 4 \uf02b x \uf02b 5 \uf02b x \uf02b 7 2002 1999 1998 1996 \uf0db x \uf02b1 \uf02b1 \uf03e \uf0e6 x\uf02b4 \uf02b1\uf0f6\uf0f8\uf0f7 \uf02b \uf0e6 x\uf02b5 \uf02b1\uf0f7\uf0f6\uf0f8 \uf02b \uf0e6 x\uf02b7 \uf02b1\uf0f8\uf0f7\uf0f6 2002 \uf0e8\uf0e7 1999 \uf0e8\uf0e7 1998 \uf0e7\uf0e8 1996 \uf0db x \uf02b 2003 \uf03e x \uf02b 2003 \uf02b x \uf02b 2003 \uf02b x \uf02b 2003 2002 1999 1998 1996 \uf0db \uf028x \uf02b 2003\uf029 \uf0e6 1 \uf02d1 \uf02d 1 \uf02d1 \uf0f6 \uf03e 0 \uf0e8\uf0e7 2002 1999 1998 1996 \uf0f7\uf0f8 Do 1 \uf02d 1 \uf02d 1 \uf02d 1 \uf03c 0 n\u00ean x + 2003 < 0 \uf0db x < \u2013 2003. 2002 1999 1998 1996 22.15. * V\u1edbi x \uf0b3 1 th\u00ec x \uf02d1 = x \u2013 1. B\u1ea5t ph\u01a3\u01a1ng tr\u00ecnh tr\u1edf th\u00e0nh x + x \u2013 1 > 5 \uf0db 2x > 6 \uf0db x > 3 (th\u1ecfa m\u00e3n). V\u1eady * V\u1edbi x < 1 th\u00ec x \uf02d1 = 1 \u2013 x B\u1ea5t ph\u01a3\u01a1ng tr\u00ecnh tr\u1edf th\u00e0nh x + 1 \u2013 x > 5 \uf0db 0x > 4 v\u00f4 nghi\u1ec7m. nghi\u1ec7m c\u1ee7a b\u1ea5t ph\u01a3\u01a1ng tr\u00ecnh l\u00e0 x > 3 Chuy\u00ean \u0111\u1ec1 23. B\u1ea4T PH\u01a2\u01a0NG TR\u00ccNH D\u1ea0NG T\u00cdCH, TH\u01a2\u01a0NG 23.1. Bi\u1ebfn \u0111\u1ed5i th\u00e0nh x2 + x \u2013 6 \uf0a3 0 \uf0db (x \u2013 2)(x + 3) \uf0a3 0 C\u00e1ch 1: L\u00fd lu\u1eadn x \u2013 2 \uf0a3 0 v\u00e0 x + 3 \uf0b3 0 (do x + 3 > x \u2013 2 , \uf022 x ) C\u00e1ch 2 : L\u1eadp b\u1ea3ng x\u00e9t d\u1ea5u . 324","Website: tailieumontoan.com Ta \u0111\u1ec1u c\u00f3 k\u1ebft qu\u1ea3 \u2013 3 \uf0a3 x \uf0a3 2. -3 0 2x Bi\u1ec3u di\u1ec5n nghi\u1ec7m tr\u00ean tr\u1ee5c s\u1ed1: 23.2. a) L\u1eadp b\u1ea3ng x\u00e9t d\u1ea5u. Nghi\u1ec7m l\u00e0 x < \uf02d 8 ; 2 \uf03c x \uf03c 2 ho\u1eb7c x > 30 19 9 3 4 b)Nh\u1eadn x\u00e9t : 3x2 \u2013 25x \u2013 50 = (3x + 5)(x \u2013 10) = \u2013 (10 \u2013 x)(3x + 5) M\u1eb7t kh\u00e1c 100 \u2013 x2= (10 \u2013 x)(10 + x) . Do \u0111\u00f3 ta bi\u1ebfn \u0111\u1ed5i BPT \uf0db (10 \u2013 x)(5x \u2013 2001) \u2013 (10 \u2013 x)(3x + 5) \u2013 (10 \u2013 x)(10 + x) < 0 \uf0db (10 \u2013 x)(x \u2013 2016) < 0 Gi\u1ea3i b\u1ea5t ph\u01a3\u01a1ng tr\u00ecnh \u0111\u01a3\u1ee3c \uf0e9x \uf03c 10 \uf0ea\uf0ebx \uf03e 2016 23.3. \u0110\u00e2y l\u00e0 c\u00e1c b\u1ea5t ph\u01a3\u01a1ng tr\u00ecnh b\u1eadc ba v\u00e0 b\u1ed1n. Ta chuy\u1ec3n v\u1ebf r\u1ed3i s\u1eed d\u1ee5ng h\u1ec7 qu\u1ea3 \u0111\u1ecbnh l\u00fd B\u00e9zout (nh\u1ea9m nghi\u1ec7m) \u0111\u1ec3 ph\u00e2n t\u00edch v\u1ebf tr\u00e1i th\u00e0nh nh\u00e2n t\u1eed. a) BPT \uf0db (x \u2013 2)(x \u2013 3)(x \u2013 4) < 0 L\u1eadp b\u1ea3ng x\u00e9t d\u1ea5u t\u00ecm \u0111\u01a3\u1ee3c nghi\u1ec7m : \uf0e93 \uf03c x \uf03c 4 \uf0ea\uf0ebx \uf03c 2 0 2 34 x b) Chuy\u1ec3n v\u1ebf v\u00e0 bi\u1ebfn \u0111\u1ed5i BPT \uf0db (x + 1)(x + 2)(x \u2013 3)(x \u2013 4) \uf0b3 0 \uf0e9x \uf0a3 \uf02d2 L\u1eadp b\u1ea3ng x\u00e9t d\u1ea5u t\u00ecm \u0111\u01a3\u1ee3c nghi\u1ec7m l\u00e0: \uf0ea\uf0ea\uf02d1 \uf0a3 x \uf0a3 3 . Bi\u1ec3u di\u1ec5n nghi\u1ec7m : \uf0ea\uf0ebx \uf0b3 4 -2 -1 0 34 x 23.4. a) Nh\u00e2n 4 v\u00e0o nh\u00e2n t\u1eed th\u1ee9 nh\u1ea5t, nh\u00e2n 2 v\u00e0o nh\u00e2n t\u1eed th\u1ee9 hai v\u00e0 nh\u00e2n 8 v\u00e0o v\u1ebf ph\u1ea3i ta \u0111\u01a3\u1ee3c : BPT \uf0db (8x + 4)(8x + 6)(8x + 5)2 \uf0a3 72 \u0110\u1eb7t 8x + 5 = y ta c\u00f3 (y \u2013 1)(y + 1)y2 \uf0a3 72 \uf0db (y2 \u2013 1)y2 \u2013 72 \uf0a3 0 \uf0db y4 \u2013 y2 \u2013 72 \uf0a3 0 \uf0db (y2 \u2013 9)(y2 + 8) \uf0a3 0 Do y2 + 8 = (8x + 5)2 + 8 > 0 ; \uf022 x n\u00ean y2 \u2013 9 \uf0a3 0 Hay (y \u2013 3)(y + 3) \uf0a3 0. Thay y = 8x + 5 v\u00e0o ta c\u00f3 : (8x + 2)(8x + 8) \uf0a3 0 325","Website: tailieumontoan.com Gi\u1ea3i \u0111\u01a3\u1ee3c \uf02d1 \uf0a3 x \uf0a3 \uf02d 1 (B\u1ea1n \u0111\u1ecdc t\u1ef1 bi\u1ec3u di\u1ec5n nghi\u1ec7m tr\u00ean tr\u1ee5c s\u1ed1) 4 b) Nh\u00e2n 2 v\u00e0o nh\u00e2n t\u1eed th\u1ee9 nh\u1ea5t, nh\u00e2n 2 v\u00e0o nh\u00e2n t\u1eed th\u1ee9 hai v\u00e0 nh\u00e2n 4 v\u00e0o v\u1ebf ph\u1ea3i ta \u0111\u01a3\u1ee3c : BPT \uf0db (4x \u2013 2)(4x \u2013 4)(4x \u2013 5)(4x \u2013 7) \uf0b3 72 \uf0db [(4x \u2013 2)(4x \u2013 7)][(4x \u2013 4)(4x \u2013 5)] \uf0b3 72 \uf0db (16x2 \u2013 36x + 14) (16x2 \u2013 36x + 20) \uf0b3 72 \u0110\u1eb7t 16x2 \u2013 36x + 17 = y ta c\u00f3 (y \u2013 3)(y + 3) \u201372 \uf0b3 0 \uf0db y2 \u2013 81 \uf0b3 0 \uf0db (y \u2013 9)(y + 9) \uf0b3 0 Do y + 9 = 16x2 \u2013 36x + 26 = (4x)2 \u2013 2.4x. 9 + 81 + 23 24 4 = \uf0e6 4x \uf02d 9 \uf0f62 \uf02b 23 \uf03e 0, \uf022x t\u1eeb \u0111\u00f3 ta c\u00f3 y \u2013 9 \uf0b3 0 \uf0e8\uf0e7 2 \uf0f7\uf0f8 4 Hay 16x2 \u2013 36x + 8 \uf0b3 0 \uf0db 4x2 \u2013 9x + 2 \uf0b3 0 \uf0db (4x \u2013 1)(x \u2013 2) \uf0b3 0 Gi\u1ea3i b\u1ea5t ph\u01a3\u01a1ng tr\u00ecnh n\u00e0y \u0111\u01a3\u1ee3c \uf0e9\uf0eax \uf0a3 1 . (B\u1ea1n \u0111\u1ecdc t\u1ef1 bi\u1ec3u di\u1ec5n nghi\u1ec7m) \uf0ea \uf0b3 4 \uf0ebx 2 c) BPT \uf0db (x \u2013 1)(x \u2013 2)(2x \u2013 3)(2x \u2013 5) > 30 \uf0db (2x \u2013 2)(2x \u2013 4)(2x \u2013 3)(2x \u2013 5) > 120 \uf0db (4x2 \u2013 14x + 10) (4x2 \u2013 14x + 12) \u2013 120 > 0 \u0110\u1eb7t 4x2 \u2013 14x + 11 = y ta c\u00f3 (y \u2013 1)(y + 1) \u2013 120 > 0 \uf0db y2 \u2013 112 > 0 \uf0db (y \u2013 11)(y + 11) > 0 Do y + 11 = 4x2 \u2013 14x + 22 = (2x)2 \u2013 2.2x. 7 + 49 + 39 24 4 = \uf0e6 2x \uf02d 7 \uf0f62 \uf02b 39 \uf03e 0 , \uf022x \uf0e7\uf0e8 2 \uf0f7\uf0f8 4 Do \u0111\u00f3 y \u2013 11 > 0 hay 4x2 \u2013 14x > 0 \uf0db 2x(2x \u2013 7) > 0. Gi\u1ea3i b\u1ea5t ph\u01a3\u01a1ng tr\u00ecnh \u0111\u01a3\u1ee3c \uf0e9x \uf03e 3,5 (B\u1ea1n \u0111\u1ecdc t\u1ef1 bi\u1ec3u di\u1ec5n nghi\u1ec7m ) \uf0ea\uf0ebx \uf03c 0 . 23.5. a) BPT \uf0db (x4 \u2013 4)(x4 \u2013 8) \uf0a3 96 \uf0db x8 \u2013 12x4 + 32 \u2013 96 \uf0a3 0 \uf0db x8 + 4x4 \u2013 16x4 \u2013 64 \uf0a3 0 \uf0db (x4 \u2013 16)(x4 + 4) \uf0a3 0 . Do x4 + 4 > 0 \uf022 x n\u00ean x4 \u2013 16 \uf0a3 0 \uf0db (x + 2)(x \u2013 2)(x2 + 4) \uf0a3 0 Do x2 + 4 > 0 \uf022 x n\u00ean (x + 2)(x \u2013 2) \uf0a3 0 \uf0db \u2013 2 \uf0a3 x \uf0a3 2 . *Ch\u00fa \u00fd : C\u00e2u a) c\u00f3 th\u1ec3 d\u00f9ng ph\u01a3\u01a1ng ph\u00e1p \u0111\u1eb7t bi\u1ebfn ph\u1ee5 : \u0110\u1eb7t x4 \u2013 6 = y ta c\u00f3 (y + 2)(y \u2013 2) \uf0a3 96 \uf0db y2 \u2013 4 \u2013 96 \uf0a3 0 326","Website: tailieumontoan.com \uf0db y2 \u2013 100 \uf0a3 0 \uf0db (y \u2013 10)(y + 10) \uf0a3 0 Do y + 10 = x4 \u2013 6 + 10 = x4 + 4 > 0 \uf022 x n\u00ean y \u2013 10 \uf0a3 0 hay x4 \u2013 16 \uf0a3 0 r\u1ed3i gi\u1ea3i nh\u01a3 tr\u00ean ta \u0111\u01a3\u1ee3c \u2013 2 \uf0a3 x \uf0a3 2 . b) \u0110\u1ec3 \u00fd r\u1eb1ng x4 + 4 = x4 + 4x2+ 4 \u2013 4x2 = (x2 + 2)2 \u2013 (2x)2 = (x2 + 2x + 2) (x2 \u2013 2x + 2) Do \u0111\u00f3 c\u00f3 (x4 + 4) \u2013 (x2 + 2x + 2)(3x + 26) \uf0b3 0 \uf0db (x2 + 2x + 2) (x2 \u2013 2x + 2) \u2013 (x2 + 2x + 2)(3x + 26) \uf0b3 0 \uf0db (x2 + 2x + 2) (x2 \u2013 5x \u2013 24) \uf0b3 0 \uf0db (x2 + 2x + 2) (x \u2013 8)(x + 3) \uf0b3 0 Do x2 + 2x + 2 = (x + 1)2 + 1 > 0 \uf022x n\u00ean ta ch\u1ec9 x\u00e9t (x \u2013 8)(x + 3) \uf0b3 0 \uf0db \uf0e9x \uf0a3 \uf02d3 \uf0ea\uf0ebx \uf0b3 8 c) BPT \uf0db (x3 \u2013 27)(x2 + x \u2013 6) > 0 \uf0db (x \u2013 3)( x2 + 3x + 9)(x \u2013 2)(x +3) > 0 Ta c\u00f3 x2 + 3x + 9 = \uf0e6 x \uf02b 3 \uf0f62 \uf02b 27 >0 \uf022x ; V\u1eady (x \u2013 3)(x +3)(x \u2013 2) >0 \uf0e7\uf0e8 2 \uf0f8\uf0f7 4 Gi\u1ea3i b\u1ea5t ph\u01a3\u01a1ng tr\u00ecnh ta c\u00f3 nghi\u1ec7m : \uf0e9\uf02d3 \uf03c x \uf03c 2 \uf0eb\uf0eax \uf03e 3 23.6. \u0110KX\u0110 x \uf0b9 \uf02d 389 . L\u1eadp b\u1ea3ng x\u00e9t d\u1ea5u : 14 Nghi\u1ec7m c\u1ee7a b\u1ea5t ph\u01a3\u01a1ng tr\u00ecnh l\u00e0 \uf02d 389 \uf03c x \uf03c 9 14 2 23.7. a) \u0110KX\u0110 x \uf0b9 4 ; BPT \uf0db 1\uf02b 5x \uf02d 2 \uf0b3 0 \uf0db 9 \uf02b 3x \uf0b3 0 \uf0db \uf0e9x \uf03e 4 x\uf02d4 x\uf02d4 \uf0ea\uf0ebx \uf0a3 \uf02d3 b) \u0110KX\u0110 x \uf0b9 \uf0b1 2 ; BPT \uf0db 11x \uf02b 2 \uf0a3 0 . (x \uf02b 2)(x \uf02d 2) \uf0e9x \uf03c \uf02d2 \uf0ea L\u1eadp b\u1ea3ng x\u00e9t d\u1ea5u ta t\u00ecm \u0111\u01a3\u1ee3c \uf0ea\uf02d 2 \uf0a3 x \uf03c 2 . c) \u0110KX\u0110 x \uf0b9 8 v\u00e0 x \uf0b9 6 \uf0eb 11 BPT \uf0db \uf02dx \uf02b10 < 0 \uf0db \uf0e96 \uf03c x \uf03c 8 (x \uf02d 8)(x \uf02d 6) \uf0ea\uf0ebx \uf03e 10 d) \u0110KX\u0110 x \uf0b9 3 v\u00e0 x \uf0b9 1 ; BPT \uf0db (x \uf02b 3)(x\uf02d 2) \uf03e 0 (x \uf02d1)(x \uf02d 3) 327","Website: tailieumontoan.com L\u1eadp b\u1ea3ng x\u00e9t d\u1ea5u , nghi\u1ec7m l\u00e0 \uf0e9x \uf03c \uf02d3 \uf0ea\uf0ea1 \uf03c x \uf03c 2 \uf0ea\uf0ebx \uf03e 3 23.8. 3\uf03c x \uf02b 3 \uf03c 5 \uf0db \uf0ec x \uf02b 3 \uf03e 3 \uf0db \uf0ec5 \uf03c x \uf03c 9 \uf0db 7<x<9 x \uf02d 5 \uf0ef\uf0ef x \uf02d 5 \uf03c 5 \uf0ef\uf0ed\uf0e9x \uf03e 7 \uf0ed x \uf02b 3 \uf0ee\uf0ef\uf0ea\uf0ebx \uf03c 5 \uf0ef x \uf02d 5 \uf0ef\uf0ee 23.9. \u0110KX\u0110 x \uf0b9 3 ; x \uf0b9 \uf0b1 1 ; R\u00fat g\u1ecdn : \uf028 x \uf02d 2016\uf029\uf0282x2 \uf02d 4x \uf02b10\uf029 \uf028x \uf02d 2016\uf029\uf028x2 \uf02d 2x \uf02b 5\uf029 A\uf03d \uf03d 8 \uf02d 8x 4(1\uf02d x) Do x2 \u2013 2x + 5 = (x \u2013 1)2 + 4 > 0 \uf022 x n\u00ean A\uf0a30 \uf0db x \uf02d 2016 \uf0a3 0 . Gi\u1ea3i \u0111\u01a3\u1ee3c \uf0e9x \uf0b3 2016 v\u00e0 x \uf0b9 \u20131 4(1\uf02d x) \uf0eb\uf0eax \uf03c 1 23.10. \u0110KX\u0110 : x \uf0b9 \uf0b1 3 ; x \uf0b9 \u20132 ; R\u00fat g\u1ecdn \u0111\u01a3\u1ee3c 2\uf02dx . B= x\uf02b3 B \uf0b3 2015 \uf0db 2 \uf02d x \uf0b3 2015 \uf0db 2 \uf02d x \uf02d 2015 \uf0b3 0 \uf0db \uf02d2016x \uf02d 6043 \uf0b3 0 x\uf02b3 x\uf02b3 x\uf02b3 Gi\u1ea3i b\u1ea5t ph\u01a3\u01a1ng tr\u00ecnh n\u00e0y \u0111\u01a3\u1ee3c . \uf02d3 \uf03c x \uf0a3 \uf02d 6043 2016 23.11. V\u1edbi x \uf0b9 2 ta c\u00f3 3 = (5 \u2013 m)(x \u2013 2) \uf0db x(m \u2013 5) = 2m \u2013 13 * V\u1edbi m = 5 ph\u01a3\u01a1ng tr\u00ecnh tr\u1edf th\u00e0nh 0x = \u2013 3 v\u00f4 nghi\u1ec7m. * V\u1edbi m \uf0b9 5 th\u00ec \uf0db x = 2m \uf02d13 m\uf02d5 \u0110\u1ec3 x \uf0b3 0 ta ph\u1ea3i c\u00f3 2m \uf02d13 \uf0b3 0 \uf0db \uf0e9m \uf0b3 6,5 m\uf02d5 \uf0eb\uf0eam \uf03c 5 23.12. Ta c\u00f3 \uf0e6 1 \uf02d1\uf0f8\uf0f7\uf0f6 \uf0e6 1 \uf02d1\uf0f6\uf0f7\uf0f8 \uf0e6 1 \uf02d1\uf0f8\uf0f6\uf0f7 \uf0e6 1 \uf02d1\uf0f7\uf0f6\uf0f8 \uf0e6 1 \uf02d1\uf0f8\uf0f6\uf0f7 \uf0e6 1 \uf02d1\uf0f7\uf0f8\uf0f6 \uf03d \uf0e7\uf0e8 3 \uf0e7\uf0e8 6 \uf0e7\uf0e8 10 \uf0e7\uf0e8 15 \uf0e8\uf0e7 21 \uf0e7\uf0e8 28 \uf0e6 2 \uf02d 1\uf0f8\uf0f7\uf0f6 \uf0e8\uf0e6\uf0e7 2 \uf02d1\uf0f6\uf0f7\uf0f8 \uf0e6 2 \uf02d 1\uf0f6\uf0f8\uf0f7 \uf0e7\uf0e8\uf0e6 2 \uf02d1\uf0f6\uf0f8\uf0f7 \uf0e6 2 \uf02d1\uf0f8\uf0f7\uf0f6 \uf0e6 2 \uf02d1\uf0f6\uf0f7\uf0f8 \uf03d \uf0e8\uf0e7 2.3 3.4 \uf0e8\uf0e7 4.5 5.6 \uf0e7\uf0e8 6.7 \uf0e7\uf0e8 7.8 \uf02d4 . \uf02d10 . \uf02d18 . \uf02d28 \uf02d40 . \uf02d54 \uf03d 1.4 . 2.5 . 3.6 . 4.7 5.8 . 6.9 \uf03d 3 2.3 3.4 4.5 5.6 6.7 7.8 2.3 3.4 4.5 5.6 6.7 7.8 7 328","Website: tailieumontoan.com Do \u0111\u00f3 b\u1ea5t ph\u01a3\u01a1ng tr\u00ecnh tr\u1edf th\u00e0nh 3 x \uf03e x2 \uf02d 4 \uf0db x2 \u2013 3x \u2013 28 < 0 77 \uf0db (x \u2013 7)(x + 4) < 0 Gi\u1ea3i b\u1ea5t ph\u01a3\u01a1ng tr\u00ecnh n\u00e0y ta \u0111\u01a3\u1ee3c \u2013 4 < x < 7. 23.13. X\u00e9t 1 \uf02b 1 \uf02b ...\uf02b 1 \uf03d 1 \uf02d 1 \uf02b 1 \uf02d 1 \uf02b ...\uf02b 1 \uf02d 1 = 1.2 3.4 99.100 1 2 3 4 99 100 1 \uf02b 1 \uf02b 1 \uf02b 1 \uf02b ... \uf02b 1 \uf02b 1 \uf02d 2 \uf0e7\uf0e8\uf0e6 1 \uf02b 1 \uf02b ... \uf02b 1 \uf0f6 \uf03d 1 2 3 4 99 100 2 4 100 \uf0f7\uf0f8 1 \uf02b 1 \uf02b 1 \uf02b 1 \uf02b ... \uf02b 1 \uf02b 1 \uf02d \uf0e6\uf0e8\uf0e71\uf02b 1 \uf02b 1 \uf02b ... \uf02b 1 \uf0f6 \uf03d 1 \uf02b 1 \uf02b ... \uf02b 1 \uf02b 1 1 2 3 4 99 100 2 3 50 \uf0f8\uf0f7 51 52 99 100 V\u1eady x2 + x + 1945 > 1975 \uf0db x2 + x \u2013 30 > 0 \uf0db (x \u2013 5)(x + 6) > 0 . Gi\u1ea3i b\u1ea5t ph\u01a3\u01a1ng tr\u00ecnh \u0111\u01a3\u1ee3c \uf0e9x \uf03c \uf02d6 \uf0ea\uf0ebx \uf03e 5 23.14. V\u1edbi a \uf0b9 \uf02d1 th\u00ec BPT \uf0db x \uf02d 1 \uf03c x \uf02d 1 \uf02b 2a \uf02d 2ax \uf02d a2 a2 \uf02da \uf02b1 a \uf02b1 a2 \uf02da \uf02b1 a2 \uf02da \uf02b1 1\uf02b a3 \uf0db a2 1 \uf02b1 \uf02d a 1 \uf02d \uf02da2 \uf02d 2ax \uf02b 2a \uf03c 0 \uf02da \uf02b1 (a \uf02b1)(a2 \uf02d a \uf02b1) \uf0db a \uf02b1\uf02d a2 \uf02b a \uf02d1\uf02b a2 \uf02b 2ax \uf02d 2a \uf03c0 \uf0db 2ax \uf03c0 (a \uf02b1)(a2 \uf02d a \uf02b1) (a \uf02b1)(a2 \uf02d a \uf02b1) Do a2 \u2013 a + 1 = \uf0e6 a \uf02d 1 \uf0f62 \uf02b 3 \uf03e 0 , \uf022a n\u00ean ta ch\u1ec9 x\u00e9t 2ax \uf03c 0 \uf0e8\uf0e7 2 \uf0f8\uf0f7 4 a \uf02b1 X\u00e9t d\u1ea5u c\u1ee7a a c\u00f3: N\u1ebfu a < \u2013 1 ho\u1eb7c a > 0 th\u00ec a > 0 nghi\u1ec7m l\u00e0 x < 0. a \uf02b1 a \uf02b1 N\u1ebfu \u20131 < a < 0 th\u00ec a < 0 nghi\u1ec7m l\u00e0 x > 0 a \uf02b1 N\u1ebfu a = 0 th\u00ec b\u1ea5t ph\u01a3\u01a1ng tr\u00ecnh tr\u1edf th\u00e0nh 0x < 0 v\u00f4 nghi\u1ec7m 23.15. Ta c\u00f3 A = \uf0e6 6 \uf02b1\uf0f6\uf0f7\uf0f8 \uf0e6 6 \uf02b1\uf0f8\uf0f7\uf0f6 \uf0e6 6 \uf02b1\uf0f8\uf0f6\uf0f7...\uf0e6\uf0e8\uf0e7 6 \uf02b1\uf0f7\uf0f6\uf0f8 \uf0e7\uf0e8 1.8 \uf0e8\uf0e7 2.9 \uf0e8\uf0e7 3.10 13.20 14 . 24 . 36 ..... 266 \uf03d 2.7 . 3.8 . 4.9 .....14.19 = 1.8 2.9 3.10 13.20 1.8 2.9 3.10 13.20 2.3.4.....13.14 . 7.8.9.....18.19 \uf03d 49 1.2.3.....12.13 8.9.10.....19.20 10 B = 3 . 8 . 15 ..... 99 \uf03d 1.3 . 2.4 . 3.5 ..... 9.11 2.2 3.3 4.4 10.10 2.2 3.3 4.4 10.10 329","Website: tailieumontoan.com 1.2.3.....8.9 . 3.4.5.....10.11 \uf03d 11 2.3.4.....9.10 2.3.4.....9.10 20 11 \uf03c x \uf02d 2 \uf03c 49 \uf0db 33 \uf03c 2x \uf02d 4 \uf03c 294 \uf0db 33 < 2x \u2013 4 < 294 20 30 10 60 60 60 \uf0db 37 < 2x < 298 \uf0db 18,5 < x < 149 23.16. x \uf02d 3 \uf03c 3 \uf0db x \uf02d 3 \uf02d 3 \uf03c 0 \uf0db x \uf02d 3 \uf02d 3x \uf02b 3 \uf03c 0 \uf0db \uf02d2x \uf03c 0 \uf0db 2x \uf03e 0 x \uf02d1 x \uf02d1 x \uf02d1 x \uf02d1 x \uf02d1 L\u1eadp b\u1ea3ng x\u00e9t d\u1ea5u : x 01 2x \u2013 0 + | + + x\u2013 1 \u2013| \u20130 + VT + || \u2013 || V\u1eady x > 1 ; x < 0 l\u00e0 nghi\u1ec7m c\u1ee7a b\u1ea5t ph\u01a3\u01a1ng tr\u00ecnh. 23.17. 3\uf02b x2 \uf02d 4 \uf02d 5 \uf03c 7 \uf02b 9 x2 \uf02b 6 x2 \uf02b1 x2 \uf02b3 x2 \uf02b5 \uf0db x2 \uf02d 4 \uf02b \uf0e6\uf0e7\uf0e81 \uf02d 5 \uf0f6 \uf02b \uf0e7\uf0e6\uf0e81 \uf02d 7 3 \uf0f6 \uf02b \uf0e7\uf0e6\uf0e81\uf02d 9 5 \uf0f6 \uf03c 0 x2 \uf02b 6 x2 \uf02b1 \uf0f7\uf0f8 x2 \uf02b \uf0f7\uf0f8 x2 \uf02b \uf0f7\uf0f8 \uf0db x2 \uf02d 4 \uf02b x2 \uf02d4 \uf02b x2 \uf02d 4 \uf02b x2 \uf02d 4 \uf03c 0 x2 \uf02b 6 x2 \uf02b1 x2 \uf02b 3 x2 \uf02b 5 \uf028 \uf029\uf0db \uf0e6 1 1 1 1 \uf0f6 x2 \uf02d 4 \uf0e8\uf0e7 x2 \uf02b 6 \uf02b x2 \uf02b1 \uf02b x2 \uf02b 3 \uf02b x2 \uf02b 5 \uf0f7\uf0f8 \uf03c 0 \uf0db (x \u2013 2)(x + 2) < 0 \uf0db \u20132<x<2 do x 1 6 \uf02b x 1 1 \uf02b x 1 3 \uf02b x 1 5 \uf03e 0 . 2\uf02b 2\uf02b 2\uf02b 2\uf02b Chuy\u00ean \u0111\u1ec1 24: PH\u01a2\u01a0NG TR\u00ccNH- B\u1ea4T PH\u01a2\u01a0NG TR\u00ccNH CH\u1ee8A D\u1ea4U GI\u00c1 TR\u1eca TUY\u1ec6T \u0110\u1ed0I 24.1. a) Bi\u1ebfn \u0111\u1ed5i PT \uf0db 5x \uf02d12 \uf03d 10x \uf02d 32 Ta c\u00f3 V\u00ec 5x \uf02d12 \uf0b3 0 n\u00ean 10x \u2013 32 \uf0b3 0 \uf0db x \uf0b3 3,2. Khi \u1ea5y 5x \uf02d12 \uf03d 5x \uf02d12 . Ph\u01a3\u01a1ng tr\u00ecnh tr\u1edf th\u00e0nh 5x \u2013 12 = 10x \u2013 32 ta t\u00ecm \u0111\u01a3\u1ee3c x = 4 (th\u1ecfa m\u00e3n) ; V\u1eady nghi\u1ec7m c\u1ee7a ph\u01a3\u01a1ng tr\u00ecnh l\u00e0 x = 4. b) Bi\u1ebfn \u0111\u1ed5i PT \uf0db 5 3x \uf02d 4 \uf03d 22 \uf02d 6x 330","Website: tailieumontoan.com X\u00e9t v\u1edbi x \uf0b3 4 ta t\u00ecm \u0111\u01a3\u1ee3c x = 2 ; X\u00e9t v\u1edbi x \uf03c 4 ta t\u00ecm \u0111\u01a3\u1ee3c x = \uf02d 2 3 39 24.2. a) PT \uf0db 2x \uf02d 3 \uf03d 2 \uf0db \uf0e9 2x \uf03d 5 \uf0db \uf0e9x \uf03d \uf0b12, 5 \uf0ea \uf0ea\uf0ebx \uf03d \uf0b10, 5 \uf0eb\uf0ea 2x \uf03d 1 b) *V\u1edbi x \uf0b3 2 th\u00ec x \uf02d 2 \uf03d x \uf02d 2 Ph\u01a3\u01a1ng tr\u00ecnh tr\u1edf th\u00e0nh \uf0db x \uf02d 2 = 2x \u2013 6 \uf0db x = 2x \u2013 4 V\u1edbi x \uf0b3 0 ta c\u00f3 x = 2x \u2013 4 \uf0db x = 4 (th\u1ecfa m\u00e3n) V\u1edbi x < 0 ta c\u00f3 \u2013 x = 2x \u2013 4 \uf0db x = 4 (lo\u1ea1i 3 *V\u1edbi x \uf03c 2 th\u00ec x \uf02d 2 \uf03d 2 \uf02d x Ph\u01a3\u01a1ng tr\u00ecnh tr\u1edf th\u00e0nh \uf0db 2 \uf02d x = 2x \u2013 6 \uf0db x = 8 \u20132x V\u1edbi x \uf0b3 0 ta c\u00f3 x = 8 \u2013 2x \uf0db x = 8 (lo\u1ea1i v\u00ec x \uf03c 2 ) V\u1edbi x < 0 3 ta c\u00f3 \u2013 x = 8 \u2013 2x \uf0db x = 8 (lo\u1ea1i) Ph\u01a3\u01a1ng tr\u00ecnh c\u00f3 nghi\u1ec7m duy nh\u1ea5t l\u00e0 x = 4 24.3. L\u1eadp b\u1ea3ng x\u00e9t gi\u00e1 tr\u1ecb tuy\u1ec7t \u0111\u1ed1i r\u1ed3i gi\u1ea3i c\u00e1c ph\u01a3\u01a1ng tr\u00ecnh. a) T\u1eadp nghi\u1ec7m l\u00e0 \uf02d 5 \uf0a3 x \uf0a3 5 44 b) B\u1ea3ng x\u00e9t gi\u00e1 tr\u1ecb tuy\u1ec7t \u0111\u1ed1i : x \u20132 3 2x \u2013 6 5 2x \uf02d 6 6 \u2013 2x | 6 \u2013 2x 0 5\u2013x | 2x \u2013 6 x\uf02d5 x+2 0 x\u20135 x\uf02b2 5\u2013x | 5\u2013x | 0x \u2013 3 V\u1ebf tr\u00e1i | x+2 \u2013x\u20132 0 x+2 | | 2x \u2013 13 \u2013 2x + 13 | \u2013 4x + 9 | * V\u1edbi x < \u20132 PT \uf0db \u2013 2x + 13 = 5 \uf0db x = 4 (lo\u1ea1i) * V\u1edbi \uf02d2 \uf0a3 x \uf03c 3 PT \uf0db \u2013 4x + 9 = 5 \uf0db \u20134x < \u2013 4 \uf0db x = 1 * V\u1edbi 3 \uf0a3 x < 5 PT \uf0db 0x \u2013 3 = 5 \uf0db 0x = 8 (v\u00f4 nghi\u1ec7m) * V\u1edbi x \uf0b3 2,5 PT \uf0db 2x \u2013 13 = 5 \uf0db 2x = 18 \uf0db x = 9 T\u1eadp nghi\u1ec7m l\u00e0 S = \uf07b1; 9\uf07d c) L\u1eadp b\u1ea3ng x\u00e9t GTT\u0110. Nghi\u1ec7m l\u00e0 x = \u2013 0,25 ; x = 0,5. 24. 4. a) PT \uf0db \uf0e9x2 \uf02d 2x \uf02d1 \uf03d 2 \uf0db \uf0e9(x \uf02d 3)(x+1) \uf03d 0 \uf0ea \uf0eb x 2 \uf02d 2x \uf02d1 \uf03d \uf02d2 \uf0eb\uf0ea(x \uf02d1)2 \uf03d 0 331","Website: tailieumontoan.com T\u1eadp nghi\u1ec7m : S = \uf07b\uf02d1;3;1\uf07d . b) L\u01a3u \u00fd : x \uf0b3 0 . T\u1eadp nghi\u1ec7m S = \uf07b2;3\uf07d . c) V\u1ebf tr\u00e1i 4x \u2013 x2 = 4 \u2013 (4 \u2013 4x + x2) = 4 \u2013 (2 \u2013 x)2 \uf0a3 4 V\u1ebf ph\u1ea3i : \u00e1p d\u1ee5ng b\u1ea5t \u0111\u1eb3ng th\u1ee9c a \uf02b b \uf0b3 a \uf02b b ta c\u00f3 V\u1ebf tr\u00e1i : x \uf02d1 \uf02b x \uf02d 5 \uf03d x \uf02d1 \uf02b 5 \uf02d x \uf0b3 x \uf02d1\uf02b 5 \uf02d x \uf03d 4 Suy ra v\u1ebf ph\u1ea3i b\u1eb1ng v\u1ebf tr\u00e1i b\u1eb1ng 4 \uf0de x = 2. d) \u00c1p d\u1ee5ng b\u1ea5t \u0111\u1eb3ng th\u1ee9c : a \uf02d b \uf0a3 a \uf02d b ta c\u00f3 : V\u1ebf tr\u00e1i x2 \uf02d 25 \uf02d x2 \uf02d 9 \uf0a3 x2 \uf02d 25 \uf02d x2 \uf02b 9 \uf03d16 M\u1eb7t kh\u00e1c v\u1ebf ph\u1ea3i x2\u2013 2x + 17 = (x \u2013 1)2 + 16 \uf0b3 16 Suy ra v\u1ebf ph\u1ea3i b\u1eb1ng v\u1ebf tr\u00e1i v\u00e0 b\u1eb1ng 16 \uf0de x = 1. 24. 5. L\u1eadp b\u1ea3ng x\u00e9t gi\u00e1 tr\u1ecb tuy\u1ec7t \u0111\u1ed1i: x 25 x\uf02d2 2\u2013x 0 x\u20132 | x\u20132 x\uf02d5 5\u2013x | 5\u2013x 0 x\u20135 V\u1ebf tr\u00e1i 7 \u2013 2x | 0x + 3 | 2x \u2013 7 * V\u1edbi x < 2 th\u00ec (1) \uf0db 7 \u2013 2x = m \uf0db x \uf03d 7 \uf02d m l\u00e0 nghi\u1ec7m n\u1ebfu 2 7 \uf02d m \uf03c 2 \uf0db m > 3. 2 * V\u1edbi 2 \uf0a3 x \uf0a3 5 th\u00ec (1) \uf0db 0x = m \u2013 3 v\u00f4 s\u1ed1 nghi\u1ec7m n\u1ebfu m = 3. * V\u1edbi x > 5 th\u00ec (1) \uf0db 2x \u2013 7 = m \uf0db x \uf03d m \uf02b 7 l\u00e0 nghi\u1ec7m n\u1ebfu 2 m \uf02b 7 \uf03e 5 \uf0db m > 3. 2 V\u1eady N\u1ebfu m > 3 th\u00ec (1) c\u00f3 hai nghi\u1ec7m l\u00e0 x \uf03d 7 \uf02d m v\u00e0 x \uf03d m \uf02b 7 . 22 N\u1ebfu m = 3 th\u00ec (1) c\u00f3 v\u00f4 s\u1ed1 nghi\u1ec7m 2 \uf0a3 x \uf0a3 5 . N\u1ebfu m < 3 th\u00ec (1) v\u00f4 nghi\u1ec7m. \uf0e92 x \uf02d 5 \uf03d x \uf02d 3 \uf02b 4 \uf0e92 x \uf02d x \uf02d 3 \uf03d 9 24.6. PT \uf0db \uf0ea \uf0db\uf0ea \uf0eb\uf0ea2 x \uf02d 5 \uf03d \uf02d4 \uf02d x \uf02d 3 \uf0eb\uf0ea2 x \uf02b x \uf02d 3 \uf03d 1 L\u1eadp b\u1ea3ng x\u00e9t gi\u00e1 tr\u1ecb tuy\u1ec7t \u0111\u1ed1i t\u00ecm \u0111\u01a3\u1ee3c t\u1eadp nghi\u1ec7m l\u00e0 S = {\u2013 12 ; 6} 24.7. \u0110\u1eb7t x \uf02d 5 = t ( t \uf0b3 0). Ph\u01a3\u01a1ng tr\u00ecnh tr\u1edf th\u00e0nh 2t \uf02d 9 \uf02b 2t \uf02d11 \uf03d12 . 332","Website: tailieumontoan.com L\u1eadp b\u1ea3ng x\u00e9t gi\u00e1 tr\u1ecb tuy\u1ec7t \u0111\u1ed1i t\u00ecm \u0111\u01a3\u1ee3c t = 2 v\u00e0 t = 8. V\u1edbi t = 2 \uf0db x\uf02d5 \uf03d2 \uf0db \uf0e9x \uf03d 7 \uf0eb\uf0eax \uf03d 3 V\u1edbi t = 8 \uf0db x\uf02d5 \uf03d8 \uf0db \uf0e9x \uf03d 13 \uf0ea\uf0ebx \uf03d \uf02d3 24.8. a) BPT \uf0db \uf02d14 \uf03c x2 \uf02d 4x \uf02b 2 \uf03c 14 \uf0db \uf0ec\uf0efx2 \uf02d 4x \uf02d 12 \uf03c 0 \uf0ed \uf0ee\uf0ef x 2 \uf02d 4x \uf02b 16 \uf03e 0 * x2 \u2013 4 x + 16 = \uf028 x \uf02d 2\uf0292 + 12 > 0 \uf022 x * x2 \u2013 4x \u2013 12 = (x \u2013 6)(x + 2) < 0 \uf0db \u2013 2 < x < 6 Nghi\u1ec7m c\u1ee7a b\u1ea5t ph\u01a3\u01a1ng tr\u00ecnh l\u00e0 \u2013 2 < x < 6 b) x \uf02d 3 \uf03c x2 \uf02b 2x \uf02b 3 \uf0db \uf02dx2 \uf02d 2x \uf02d 3 \uf03c x \uf02d 3 \uf03c x2 \uf02b 2x \uf02b 3 \uf0db \uf0ef\uf0ecx2 \uf02b 3x \uf03e 0 \uf0ed \uf0ef\uf0ee x 2 \uf02b x \uf02b 6 \uf03e 0 * x2 + x + 6 = \uf0e6 x \uf02b 1 \uf0f62 \uf02b 23 \uf03e 0 , \uf022x > 0 \uf0e8\uf0e7 2 \uf0f8\uf0f7 4 * x2 + 3x = x(x + 3) > 0 \uf0db \uf0e9x \uf03e 0 \uf0eb\uf0eax \uf03c \uf02d3 Nghi\u1ec7m c\u1ee7a b\u1ea5t ph\u01a3\u01a1ng tr\u00ecnh l\u00e0 \uf0e9x \uf03e 0 \uf0ea\uf0ebx \uf03c \uf02d3 c) BPT \uf0db \uf0ec6x \uf02d15 \uf0a3 2x \uf02b 5 \uf0db 1, 25 \uf0a3 x \uf0a3 5 \uf0ed\uf0ee6x \uf02d15 \uf0b3 \uf02d2x \uf02d 5 \uf0e92(5x \uf02d1) \uf03e 5x \uf02b1 \uf0e95x \uf03e 3 \uf0e9x \uf03e 0,6 \uf0ea\uf0eb2(5x \uf02d1) \uf03c \uf02d5x \uf02d1 \uf0ea\uf0eb15x \uf03c 1 \uf0ea 24.9. a) BPT \uf0db \uf0db \uf0db \uf0eax \uf03c 1 \uf0eb 15 \uf0e9x \uf03e 0,6 \uf0ea Nghi\u1ec7m b\u1ea5t ph\u01a3\u01a1ng tr\u00ecnh l\u00e0 \uf0eax \uf03c 1 \uf0eb 15 b) V\u1edbi x \uf0b9 2 b\u1ea5t ph\u01a3\u01a1ng tr\u00ecnh \u0111\u00e3 cho t\u01a3\u01a1ng \u0111\u01a3\u01a1ng v\u1edbi: \uf0e9 1 \uf02b1 \uf03e 3 \uf0e9 1 \uf02d 2 \uf03e 0 \uf0e95\uf02d 2x \uf03e 0 \uf0ea\uf0e92 \uf03c x \uf03c 5 \uf0ea \uf02d \uf0ea \uf02d \uf0ea \uf03c 0 \uf0ea 2 \uf0ea x 2 \uf0db \uf0ea x 2 \uf0db \uf0ea x\uf02d2 \uf0db \uf0ea 4x \uf02d7 \uf0ea 1 \uf02b1 \uf03c \uf02d3 \uf0ea 1 \uf02b 4 \uf03c 0 \uf0ea7 \uf03c x \uf03c 2 \uf0ea\uf0eb x \uf02d 2 \uf0ea\uf0eb x \uf02d 2 \uf0eb\uf0ea x \uf02d 2 \uf0eb\uf0ea 4 H\u1ee3p nghi\u1ec7m \u0111\u01a3\u1ee3c 7 \uf03c x \uf03c 5 tr\u1eeb x = 2 . 42 c) V\u1edbi x \uf0b9 \u20133. T\u01a3\u01a1ng t\u1ef1 (b) ho\u1eb7c bi\u1ebfn \u0111\u1ed5i BPT \uf0db x \uf02d 3 \uf03e 2 . x\uf02b3 T\u00ecm \u0111\u01a3\u1ee3c \u2013 9 < x <\u2013 1 tr\u1eeb x = \u20133. 333","Website: tailieumontoan.com d) Ta c\u00f3 \uf02dx2 \uf02d 2x \uf02d 2016 \uf03d x2 \uf02b 2x+2016 (do \uf02d x2 \uf02d 2x \uf02d 2016 \uf03c 0 ; \uf022x) v\u00e0 x2 + 2018 > 0 , \uf022 x n\u00ean BPT \uf0db x2 + 2x + 2016 \uf0b3 x2 + 2018 \uf0db 2x \uf0b3 2 \uf0db x \uf0b3 1. 24.10. a) 2 \uf02b 5 \uf02b 8 \uf02b ... \uf02b 89 \uf03d (2 \uf02b 89).30 \uf03d 91.15 . 2 Do \u0111\u00f3 BPT \uf0db 2x \uf02d 3 \uf0b3 15 \uf0db \uf0e92x \uf02d 3 \uf0b3 15 \uf0db \uf0e9x \uf0b3 9 \uf0ea\uf0eb2x \uf02d 3 \uf0a3 \uf02d15 \uf0eb\uf0eax \uf0a3 \uf02d6 b) BPT \uf0db x2 \uf02d 4 \uf0b3 2 \uf02d x \uf0db \uf0e9x2 \uf02d 4 \uf0b3 2 \uf02d x \uf0e9(x \uf02d 2)(x \uf02b 3) \uf0b3 0 \uf0e9\uf0e9x \uf0b3 2 \uf0ea\uf0eb(x \uf02d 2)(x \uf02b1) \uf0a3 0 \uf0ea \uf0eb\uf0eax \uf0ea x 2 \uf02d 4 \uf0a3 x \uf02d 2 \uf0db \uf0db \uf0ea \uf0a3 \uf02d3 \uf0eb \uf0eb\uf0ea\uf02d1 \uf0a3 x \uf0a3 2 T\u1ed5ng h\u1ee3p nghi\u1ec7m: \uf0e9x \uf0b3 \uf02d1 \uf0ea\uf0ebx \uf0a3 \uf02d3 c) BPT \uf0db \uf0e94x \uf02b 5 \uf03e x2 \uf02d 2x \uf02b 5 \uf0db \uf0e9x2 \uf02d 6x \uf03c 0 (1c) \uf0ea \uf0ea \uf0eb4x \uf02b 5 \uf03c \uf02d( x 2 \uf02d 2x \uf02b 5) \uf0ebx 2 \uf02b 2x \uf02b10 \uf03c 0 (2c) (1c) \uf0db x(x \u2013 6) < 0 \uf0db 0 < x < 6 (2c) v\u00f4 nghi\u1ec7m Nghi\u1ec7m c\u1ee7a b\u1ea5t ph\u01a3\u01a1ng tr\u00ecnh l\u00e0 0 < x < 6 d) Ta c\u00f3 \uf02d2. 3x \uf02d 5 \uf0a3 0 ; \uf022x v\u00e0 x2 \uf02d x \uf02b1 \uf03d \uf0e6 x \uf02d 1 \uf0f62 \uf02b 3 \uf03e 0 ; \uf022x \uf0e8\uf0e7 2 \uf0f7\uf0f8 4 N\u00ean b\u1ea5t ph\u01a3\u01a1ng tr\u00ecnh v\u00f4 nghi\u1ec7m 24.11. a) B\u00ecnh ph\u01a3\u01a1ng hai v\u1ebf . Ho\u1eb7c l\u1eadp b\u1ea3ng x\u00e9t gi\u00e1 tr\u1ecb tuy\u1ec7t \u0111\u1ed1i. Nghi\u1ec7m c\u1ee7a (1) l\u00e0 x < \u20131. b) L\u1eadp b\u1ea3ng x\u00e9t gi\u00e1 tr\u1ecb tuy\u1ec7t \u0111\u1ed1i : x 56 x\uf02d5 5\u2013x 0 x\u20135 | x\u20135 x\uf02d6 6\u2013x | 6\u2013x 0 x\u20136 V\u1ebf tr\u00e1i 11 \u2013 2x | 0x + 1 | 2x \u2013 11 * V\u1edbi x < 5 th\u00ec (2) \uf0db 11 \u2013 2x > 3x \u201311 \uf0db x < 4,4 * V\u1edbi 5 \uf0a3 x \uf0a3 6 th\u00ec (2) \uf0db 0x + 1 > 3x \u2013 11 \uf0db x < 4 (lo\u1ea1i) * V\u1edbi x > 6 th\u00ec (2) \uf0db 2x \u2013 11 > 3x \u2013 11 \uf0db x < 0 (lo\u1ea1i) 334","Website: tailieumontoan.com V\u1eady nghi\u1ec7m c\u1ee7a (2) l\u00e0 x < 4,4. \uf0e9 x \uf02d 4 \uf02d 6 \uf03e 10 \uf0e9 x \uf02d 4 \uf03e 16 24. 12. a) BPT \uf0db x \uf02d 4 \uf02d 6 \uf03e 10 \uf0db \uf0ea \uf0db\uf0ea \uf0ea\uf0eb x \uf02d 4 \uf02d 6 \uf03c \uf02d10 \uf0eb\uf0ea x \uf02d 4 \uf03c \uf02d4 (loai) \uf0db \uf0e9x \uf02d 4 \uf03e 16 \uf0db \uf0e9x \uf03e 20 \uf0ea\uf0ebx \uf02d 4 \uf03c \uf02d16 \uf0eb\uf0eax \uf03c \uf02d12 b) BPT \uf0db 2x \uf02d 3 \uf02d11 \uf03c 1 \uf0db \uf02d1 \uf03c 2x \uf02d 3 \uf02d11 \uf03c 1 \uf0db \uf0ec\uf0ef 2x \uf02d 3 \uf03c 12 \uf0db \uf0ec\uf02d12 \uf03c 2x \uf02d3\uf03c 12 \uf0ec \uf02d4, 5\uf03c x \uf03c 7, 5 \uf0ed 2x \uf02d 2 \uf03e 10 \uf0ed\uf0ef\uf0e92x \uf02d 2\uf03e 10 \uf0db \uf0ef\uf0ed\uf0e9 x \uf03e 6 \uf0ee\uf0ef \uf0ee\uf0ef\uf0ea\uf0eb2x \uf02d 2\uf03c \uf02d10 \uf0ee\uf0ef\uf0eb\uf0ea x \uf03c \uf02d 4 H\u1ee3p nghi\u1ec7m ta \u0111\u01a3\u1ee3c nghi\u1ec7m c\u1ee7a BPT l\u00e0 \uf0e96 \uf03c x \uf03c 7,5 \uf0eb\uf0ea\uf02d4,5 \uf03c x \uf03c \uf02d4 24.13.L\u1eadp b\u1ea3ng x\u00e9t GTT\u0110 r\u1ed3i x\u00e9t c\u00e1c kho\u1ea3ng : *N\u1ebfu x <\u2013 2 th\u00ec PT \uf0db \u2013 x \u2013 x \u2013 1\u2013 x \u2013 2 = 7 \uf0db x = \uf02d 10 3 *N\u1ebfu \u2013 2 \uf0a3 x < \u2013 1 th\u00ec PT \uf0db \u2013 x \u2013 x \u2013 1+ x + 2 = 7 \uf0db x = \u2013 6 (lo\u1ea1i) *N\u1ebfu \u2013 1 \uf0a3 x \uf0a3 0 th\u00ec PT \uf0db \u2013 x + x + 1+ x + 2 = 7 \uf0db x = 4 (lo\u1ea1i) *N\u1ebfu x > 0 th\u00ec PT \uf0db x + x + 1+ x + 2 = 7 \uf0db x= 4 . 3 Ph\u01a3\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m l\u00e0 x = \uf02d 10 v\u00e0 x = 4 . 33 24.14. x2 \uf02d1 \uf02b x2 \uf02d 4 \uf03d x2 \uf02d1 \uf02b 4 \uf02d x2 \uf0b3 x2 \u2013 1 + 4 \u2013 x2 = 3 \uf0ec\uf02d2 \uf0a3 x \uf0a3 2 D\u1ea5u \u201c=\u201d x\u1ea3y ra \uf0db (x2 \uf02d1)(4 \uf02d x2) \uf0b3 0 \uf0db 1 \uf0a3 x2 \uf0a3 4 \uf0db 1 \uf0a3 x \uf0a3 2 \uf0db \uf0ef\uf0ed\uf0e9x \uf0b3 1 \uf0ee\uf0ef\uf0ea\uf0ebx \uf0a3 \uf02d1 \uf0db \uf0e91 \uf0a3 x \uf0a3 2 \uf02d1 . Nghi\u1ec7m ph\u01a3\u01a1ng tr\u00ecnh l\u00e0 1\uf0a3 x \uf0a3 2 ; \uf02d2 \uf0a3 x \uf0a3 \uf02d1 \uf0eb\uf0ea\uf02d2 \uf0a3 x \uf0a3 24.15. L\u1eadp b\u1ea3ng x\u00e9t GTT\u0110 r\u1ed3i x\u00e9t c\u00e1c kho\u1ea3ng : * V\u1edbi x \uf0a3 0 ph\u01a3\u01a1ng tr\u00ecnh th\u00e0nh \u2013 x \u2013 2 + x = 2 \uf0db 0x = 4 v\u00f4 nghi\u1ec7m. * V\u1edbi 0 \uf03c x \uf0a3 2 ph\u01a3\u01a1ng tr\u00ecnh th\u00e0nh x \u2013 2 + x = 2 \uf0db x = 2 (nh\u1eadn) * V\u1edbi x > 2 ph\u01a3\u01a1ng tr\u00ecnh th\u00e0nh x \u2013 x + 2 = 2 \uf0db 0x + 2 = 2 v\u00f4 s\u1ed1 nghi\u1ec7m. V\u1eady nghi\u1ec7m c\u1ee7a ph\u01a3\u01a1ng tr\u00ecnh l\u00e0 x \uf0b3 2 24.16. \u0110\u1eb7t y = x \uf02d1 th\u00ec y \uf0b3 0. Ph\u01a3\u01a1ng tr\u00ecnh tr\u1edf th\u00e0nh : y2 + 2y \u2013 8 = 0 \uf0db (y \u2013 2)(y + 4) = 0 \uf0db y = 2 ho\u1eb7c y = \u2013 4 (lo\u1ea1i) . V\u1eady y = 2 \uf0db x \uf02d1 =2 \uf0db \uf0e9x \uf02d1 \uf03d 2 \uf0db \uf0e9x \uf03d 3 \uf0ea\uf0ebx \uf02d1 \uf03d \uf02d2 \uf0eb\uf0eax \uf03d \uf02d1 Nghi\u1ec7m c\u1ee7a ph\u01a3\u01a1ng tr\u00ecnh l\u00e0 x = 3 v\u00e0 x = \u20131. 335","Website: tailieumontoan.com 24.17. * N\u1ebfu x \uf0b3 \uf02d2,5 th\u00ec 2x \uf02b 5 \uf03d 2x+5 khi \u1ea5y 2x+5 \uf03d x2 \uf02b 3x \uf02d1 \uf0db x2 + x \u2013 6 = 0 \uf0db (x + 3)(x \u2013 2) = 0 \uf0db x = \u2013 3 ho\u1eb7c x = 2 . Lo\u1ea1i x = \u2013 3 * N\u1ebfu x \uf03c \uf02d2,5 th\u00ec 2x \uf02b 5 \uf03d \uf02d2x \uf02d 5 khi \u1ea5y \uf02d2x \uf02d5 \uf03d x 2 \uf02b 3x \uf02d1 \uf0db x2 + 5x + 4 = 0 \uf0db (x + 4)(x + 1) = 0 \uf0db x = \u2013 4 ho\u1eb7c x = \u2013 1. Lo\u1ea1i x = \u2013 1. Nghi\u1ec7m c\u1ee7a ph\u01a3\u01a1ng tr\u00ecnh l\u00e0 x = 2 v\u00e0 x = \u2013 4. 24.18. Ta c\u00f3 ab \uf03d a . b n\u00ean x \uf02b1 \uf02b x \uf02d1 \uf03d1\uf02b x2 \uf02d1 \uf0db x \uf02b1. x \uf02d1 \uf02d x \uf02b1 \uf02d x \uf02d1 \uf02b1\uf03d 0 \uf0db \uf028 x \uf02b1 \uf02d1\uf029.\uf028 x \uf02d1 \uf02d1\uf029 \uf03d 0 \uf0db \uf0e9 x \uf02b 1 \uf02d1 \uf03d 0 (1) \uf0ea \uf0eb\uf0ea x \uf02d1 \uf02d1 \uf03d 0 (2) * (1) \uf0db x+1= \uf0b1 1\uf0db \uf0e9x \uf03d 0 ; (2) \uf0db x\u20131= \uf0b1 1 \uf0db \uf0e9x \uf03d 0 \uf0ea\uf0ebx \uf03d \uf02d2 \uf0ea\uf0ebx \uf03d 2 T\u1eadp nghi\u1ec7m c\u1ee7a ph\u01a3\u01a1ng tr\u00ecnh l\u00e0 S = \uf07b\uf02d2; 0; 2\uf07d 24.19. * Khi x = 2005 ; x = 2006 th\u00ec v\u1ebf tr\u00e1i v\u00e0 v\u1ebf ph\u1ea3i \u0111\u1ec1u c\u00f9ng s\u1ed1 tr\u1ecb l\u00e01. Do \u0111\u00f3 x = 2005 v\u00e0 x = 2006 l\u00e0 nghi\u1ec7m c\u1ee7a ph\u01a3\u01a1ng tr\u00ecnh. *V\u1edbi x < 2005 th\u00ec x \uf02d 2005 \uf03e 0 v\u00e0 x \uf02d 2006 \uf03e 1 Do \u0111\u00f3 x \uf02d 2005 2006 \uf02b x \uf02d 2006 2006 \uf03e 1 \uf0de ph\u01a3\u01a1ng tr\u00ecnh v\u00f4 nghi\u1ec7m. *V\u1edbi x > 2006 th\u00ec x \uf02d 2005 \uf03e 1 v\u00e0 x \uf02d 2006 \uf03e 0 Do \u0111\u00f3 x \uf02d 2005 2006 \uf02b x \uf02d 2006 2006 \uf03e 1 \uf0de ph\u01a3\u01a1ng tr\u00ecnh v\u00f4 nghi\u1ec7m. * V\u1edbi 2005 < x < 2006 th\u00ec 0 < x \u2013 2005 < 1 v\u00e0 \u2013 1 < x \u2013 2006 < 0 \uf0de x \uf02d 2005 2006 \uf03c x \uf02d 2005 \uf03d x \uf02d 2005 v\u00e0 x \uf02d 2006 2006 \uf03c x \uf02d 2006 \uf03d 2006 \uf02d x \uf0de x \uf02d 2005 2006 \uf02b x \uf02d 2006 2006 \uf03c x \uf02d 2005 \uf02b 2006 \uf02d x \uf03d1 \uf0de ph\u01a3\u01a1ng tr\u00ecnh v\u00f4 nghi\u1ec7m. V\u1eady nghi\u1ec7m c\u1ee7a ph\u01a3\u01a1ng tr\u00ecnh l\u00e0 x = 2005 v\u00e0 x = 2006. Chuy\u00ean \u0111\u1ec1 25: GI\u00c1 TR\u1eca NH\u1ece NH\u1ea4T V\u00c0 L\u1edaN NH\u1ea4T 25. 1. a) A(x) = 4(x + 1)2 + 11 \uf0b3 11, \uf022 x . V\u1eady min A(x) = 11 \uf0db x= \u2013 1. b) A(y) = 2y2 \u2013 16y \u2013 2 = 2(y \u2013 4)2 \u2013 34 \uf0b3 \u2013 34 , \uf022 y. V\u1eady min A(y) = \u2013 34 \uf0db y = 4. c) A(z) = 6z2 + 12z + 16 = 6(z + 1)2 + 10 \uf0b3 10, \uf022 z. V\u1eady min A(z) = 10 \uf0db z = \u2013 1. 25.2. a) B(x) = 24 \u2013 (x2 \u2013 6x + 9) = 24 \u2013 (x \u2013 3)2 \uf0a3 24 , \uf022 x . V\u1eady max B(x) = 24 \uf0db x= 3. ; 336","Website: tailieumontoan.com b) B(y) = \u2013 2y2 \u2013 4y + 10 = 12 \u2013 2(y + 1)2 \uf0a3 12 , \uf022 y; V\u1eady max B(y) = 12 \uf0db y = \u2013 1; c) R\u00fat g\u1ecdn \uf0e61 \uf02d1\uf0f7\uf0f6\uf0f8 \uf0e6 1 \uf02d1\uf0f8\uf0f7\uf0f6 \uf0e6 1 \uf02d 1\uf0f8\uf0f6\uf0f7 ... \uf0e6 1 \uf02d1\uf0f7\uf0f8\uf0f6 = \uf02d 11 (b\u1ea1n \u0111\u1ecdc t\u1ef1 r\u00fat g\u1ecdn . L\u01a3u \u00fd 1 \uf02d1 \uf03d \uf02d 1.3 ; \uf0e7\uf0e8 22 \uf0e8\uf0e7 32 \uf0e7\uf0e8 42 \uf0e8\uf0e7 102 20 22 2.2 1 \uf02d1 \uf03d \uf02d 2.4 ; .... ; 1 \uf02d1 \uf03d \uf02d 9.11 ). 32 3.3 102 10.10 Do \u0111\u00f3 B(z) = \u2013 20(z2 \u2013 2z + 3) = \u2013 40 \u2013 20(z \u2013 1)2 \uf0a3 \u2013 40 , \uf022 z; V\u1eady max B(z) = \u2013 40 \uf0db z = 1. 25.3. a) C = (x2 \u2013 8x + 15) (x2 \u2013 8x + 17) \u0110\u1eb7t x2 \u2013 8x + 16 = y ta c\u00f3 C = (y \u2013 1)(y + 1) = y2 \u2013 1 \uf0b3 \u2013 1 , \uf022 y. V\u1eady min C = \u2013 1 \uf0db y = 0 \uf0db (x \u2013 4)2 = 0 \uf0db x = 4. b) D = (1 \u2013 x)(x \u2013 5)(x2 \u2013 6x + 11) = \u2013 (x2 \u2013 6x + 5) )(x2 \u2013 6x + 11) \u0110\u1eb7t x2 \u2013 6x + 8 = y ta c\u00f3 D = \u2013 (y \u2013 3)(y + 3) = 9 \u2013 y2 \uf0a3 9, \uf022 y V\u1eady maxD = 9 \uf0db y = 0 \uf0db x2 \u2013 6x +8=0 \uf0db (x \u2013 2)(x \u2013 4) = 0 \uf0db \uf0e9x \uf03d 2 \uf0eb\uf0eax \uf03d 4 c) E = (x + 6)(x + 3)(x + 2)(x \u2013 1) + 1= (x2 + 5x \u2013 6)( x2 + 5x + 6) + 1 \u0110\u1eb7t x2 + 5x = y ta c\u00f3 E = (y \u2013 6)(y + 6) + 1 = y2 \u2013 36 + 1 \uf0b3 \u201335 , \uf022 y V\u1eady min E = \u2013 35 \uf0db y = 0 \uf0db x2 + 5x = (x + 5)x = 0 \uf0db x = 0 ; x = \u20135. d) \u0110\u1eb7t x \u2013 2015 = y th\u00ec F = 2018 \u2013 \uf0e9\uf0eb\uf028 y \uf02b 1\uf0294 \uf02b \uf028 y \uf02d 1\uf0294 \uf0f9 \uf0fb \u00c1p d\u1ee5ng h\u1eb1ng \u0111\u1eb3ng th\u1ee9c (a + b)4 = a4 + 4a3b + 6a2b2 + 4ab3 + b4 ta c\u00f3 F = 2018 \u2013 2(y4 + 6y2 + 1) = 2016 \u2013 2(y4 + 6y2) \uf0a3 2016 , \uf022 y V\u1eady max F = 2016 \uf0db y = 0 \uf0db x = 2015. 25.4. a) M(x, y) = ( x \u2013 y)2 + 3(y + 2)2 + 10 \uf0b3 10 , \uf022 x,y . Do \u0111\u00f3 min M(x,y) = 10 \uf0db (x = \u20132; y = \u20132) . b) N(x,y) = 2015 \u2013 (x + y + 1)2 \u2013 2(y \u2013 2)2 \uf0a3 2015, \uf022 x,y Do \u0111\u00f3 maxN(x,y) = 2015 \uf0db (x = \u2013 3; y = 2). c) P(x,y,z) = 15 \u2013 (x \u2013 1)2 \u2013 (y \u2013 2 )2 \u2013 (z \u2013 3)2 \uf0a3 15, \uf022 x,y,z. Do \u0111\u00f3 maxP(x,y,z) = 15 \uf0db (x = 1; y = 2 ; z = 3). d) Q(x,y,z,t) = (x + y + z)2 + (x + t)2 + (y \u2013 2)2 + (t \u2013 3)2 + 100 \uf0b3 100 , \uf022 x,y,z,t . Do \u0111\u00f3 min Q(x,y,z,t) = 100 \uf0db (x = \u20133; y = 2 ; z = 1; t = 3) . 25.5. a) R = \uf028x1 \uf02d 2\uf0292 \uf02b \uf028x2 \uf02d 2\uf0292 \uf02b \uf028x3 \uf02d 2\uf0292 \uf02b...\uf02b \uf028x10 \uf02d 2\uf0292 \uf02d 40 \uf0b3 \uf02d40 , \uf022xi(i \uf03d1,2,...,10) min R = \u2013 40 \uf0db (x1 = x2 = ...= x10 = 2) b) Ta c\u00f3 i2xi2 \uf02d 2ixi \uf02b1 \uf03d \uf028ixi \uf02d1\uf0292 . Do \u0111\u00f3 337","Website: tailieumontoan.com S = (x1 \uf02d1)2 \uf02b (2x2 \uf02d1)2 \uf02b (3x3 \uf02d1)2 \uf02b... \uf02b (nx n \uf02d1) 2 \uf02b n \uf0b3 n , \uf022x i(i \uf03d1, 2,..., n) Do \u0111\u00f3 min S = n \uf0db x1 \uf03d 1; x2 \uf03d 1; x3 \uf03d 1 ;...; xn \uf03d 1 . 2 3 n c) Ta c\u00e3 xi2 \uf02d 2ixi \uf02b i2 \uf03d \uf028 xi \uf02d i\uf0292 ( i = 1,2,3,...,n). Do \u0111\u00f3 : T = 100 \u2013 (x1 \u2013 1)2 \u2013 (x2 \u2013 2)2 \u2013 (x3 \u2013 3)2 \u2013 ... \u2013 (xn \u2013 n)2 \uf0a3 100, \uf022 xi Do \u0111\u00f3 maxT = 100 \uf0db (x1 = 1; x2 = 2 ; x3 = 3;...; xn = n). 25.6. a) A = (4x 200 \uf02b 20 \uf0a3 200 \uf03d 10 , \uf022x \uf02d1)2 20 V\u1eady maxA= 10 \uf0db x = 0,25 . b) B = (x2 \uf02d50 \uf03d \uf02d50 \uf02b2 \uf0b3 \uf02d50 \uf03d \uf02d 25 \uf022x \uf02d 4x \uf02b 4) \uf02b 2 2 \uf028x \uf02d 2\uf0292 V\u1eady min B = \u2013 25 \uf0db x = 2 . c) E = \uf028 \uf02d1\uf0292 \uf02b 2015 \uf02b \uf0a3 2015 \uf022 x, y 2016 x \uf028 y \uf02d1\uf0292 2016 V\u1eady max E = 2015 \uf0db \uf0ecx \uf03d1 . 2016 \uf0ee\uf0edy \uf03d 1 25.7: a) D = 4( x 2 \uf02b 2) \uf02b (x2 \uf02d 2x \uf02b 1) \uf03d 4\uf02b \uf028 x \uf02d1\uf0292 \uf0b3 4 , \uf022x x2 \uf02b 2 x2 \uf02b 2 V\u1eady min D = 4 \uf0db x = 1 . b) E = 5(x2 \uf02b 5) \uf02b1 \uf03d 5 \uf02b 1 .Do \uf022 x ta c\u00f3 x2 + 5 \uf0b3 5 \uf0db 1 \uf0a3 1 \uf0db 5 \uf02b 1 \uf0a3 26 , \uf022x x2 \uf02b5 x2 \uf02b \uf0de max E = 5,2 \uf0db x = 0 . 5 x2 \uf02b5 5 x2 \uf02b5 5 c) F \uf03d 2 \uf02b 2(x \uf02d 2)2 \uf0b3 2 , \uf022x . V\u1eady min F = 2 \uf0db x = 2. x2 \uf02b4 d) Q = 4\uf02b (x \uf02b 6 \uf02b 4 \uf0a3 4\uf02b 6 , \uf022x \uf0de max Q = 5,5 \uf0db x = \u20132 . 2)2 4 25.8. a) V\u1edbi x \uf0b9 \uf02d1; f(x) = 3(x \uf02b1) \uf02d 3 \uf03d 3\uf02d 3 . \u0110\u1eb7t y\uf03d 1 (x \uf02b1)2 x \uf02b1 (x \uf02b1)2 x \uf02b1 ta c\u00f3 f(x) = 3y \u2013 3y2 = \uf02d3\uf0e8\uf0e6\uf0e7 y2 \uf02d 2.y. 1 \uf02b 1 \uf02d 1 \uf0f6 \uf03d \uf02d3\uf0e7\uf0e6\uf0e8 y \uf02d 1 \uf0f62 \uf02b 3 \uf0a3 3 , \uf022y 2 4 4 \uf0f8\uf0f7 2 \uf0f8\uf0f7 4 4 V\u1eady max f(x) = 3 \uf0db y = 1 hay x = 1 . 42 b) g(x) = (x 1 \uf02d x 2 \uf02b3 \uf03d y2 \uf02d 2y \uf02b1\uf02b 2 \uf03d \uf028 y \uf02d1\uf0292 \uf02b 2 \uf0b3 2 \uf022y v\u1edbi y \uf03d 1 v\u00e0 v\u1edbi x \uf0b9 2 . V\u1eady \uf02d 2)2 \uf02d2 x\uf02d2 min g(x) = 2 \uf0db y = 1 hay x = 3 . 25.9. a) Ta ch\u1ee9ng minh A \uf0b3 6 , \uf022 x. Th\u1eadt v\u1eady x2 \u2013 6x + 15 \uf0b3 6 \uf0db x2\u2013 6x + 9 \uf0b3 0 \uf0db (x \u2013 3)2 \uf0b3 0 \u0111\u00fang \uf022 x. 338","D\u1ea5u \u201c=\u201d x\u1ea3y ra \uf0db x = 3. Website: tailieumontoan.com b) Ta ch\u1ee9ng minh B \uf0a3 8 , \uf022 x. Th\u1eadt v\u1eady: \uf022 x ta c\u00f3 \uf02dx2 \uf02b 4x \uf02b 4 \uf0a3 8 \uf0db \uf02dx2 \uf02b 4x \uf02b 4 \uf02d 8(x2 \uf02d 4x \uf02b 5) \uf0a30 \uf0db \uf02d9(x \uf02d 2)2 \uf0a3 0 hi\u1ec3n nhi\u00ean \u0111\u00fang . D\u1ea5u \u201c=\u201d x\u1ea3y ra x2 \uf02d 4x \uf02b 5 x2 \uf02d 4x \uf02b 5 (x \uf02d 2)2 \uf02b1 \uf0db x = 2. c) X\u00e9t C\u20131= 1\uf02b 2y \uf02d1 \uf03d 1\uf02b 2y \uf02d 2 \uf02d y2 \uf03d \uf02d(y \uf02d1)2 \uf0a3 0 , \uf022 y . Nh\u01a3 v\u1eady 2 \uf02b y2 2 \uf02b y2 2 \uf02b y2 C \uf0a3 1, \uf022 y , d\u1ea5u \u201c=\u201d x\u1ea3y ra \uf0db y = 1 ngh\u0129a l\u00e0 maxC = 1 \uf0db y =1. X\u00e9t C + 1 = 1\uf02b 2y \uf02b 1 \uf03d 2 \uf02b 4y \uf02b 2 \uf02b y2 \uf03d (y \uf02b 2)2 \uf0b3 0 , \uf022y. Nh\u01a3 v\u1eady 2 2 \uf02b y2 2 2 \uf02b y2 2 \uf02b y2 C \uf0b3 \u2013 0,5 , \uf022 y , d\u1ea5u \u201c=\u201d x\u1ea3y ra \uf0db y = \u20132 ngh\u0129a l\u00e0 minC = \u2013 0,5 \uf0db y = \u20132 25.10. a)V\u1edbi x > 4 th\u00ec A < 0 . V\u1edbi x \uf0ceZ X\u00e9t x < 4 th\u00ec m\u1eabu 4 \u2013 x l\u00e0 s\u1ed1 nguy\u00ean d\u01a3\u01a1ng.Ph\u00e2n s\u1ed1 A c\u00f3 t\u1eed v\u00e0 m\u1eabu \u0111\u1ec1u d\u01a3\u01a1ng, t\u1eed b\u1eb1ng 30 kh\u00f4ng \u0111\u1ed5i n\u00ean A l\u1edbn nh\u1ea5t \uf0db m\u1eabu ( 4 \u2013 x) l\u00e0 s\u1ed1 nguy\u00ean d\u01a3\u01a1ng nh\u1ecf nh\u1ea5t. Do \u0111\u00f3 4 \u2013 x = 1 \uf0db x = 3 khi \u0111\u00f3 A = 30. V\u1eady maxA = 30 \uf0db x = 3 b) V\u1edbi x \uf0b9 3 th\u00ec B \uf03d x \uf02d 26 \uf03d (x \uf02d 3) \uf02d 23 \uf03d 1\uf02d 23 \uf03d 1\uf02b 23 x \uf02d3 x \uf02d3 x \uf02d3 3\uf02dx B l\u1edbn nh\u1ea5t khi 23 l\u1edbn nh\u1ea5t . N\u1ebfu x > 3 th\u00ec 23 < 0 3\uf02dx 3\uf02dx N\u1ebfu x < 3 th\u00ec 23 > 0 n\u00ean 23 l\u1edbn nh\u1ea5t \uf0db (3 \u2013 x) nh\u1ecf nh\u1ea5t 3\uf02dx 3\uf02dx \uf0ec3 \uf02d x \uf03e 0 \uf0ef \uf0ed 3 \uf02d x nh\u1ecf nh\u1ea5t \uf0de 3 \u2013 x = 1 hay x = 2 . \uf0ee\uf0ef(3 \uf02d x) \uf0ce Z Khi \u0111\u00f3 max B = 24 \uf0db x = 2. c) V\u1edbi x \uf0b9 1945 th\u00ec C = 1975 \uf02d x \uf03d 30 \uf02d (x \uf02d1945) \uf03d 30 \uf02d1 x \uf02d1945 x \uf02d1945 x \uf02d1945 \u0110\u1eb7t E = 30 Ta c\u00f3 C nh\u1ecf nh\u1ea5t \uf0db E nh\u1ecf nh\u1ea5t . x \uf02d1945 *V\u1edbi x > 1945 th\u00ec E > 0 *V\u1edbi x < 1945 th\u00ec E < 0 n\u00ean C nh\u1ecf nh\u1ea5t \uf0db s\u1ed1 \u0111\u1ed1i c\u1ee7a E l\u1edbn nh\u1ea5t \uf0db 30 l\u1edbn nh\u1ea5t . Do 1945 \u2013 1945 \uf02d x x > 0 n\u00ean 30 l\u1edbn nh\u1ea5t \uf0db 1945 \uf02d x (1945 \u2013 x) nh\u1ecf nh\u1ea5t 339","\uf0ec1945 \uf02d x \uf03e 0 Website: tailieumontoan.com \uf0ed\uf0ef(1945 \uf02d x) nh\u1ecf nh\u1ea5t \uf0ef\uf0ee1945 \uf02d x \uf0ce Z \uf0de 1945 \u2013 x = 1. Khi \u0111\u00f3 C = \u2013 31 . V\u1eady min C = \u2013 31 \uf0db x = 1944. 25.11. a) D = (9x2 \uf02b 2) \uf02d (9x 2 \uf02d 6x \uf02b1) \uf03d 1\uf02d (3x \uf02d1)2 \uf0a3 1 \uf022 x. 9x2 \uf02b 2 9x2 \uf02b 2 Do \u0111\u00f3 maxD = 1 \uf0db x = 1 3 D= 12x \uf02b 2 \uf03d (9x2 \uf02b12x \uf02b 4) \uf02d (9x2 \uf02b 2) \uf03d (3x \uf02b 2)2 \uf02d 1 \uf0b3 \uf02d 1 \uf022 x. 2(9x2 \uf02b 2) 2(9x2 \uf02b 2) 2(9x2 \uf02b 2) 2 2 Do \u0111\u00f3 minD = \uf02d 1 \uf0db x = \uf02d2 2 3 b) E = 2(x2 \uf02b 2x \uf02b 3) \uf02d x 2 \uf03d 2 \uf02d x 2 \uf0a3 2 , \uf022x x2 \uf02b 2x \uf02b 3 x2 \uf02b 2x \uf02b 3 Do \u0111\u00f3 maxE = 2 \uf0db x = 0 E= 2x2 \uf02b 8x+12 (x2 \uf02b 2x \uf02b 3) \uf02b (x2 \uf02b 6x \uf02b 9) \uf03d 1 \uf02b (x \uf02b 3)2 \uf0b3 1 , \uf022x 2(x2 \uf02b 2x \uf02b 3) = 2(x2 \uf02b 2x \uf02b 3) 2 2(x2 \uf02b 2x \uf02b 2 3) Do minE = 1 \uf0db x = \u2013 3. 2 c) G= 2(x2 \uf02b x \uf02b1) \uf03d 2 \uf0a32,\uf022x \uf0b3 0. (x2 \uf02b x \uf02b1) \uf02b x 1\uf02b x \uf02bx x2 \uf02b1 V\u1eady max G = 2 \uf0db x = 0. *X\u00e9t v\u1edbi x > 0 th\u00ec G \uf03d 2x2 \uf02b 4x \uf02b 2 \uf02d 2x \uf03d 2 \uf02d 2x \uf03d 2 \uf02d 2 Do x \uf02b 1 \uf0b3 2 n\u00ean x2 \uf02b 2x \uf02b1 \uf02b 2x 2\uf02b x x2 \uf02b1 x \uf02b 1 x 2\uf02d 2 \uf0b3 3 , \uf022 x > 0. V\u1eady min G = 1,5 \uf0db x=1. x\uf02b 1 2 \uf02b2 x d) Ta c\u00f3 2xy \uf0a3 x2 \uf02b y2 \uf0de (x + y)2 \uf0a3 2(x2 \uf02b y2 ) = 100 \uf0de x \uf02b y \uf0a3 10 \uf0de \u2013 10 \uf0a3 x \uf02b y \uf0a310 V\u1eady max K = 10 \uf0db x = y = 5 ; minK = \u2013 10 \uf0db x = y = \u2013 5 25.12. \u00c1p d\u1ee5ng tr\u1ef1c ti\u1ebfp \u0111\u1ecbnh l\u00fd v\u1ec1 c\u1ef1c tr\u1ecb. 340","Website: tailieumontoan.com 25.13. a) B = 2x2 \uf02b 25x \uf02b 72 \uf03d 2x \uf02b 72 \uf02b 25 xx Ta c\u00f3 v\u1edbi x > 0 th\u00ec 2x v\u00e0 72 l\u00e0 hai s\u1ed1 d\u01a3\u01a1ng c\u00f3 t\u00edch b\u1eb1ng 144 kh\u00f4ng \u0111\u1ed5i x n\u00ean t\u1ed5ng c\u1ee7a ch\u00fang nh\u1ecf nh\u1ea5t khi v\u00e0 ch\u1ec9 khi hai s\u1ed1 \u0111\u00f3 b\u1eb1ng nhau t\u1ee9c l\u00e0 : 2x \uf03d 72 \uf0db x2 = 36 . Nghi\u1ec7m x = 6 th\u1ecfa m\u00e3n \u0111i\u1ec1u ki\u1ec7n c\u1ee7a b\u00e0i . x V\u1eady min B = 49 \uf0db x = 6 . b) C = (x2 \uf02b 2x \uf02b1) \uf02b 4 \uf02d 2(x \uf02b1) \uf03d (x \uf02b 1) \uf02b 4 \uf02d 2 x \uf02b1 x \uf02b1 Ta c\u00f3 v\u1edbi x \uf0b3 0 , 2 s\u1ed1 d\u01a3\u01a1ng x + 1 v\u00e0 4 c\u00f3 t\u00edch b\u1eb1ng 4 kh\u00f4ng \u0111\u1ed5i x \uf02b1 N\u00ean C nh\u1ecf nh\u1ea5t \uf0db x \uf02b1\uf03d 4 \uf0db (x + 1)2 = 4 . Nghi\u1ec7m x = 1 th\u1ecfa m\u00e3n \u0111i\u1ec1u ki\u1ec7n \u0111\u1ea7u b\u00e0i . V\u1eady min C x \uf02b1 =2 \uf0db x=1 . c) T\u1ed5ng (x2 \u2013 5x \u2013 20) + (28 \u2013 x2 + 5x) = 8 kh\u00f4ng \u0111\u1ed5i n\u00ean t\u00edch c\u1ee7a ch\u00fang l\u1edbn nh\u1ea5t khi hai s\u1ed1 \u0111\u00f3 b\u1eb1ng nhau x2 \u2013 5x \u2013 20 = 28 \u2013 x2 + 5x \uf0db x2 \u2013 5x \u2013 24 = 0 \uf0db (x + 3)(x \u2013 8) = 0 \uf0db x = \u2013 3 ; x = 8. V\u1eady max D = 4.4 = 16 \uf0db \uf0e9x \uf03d \uf02d3 \uf0eb\uf0eax \uf03d 8 . 25.14. a) Ta \u0111\u00e3 bi\u1ebft x \uf02b 1 \uf0b3 2 ,\uf022x \uf03e 0 . Do \u0111\u00f3 a \uf02b b \uf0b3 2 (1). x ba Do vai tr\u00f2 c\u1ee7a a, b, c l\u00e0 nh\u01a3 nhau n\u00ean ta gi\u1ea3 s\u1eed a \uf0b3 b \uf0b3 c \uf03e 0. Ta c\u00f3 a \u2013 c \uf0b3 0 v\u00e0 b(a \u2013 c) \uf0b3 c(a \u2013 c) \uf0de ab \u2013 bc + c2 \uf0b3 ac \uf0de b \uf02d b \uf02b c \uf0b31 (2). T\u1eeb (1) v\u00e0 (2) \uf0de a \uf02b b \uf02b c \uf0b3 3 \uf0de G = 2020 \u2013 \uf0e6 a \uf02b b \uf02b c \uf0f6 \uf0a3 2017. caa b c a \uf0e7\uf0e8 b c a \uf0f7\uf0f8 V\u1eady maxG = 2017 \uf0db a = b = c v\u00e0 a,b, c > 0. b) H\uf03d 4 \uf02b \uf0e6 a \uf02b b \uf0f6 \uf02b \uf0e6 a \uf02b c \uf0f6 \uf02b \uf0e6 a \uf02b d \uf0f6 \uf02b \uf0e6 b \uf02b c \uf0f6 \uf02b \uf0e6 b \uf02b d \uf0f6 \uf02b \uf0e6 c \uf02b d \uf0f6 \uf02b 4 \uf0b3 8 \uf02b 2.6 \uf03d 20 \uf0e8\uf0e7 b a \uf0f8\uf0f7 \uf0e7\uf0e8 c a \uf0f7\uf0f8 \uf0e8\uf0e7 d a \uf0f7\uf0f8 \uf0e7\uf0e8 c b \uf0f7\uf0f8 \uf0e7\uf0e8 d b \uf0f7\uf0f8 \uf0e8\uf0e7 d c \uf0f8\uf0f7 V\u1eady min H = 20 \uf0db a = b = c = d v\u00e0 a, b, c, d > 0 . 25.15. a) K = \uf0e6 x \uf02b y\uf02b z\uf0f6 + \uf0e6 x \uf02b y \uf02b z \uf0f6 \uf0e7 y z \uf0f7 \uf0e7 \uf02b \uf02b \uf02b \uf0f7 \uf0e8 x \uf0f8 \uf0e8 y z z x x y \uf0f8 Ta c\u00f3 x \uf02b y \uf02b z \uf0b3 3 (xem b\u00e0i t\u1eadp 25.14) v\u00e0 x \uf02b y \uf02b z \uf0b3 3 (xem v\u00ed d\u1ee5 8 chuy\u00ean \u0111\u1ec1 20) \uf0de K yzx y\uf02bz z\uf02bx x\uf02by 2 \uf0b33\uf02b 3 \uf03d 9 22 341","Website: tailieumontoan.com V\u1eady min K = 4,5 \uf0db x = y = z v\u00e0 x, y, z > 0 . b) Bi\u1ebfn \u0111\u1ed5i L = x \uf02b y \uf02b z \uf02b y \uf02b z \uf02b z \uf02b x \uf02b x \uf02b y y\uf02bz z\uf02bx x\uf02by x y z = x\uf02by\uf02b x z y \uf02b \uf0e6 x \uf02b y \uf0f6 \uf02b \uf0e6 x \uf02b z \uf0f6 \uf02b \uf0e6 y \uf02b z \uf0f6 \uf0b3 3 \uf02b2\uf02b2\uf02b2 y\uf02bz z\uf02bx \uf02b \uf0e7 y x \uf0f7 \uf0e7\uf0e8 z x \uf0f8\uf0f7 \uf0e7 z y \uf0f7 2 \uf0e8 \uf0f8 \uf0e8 \uf0f8 V\u1eady min L = 7,5 \uf0db x = y = z v\u00e0 x, y, z > 0 . 25.16 : a) a + b = 4 \uf0de 16 = a2 + b2 + 2ab = 2(a2 + b2 ) \u2013 (a \u2013 b)2 \uf0de 16 \uf0a3 2(a2 + b2 ) \uf0de a2 + b2 \uf0b3 8 . V\u1eady min D = 8 \uf0db a = b = 2 . b) Ta c\u00f3 3(a2 + b2 + c2) \uf0b3 (a + b + c)2 (xem b\u00e0i t\u1eadp 21.1) Do \u0111\u00f3 3E \uf0b3 (a + b + c)2 = 9. V\u1eady min E = 3 \uf0db a = b = c = 1 c) F = a3 + b3 + 2ab = (a + b)(a2 \u2013 ab + b2) + 2ab . Do a + b = 2 n\u00ean F = 2(a2 \u2013 ab + b2) + 2ab = 2a2 + 2b2 = 2a2 + 2(2 \u2013 a)2 = 4a2 \u2013 8a + 8 = 4\uf028a \uf02d1\uf0292 \uf02b 4 \uf0b3 4 ,\uf022a V\u1eady min F = 4 \uf0db a = b = 1. 25.17. a) a + 2b =2 \uf0de a = 2 \u2013 2b \uf0de G = 2ab = 4(1 \u2013 b).b = \u20134(b2 \u2013 b) = \uf02d4 \uf0e6 b \uf02d 1 \uf0f62 \uf02b 1\uf0a3 1 ,\uf022 b V\u1eady max G = 1 \uf0db b = 1 v\u00e0 a = 1 . \uf0e8\uf0e7 2 \uf0f7\uf0f8 2 b) \u0110\u1eb7t a + 1 = x ; b + 1 = y ; c + 1 = z th\u00ec x + y + z = a + b + c + 3 \uf0a3 6 n\u00ean 1 \uf0b3 1 v\u00e0 . x\uf02by\uf02bx 6 Ta c\u00f3 (x + y + z) \uf0e6\uf0e8\uf0e7\uf0e7 1 \uf02b 1 \uf02b 1 \uf0f6\uf0f8\uf0f7\uf0f7 \uf0b3 9 (xem v\u00ed d\u1ee5 7 chuy\u00ean \u0111\u1ec1 21) \uf0de 1 \uf02b 1\uf02b 1 \uf0b3 9 \uf0b3 9 \uf03d 3 \uf0de x y z x y z x\uf02by\uf02bz 6 2 1\uf02d \uf0e6 1 \uf02b 1 \uf02b 1 \uf0f6 \uf0a3 1\uf02d 3 \uf03d \uf02d 1 \uf0e7 x y z \uf0f7 2 2 \uf0e8 \uf0f8 maxH = \uf02d 1 \uf0db x \uf03d y \uf03d z \uf03d 2 \uf0db a = b = c = 1. 2 25.18. a) S\u1eed d\u1ee5ng b\u1ea5t \u0111\u1eb3ng th\u1ee9c a \uf02b b \uf0b3 a \uf02b b . D\u1ea5u \u201c=\u201d x\u1ea3y ra \uf0db ab \uf0b3 0 . L= 5x \uf02d 2010 \uf02b 5x \uf02d 2020 = 5x \uf02d 2010 \uf02b 2020 \uf02d 5x \uf0b3 5x \uf02d 2010 \uf02b 2020 \uf02d 5x =10 V\u1eady L \uf0b3 10 . D\u1ea5u \u201c=\u201dx\u1ea3y ra \uf0db (2020 \u2013 5x)(5x \u2013 2010) \uf0b3 0 \uf0db 402 \uf0a3 x \uf0a3 404. Do \u0111\u00f3 min L = 10 \uf0db 402 \uf0a3 x \uf0a3 404. (c\u00f3 th\u1ec3 l\u1eadp b\u1ea3ng x\u00e9t gi\u00e1 tr\u1ecb tuy\u1ec7t \u0111\u1ed1i \u0111\u1ec3 gi\u1ea3i) b) \u00a7\u00c6t M1 \uf03d x \uf02d 2015 \uf02b x \uf02d 2018 ; M2 \uf03d x \uf02d 2016 \uf02b x \uf02d 2017 Gi\u1ea3i t\u01a3\u01a1ng t\u1ef1 a) ta c\u00f3 min M1 = 3 \uf0b3 \uf0db 2015 \uf0a3 x \uf0a3 2018 min M2 = 1 \uf0b3 \uf0db 2016 \uf0a3 x \uf0a3 2017 V\u00cby min M = 4 \uf0db 2016 \uf0a3 x \uf0a3 2017 . 342","Website: tailieumontoan.com c) \u0110\u1eb7t 19x \uf02d 8 = y th\u00ec N = y2 \u2013 10y + 25 + 1945 = (y \u2013 5)2 + 1945\uf0b3 1945. V\u1eady min N = 1945 \uf0db y = 5 \uf0db 19x \uf02d 8 = 5 \uf0db x = 13 ; x = 3 . 19 19 25. 19. a) maxP = 8 \uf0db y =\u2013 3 b) \uf022y ta c\u00f3 7y \uf02d 5 \uf0b3 0 \uf0db 7y \uf02d 5 \uf02b 60 \uf0b3 60 \uf0db 1 \uf0a3 1 7y \uf02d 5 \uf02b 60 60 \uf0db 2014 \uf0a3 2014 \uf0db 2014 \uf02d 1954\uf0a3 2014\uf02d 1954\uf03d1 7y \uf02d 5 \uf02b 60 60 7y\uf02d 5 \uf02b 60 60 60 60 V\u1eady maxQ = 1 \uf0db y = 5 . 7 c) V\u1edbi x \uf0a3 \uf02d5 th\u00ec T = \u2013 x \u2013 5 + x + 2 = \u20133 V\u1edbi \u2013 5 < x < \uf02d2 th\u00ec T = x + 5 + x + 2 = 2x + 7 Do \u2013 5 < x < \uf02d2 n\u00ean \u2013 10 < 2x < \uf02d4 \uf0de \u2013 3 < T < 3 V\u1edbi x \uf0b3 \u2013 2 th\u00ec T = x + 5 \u2013 x \u2013 2 = 3 V\u1eady max T = 3 \uf0db x \uf0b3 \u2013 2 25.20. \u0110\u1eb7t S1 = z \uf02d1 \uf02b z \uf02d100 ; S2 = z \uf02d 2 \uf02b z \uf02d 99 ; S3 = z \uf02d 3 \uf02b z \uf02d 98 ; ...; S50 = z \uf02d 50 \uf02b z \uf02d 51 . T\u01a3\u01a1ng t\u1ef1 b\u00e0i 24.18 a) Ta c\u00f3 : min S1 = 99 \uf0db 1 \uf0a3 z \uf0a3 100 min S2 = 97 \uf0db 2 \uf0a3 z \uf0a3 99 min S3 = 95 \uf0db 3 \uf0a3 z \uf0a3 98 .......................................................... min S49 = 3 \uf0db 49 \uf0a3 z \uf0a3 52 min S50 = 1 \uf0db 50 \uf0a3 z \uf0a3 51 Ta c\u00f3 1 + 3 + 5 + ...+ 97 + 99 = (1 + 99).50 : 2 = 2500 V\u1eady minS = min S1+ min S2+ min S3+... min S49+ min S50 = 2500 \uf0db 50 \uf0a3 z \uf0a3 51. 25.21. \u00c1p d\u1ee5ng b\u1ea5t \u0111\u1eb3ng th\u1ee9c a \uf02b b \uf0b3 a \uf02b b , d\u1ea5u \u201c=\u201d \uf0db ab \uf0b3 0 Ta c\u00f3 y \uf03d x2 \uf02b x \uf02b16 \uf02b 6 \uf02d x2 \uf02d x \uf0b3 x2 \uf02b x \uf02b16 \uf02b 6 \uf02d x2 \uf02d x \uf03d 22 . D\u1ea5u \u201c=\u201d x\u1ea3y ra \uf0db (x2 \uf02b x \uf02b16)(6 \uf02d x2 \uf02d x) \uf0b3 0 \uf0db 6 \u2013 x2 \u2013 x \uf0b3 0 do x2 + x + 16 = \uf0e6 x \uf02b 1 \uf0f62 \uf02b 63 \uf03e 0 , \uf022x. \uf0e8\uf0e7 2 \uf0f8\uf0f7 4 Hay l\u00e0 x2+ x \u2013 6 \uf0a3 0 \uf0db (x + 3)(x \u2013 2) \uf0a3 0 \uf0db \u2013 3 \uf0a3 x \uf0a3 2 343","Website: tailieumontoan.com V\u1eady min y = 22 \uf0db \u2013 3 \uf0a3 x \uf0a3 2. 25.22. Ta c\u00f3 2A = 8 \u2013 [(x2 \u2013 2xy + y2) +(x2 \u2013 4x + 4) +(y2 \u2013 4y + 4)] = 8 \u2013 [(x \u2013 y)2 + (x \u2013 2)2 +(y \u2013 2)2] \uf0a3 8 \uf0de max A = 4 \uf0db x = y = 2. 25.23. P = (x3 \uf02b y3) \uf02d (x2 \uf02b y2 ) \uf03d x2 (x \uf02d1) \uf02b y2(y\uf02d1) \uf03d x2 \uf02b y2 (x \uf02d1)(y \uf02d1) (x \uf02d1)(y \uf02d1) y \uf02d1 x \uf02d1 \u0110\u1eb7t x \u2013 1 = a v\u00e0 y \u2013 1 = b, do x > 1 v\u00e0 y > 1 n\u00ean a > 0 v\u00e0 b > 0 \u0111\u1ed3ng th\u1eddi x = a + 1 v\u00e0 y = b + 1. Khi \u1ea5y P = (a \uf02b1)2 \uf02b (b\uf02b1)2 . ba \u00c1p d\u1ee5ng b\u1ea5t \u0111\u1eb3ng th\u1ee9c (x + y)2 \uf0b3 4xy v\u00e0 x + 1 \uf0b3 2 (v\u1edbi x > 0) ta c\u00f3 x (a + 1)2 \uf0b3 4a ; (b + 1)2 \uf0b3 4b ; N\u00ean P \uf0b3 4a \uf02b 4b \uf03d 4 \uf0e6 a \uf02b b \uf0f6 \uf0b3 8 b a \uf0e8\uf0e7 b a \uf0f7\uf0f8 V\u1eady min P = 2 \uf0db a = b = 1 hay x = y = 2. 25.24. Ta c\u00f3 P = (4x2 + 9y2 + 64 \u2013 12xy + 32x \u2013 48y) + (x2 \u2013 8x + 16) + 2 = ( 2x \u2013 3y + 8)2 + (x \u2013 4)2 + 2 \uf0b3 2 D\u1ea5u \u201c=\u201d x\u1ea3y ra \uf0db \uf0ec2x \uf02d 3y \uf02b 8 \uf03d 0 \uf0db \uf0ef\uf0ecx \uf03d 4 . V\u1eady min P = 2 \uf0db \uf0ef\uf0ecx \uf03d 4 \uf0ee\uf0edx \uf02d 4 \uf03d 0 \uf0ed \uf03d 16 \uf0ed \uf03d 16 \uf0ee\uf0efy 3 \uf0ee\uf0efy 3 25.25. S\u1eed d\u1ee5ng b\u1ea5t \u0111\u1eb3ng th\u1ee9c (a + b)2 \uf0b3 4ab Ta c\u00f3 : (x + 2010)2 \uf0b3 4. 2010.x \uf0de N= x \uf0a31 (x \uf02b 2010 )2 8040 V\u1eady max N = 1 , \u0111\u1ea1t \u0111\u01a3\u1ee3c khi x = 2010 . 8040 25.26. B = 3\uf02b x2 1 \uf03d 3\uf02b 1 \uf02b2 \uf0a3 7 \uf02b 2x \uf02b 3 (x \uf02b1)2 2 \uf0de max B = 3,5 \uf0db x = \u2013 1 25.27. X\u00e9t 4P = (4x2 + y2 + 4 + 4xy \u2013 8x \u2013 4y) + 3y2 \u2013 8y + 8036 = (2x + y \u2013 2)2 + 3\uf0e7\uf0e8\uf0e6 y2 \uf02d 8 y \uf02b 8036 \uf0f6 = (2x + y \u2013 2)2+ 3\uf0e8\uf0e7\uf0e6 y \uf02d 4 \uf0f62 \uf02b 24092 \uf0b3 24092 33 \uf0f8\uf0f7 3 \uf0f7\uf0f8 3 3 \uf0de min(4P) = 24092 \uf0de min P = 6023 \uf0db y \uf03d 4 ; x \uf03d 1 . 3 3 33 25.28. a) M = 2 \uf0db 10x \uf02b y \uf03d 2 \uf0db 10x + y = 2x + 2y \uf0db y = 8x x\uf02by V\u00ec x; y l\u00e0 c\u00e1c ch\u1eef s\u1ed1 n\u00ean 1\uf0a3 x \uf0a3 9 ; 0 \uf0a3 y \uf0a3 9 \uf0de x = 1; y = 8 ; n = 18. 344","Website: tailieumontoan.com b) M = x \uf02b y \uf02b 9x \uf03d1\uf02b 9x \uf03d 1\uf02b 9 . M nh\u1ecf nh\u1ea5t \uf0db y l\u1edbn nh\u1ea5t x\uf02by x\uf02by 1\uf02b x y x \uf0db y l\u1edbn nh\u1ea5t v\u00e0 x nh\u1ecf nh\u1ea5t \uf0de y = 9 ; x = 1 v\u00e0 n = 19 \uf0de minM = 19 . 10 25.29. Do vai tr\u00f2 a, b, c nh\u01a3 nhau. Gi\u1ea3 s\u1eed a = max\uf07ba, b, c\uf07d khi \u0111\u00f3 2 \uf0a3 a \uf0a3 4 . Ta c\u00f3 P \uf03d a2 \uf02b b2 \uf02b c2 \uf02b (a \uf02b b \uf02b c)2 \uf03d a2 \uf02b b2 \uf02b c2 \uf02b 36 . M\u1eb7t kh\u00e1c v\u00ec bc \uf0b3 0 n\u00ean a2 + b2 + c2= a2 + (b + c)2 \u2013 2bc \uf0a3 a2+(6 22 \u2013 a)2 = 2a2 \u2013 12a + 36 \uf0eca \uf0b3 b;a \uf0b3 c; bc \uf03d 0 = 2\uf05b(a \uf02d 2)(a\uf02d 4)\uf02b 10\uf05d \uf0a3 20 \uf0de max(a2 + b2 + c2) = 20 \uf0db \uf0ed\uf0ef(a \uf02d 2)(a \uf02d 4) \uf03d 0 \uf0ef\uf0eea \uf02b b \uf02b c \uf03d 6 \uf0db (a; b; c) = (4; 2; 0) ho\u1eb7c (4; 0; 2). Khi \u0111\u00f3 max P = 28 \uf0db (a; b; c) = (4; 2; 0) v\u00e0 c\u00e1c ho\u00e1n v\u1ecb c\u1ee7a n\u00f3. Chuy\u00ean \u0111\u1ec1 26: \u0110\u1ed2NG D\u01a2 TH\u1ee8C 26.1. V\u1edbi nh\u1eefng b\u00e0i to\u00e1n d\u1ea1ng n\u00e0y, ph\u01a3\u01a1ng ph\u00e1p chung l\u00e0 t\u00ednh to\u00e1n \u0111\u1ec3 \u0111i \u0111\u1ebfn a \uf0ba b (mod m) v\u1edbi b l\u00e0 s\u1ed1 c\u00f3 tr\u1ecb tuy\u1ec7t \u0111\u1ed1i nh\u1ecf nh\u1ea5t c\u00f3 th\u1ec3 \u0111\u01a3\u1ee3c (t\u1ed1t nh\u1ea5t l\u00e0 b = \uf0b1 1) t\u1eeb \u0111\u00f3 t\u00ednh \u0111\u01a3\u1ee3c thu\u1eadn l\u1ee3i an \uf0ba bn (mod m) a) 8! = 1.2.3.4.5.6.7.8. Ta c\u00f3 3.4 = 12 \uf0ba 1 (mod 11) ; 2.6 = 12 \uf0ba 1 (mod 11) ; 7.8 \uf0ba 1 (mod 11) V\u1eady 8! \uf0ba 5 (mod 11) \uf0de 8! \u2013 1 \uf0ba 4 (mod 11). S\u1ed1 d\u01a3 trong ph\u00e9p chia 8! \u2013 1 cho 11 l\u00e0 4. b) 2014 \uf0ba \u2013 1 (mod 5) \uf0de 20142015 \uf0ba \u2013 1 (mod 5) 2016 \uf0ba 1 (mod 5) \uf0de 20162015 \uf0ba 1 (mod 5) ; 2018 \uf0ba 3 (mod 5) 20142015 + 20162015 + 2018 \uf0ba 3 (mod 5). c) 23 \uf0ba 1 (mod 7) \uf0de 250 = (23)16. 4 \uf0ba 4 (mod 7) 41 \uf0ba \u20131 (mod 7) \uf0de 4165 \uf0ba (\u20131)65 \uf0ba \u20131 (mod 7) 250 + 4165 \uf0ba 4 \u2013 1 \uf0ba 3 (mod 7). d) 15 \uf0ba 1 (mod 4); 35 \uf0ba \u2013 1 (mod 4) ; 55 \uf0ba 1 (mod 4) ; ...; 975 \uf0ba 1 (mod 4); 995 \uf0ba \u2013 1 (mod 4). \u0110\u00e1p s\u1ed1 : D\u01a3 0 . 26.2. a) 1532 \uf0ba 2 (mod 9) \uf0de 15325 \uf0ba 25 \uf0ba 5 (mod 9) \uf0de 15325 \u2013 4 \uf0ba 1 (mod 9) b) 25 = 32 \uf0ba 7 (mod 25) \uf0de 210 = (25)2 \uf0ba 72 \uf0ba \u2013 1 (mod 25). 22000 = (210)200 \uf0ba (\u2013 1)200 \uf0ba 1 (mod 25). c) 2014 = 155.13 \u2013 1 n\u00ean 2014 \uf0ba \u2013 1 (mod 13); 20152016 = 2k + 1 (k\uf0ce N) \uf0de 201420152016 \uf0ba (\u2013 1)2k+1 \uf0ba \u2013 1 (mod 13). \u0110\u00e1p s\u1ed1 : d\u01a3 12. 345","Website: tailieumontoan.com 26.3. a) Ta c\u00f3 352 = 1225 = 425.3 \u2013 50 \uf0ba \u201350(mod 425) 353 = 352. 35 \uf0ba \u201350. 35 \uf0ba \u2013 1750 \uf0ba \u201350(mod 425) 354 = (352)2 \uf0ba (\u2013 50)2 \uf0ba 2500 \uf0ba \u201350(mod 425) T\u01a3\u01a1ng t\u1ef1 v\u1edbi 358 ; 3516 ; 3532 . T\u1eeb \u0111\u00f3 c\u00f3 A \uf0ba \u2013100(mod 425). Hay s\u1ed1 d\u01a3 trong ph\u00e9p chia A cho 425 l\u00e0 325. b) Ta c\u00f3 105 = 7.14285 + 5 \uf0ba 5(mod 7); 106 = 5.10 \uf0ba 1(mod 7); 10n \u2013 4 = 99...96 \uf0ba 0 (mod 2) v\u00e0 99...96 \uf0ba 0(mod 3) \uf0de 10n \u2013 4 \uf0ba 0(mod 6) n \uf02d1 so\u02c6 \uf0a29 n \uf02d1 so\u02c6 \uf0a29 \uf0de 10n \uf0ba 4(mod 6) v\u00e0 10n = 6k + 4 (k, n \uf0ce N*). \uf028 \uf029Do \u0111\u00f3 1010n \uf03d 106k\uf02b4 \uf03d 106 k .104 \uf0ba 104 (mod 7) V\u1eady B \uf0ba 104 +104 +104 +... +104 \uf0ba 10. 104 \uf0ba 105 \uf0ba 5(mod 7). 26. 4. a) Ta t\u00ecm d\u01a3 trong ph\u00e9p chia s\u1ed1 \u0111\u00f3 cho 10. V\u00ec 42 \uf0ba 6(mod 10) n\u00ean 432 = 49 = (42)4.4 \uf0ba 6.4 \uf0ba 4(mod 10) \uf0de ch\u1eef s\u1ed1 t\u1eadn c\u00f9ng l\u00e0 4. b) Ta t\u00ecm d\u01a3 trong ph\u00e9p chia s\u1ed1 \u0111\u00f3 cho 100. Theo v\u00ed d\u1ee5 3 chuy\u00ean \u0111\u1ec1 26 ta \u0111\u00e3 c\u00f3 31000 \uf0ba 01 (mod 100) ngh\u0129a l\u00e0 hai ch\u1eef s\u1ed1 sau c\u00f9ng c\u1ee7a 31000 l\u00e0 01. S\u1ed1 31000 l\u00e0 b\u1ed9i s\u1ed1 c\u1ee7a 3 n\u00ean ch\u1eef s\u1ed1 h\u00e0ng tr\u0103m c\u1ee7a n\u00f3 khi chia cho 3 ph\u1ea3i c\u00f3 s\u1ed1 d\u01a3 l\u00e0 2 \u0111\u1ec3 chia ti\u1ebfp th\u00ec 201 chia h\u1ebft cho 3 ( n\u1ebfu s\u1ed1 d\u01a3 l\u00e0 0 hay 1 th\u00ec 001; 101 \u0111\u1ec1u kh\u00f4ng chia h\u1ebft cho 3). V\u1eady s\u1ed1 3999 = 31000 : 3 c\u00f3 hai ch\u1eef s\u00f4 t\u1eadn c\u00f9ng b\u1eb1ng 201 : 3 = 67. c) Ta t\u00ecm d\u01a3 trong ph\u00e9p chia s\u1ed1 \u0111\u00f3 cho 1000. Do 1000 = 125.8 tr\u01a3\u1edbc h\u1ebft ta t\u00ecm s\u1ed1 d\u01a3 c\u1ee7a 2512 cho 125. T\u1eeb h\u1eb1ng \u0111\u1eb3ng th\u1ee9c: (a + b)5 = a5 + 5a4b + 10a3b2 + 10a2b3 + 5ab4 + b5 ta c\u00f3 nh\u1eadn x\u00e9t n\u1ebfu a 25 th\u00ec (a + b)5 \uf0ba b5 (mod 125). V\u00ec 210 = 1024 \uf0ba \u2013 1 (mod 25) n\u00ean 210 = 25k \u2013 1 (k \uf0ce N). T\u1eeb nh\u1eadn x\u00e9t tr\u00ean ta c\u00f3 250 = (210)5 = (25k \u2013 1)5 \uf0ba \u2013 1 (mod 125) V\u00ec v\u1eady 2512 = (250)10. 212 \uf0ba (\u2013 1)10. 212 \uf0ba 212 (mod 125). Do 212 = 210 . 22 = 1024. 4 \uf0ba 24.4 \uf0ba 96 (mod 125). V\u1eady 2512 \uf0ba 96 (mod 125). Hay 2512 = 125m + 96, m\uf0ceN . Do 2512 8 ; 96 8 n\u00ean m 8 \uf0de m = 8n (n \uf0ce N). 2512 = 125. 8n + 96 = 1000n + 96. V\u1eady ba ch\u1eef s\u1ed1 t\u1eadn c\u00f9ng c\u1ee7a s\u1ed1 2512 l\u00e0 096. 26.5. \u0110\u1ec3 ch\u1ee9ng t\u1ecf a m ta ch\u1ee9ng minh a \uf0ba 0 (mod m) a) 41 = 42 \u2013 1 \uf0ba \u2013 1 (mod 7). Do \u0111\u00f3 412015 \uf0ba (\u2013 1)2015 \uf0ba \u2013 1 (mod 7) Hay 412015 \uf0ba 6 (mod 7) \uf0de 412015 \u2013 6 \uf0ba 0 (mod 7) b) Ta c\u00f3 24 = 16 \uf0ba 1 (mod 15) \uf0de 24n \uf0ba 1 (mod 15) \uf0de 24n \u2013 1 \uf0ba 0 (mod 15) Do \u0111\u00f3 24n+1 \u2013 2 = 2(24n \u2013 1) \uf0ba 0 (mod 15). c) Ta c\u00f3 33 = 27 \uf0ba 1 (mod 13) ; 376 = (33)25.3 \uf0ba 3 (mod 13) Ta c\u00f3 24 \uf0ba 3 (mod 13) \uf0de 26 \uf0ba 12 \uf0ba \u2013 1 (mod 13) 276 = (26)12. 24 \uf0ba 3 (mod 13) 346","Website: tailieumontoan.com Do \u0111\u00f3 376 \u2013 276 \uf0ba 0 (mod 13) hay 376 \u2013 276 13 d) 341 = 11 . 31 * Ta c\u00f3 25 = 32 \uf0ba \u20131(mod 11) ; 20 = 22 \u2013 2 \uf0ba \u2013 2 (mod 11) Do \u0111\u00f3 2015 \uf0ba (\u2013 2)15 \uf0ba \u2013(25)3 \uf0ba 1(mod 11) * 2015 = (25)3. (53)5 \uf0ba 1(mod 31) do 25 \uf0ba 1(mod 31) v\u00e0 53 \uf0ba 1(mod 31) Do \u0111\u00f3 2015 \uf0ba 1 (mod 11.31) hay 2015 \uf0ba 1 (mod 341) \uf0de 2015 \u2013 1 341 26.6. 1890 \uf0ba 0 (mod 7) ; 1945 \uf0ba \u2013 1 (mod 7) ; 2017 \uf0ba 1 (mod 7) 189079 \uf0ba 0 (mod 7) ; 19452015 \uf0ba \u2013 1 (mod 7) ; 20172018 \uf0ba 1 (mod 7) \uf0de \u0111pcm. 26.7. a)Ta c\u00f3 5555 = 793.7 + 4 \uf0ba 4(mod 7); 2222 = 318.7 \u2013 4 \uf0ba \u2013 4(mod 7) \uf0de 55552222 + 22225555 \uf0ba 42222 + (\u2013 4)5555 \uf0ba \u2013 42222(43333\u2013 1) (mod 7) \uf0e9\uf028 \uf029Do 43333 \u2013 1 =431111 \uf02d1\uf0fa\uf0fb\uf0f9; 43 = 64 \uf0ba 1 (mod 7) n\u00ean (43)1111 \uf0ba 1 (mod 7) \uf0ea\uf0eb Hay 43333 \u2013 1 \uf0ba 0 (mod 7) . Do \u0111\u00f3 55552222 + 22225555 \uf0ba 0 (mod 7) v\u00e0 155541111 = (2. 7777)1111 = 21111. 77771111 \uf0ba 0 (mod 7) \uf0de \u0111pcm. b) Ta c\u00f3 102 = 2.3.17. Ta c\u00f3 (220 + 119 + 69)102 \uf0ba 0 (mod 102) *220 \uf0ba 0 (mod 2) ; 119 \uf0ba \u2013 1 (mod 2) ; 69 \uf0ba 1 (mod 2) \uf0de M \uf0ba 0 (mod 2) *220 \uf0ba 1 (mod 3) ; 119 \uf0ba \u2013 1 (mod 3) ; 69 \uf0ba 0 (mod 3) \uf0de M \uf0ba 0 (mod 3) *220 \uf0ba \u20131(mod 17);119 \uf0ba 0 (mod 17) ; 69 \uf0ba 1(mod 17) \uf0de M \uf0ba 0 (mod 17) (\u0110\u1ec3 \u00fd 11969 v\u00e0 69220 l\u00e0 c\u00e1c s\u1ed1 l\u1ebb) ; \uf0de M \uf0ba 0 (mod 2.3.17). Hay M 102 26.8. \u0110\u1eb7t A = 52n-1 . 2n+1 + 22n-1 . 3n+1 . Ta c\u00f3 A 2, \uf022 n \uf0ce N* ; Ta c\u00f3 A = 2n (52n-1 . 2 + 2n-1 . 3n+1) = 2n (25n-1 . 10 + 6n-1 . 9) Do 25 \uf0ba 6 (mod 19) \uf0de A \uf0ba 2n (6n-1 .10 + 6n-1 . 9) \uf0ba 2n.6n-1 . 19 \uf0ba 0 (mod 19) Hay A 19. M\u00e0 (2 ; 19) = 1 \uf0de A 19. 2 \uf0de A 38. 26.9. Ta c\u00f3 a = anan\uf02d1...a1a0 = an.10n + an-1.10n-1 + ...+ a1.10 + a0 . a) Ta c\u00f3 10 \uf0ba 1(mod 9) do \u0111\u00f3 ai. 10i \uf0ba ai (mod 9) , i = 1; 2; 3; ...; n Do \u0111\u00f3 a \uf0ba (an + an-1+ ...+ a1 + a0) (mod 9). V\u1eady a 9 \uf0db an + an-1+ ...+ a1 + a0 \uf0ba 0 (mod 9) \uf0db an + an-1+ ...+ a1 + a0 9. b) Ta c\u00f3 102 = 100 \uf0ba 0 (mod 25) \uf0de ai. 10i \uf0ba 0 (mod 25) , i = 2; 3; ...; n. \uf0de a \uf0ba (a1.10 + a0) (mod 25). V\u1eady a 25 \uf0db a1. 10 + a0 \uf0ba 0 (mod 25) \uf0db a1a0 25. c) Do 10 \uf0ba \u2013 1 (mod 11) \uf0de ai. 10i \uf0ba ai .(\u2013 1)i (mod 11) a \uf0ba (a0 + a2 + a4 + ...) \u2013 (a1 + a3 + a5 + ...) (mod 11) Do \u0111\u00f3 a 11 \uf0db (a0 + a2 + a4 + ...) \u2013 (a1 + a3 + a5 + ...) \uf0ba 0 (mod 11) T\u1ee9c l\u00e0 hi\u1ec7u c\u1ee7a t\u1ed5ng c\u00e1c ch\u1eef s\u1ed1 \u1edf v\u1ecb tr\u00ed l\u1ebb v\u00e0 t\u1ed5ng c\u00e1c ch\u1eef s\u1ed1 \u1edf v\u1ecb tr\u00ed ch\u1eb5n b\u1eb1ng 0. 347","Website: tailieumontoan.com d) Ta c\u00f3 103 = 1000 \uf0ba 0 (mod 8) \uf0de ai. 10i \uf0ba 0 (mod 8) , i = 3; 4; ...; n. \uf0de a \uf0ba (a2. 102 + a1.10 + a0) (mod 8). V\u1eady a 8 \uf0db a2. 102 + a1. 10 + a0 \uf0ba 0 (mod 8) \uf0db a2a1a0 8. 26.10. Theo \u0111\u1ecbnh l\u00fd Fermat b\u00e9, do 11 l\u00e0 s\u1ed1 nguy\u00ean t\u1ed1 n\u00ean ta c\u00f3 210 \uf0ba 1 (mod 11) \uf0de 210n \uf0ba 1 (mod 11) \uf0de 210n + 1 = 2. 210n \uf0ba 2 (mod 22) \uf0de 210n + 1 = 22k + 2 (k \uf0ce N) Do 23 l\u00e0 s\u1ed1 nguy\u00ean t\u1ed1 ta c\u0169ng c\u00f3 222 \uf0ba 1 (mod 23) \uf0de 2210n\uf02b1 \uf03d 222k\uf02b2 \uf03d 4.222k \uf0ba 4 (mod 23) \uf0de 2210n\uf02b1 \uf02b19 \uf0ba 4 + 19 \uf0ba 0 (mod 23) T\u1ee9c l\u00e0 A 23. M\u00e0 A > 23, \uf022n \uf0b31 n\u00ean A l\u00e0 h\u1ee3p s\u1ed1. 26.11. Theo \u0111\u1ecbnh l\u00fd Wilson : V\u1edbi m\u1ecdi s\u1ed1 nguy\u00ean t\u1ed1 p th\u00ec (p \u2013 1)! \uf0ba \u20131 (mod p). Do 13 nguy\u00ean t\u1ed1 n\u00ean 12! \uf0ba \u20131 (mod 13) \uf0de \uf02812!\uf02913 \uf0ba (\u20131)13 \uf0ba \u20131 (mod 13). Ta c\u00f3 2016 = 13.155 + 1 \uf0ba 1 (mod 13) \uf0de 20162015 \uf0ba 1 (mod 13). Do \u0111\u00f3 B = \uf02812!\uf02913 + 20162015 \uf0ba 0 (mod 13). Hay B 13. 26.12. a) Theo \u0110\u1ecbnh l\u00fd Fermat b\u00e9 , do 7 l\u00e0 s\u1ed1 nguy\u00ean t\u1ed1 n\u00ean 26 \uf0ba 1 (mod 7). Ta c\u00f3 4 \uf0ba 1 (mod 3) \uf0de 4n \uf0ba 1 (mod 3) \uf0de 2.4n \uf0ba 2 (mod 6) . Ngh\u0129a l\u00e0 22n + 1 = 2(22)n = 2. 4n \uf0ba 2 (mod 6) \uf0de 22n + 1 = 6k + 2 , (k\uf0ce N) M\u1eb7t kh\u00e1c 23n = (23)n = 8n \uf0ba 1 (mod 7) \uf0de 3. 23n \uf0ba 3 (mod 7). Do \u0111\u00f3 222n\uf02b1 \uf02b 3.23n \uf0ba 26k + 2 + 3 \uf0ba 22. (26)k + 3 \uf0ba 22.1 + 3 \uf0ba 0 (mod 7). b) Do 11 l\u00e0 s\u1ed1 nguy\u00ean t\u1ed1 n\u00ean 210 \uf0ba 1 (mod 11) Ta c\u00f3 16 \uf0ba 1 (mod 5) \uf0de 16n \uf0ba 1 (mod 5) \uf0de 2.16n \uf0ba 2 (mod 10). Ngh\u0129a l\u00e0 24n + 1 = 2(24)n = 2.16n \uf0ba 2 (mod 10) \uf0de 24n + 1 = 10k + 2 , (k\uf0ce N) M\u1eb7t kh\u00e1c 12 \uf0ba 1 (mod 11) \uf0de 125n+ 1 \uf0ba 1 (mod 11) \uf0de 2. 125n +1 \uf0ba 2 (mod 11) ; Do 102 \uf0ba 1 (mod 11) \uf0de 102n \uf0ba 1 (mod 11) \uf0de 5.102n \uf0ba 5 (mod 11). V\u00ec th\u1ebf 224n\uf02b1 \uf02b 2.125n\uf02b1 \uf02b 5.102n \uf0ba 210k + 2 + 2 + 5 \uf0ba 22 + 7 \uf0ba 0 (mod 11). 26.13. a) Ta c\u00f3 72 = 8.9 v\u00e0 (8; 9) = 1. *63 \uf0ba 0 (mod 9); khi n = 2 th\u00ec 3n \uf0ba 0 (mod 9) do \u0111\u00f3 3n + 63 \uf0ba 0 (mod 9). *M\u1eb7t kh\u00e1c, v\u1edbi n = 2k (k\uf0ce N*) th\u00ec 3n \u2013 1 = 32k \u2013 1 = 9k \u2013 1 \uf0ba 1k \u2013 1 \uf0ba 0 (mod 8) do \u0111\u00f3 3n + 63 = 3n \u2013 1 + 64 \uf0ba 0 (mod 8). V\u1eady v\u1edbi n = 2k (k\uf0ceN*) th\u00ec 3n + 63 72 . b) Ta c\u00f3 323 = 17 . 19 v\u00e0 (17; 19) = 1. *A = (20n \u2013 1) + (16n \u2013 3n) = P + Q. Ta c\u00f3 20n \uf0ba 1(mod 19) \uf0de P \uf0ba 0 (mod 19). N\u1ebfu n = 2k (k\uf0ceN*) th\u00ec Q = 162k \u2013 32k \uf0ba (\u2013 3)2k \u2013 32k \uf0ba 32k \u2013 32k \uf0ba 0 (mod 19) \uf0de A = P + Q \uf0ba 0 (mod 19) * A = (20n \u2013 3n ) + (16n \u20131) = P\u2019 + Q\u2019 348","Website: tailieumontoan.com 20n \uf0ba 3n (mod 17). Do \u0111\u00f3 P\u2019 = 20n \u2013 3n \uf0ba 0 (mod 17). N\u1ebfu n = 2k (k\uf0ceN*) th\u00ec Q\u2019 = 162k \u2013 1 = (\u2013 1)2k \u2013 1 \uf0ba 1 \u2013 1 \uf0ba 0 (mod 17) \uf0de A = P\u2019 + Q\u2019 \uf0ba 0 (mod 17). Do (17 ; 19) = 1 n\u00ean A \uf0ba 0 (mod 17. 19). V\u1eady v\u1edbi n = 2k (k\uf0ceN*) th\u00ec A = 20n + 16n \u2013 3n \u2013 1 323 . 26.14. Theo \u0111\u1ecbnh l\u00fd Fermat b\u00e9 ta c\u00f3 2p \uf0ba 2 (mod p) n\u00ean n\u1ebfu 2p \uf0ba \u2013 1 (mod p) th\u00ec ta c\u00f3 3 \uf0ba 0 (mod p) \uf0de p = 3. M\u1eb7t kh\u00e1c khi p = 3 th\u00ec 23 + 1 = 9 \uf0ba 0 (mod 3) . V\u1eady p = 3 l\u00e0 s\u1ed1 c\u1ea7n t\u00ecm. 26.15. V\u1edbi p = 3 th\u00ec p2 + 20 = 29 l\u00e0 s\u1ed1 nguy\u00ean t\u1ed1. V\u1edbi p \uf0b9 3 th\u00ec p2 \uf0ba 1 (mod 3) n\u00ean p2 + 20 \uf0ba 21 \uf0ba 0 (mod 3). V\u1eady p2 + 20 3 m\u1eb7t kh\u00e1c p2 + 20 > 3 n\u00ean p2 + 20 l\u00e0 h\u1ee3p s\u1ed1 . V\u1eady ch\u1ec9 c\u00f3 1 s\u1ed1 nguy\u00ean t\u1ed1 c\u1ea7n t\u00ecm l\u00e0 p = 3. 26.16. V\u1edbi a, b \uf0ceN*. N\u1ebfu ab p th\u00ec s\u1ed1 abp \u2013 bap p N\u1ebfu ab \uf02f p th\u00ec (a, p) = (b, p) = 1. Do \u0111\u00f3 ap-1 \uf0ba bp-1 \uf0ba 1 (mod p) \uf0de ap-1 \u2013 bp-1 \uf0ba 0 (mod p) \uf0de ab(ap-1 \u2013 bp-1) \uf0ba 0 (mod p) \uf0de abp \u2013 bap \uf0ba 0 (mod p) hay abp \u2013 bap p , \uf022 a, b \uf0ceN*. 26.17. a) Gi\u1ea3 s\u1eed a, b, c \uf0ce Z m\u00e0 a2 + b2 + c2 \uf0ba 7 (mod 8). Ta c\u00f3 a \uf0ba 0; \uf0b1 1; \uf0b1 2; \uf0b1 3; 4 (mod 8) \uf0de a2 \uf0ba 0; 1; 4 (mod 8) \uf0de b2 + c2 \uf0ba 7 ; 6 ; 3 (mod 8). \u0110i\u1ec1u n\u00e0y v\u00f4 l\u00fd v\u00ec b2 \uf0ba 0; 1; 4 (mod 8) v\u00e0 c2 \uf0ba 0; 1; 4 (mod 8) \uf0de b2 + c2 \uf0ba 0 ; 1 ; 2; 4; 5 (mod 8). V\u1eady a2 + b2 + c2 \uf0ba\uf02f 7 (mod 8). b) \u00c1p d\u1ee5ng c\u00e2u a) ta c\u00f3 v\u1edbi x , y , z \uf0ce Z 4x2 + y2 + 9z2 = (2x)2 + y2 + (3z)2 \uf0ba\uf02f 7 (mod 8). M\u00e0 2015 = 8. 251 + 7 \uf0ba 7 (mod 8) V\u1eady ph\u01a3\u01a1ng tr\u00ecnh \u0111\u00e3 cho kh\u00f4ng c\u00f3 nghi\u1ec7m nguy\u00ean. 26.18. Ta c\u00f3 2011 \uf0ba 11 (mod 100) ; 112 \uf0ba 21 (mod 100) ; 113 \uf0ba 31 (mod 100); 115 \uf0ba 21.31 \uf0ba 51 (mod 100) \uf0de 1110 \uf0ba 512 \uf0ba 1 (mod 100). Ta c\u00f3 20102009 \uf0ba 0 (mod 10) \uf0de 20102009 = 10k (k \uf0ce Z) \uf0de 201120102009 = 201110k \uf0ba 1110k \uf0ba (1110)k \uf0ba 1 (mod 100). Do \u0111\u00f3 hai ch\u1eef s\u1ed1 t\u1eadn c\u00f9ng l\u00e0 s\u1ed1 01. 26.19. B\u00e0i to\u00e1n c\u00f3 nhi\u1ec1u c\u00e1ch gi\u1ea3i. Sau \u0111\u00e2y l\u00e0 c\u00e1ch gi\u1ea3i theo \u0111\u1ed3ng d\u01a3 th\u1ee9c: * Ta c\u00f3 \uf022 n \uf0ceN* th\u00ec n5 \u2013 n \uf0ba 0 (mod 30) (v\u00ed d\u1ee5 8 chuy\u00ean \u0111\u1ec1 26 \u0111\u00e3 ch\u1ee9ng minh) A = (a2012 \u2013 a2008) + (b2012 \u2013 b2008) + (c2012 \u2013 c2008) A = a2007 (a5 \u2013 a) + b2007 (b5 \u2013 b) + c2007 (c5 \u2013 c) Ta c\u00f3 a5 \u2013 a \uf0ba 0 (mod 30) \uf0de a2007 (a5 \u2013 a) \uf0ba 0 (mod 30) T\u01a3\u01a1ng t\u1ef1 b2007 (b5 \u2013 b) \uf0ba 0 (mod 30) ; c2007 (c5 \u2013 c) \uf0ba 0 (mod 30) V\u1eady A \uf0ba 0 (mod 30) . Hay A 30 . 26.20. Gi\u1ea3 s\u1eed t\u1ed3n t\u1ea1i b\u1ed9 ba s\u1ed1 nguy\u00ean (x; y ; z) th\u1ecfa m\u00e3n x4 + y4 = 7z4 + 5 349"]
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