["B\u00e0i 3: T\u00ecm m\u1ed9t s\u1ed1 t\u1ef1 nhi\u00ean c\u00f3 3 ch\u1eef s\u1ed1, bi\u1ebft r\u1eb1ng s\u1ed1 \u0111\u00f3 g\u1ea5p 5 l\u1ea7n t\u00edch c\u00e1c ch\u1eef s\u1ed1 c\u1ee7a n\u00f3 Gi\u1ea3i C\u00e1ch 1: G\u1ecdi s\u1ed1 ph\u1ea3i t\u00ecm l\u00e0 abc. Theo b\u00e0i ra ta c\u00f3 : abc = 5 x a x b x c. V\u00ec a x 5 x b x c chia h\u1ebft cho 5 n\u00ean abc chia h\u1ebft cho 5. V\u1eady c = 0 ho\u1eb7c 5, nh\u01b0ng c kh\u00f4ng th\u1ec3 b\u1eb1ng 0, v\u1eady c = 5. S\u1ed1 ph\u1ea3i t\u00ecm c\u00f3 d\u1ea1ng ab5. Thay v\u00e0o ta c\u00f3: 100 x a + 10 x b + 5 = 25 x a x b. 20 x a + 2 x b +1 = 5 x a x b. V\u00ec a x 5 x b chia h\u1ebft cho 5 n\u00ean 2 x b + 1 chia h\u1ebft cho 5. V\u1eady 2 x b c\u00f3 t\u1eadn c\u00f9ng b\u1eb1ng 4 ho\u1eb7c 9, nh\u01b0ng 2 x b l\u00e0 s\u1ed1 ch\u1eb5n n\u00ean b = 2 ho\u1eb7c 7. - Tr\u01b0\u1eddng h\u1ee3p b = 2 ta c\u00f3 a25 = 5 x a x 2. V\u1ebf tr\u00e1i l\u00e0 s\u1ed1 l\u1ebb m\u00e0 v\u1ebf ph\u1ea3i l\u00e0 s\u1ed1 ch\u1eb5n. V\u1eady tr\u01b0\u1eddng h\u1ee3p b = 2 b\u1ecb lo\u1ea1i. - Tr\u01b0\u1eddng h\u1ee3p b = 7 ta c\u00f3 20 x a + 15 = 35 x a. T\u00ednh ra ta \u0111\u01b0\u1ee3c a = 1. Th\u1eed l\u1ea1i: 175 = 5 x 7 x 5. V\u1eady s\u1ed1 ph\u1ea3i t\u00ecm l\u00e0 175. C\u00e1ch 2: T\u01b0\u01a1ng t\u1ef1 cach 1 ta c\u00f3: ab5 = 25 x a x b V\u1eady ab5 chia h\u1ebft cho 25, suy ra b = 2 ho\u1eb7c 7. M\u1eb7t kh\u00e1c, ab5 l\u00e0 s\u1ed1 l\u1ebb cho n\u00ean a, b ph\u1ea3i l\u00e0 s\u1ed1 l\u1ebb suy ra b = 7. Ti\u1ebfp theo t\u01b0\u01a1ng t\u1ef1 c\u00e1ch 1 ta t\u00ecm \u0111\u01b0\u1ee3c a = 1. S\u1ed1 ph\u1ea3i t\u00ecm l\u00e0 175. * Lo\u1ea1i 4: So s\u00e1nh t\u1ed5ng ho\u1eb7c \u0111i\u1ec1n d\u1ea5u B\u00e0i 1: Cho A = abc + ab + 1997 B = 1ab9 + 9ac + 9b So s\u00e1nh A v\u00e0 B Gi\u1ea3i Ta th\u1ea5y: B = 1009 + ab0 + 900 + ac + 90 + b = 1999 + ab0 + a0 + c + b = 1999 + abc + ab . . .-> A < B B\u00e0i 2: So s\u00e1nh t\u1ed5ng A v\u00e0 B. A = abc +de + 1992 B = 19bc + d1 + a9e Gi\u1ea3i Ta th\u1ea5y: B = 1900 + bc + d0 + 1 + a00 + e + 90 = abc + de + 1991 T\u1eeb \u0111\u00f3 ta suy ra A > B. B\u00e0i 3: \u0110i\u1ec1n d\u1ea5u 1a26 + 4b4 +5bc [ ] abc + 1997 abc + m000 [ ] m0bc + a00 x5 + 5x [ ] xx +56 2. D\u1ea1ng 2: K\u0129 thu\u1eadt t\u00ednh v\u00e0 quan h\u1ec7 gi\u1eefa c\u00e1c ph\u00e9p t\u00ednh 50","B\u00e0i 1: T\u1ed5ng c\u1ee7a hai s\u1ed1 g\u1ea5p \u0111\u00f4i s\u1ed1 th\u1ee9 nh\u1ea5t. T\u00ecm th\u01a3\u01a1ng c\u1ee7a 2 s\u1ed1 \u0111\u00f3. Gi\u1ea3i Ta c\u00f3: STN + ST2 = T\u1ed5ng. M\u00e0 t\u1ed5ng g\u1ea5p \u0111\u00f4i STN n\u00ean STN = ST2 suy ra th\u01b0\u01a1ng c\u1ee7a 2 s\u1ed1 \u0111\u00f3 b\u1eb1ng 1 B\u00e0i 2: M\u1ed9t ph\u00e9p chia c\u00f3 th\u01a3\u01a1ng l\u00e0 6 v\u00e0 s\u1ed1 d\u01a3 l\u00e0 3, t\u1ed5ng c\u1ee7a s\u1ed1 b\u1ecb chia, s\u1ed1 chia v\u00e0 s\u1ed1 d\u01a3 b\u1eb1ng 195. T\u00ecm s\u1ed1 b\u1ecb chia v\u00e0 s\u1ed1 chia. Gi\u1ea3i G\u1ecdi s\u1ed1 b\u1ecb chia l\u00e0 A, s\u1ed1 chia l\u00e0 B Ta c\u00f3: A: B = 6 (d\u01b0 3) hay A = B x 6 + 3 V\u00e0: A + B + 3 = 195 -> A + B = 1995 \u2013 3 = 1992 B = (1992 \u2013 3): (6 + 1) = 27 A = 27 x 6 + 3 = 165 B\u00e0i 3: Hi\u1ec7u c\u1ee7a 2 s\u1ed1 l\u00e0 33, l\u1ea5y s\u1ed1 l\u1edbn chia cho s\u1ed1 nh\u1ecf \u0111\u01a3\u1ee3c th\u01a3\u01a1ng l\u00e0 3 v\u00e0 s\u1ed1 d\u01a3 l\u00e0 3. T\u00ecm 2 s\u1ed1 \u0111\u00f3. Gi\u1ea3i S\u1ed1 b\u00e9 l\u00e0: (33 \u2013 3): 2 = 15 S\u1ed1 l\u1edbn l\u00e0: 33 + 15 = 48 \u0110\u00e1p s\u1ed1: SL 48 ; SB 15. CHUY\u00caN \u0110\u1ec0 29. D\u00c3Y S\u1ed0 D\u1ea1ng 1. QUY LU\u1eacT VI\u1ebeT D\u00c3Y S\u1ed0: A. L\u00dd THUY\u1ebeT Tr\u01b0\u1edbc h\u1ebft ta c\u1ea7n x\u00e1c \u0111\u1ecbnh quy lu\u1eadt c\u1ee7a d\u00e3y s\u1ed1. Nh\u1eefng quy lu\u1eadt th\u01b0\u1eddng g\u1eb7p l\u00e0: + M\u1ed7i s\u1ed1 h\u1ea1ng (k\u1ec3 t\u1eeb s\u1ed1 h\u1ea1ng th\u1ee9 hai) b\u1eb1ng s\u1ed1 h\u1ea1ng \u0111\u1ee9ng tr\u01b0\u1edbc n\u00f3 c\u1ed9ng (ho\u1eb7c tr\u1eeb) v\u1edbi 1 s\u1ed1 t\u1ef1 nhi\u00ean d; + M\u1ed7i s\u1ed1 h\u1ea1ng (k\u1ec3 t\u1eeb s\u1ed1 h\u1ea1ng th\u1ee9 hai) b\u1eb1ng s\u1ed1 h\u1ea1ng \u0111\u1ee9ng tr\u01b0\u1edbc n\u00f3 nh\u00e2n (ho\u1eb7c chia) v\u1edbi 1 s\u1ed1 t\u1ef1 nhi\u00ean q kh\u00e1c 0; + M\u1ed7i s\u1ed1 h\u1ea1ng (k\u1ec3 t\u1eeb s\u1ed1 h\u1ea1ng th\u1ee9 ba) b\u1eb1ng t\u1ed5ng hai s\u1ed1 h\u1ea1ng \u0111\u1ee9ng tr\u01b0\u1edbc n\u00f3 + M\u1ed7i s\u1ed1 h\u1ea1ng (k\u1ec3 t\u1eeb s\u1ed1 h\u1ea1ng th\u1ee9 t\u01b0) b\u1eb1ng t\u1ed5ng c\u1ee7a s\u1ed1 h\u1ea1ng \u0111\u1ee9ng tr\u01b0\u1edbc n\u00f3 c\u1ed9ng v\u1edbi s\u1ed1 t\u1ef1 nhi\u00ean d c\u1ed9ng v\u1edbi s\u1ed1 th\u1ee9 t\u1ef1 c\u1ee7a s\u1ed1 h\u1ea1ng \u1ea5y + S\u1ed1 h\u1ea1ng \u0111\u1ee9ng sau b\u1eb1ng s\u1ed1 h\u1ea1ng \u0111\u1ee9ng tr\u01b0\u1edbc nh\u00e2n v\u1edbi s\u1ed1 th\u1ee9 t\u1ef1 v . . . v. . . B. B\u00c0I T\u1eacP V\u1eacN D\u1ee4NG 1. Lo\u1ea1i 1: D\u00e3y s\u1ed1 c\u00e1ch \u0111\u1ec1u: B\u00e0i 1: Vi\u1ebft ti\u1ebfp 3 s\u1ed1: a, 5, 10, 15, ... 51b, 3, 7, 11, ...","Gi\u1ea3i a, V\u00ec: 10 \u2013 5 = 5 15 \u2013 10 = 5 D\u00e3y s\u1ed1 tr\u00ean 2 s\u1ed1 h\u1ea1ng li\u1ec1n nhau h\u01a1n k\u00e9m nhau 5 \u0111\u01a1n v\u1ecb. V\u1eady 3 s\u1ed1 ti\u1ebfp theo l\u00e0: 15 + 5 = 20 20 + 5 = 25 25 + 5 = 30 D\u00e3y s\u1ed1 m\u1edbi l\u00e0: 5, 10, 15, 20, 25, 30. b, 7 \u2013 3 = 4 11 \u2013 7 = 4 D\u00e3y s\u1ed1 tr\u00ean 2 s\u1ed1 h\u1ea1ng li\u1ec1n nhau h\u01a1n k\u00e9m nhau 4 \u0111\u01a1n v\u1ecb. V\u1eady 3 s\u1ed1 ti\u1ebfp theo l\u00e0: 11 + 4 = 15 15 + 4 = 19 19 + 4 = 23 D\u00e3y s\u1ed1 m\u1edbi l\u00e0: 3, 7, 11, 15, 19, 23. D\u00e3y s\u1ed1 c\u00e1ch \u0111\u1ec1u th\u00ec hi\u1ec7u c\u1ee7a m\u1ed7i s\u1ed1 h\u1ea1ng v\u1edbi s\u1ed1 li\u1ec1n tr\u01b0\u1edbc lu\u00f4n b\u1eb1ng nhau 1. Lo\u1ea1i 2: D\u00e3y s\u1ed1 kh\u00e1c: B\u00e0i 1: Vi\u1ebft ti\u1ebfp 3 s\u1ed1 h\u1ea1ng v\u00e0o d\u00e3y s\u1ed1 sau: a, 1, 3, 4, 7, 11, 18, ... b, 0, 2, 4, 6, 12, 22, ... c, 0, 3, 7, 12, ... d, 1, 2, 6, 24, ... Gi\u1ea3i a, Ta nh\u1eadn x\u00e9t: 4 = 1 + 3 7=3+4 11 = 4 + 7 18 = 7 + 11 ... T\u1eeb \u0111\u00f3 r\u00fat ra quy lu\u1eadt c\u1ee7a d\u00e3y s\u1ed1 l\u00e0: M\u1ed7i s\u1ed1 h\u1ea1ng (K\u1ec3 t\u1eeb s\u1ed1 h\u1ea1ng th\u1ee9 ba) b\u1eb1ng t\u1ed5ng c\u1ee7a hai s\u1ed1 h\u1ea1ng \u0111\u1ee9ng tr\u01b0\u1edbc n\u00f3. Vi\u1ebft ti\u1ebfp ba s\u1ed1 h\u1ea1ng, ta \u0111\u01b0\u1ee3c d\u00e3y s\u1ed1 sau: 1, 3, 4, 7, 11, 18, 29, 47, 76,... b, T\u01b0\u01a1ng t\u1ef1 b\u00e0i a, ta t\u00ecm ra quy lu\u1eadt c\u1ee7a d\u00e3y s\u1ed1 l\u00e0: M\u1ed7i s\u1ed1 h\u1ea1ng (k\u1ec3 t\u1eeb s\u1ed1 h\u1ea1ng th\u1ee9 t\u01b0) b\u1eb1ng t\u1ed5ng c\u1ee7a 3 s\u1ed1 h\u1ea1ng \u0111\u1ee9ng tr\u01b0\u1edbc n\u00f3. Vi\u1ebft ti\u1ebfp ba s\u1ed1 h\u1ea1ng, ta \u0111\u01b0\u1ee3c d\u00e3y s\u1ed1 sau. 0, 2, 4, 6, 12, 22, 40, 74, 136, ... c, Ta nh\u1eadn x\u00e9t: S\u1ed1 h\u1ea1ng th\u1ee9 hai l\u00e0: 3=0+1+2 S\u1ed1 h\u1ea1ng th\u1ee9 ba l\u00e0: 7=3+1+3 S\u1ed1 h\u1ea1ng th\u1ee9 t\u01b0 l\u00e0: 12 = 7 + 1 + 4 ... T\u1eeb \u0111\u00f3 r\u00fat ra quy lu\u1eadt c\u1ee7a d\u00e3y l\u00e0: M\u1ed7i s\u1ed1 h\u1ea1ng (k\u1ec3 t\u1eeb s\u1ed1 h\u1ea1ng th\u1ee9 hai) b\u1eb1ng t\u1ed5ng c\u1ee7a s\u1ed1 h\u1ea1ng \u0111\u1ee9ng tr\u01b0\u1edbc n\u00f3 c\u1ed9ng v\u1edbi 1 v\u00e0 c\u1ed9ng 52v\u1edbi s\u1ed1 th\u1ee9 t\u1ef1 c\u1ee7a s\u1ed1 h\u1ea1ng \u1ea5y.","Vi\u1ebft ti\u1ebfp ba s\u1ed1 h\u1ea1ng ta \u0111\u01b0\u1ee3c d\u00e3y s\u1ed1 sau. 0, 3, 7, 12, 18, 25, 33, ... d, Ta nh\u1eadn x\u00e9t: S\u1ed1 h\u1ea1ng th\u1ee9 hai l\u00e0 2=1x2 S\u1ed1 h\u1ea1ng th\u1ee9 ba l\u00e0 6=2x3 s\u1ed1 h\u1ea1ng th\u1ee9 t\u01b0 l\u00e0 24 = 6 x 4 ... T\u1eeb \u0111\u00f3 r\u00fat ra quy lu\u1eadt c\u1ee7a d\u00e3y s\u1ed1 l\u00e0: M\u1ed7i s\u1ed1 h\u1ea1ng (k\u1ec3 t\u1eeb s\u1ed1 h\u1ea1ng th\u1ee9 hai) b\u1eb1ng t\u00edch c\u1ee7a s\u1ed1 h\u1ea1ng \u0111\u1ee9ng li\u1ec1n tr\u01b0\u1edbc n\u00f3 nh\u00e2n v\u1edbi s\u1ed1 th\u1ee9 t\u1ef1 c\u1ee7a s\u1ed1 h\u1ea1ng \u1ea5y. Vi\u1ebft ti\u1ebfp ba s\u1ed1 h\u1ea1ng ta \u0111\u01b0\u1ee3c d\u00e3y s\u1ed1 sau: 1, 2, 6, 24, 120, 720, 5040, ... B\u00e0i 2: T\u00ecm s\u1ed1 h\u1ea1ng \u0111\u1ea7u ti\u00ean c\u1ee7a c\u00e1c d\u00e3y s\u1ed1 sau: a, . . ., 17, 19, 21 b, . . . , 64, 81, 100 Bi\u1ebft r\u1eb1ng m\u1ed7i d\u00e3y c\u00f3 10 s\u1ed1 h\u1ea1ng. Gi\u1ea3i a, Ta nh\u1eadn x\u00e9t: S\u1ed1 h\u1ea1ng th\u1ee9 m\u01b0\u1eddi l\u00e0 21 = 2 x 10 + 1 S\u1ed1 h\u1ea1ng th\u1ee9 ch\u00edn l\u00e0: 19 = 2 x 9 + 1 S\u1ed1 h\u1ea1ng th\u1ee9 t\u00e1m l\u00e0: 17 = 2 x 8 + 1 ... T\u1eeb \u0111\u00f3 suy ra quy lu\u1eadt c\u1ee7a d\u00e3y s\u1ed1 tr\u00ean l\u00e0: M\u1ed7i s\u1ed1 h\u1ea1ng c\u1ee7a d\u00e3y b\u1eb1ng 2 x th\u1ee9 t\u1ef1 c\u1ee7a s\u1ed1 h\u1ea1ng trong d\u00e3y r\u1ed3i c\u1ed9ng v\u1edbi 1. V\u1eady s\u1ed1 h\u1ea1ng \u0111\u1ea7u ti\u00ean c\u1ee7a d\u00e3y l\u00e0 2x1+1=3 b, T\u01b0\u01a1ng t\u1ef1 nh\u01b0 tr\u00ean ta r\u00fat ra quy lu\u1eadt c\u1ee7a d\u00e3y l\u00e0: M\u1ed7i s\u1ed1 h\u1ea1ng b\u1eb1ng s\u1ed1 th\u1ee9 t\u1ef1 nh\u00e2n s\u1ed1 th\u1ee9 t\u1ef1 c\u1ee7a s\u1ed1 h\u1ea1ng \u0111\u00f3. V\u1eady s\u1ed1 h\u1ea1ng \u0111\u1ea7u ti\u00ean c\u1ee7a d\u00e3y l\u00e0: 1x1=1 B\u00e0i 3: L\u00fac 7 gi\u1edd s\u00e1ng, M\u1ed9t ng\u01a3\u1eddi xu\u1ea5t ph\u00e1t t\u1eeb A, \u0111i xe \u0111\u1ea1p v\u1ec1 B. \u0110\u1ebfn 11 gi\u1edd tr\u01a3a ng\u01a3\u1eddi \u0111\u00f3 d\u1eebng l\u1ea1i ngh\u1ec9 \u0103n tr\u01a3a m\u1ed9t ti\u1ebfng, sau \u0111\u00f3 l\u1ea1i \u0111i ti\u1ebfp v\u00e0 3 gi\u1edd chi\u1ec1u th\u00ec v\u1ec1 \u0111\u1ebfn B. Do ng\u01a3\u1ee3c gi\u00f3, cho nen t\u1ed1c \u0111\u1ed9 c\u1ee7a ng\u01a3\u1eddi \u0111\u00f3 sau m\u1ed7i gi\u1edd l\u1ea1i gi\u1ea3m \u0111i 2 km. T\u00ecm t\u1ed1c \u0111\u1ed9 c\u1ee7a ng\u01a3\u1eddi \u0111\u00f3 khi xu\u1ea5t ph\u00e1t, bi\u1ebft r\u1eb1ng t\u1ed1c \u0111\u1ecd \u0111i trong ti\u1ebfng cu\u1ed1i qu\u00e3ng \u0111\u01a3\u1eddng l\u00e0 10 km\/ gi\u1edd. Gi\u1ea3i Th\u1eddi gian ng\u01b0\u1eddi \u0111\u00f3 \u0111i tr\u00ean \u0111\u01b0\u1eddng l\u00e0: (11 \u2013 7) + (15 \u2013 12) = 7 (gi\u1edd) Ta nh\u1eadn x\u00e9t: T\u1ed1c \u0111\u1ed9 ng\u01b0\u1eddi \u0111\u00f3 \u0111i trong ti\u1ebfng th\u1ee9 7 l\u00e0: 10 (km\/gi\u1edd) = 10 + 2 x 0 T\u1ed1c \u0111\u1ed9 ng\u01b0\u1eddi \u0111\u00f3 \u0111i trong ti\u1ebfng th\u1ee9 6 53l\u00e0:","12 (km\/gi\u1edd) = 10 + 2 x 1 T\u1ed1c \u0111\u1ed9 ng\u01b0\u1eddi \u0111\u00f3 \u0111i trong ti\u1ebfng th\u1ee9 5 l\u00e0: 14 (km\/gi\u1edd) = 10 + 2 x 2 ... T\u1eeb \u0111\u00f3 r\u00fat ra t\u1ed1c \u0111\u1ed9 ng\u01b0\u1eddi \u0111\u00f3 l\u00fac xu\u1ea5t ph\u00e1t (trong ti\u1ebfng th\u1ee9 nh\u1ea5t) l\u00e0: 10 + 2 x 6 = 22 (km\/gi\u1edd) B\u00e0i 4: \u0110i\u1ec1n c\u00e1c s\u1ed1 th\u00edch h\u1ee3p v\u00e0o \u00f4 tr\u1ed1ng, sao cho t\u1ed5ng c\u00e1c s\u1ed1 \u1edf 3 \u00f4 li\u00ean ti\u1ebfp \u0111\u1ec1u b\u1eb1ng 1996: Gi\u1ea3i Ta \u0111\u00e1nh s\u1ed1 c\u00e1c \u00f4 theo th\u1ee9 t\u1ef1 nh\u01b0 sau: Theo \u0111i\u1ec1u ki\u1ec7n c\u1ee7a \u0111\u1ea7u b\u00e0i ta c\u00f3: 496 + \u00f47 + \u00f4 8 = 1996 \u00f47 + \u00f48 + \u00f49 = 1996 V\u1eady \u00f49 = 496. T\u1eeb \u0111\u00f3 ta t\u00ednh \u0111\u01b0\u1ee3c \u00f48 = \u00f45 = \u00f42 = 1996 \u2013 (496 + 996) = 504; \u00f47 = \u00f44 = \u00f41 = 996 v\u00e0 \u00f43 = \u00f46 = 496 \u0110i\u1ec1n v\u00e0o ta \u0111\u01b0\u1ee3c d\u00e3y s\u1ed1: D\u1ea1ng 2. X\u00e1c \u0111\u1ecbnh s\u1ed1 a c\u00f3 thu\u1ed9c d\u00e3y \u0111\u00e3 cho hay kh\u00f4ng: A. L\u00dd THUY\u1ebeT - X\u00e1c \u0111\u1ecbnh quy lu\u1eadt c\u1ee7a d\u00e3y. - Ki\u1ec3m tra s\u1ed1 a c\u00f3 tho\u1ea3 m\u00e3n quy lu\u1eadt \u0111\u00f3 hay kh\u00f4ng. B. B\u00c0I T\u1eacP V\u1eacN D\u1ee4NG Em h\u00e3y cho bi\u1ebft: a, C\u00e1c s\u1ed1 50 v\u00e0 133 c\u00f3 thu\u1ed9c d\u00e3y 90, 95, 100,. .. hay kh\u00f4ng? b, S\u1ed1 1996 thu\u1ed9c d\u00e3y 3, 6, 8, 11,. .. hay kh\u00f4ng? c, S\u1ed1 n\u00e0o trong c\u00e1c s\u1ed1 666, 1000, 9999 thu\u1ed9c d\u00e3y 3, 6, 12, 24,. ..? Gi\u1ea3i th\u00edch t\u1ea1i sao? Gi\u1ea3i a, C\u1ea3 2 s\u1ed1 50 v\u00e0 133 \u0111\u1ec1u kh\u00f4ng thu\u1ed9c d\u00e3y \u0111\u00e3 cho v\u00ec - C\u00e1c s\u1ed1 h\u1ea1ng c\u1ee7a d\u00e3y \u0111\u00e3 cho \u0111\u1ec1u l\u1edbn h\u01a1n 50; - C\u00e1c s\u1ed1 h\u1ea1ng c\u1ee7a d\u00e3y \u0111\u00e3 cho \u0111\u1ec1u chia h\u1ebft cho 5 m\u00e0 133 kh\u00f4ng chia h\u1ebft cho 5. b, S\u1ed1 1996 kh\u00f4ng thu\u1ed9c d\u00e3y \u0111\u00e3 cho, V\u00ec m\u1ecdi s\u1ed1 h\u1ea1ng c\u1ee7a d\u00e3y khi chia cho \u0111\u1ec1u d\u01b0 2 m\u00e0 1996: 3 th\u00ec d\u01b0 1. c, C\u1ea3 3 s\u1ed1 666, 1000, 9999 \u0111\u1ec1u kh\u00f4ng thu\u1ed9c d\u00e3y 3, 6, 12, 24,. .., v\u00ec - M\u1ed7i s\u1ed1 h\u1ea1ng c\u1ee7a d\u00e3y (k\u1ec3 t\u1eeb s\u1ed1 h\u1ea1ng th\u1ee9 2) b\u1eb1ng s\u1ed1 h\u1ea1ng li\u1ec1n tr\u01b0\u1edbc nh\u00e2n v\u1edbi 2. Cho n\u00ean c\u00e1c s\u1ed1 h\u1ea1ng (k\u1ec3 t\u1eeb s\u1ed1 h\u1ea1ng th\u1ee9 3) c\u00f3 s\u1ed1 h\u1ea1ng \u0111\u1ee9ng li\u1ec1n tr\u01b0\u1edbc l\u00e0 s\u1ed1 ch\u1eb5n m\u00e0 666: 2 = 333 l\u00e0 s\u1ed1 l\u1ebb. - C\u00e1c s\u1ed1 h\u1ea1ng c\u1ee7a d\u00e3y \u0111\u1ec1u chia h\u1ebft cho 3 m\u00e0 1000 kh\u00f4ng chia h\u1ebft cho 3 - C\u00e1c s\u1ed1 h\u1ea1ng c\u1ee7a d\u00e3y (k\u1ec3 t\u1eeb s\u1ed1 h\u1ea1ng th\u1ee9 hai) \u0111\u1ec1u ch\u1eb5n m\u00e0 9999 l\u00e0 s\u1ed1 l\u1ebb. D\u1ea1ng 3. T\u00ecm s\u1ed1 s\u1ed1 h\u1ea1ng c\u1ee7a d\u00e3y s\u1ed1: A. L\u00dd THUY\u1ebeT - \u1ede d\u1ea1ng n\u00e0y th\u01b0\u1eddng s\u1eed d\u1ee5ng ph\u01b0\u01a1ng ph\u00e1p gi\u1ea3i to\u00e1n kho\u1ea3ng c\u00e1ch (tr\u1ed3ng c\u00e2y).Ta c\u00f3 c\u00f4ng th\u1ee9c sau: S\u1ed1 s\u1ed1 h\u1ea1ng c\u1ee7a d\u00e3y = S\u1ed1 54kho\u1ea3ng c\u00e1ch + 1","- N\u1ebfu quy lu\u1eadt c\u1ee7a d\u00e3y l\u00e0: s\u1ed1 \u0111\u1ee9ng sau b\u1eb1ng s\u1ed1 h\u1ea1ng li\u1ec1n tr\u01b0\u1edbc c\u1ed9ng v\u1edbi s\u1ed1 kh\u00f4ng \u0111\u1ed5i th\u00ec: S\u1ed1 c\u00e1c s\u1ed1 h\u1ea1ng c\u1ee7a d\u00e3y = (S\u1ed1 cu\u1ed1i \u2013 s\u1ed1 \u0111\u1ea7u): K\/c + 1 B. B\u00c0I T\u1eacP V\u1eacN D\u1ee4NG B\u00e0i 1: Vi\u1ebft c\u00e1c s\u1ed1 l\u1ebb li\u00ean ti\u1ebfp t\u1eeb 211. S\u1ed1 cu\u1ed1i c\u00f9ng l\u00e0 971. H\u1ecfi vi\u1ebft \u0111\u01a3\u1ee3c bao nhi\u00eau s\u1ed1? Gi\u1ea3i Hai s\u1ed1 l\u1ebb li\u00ean ti\u1ebfp h\u01a1n k\u00e9m nhau 2 \u0111\u01a1n v\u1ecb S\u1ed1 cu\u1ed1i h\u01a1n s\u1ed1 \u0111\u1ea7u s\u1ed1 \u0111\u01a1n v\u1ecb l\u00e0: 971 \u2013 211 = 760 (\u0111\u01a1n v\u1ecb) 760 \u0111\u01a1n v\u1ecb c\u00f3 s\u1ed1 kho\u1ea3ng c\u00e1ch l\u00e0: 760: 2 = 380 (K\/ c) D\u00e3y s\u1ed1 tr\u00ean c\u00f3 s\u1ed1 s\u1ed1 h\u1ea1ng l\u00e0: 380 +1 = 381 (s\u1ed1) \u0110\u00e1p s\u1ed1: 381 s\u1ed1 h\u1ea1ng B\u00e0i 2: Cho d\u00e3y s\u1ed1 11, 14, 17,. .., 68. a, H\u00e3y x\u00e1c \u0111\u1ecbnh d\u00e3y tr\u00ean c\u00f3 bao nhi\u00eau s\u1ed1 h\u1ea1ng? b, N\u1ebfu ta ti\u1ebfp t\u1ee5c k\u00e9o d\u00e0i c\u00e1c s\u1ed1 h\u1ea1ng c\u1ee7a d\u00e3y s\u1ed1 th\u00ec s\u1ed1 h\u1ea1ng th\u1ee9 1 996 l\u00e0 s\u1ed1 m\u1ea5y? Gi\u1ea3i a, Ta c\u00f3: 14 \u2013 11 = 3 17 \u2013 14 = 3 V\u1eady quy lu\u1eadt c\u1ee7a d\u00e3y l\u00e0: m\u1ed7i s\u1ed1 h\u1ea1ng \u0111\u1ee9ng sau b\u1eb1ng s\u1ed1 h\u1ea1ng \u0111\u1ee9ng tr\u01b0\u1edbc c\u1ed9ng v\u1edbi 3. S\u1ed1 c\u00e1c s\u1ed1 h\u1ea1ng c\u1ee7a d\u00e3y l\u00e0: ( 68 \u2013 11 ): 3 + 1 = 20 (s\u1ed1 h\u1ea1ng) b, Ta nh\u1eadn x\u00e9t: S\u1ed1 h\u1ea1ng th\u1ee9 hai: 14 = 11 + 3 = 11 + (2 \u2013 1) x 3 S\u1ed1 h\u1ea1ng th\u1ee9 ba: 17 = 11 + 6 = 11 + (3 \u2013 1) x 3 S\u1ed1 h\u1ea1ng th\u1ee9 t\u01b0 : 20 = 11 + 9 = 11 + (4 \u2013 1) x 3 V\u1eady s\u1ed1 h\u1ea1ng th\u1ee9 1 996 l\u00e0: 11 + (1 996 \u2013 1) x 3 = 5 996 \u0110\u00e1p s\u1ed1: 20 s\u1ed1 h\u1ea1ng; 5996 B\u00e0i 3: Trong c\u00e1c s\u1ed1 c\u00f3 ba ch\u1eef s\u1ed1, c\u00f3 bao nhi\u00eau s\u1ed1 chia h\u1ebft cho 4? Gi\u1ea3i Ta c\u00f3 nh\u1eadn x\u00e9t: s\u1ed1 nh\u1ecf nh\u1ea5t c\u00f3 ba ch\u1eef s\u1ed1 chia h\u1ebft cho 4l\u00e0 100 v\u00e0 s\u1ed1 l\u1edbn nh\u1ea5t c\u00f3 ba ch\u1eef s\u1ed1 chia h\u1ebft cho 4 l\u00e0 996. Nh\u01b0 v\u1eady c\u00e1c s\u1ed1 c\u00f3 ba ch\u1eef s\u1ed1 chia h\u1ebft cho 4 l\u1eadp th\u00e0nh m\u1ed9t d\u00e3y s\u1ed1 c\u00f3 s\u1ed1 h\u1ea1ng \u0111\u1ea7u l\u00e0 100, s\u1ed1 h\u1ea1ng cu\u1ed1i l\u00e0 996 v\u00e0 m\u1ed7i s\u1ed1 h\u1ea1ng c\u1ee7a d\u00e3y (K\u1ec3 t\u1eeb s\u1ed1 h\u1ea1ng th\u1ee9 hai) b\u1eb1ng s\u1ed1 h\u1ea1ng \u0111\u1ee9ng k\u1ec1 tr\u01b0\u1edbc c\u1ed9ng v\u1edbi 4. V\u1eady c\u00e1c s\u1ed1 c\u00f3 3 ch\u1eef s\u1ed1 chia h\u1ebft cho 4 l\u00e0: (996 \u2013 100): 4 + 1 = 225 (s\u1ed1) \u0110\u00e1p s\u1ed1: 225 s\u1ed1 D\u1ea1ng 4. T\u00ecm t\u1ed5ng c\u00e1c s\u1ed1 h\u1ea1ng c\u1ee7a d\u00e3y s\u1ed1: A. L\u00dd THUY\u1ebeT N\u1ebfu c\u00e1c s\u1ed1 h\u1ea1ng c\u1ee7a d\u00e3y s\u1ed1 c\u00e1ch \u0111\u1ec1u nhau th\u00ec t\u1ed5ng c\u1ee7a 2 s\u1ed1 h\u1ea1ng c\u00e1ch \u0111\u1ec1u s\u1ed1 h\u1ea1ng \u0111\u1ea7u v\u00e0 s\u1ed1 h\u1ea1ng cu\u1ed1i trong d\u00e3y \u0111\u00f3 b\u1eb1ng nhau. V\u00ec v\u1eady: T\u1ed5ng c\u00e1c s\u1ed1 h\u1ea1ng c\u1ee7a d\u00e3y = t\u1ed5ng c\u1ee7a 1 c\u1eb7p 2 s\u1ed1 h\u1ea1ng c\u00e1ch \u0111\u1ec1u s\u1ed1 h\u1ea1ng \u0111\u1ea7u v\u00e0 cu\u1ed1i x s\u1ed1 h\u1ea1ng c\u1ee7a d\u00e3y: 2 B. B\u00c0I T\u1eacP V\u1eacN D\u1ee4NG 55","B\u00e0i 1: T\u00ednh t\u1ed5ng c\u1ee7a 100 s\u1ed1 l\u1ebb \u0111\u1ea7u ti\u00ean. Gi\u1ea3i D\u00e3y c\u1ee7a 100 s\u1ed1 l\u1ebb \u0111\u1ea7u ti\u00ean l\u00e0: 1 + 3 + 5 + 7 + 9 +. . . + 197 + 199. Ta c\u00f3: 1 + 199 = 200 3 + 197 = 200 5 + 195 = 200 ... V\u1eady t\u1ed5ng ph\u1ea3i t\u00ecm l\u00e0: 200 x 100: 2 = 10 000 \u0110\u00e1p s\u1ed1: 10 000 B\u00e0i 2: Cho 1 s\u1ed1 t\u1ef1 nhi\u00ean g\u1ed3m c\u00e1c s\u1ed1 t\u1ef1 nhi\u00ean li\u00ean ti\u1ebfp t\u1eeb 1 \u0111\u1ebfn 1983 \u0111\u01a3\u1ee3c vi\u1ebft theo th\u1ee9 t\u1ef1 li\u1ec1n nhau nh\u01a3 sau: 1 2 3 4 5 6 7 8 9 10 11 12 13. . . 1980 1981 1982 1983 H\u00e3y t\u00ednh t\u1ed5ng t\u1ea5t c\u1ea3 c\u00e1c ch\u1eef s\u1ed1 c\u1ee7a s\u1ed1 \u0111\u00f3. (\u0110\u1ec1 thi h\u1ecdc sinh gi\u1ecfi to\u00e0n qu\u1ed1c n\u0103m 1983) Gi\u1ea3i C\u00e1ch 1. Ta nh\u1eadn x\u00e9t: C\u00e1c c\u1eb7p s\u1ed1: - 0 v\u00e0 1999 c\u00f3 t\u1ed5ng c\u00e1c ch\u1eef s\u1ed1 l\u00e0: 0 + 1 + 9 + 9 + 9 = 28 - 1 v\u00e0 1998 c\u00f3 t\u1ed5ng c\u00e1c ch\u1eef s\u1ed1 l\u00e0: 1 + 1 + 9 + 9 + 8 = 28 - 2 v\u00e0 1997 c\u00f3 t\u1ed5ng c\u00e1c ch\u1eef s\u1ed1 l\u00e0: 2 + 1 + 9 + 9 + 7 = 28 - 998 v\u00e0 1001 c\u00f3 t\u1ed5ng c\u00e1c ch\u1eef s\u1ed1 l\u00e0: 9 + 9 + 8 + 1 + 1 = 28 - 999 v\u00e0 1000 c\u00f3 t\u1ed5ng c\u00e1c ch\u1eef s\u1ed1 l\u00e0: 9 + 9 + 9 + 1 = 28 Nh\u01b0 v\u1eady trong d\u00e3y s\u1ed1 0, 1, 2, 3, 4, 5,. . . , 1997, 1998, 1999 Hai s\u1ed1 h\u1ea1ng c\u00e1ch \u0111\u1ec1u s\u1ed1 h\u1ea1ng \u0111\u1ea7u v\u00e0 s\u1ed1 h\u1ea1ng cu\u1ed1i \u0111\u1ec1u c\u00f3 t\u1ed5ng b\u1eb1ng 28. C\u00f3 1000 c\u1eb7p nh\u01b0 v\u1eady, do \u0111\u00f3 t\u1ed5ng c\u00e1c ch\u1eef s\u1ed1 t\u1ea1o n\u00ean d\u00e3y s\u1ed1 tr\u00ean l\u00e0: 28 x 1000 = 28 000 S\u1ed1 t\u1ef1 nhi\u00ean \u0111\u01b0\u1ee3c t\u1ea1o th\u00e0nh b\u1eb1ng c\u00e1ch vi\u1ebft li\u00ean ti\u1ebfp c\u00e1c s\u1ed1 t\u1ef1 nhi\u00ean t\u1eeb 1984 \u0111\u1ebfn 1999 l\u00e0 : V\u1eady t\u1ed5ng c\u00e1c ch\u1eef s\u1ed1 c\u1ee7a s\u1ed1 t\u1ef1 nhi\u00ean \u0111\u00e3 cho l\u00e0: 28 000 \u2013 382 = 27 618 B\u00e0i 3: Vi\u1ebft c\u00e1c s\u1ed1 ch\u1eb5n li\u00ean ti\u1ebfp: 2, 4, 6, 8,. . . , 2000 T\u00ednh t\u1ed5ng c\u1ee7a d\u00e3y s\u1ed1 tr\u00ean 56","Gi\u1ea3i D\u00e3y s\u1ed1 tr\u00ean 2 s\u1ed1 ch\u1eb5n li\u00ean ti\u1ebfp h\u01a1n k\u00e9m nhau 2 \u0111\u01a1n v\u1ecb. D\u00e3y s\u1ed1 tr\u00ean c\u00f3 s\u1ed1 s\u1ed1 h\u1ea1ng l\u00e0: (2000 \u2013 2): 2 + 1 = 1000 (s\u1ed1) 1000 s\u1ed1 c\u00f3 s\u1ed1 c\u1eb7p s\u1ed1 l\u00e0: 1000: 2 = 500 (c\u1eb7p) T\u1ed5ng 1 c\u1eb7p l\u00e0: 2 + 2000 = 2002 T\u1ed5ng c\u1ee7a d\u00e3y s\u1ed1 l\u00e0: 2002 x 500 = 100100 CHUY\u00caN \u0110\u1ec0 30. TO\u00c1N CHUY\u1ec2N \u0110\u1ed8NG A. L\u00dd THUY\u1ebeT 1. M\u1ed7i quan h\u1ec7 gi\u1eefa qu\u00e3ng \u0111\u1eddng (s), v\u1eadn t\u1ed1c (v) v\u00e0 th\u1eddi gian (t) s 1.1. V\u1eadn t\u1ed1c: v=t 1.2. Qu\u00e3ng \u0111\u1eddng: s = v x t 1.3. Th\u1eddi gian: t = s : v - V\u1edbi c\u00f9ng m\u1ed9t v\u1eadn t\u1ed1c th\u00ec qu\u00e3ng \u0111\u1eddng v\u00e0 th\u1eddi gian l\u00e0 2 \u0111\u1ea1i l\u1ee3ng t\u1ec9 l\u1ec7 thu\u1eadn v\u1edbi nhau. - V\u1edbi c\u00f9ng m\u1ed9t th\u1eddi gian th\u00ec qu\u00e3ng \u0111\u1eddng v\u00e0 v\u1eadn t\u1ed1c l\u00e0 2 \u0111\u1ea1i l\u1ee3ng t\u1ec9 l\u1ec7 thu\u1eadn v\u1edbi nhau. - V\u1edbi c\u00f9ng m\u1ed9t qu\u00e3ng \u0111\u1eddng th\u00ec v\u1eadn t\u1ed1c v\u00e0 th\u1eddi gian l\u00e0 2 \u0111\u1ea1i l\u1ee3ng t\u1ec9 l\u1ec7 ngh\u1ecbch v\u1edbi nhau. 2. B\u00e0i to\u00e1n c\u00f3 m\u1ed9t \u0111\u1ed9ng t\u1eed (ch\u1ec9 c\u00f3 m\u1ed9t v\u1eadt tham gia chuy\u1ec3n \u0111\u1ed9ng,v\u00ed d\u1ee5: \u00f4 t\u00f4, xe m\u00e1y, xe \u0111\u1ea1p, ng\u1eddi \u0111i b\u1ed9, xe l\u1eeda, \u2026) 2.1. Th\u1eddi gian \u0111i = th\u1eddi gian \u0111\u1ebfn - th\u1eddi gian kh\u1edfi h\u00e0nh - th\u1eddi gian ngh\u1ec9 (n\u1ebfu c\u00f3). 2.2. Th\u1eddi gian \u0111\u1ebfn = th\u1eddi gian kh\u1edfi h\u00e0nh + th\u1eddi gian \u0111i + th\u1eddi gian ngh\u1ec9 (n\u1ebfu c\u00f3). 2.3. Th\u1eddi gian kh\u1edfi h\u00e0nh = th\u1eddi gian \u0111\u1ebfn - th\u1eddi gian \u0111i - th\u1eddi gian ngh\u1ec9 (n\u1ebfu c\u00f3). 3. B\u00e0i to\u00e1n \u0111\u1ed9ng t\u1eed ch\u1ea1y ng\u1ee3c chi\u1ec1u 3.1. Th\u1eddi gian g\u1eb7p nhau = qu\u00e3ng \u0111\u1eddng : t\u1ed5ng v\u1eadn t\u1ed1c 3.2. T\u1ed5ng v\u1eadn t\u1ed1c = qu\u00e3ng \u0111\u1eddng : th\u1eddi gian g\u1eb7p nhau 3.3. Qu\u00e3ng \u0111\u1eddng = th\u1eddi gian g\u1eb7p nhau \uf0b4 t\u1ed5ng v\u1eadn t\u1ed1c 4. B\u00e0i to\u00e1n \u0111\u1ed9ng t\u1eed ch\u1ea1y c\u00f9ng chi\u1ec1u 4.1. Th\u1eddi gian g\u1eb7p nhau = kho\u1ea3ng c\u00e1ch ban \u0111\u1ea7u : hi\u1ec7u v\u1eadn t\u1ed1c 4.2. Hi\u1ec7u v\u1eadn t\u1ed1c = kho\u1ea3ng c\u00e1ch ban \u0111\u1ea7u : th\u1eddi gian g\u1eb7p nhau 4.3. Kho\u1ea3ng c\u00e1ch ban \u0111\u1ea7u = th\u1eddi gian g\u1eb7p nhau \uf0b4 hi\u1ec7u v\u1eadn t\u1ed1c 5. B\u00e0i to\u00e1n \u0111\u1ed9ng t\u1eed tr\u00ean d\u00f2ng n\u1edbc 5.1. V\u1eadn t\u1ed1c xu\u00f4i d\u00f2ng = v\u1eadn t\u1ed1c c\u1ee7a v\u1eadt + v\u1eadn t\u1ed1c d\u00f2ng n\u1edbc 5.2. V\u1eadn t\u1ed1c ng\u1ee3c d\u00f2ng = v\u1eadn t\u1ed1c c\u1ee7a v\u1eadt - v\u1eadn t\u1ed1c d\u00f2ng n\u1edbc 5.3. V\u1eadn t\u1ed1c c\u1ee7a v\u1eadt = (v\u1eadn t\u1ed1c xu\u00f4i d\u00f2ng + v\u1eadn t\u1ed1c ng\u1ee3c d\u00f2ng) : 2 5.4. V\u1eadn t\u1ed1c d\u00f2ng n\u1edbc = (v\u1eadn t\u1ed1c xu\u00f4i d\u00f2ng - v\u1eadn t\u1ed1c ng\u1ee3c d\u00f2ng) : 2 6. \u0110\u1ed9ng t\u1eed c\u00f3 chi\u1ec1u d\u00e0i \u0111\u00e1ng k\u1ec3 6.1. \u0110o\u00e0n t\u00e0u c\u00f3 chi\u1ec1u d\u00e0i b\u1eb1ng l ch\u1ea1y qua m\u1ed9t c\u1ed9t \u0111i\u1ec7n Th\u1eddi gian ch\u1ea1y qua c\u1ed9t \u0111i\u1ec7n = l : v\u1eadn t\u1ed1c \u0111o\u00e0n t\u00e0u 6.2. \u0110o\u00e0n t\u00e0u c\u00f3 chi\u1ec1u d\u00e0i l ch\u1ea1y qua m\u1ed9t c\u00e1i c\u1ea7u c\u00f3 chi\u1ec1u d\u00e0i d Th\u1eddi gian ch\u1ea1y qua c\u1ea7u = (l + d) : v\u1eadn t\u1ed1c \u0111o\u00e0n t\u00e0u 6.3. \u0110o\u00e0n t\u00e0u c\u00f3 chi\u1ec1u d\u00e0i l ch\u1ea1y qua m\u1ed9t \u00f4 t\u00f4 \u0111ang ch\u1ea1y ng\u1ee3c chi\u1ec1u (chi\u1ec1u d\u00e0i c\u1ee7a \u00f4 t\u00f4 l\u00e0 kh\u00f4ng \u0111\u00e1ng k\u1ec3) 57","Th\u1eddi gian \u0111i qua nhau = c\u1ea3 qu\u00e3ng \u0111\u1eddng : t\u1ed5ng v\u1eadn t\u1ed1c 6.4. \u0110o\u00e0n t\u00e0u c\u00f3 chi\u1ec1u d\u00e0i l ch\u1ea1y qua m\u1ed9t \u00f4 t\u00f4 ch\u1ea1y c\u00f9ng chi\u1ec1u (chi\u1ec1u d\u00e0i \u00f4 t\u00f4 l\u00e0 kh\u00f4ng \u0111\u00e1ng k\u1ec3) Th\u1eddi gian \u0111i qua nhau = c\u1ea3 qu\u00e3ng \u0111\u1eddng: hi\u1ec7u v\u1eadn t\u1ed1c B. B\u00c0I T\u1eacP V\u1eacN D\u1ee4NG B\u00e0i 1: M\u1ed9t t\u00e0u tu\u1ea7n ti\u1ec3u c\u00f3 v\u1eadn t\u1ed1c 40 km\/gi\u1edd, \u0111\u01a3\u1ee3c l\u1ec7nh ti\u1ebfn h\u00e0nh trinh s\u00e1t ph\u00eda tr\u01a3\u1edbc h\u1ea1m \u0111\u1ed9i theo ph\u01a3\u01a1ng ti\u1ebfn c\u1ee7a h\u1ea1m \u0111\u1ed9i v\u00e0 quay v\u1ec1 h\u1ea1m \u0111\u1ed9i sau 3 gi\u1edd. Bi\u1ebft v\u1eadn t\u1ed1c c\u1ee7a h\u1ea1m \u0111\u1ed9i \u0111i \u0111\u01a3\u1ee3c 24 km\/gi\u1edd. H\u1ecfi t\u00e0u tu\u1ea7n ti\u1ec3u \u0111\u00f3 t\u1eeb khi b\u1eaft \u0111\u1ea7u \u0111i \u0111\u01a3\u1ee3c kho\u1ea3ng c\u00e1ch bao xa \u0111\u1ec3 tr\u1edf v\u1ec1 h\u1ea1m \u0111\u1ed9i \u0111\u00fang th\u1eddi gian quy \u0111\u1ecbnh? T\u1ed5ng qu\u00e3ng \u0111\u01b0\u1eddng c\u1ee7a t\u00e0u tu\u1ea7n ti\u1ec3u v\u00e0 h\u1ea1m \u0111\u1ed9i \u0111i g\u1ea5p 2 l\u1ea7n kho\u1ea3ng c\u00e1ch c\u1ea7n thi\u1ebft c\u1ee7a t\u00e0u tu\u1ea7n ti\u1ec3u ph\u1ea3i \u0111i. T\u1ed5ng v\u1eadn t\u1ed1c c\u1ee7a t\u00e0u tu\u1ea7n ti\u1ec3u v\u00e0 h\u1ea1m \u0111\u1ed9i: 40 + 24 = 64 (km\/gi\u1edd) Hai l\u1ea7n kho\u1ea3ng c\u00e1ch \u0111\u00f3 l\u00e0: 64 x 3 = 192 (km) Kho\u1ea3ng c\u00e1ch c\u1ee7a t\u00e0u tu\u1ea7n ti\u1ec3u ph\u1ea3i \u0111i l\u00e0: 192 : 2 = 96 (km) B\u00e0i 2: Hi\u1ec7n l\u00e0 12 gi\u1edd. Sau bao l\u00e2u 2 kim \u0111\u1ed3ng h\u1ed3 s\u1ebd ch\u1eadp nhau? M\u1eb7t tr\u00f2n \u0111\u1ed3ng h\u1ed3 \u0111\u01b0\u1ee3c chia l\u00e0m 12 kho\u1ea3ng theo tr\u1ee5 s\u1ed1. M\u1ed7i gi\u1edd kim gi\u1edd ch\u1ec9 ch\u1ea1y \u0111\u01b0\u1ee3c \u0111\u00fang 1 kho\u1ea3ng, kim ph\u00fat ch\u1ea1y \u0111\u00fang 12 kho\u1ea3ng. Ta xem nh\u01b0 kim gi\u1edd ch\u1ea1y tr\u01b0\u1edbc kim ph\u00fat \u0111\u00fang 1 v\u00f2ng (12 kho\u1ea3ng. V\u00ec 2 kim g\u1eb7p nhau t\u1ea1i s\u1ed1 12). Hi\u1ec7u s\u1ed1 v\u1eadn t\u1ed1c c\u1ee7a kim ph\u00fat v\u00e0 kim gi\u1edd l\u00e0: 12 - 1 = 11 (kho\u1ea3ng\/gi\u1edd) Th\u1eddi gian \u0111\u1ec3 2 kim \u0111\u1ed3ng h\u1ed3 ch\u1eadp nhau l\u1ea7n k\u1ebf ti\u1ebfp: 2 : 11 = 1 1 (gi\u1edd) 11 B\u00e0i 3: B\u00ednh \u0111i t\u1eeb A \u0111\u1ebfn B. N\u1eeda qu\u00e3ng \u0111\u01a3\u1eddng \u0111\u1ea7u B\u00ednh \u0111i v\u1edbi v\u1eadn t\u1ed1c 60 km\/gi\u1edd. N\u1eeda qu\u00e3ng \u0111\u01a3\u1eddng c\u00f2n l\u1ea1i B\u00ednh \u0111i v\u1edbi v\u1eadn t\u1ed1c 30 km\/gi\u1edd. T\u00ednh v\u1eadn t\u1ed1c trung b\u00ecnh c\u1ee7a B\u00ednh tr\u00ean su\u1ed1t qu\u00e3ng \u0111\u01a3\u1eddng AB. Gi\u1ea3 s\u1eed qu\u00e3ng \u0111\u01b0\u1eddng AB d\u00e0i 120km. N\u1eeda qu\u00e3ng \u0111\u01b0\u1eddng AB l\u00e0: 120 : 2 = 60 (km) Th\u1eddi gian \u0111i n\u1eeda qu\u00e3ng \u0111\u01b0\u1eddng \u0111\u1ea7u: 60 : 60 = 1 (gi\u1edd) Th\u1eddi gian \u0111i n\u1eeda qu\u00e3ng \u0111\u01b0\u1eddng sau: 60 : 30 = 2 (gi\u1edd) T\u1ed5ng th\u1eddi gian \u0111i h\u1ebft qu\u00e3ng \u0111\u01b0\u1eddng: 1 + 2 = 3 (gi\u1edd) V\u1eadn t\u1ed1c trung b\u00ecnh c\u1ee7a B\u00ednh tr\u00ean su\u1ed1t qu\u00e3ng \u0111\u01b0\u1eddng:120 : 3 = 40 (km\/gi\u1edd) B\u00e0i 4: An ng\u1ed3i l\u00e0m b\u00e0i l\u00fac h\u01a1n 2 gi\u1edd m\u1ed9t ch\u00fat. Khi An l\u00e0m b\u00e0i xong th\u00ec th\u1ea5y 2 kim \u0111\u1ed3ng h\u1ed3 \u0111\u00e3 \u0111\u1ed5i ch\u1ed7 cho nhau. L\u00fac n\u00e0y h\u01a1n 3 gi\u1edd. H\u1ecfi An l\u00e0m b\u00e0i h\u1ebft bao nhi\u00eau ph\u00fat? V\u1eadn t\u1ed1c: Kim gi\u1edd m\u1ed7i gi\u1edd ch\u1ea1y 12 kho\u1ea3ng; kim gi\u1edd m\u1ed7i gi\u1edd ch\u1ea1y 1 kho\u1ea3ng. T\u1ed5ng v\u1eadn t\u1ed1c c\u1ee7a 2 kim: 12 + 1 = 13 (kho\u1ea3ng gi\u1edd) Th\u1eddi gian 2 kim \u0111\u1ed5i ch\u1ed7 cho nhau:12 : 13 = 55 5 (ph\u00fat) 13 B\u00e0i 5: M\u1ed9t con Ch\u00f3 \u0111u\u1ed5i 1 con Th\u1ecf \u1edf c\u00e1ch xa 17 b\u01a3\u1edbc c\u1ee7a Ch\u00f3. Con Th\u1ecf \u1edf c\u00e1ch hang c\u1ee7a n\u00f3 80 b\u01a3\u1edbc c\u1ee7a Th\u1ecf. Khi Th\u1ecf ch\u1ea1y \u0111\u01a3\u1ee3c 3 b\u01a3\u1edbc th\u00ec Ch\u00f3 ch\u1ea1y \u0111\u01a3\u1ee3c 1 b\u01a3\u1edbc. M\u1ed9t b\u01a3\u1edbc c\u1ee7a Ch\u00f3 b\u1eb1ng 8 b\u01a3\u1edbc c\u1ee7a Th\u1ecf. H\u1ecfi Ch\u00f3 c\u00f3 b\u1eaft \u0111\u01a3\u1ee3c Th\u1ecf kh\u00f4ng? 80 b\u01b0\u1edbc c\u1ee7a th\u1ecf b\u1eb1ng s\u1ed1 b\u01b0\u1edbc c\u1ee7a ch\u00f3 l\u00e0 : 80 : 8 = 10 ( b\u01b0\u1edbc ch\u00f3) Ch\u00f3 \u1edf c\u00e1ch hang th\u1ecf s\u1ed1 b\u01b0\u1edbc l\u00e0 : 17 + 10 = 27 ( b\u01b0\u1edbc) \u0110\u1ec3 \u0111\u1ebfn hang th\u1ecf th\u00ec ch\u00f3 ph\u1ea3i ch\u1ea1y s\u1ed1 b\u01b0\u1edbc t\u00ednh b\u1eb1ng b\u01b0\u1edbc th\u1ecf l\u00e0 : 27 x 3 = 81 ( b\u01b0\u1edbc 58","th\u1ecf) M\u00e0 th\u1ecf \u1edf c\u00e1ch hang c\u1ee7a n\u00f3 80 b\u01b0\u1edbc th\u1ecf n\u00ean th\u1ecf \u0111\u00e3 \u0111\u1ebfn tr\u01b0\u1edbc 1 b\u01b0\u1edbc v\u00e0 v\u00e0o hang. V\u00ec v\u1eady ch\u00f3 kh\u00f4ng b\u1eaft \u0111\u01a3\u1ee3c th\u1ecf. B\u00e0i 6: An \u0111i t\u1eeb A \u0111\u1ebfn B m\u1ea5t 4 gi\u1edd, B\u00ecnh \u0111i t\u1eeb B v\u1ec1 A m\u1ea5t 5 gi\u1edd. Bi\u1ebft r\u1eb1ng n\u1ebfu An v\u00e0 B\u00ecnh c\u00f9ng xu\u1ea5t ph\u00e1t c\u00f9ng m\u1ed9t l\u00fac th\u00ec sau 2 gi\u1edd 30 ph\u00fat hai ng\u01a3\u1eddi c\u00e1ch nhau 20 km. T\u00ednh \u0111\u1ed9 d\u00e0i qu\u00e3ng \u0111\u01a3\u1eddng AB. M\u1ed7i gi\u1edd An \u0111i \u0111\u01b0\u1ee3c 1\/4 qu\u00e3ng \u0111\u01b0\u1eddng; B\u00ecnh \u0111i \u0111\u01b0\u1ee3c 1\/5 qu\u00e3ng \u0111\u01b0\u1eddng. M\u1ed7i gi\u1edd c\u1ea3 2 ng\u01b0\u1eddi \u0111i \u0111\u01b0\u1ee3c: 1\/4 + 1\/5 = 9\/20 (qu\u00e3ng \u0111\u01b0\u1eddng) 2 gi\u1edd 30 ph\u00fat (2,5 gi\u1edd) c\u1ea3 2 ng\u01b0\u1eddi \u0111i \u0111\u01b0\u1ee3c: 9\/20 x 2,5 = 45\/40 (qu\u00e3ng \u0111\u01b0\u1eddng) Ph\u00e2n s\u1ed1 ch\u1ec9 20 km: 45\/40 - 1 = 5\/40 (qu\u00e3ng \u0111\u01b0\u1eddng) Qu\u00e3ng \u0111\u01b0\u1eddng AB l\u00e0: 20 : 5 x 40 = 160 (km) B\u00e0i 7: B\u00e2y gi\u1edd l\u00e0 3 gi\u1edd.H\u1ecfi sau \u00edt nh\u1ea5t l\u00e0 bao nhi\u00eau gi\u1edd n\u1eefa th\u00ec kim gi\u1edd v\u00e0 kim ph\u00fat s\u1ebd tr\u00f9ng nhau? D\u1ea1ng 2 chuy\u1ec3n \u0111\u1ed9ng c\u00f9ng chi\u1ec1u. V\u1eadn t\u1ed1c kim ph\u00fat 12 kho\u1ea3ng \/gi\u1edd V\u1eadn t\u1ed1c kim gi\u1edd 1 kho\u1ea3ng\/gi\u1edd L\u00fac 3 gi\u1edd, kim ph\u00fat \u1edf sau kim gi\u1edd 3 kho\u1ea3ng Hi\u1ec7u v\u1eadn t\u1ed1c 2 kim l\u00e0 12 \u2013 1 = 11 (kho\u1ea3ng\/g) Th\u1eddi gian kim ph\u00fat tr\u00f9ng kim gi\u1edd l\u00e0: 3 : 1 = 3\/11 (gi\u1edd) B\u00e0i 8: L\u00fac 6 gi\u1edd m\u1ed9t ng\u01a3\u1eddi xu\u1ea5t ph\u00e1t \u0111i t\u1eeb A \u0111\u1ebfn B v\u1edbi v\u1eadn t\u1ed1c 30km\/h, sau 30 ph\u00fat m\u1ed9t ng\u01a3\u1eddi kh\u00e1c \u0111i t\u1eeb B v\u1ec1 A v\u1edbi v\u1eadn t\u1ed1c 40km\/h.Bi\u1ebft h\u1ecd g\u1eb7p nhau luc 8h30. T\u00ednh \u0111\u1ed9 d\u00e0i qu\u00e3ng \u0111\u01a3\u1eddng AB? Sau 30 ph\u00fat th\u00ec ng\u01b0\u1eddi \u0111i t\u1eeb A \u0111i \u0111\u01b0\u1ee3c: 30 : 2 = 15 (km) L\u00fac n\u00e0y 2 ng\u01b0\u1eddi c\u00f9ng xu\u1ea5t ph\u00e1t l\u00fac 6g30ph. T\u1ed5ng v\u1eadn t\u1ed9c 2 ng\u01b0\u1eddi l\u00e0 bao nhi\u00eau? \u0110i bao l\u00e2u 2 ng\u01b0\u1eddi s\u1ebd g\u1eb7p nhau? Ta t\u00ednh \u0111\u01b0\u1ee3c \u0111\u1ed9 d\u00e0i qu\u00e3ng \u0111\u01b0\u1eddng. B\u00e0i 9: Hai v\u1eadn \u0111\u1ed9ng vi\u00ean \u0111ua xe \u0111\u1ea1p \u0111\u01a3\u1eddng tr\u01a3\u1eddng 10 v\u00f2ng quanh m\u1ed9t c\u00e1i h\u1ed3 h\u00ecnh tr\u00f2n c\u00f3 chu vi 10km. V\u1eadn t\u1ed1c trung b\u00ecnh c\u1ee7a ng\u01a3\u1eddi th\u1ee9 nh\u1ea5t l\u00e0 32km\/gi\u1edd; v\u1eadn t\u1ed1c c\u1ee7a ng\u01a3\u1eddi th\u1ee9 hai l\u00e0 35km\/gi\u1edd. H\u1ecfi sau 2 gi\u1edd hai ng\u01a3\u1eddi c\u00e1ch nhau bao xa? Sau 2 gi\u1edd ng\u01b0\u1eddi th\u1ee9 nh\u1ea5t \u0111i \u0111\u01b0\u1ee3c: 32 x 2 = 64 (km) Sau 2 gi\u1edd ng\u01b0\u1eddi th\u1ee9 hai \u0111i \u0111\u01b0\u1ee3c: 35 x 2 = 70 (km) Ta th\u1ea5y sau 2 gi\u1edd ng\u01b0\u1eddi th\u1ee9 hai \u0111i v\u1eeba \u0111\u00fang 7 v\u00f2ng v\u1ec1 \u0111\u1ebfn \u0111i\u1ec3m xu\u1ea5t ph\u00e1t (do 70 h\u1ebft cho 10); ng\u01b0\u1eddi th\u1ee9 nh\u1ea5t v\u1eeba qua \u0111i\u1ec3m xu\u1ea5t ph\u00e1t 6 v\u00f2ng v\u00e0 th\u00eam 4km : (64 \u2013 (10x6) = 4(km)). Nh\u01b0 v\u1eady 2 ng\u01b0\u1eddi c\u00e1ch nhau 4 km. B\u00e0i 10: M\u1ed9t ca n\u00f4 \u0111i xu\u00f4i d\u00f2ng t\u1eeb b\u1ebfn A \u0111\u1ebfn b\u1ebfn B m\u1ea5t 2 gi\u1edd; \u0111i ng\u01a3\u1ee3c d\u00f2ng t\u1eeb b\u1ebfn B \u0111\u1ebfn b\u1ebfn A m\u1ea5t 3 gi\u1edd. Bi\u1ebft v\u1eadn t\u1ed1c gi\u1eefa khi \u0111i xu\u00f4i d\u00f2ng v\u00e0 khi \u0111i ng\u01a3\u1ee3c d\u00f2ng l\u00e0 95km\/gi\u1edd. T\u00ednh \u0111\u1ed9 d\u00e0i qu\u00e3ng \u0111\u01a3\u1eddng AB? Hi\u1ec3u l\u00e0 t\u1ed5ng v\u1eadn t\u1ed1c \u0111i xu\u00f4i d\u00f2ng v\u00e0 \u0111i ng\u01b0\u1ee3c d\u00f2ng l\u00e0 95km\/gi\u1edd. C\u00f9ng qu\u00e3ng \u0111\u01b0\u1eddng th\u00ec th\u1eddi gian t\u1ec9 l\u1ec7 ngh\u1ecbch v\u1edbi v\u1eadn t\u1ed1c. T\u1ec9 s\u1ed1 v\u1eadn t\u1ed1c gi\u1eefa xu\u00f4i d\u00f2ng v\u00e0 ng\u01b0\u1ee3c d\u00f2ng l\u00e0: 3\/2. V\u1eadn t\u1ed1c xu\u00f4i d\u00f2ng l\u00e0: 95 : (3+2) x 3 = 57 (km\/gi\u1edd) Qu\u00e3ng \u0111\u01b0\u1eddng AB l\u00e0: 57 x 2 = 114 (km) 59","B\u00e0i 11: M\u1ed9t ca n\u00f4 \u0111i xu\u00f4i d\u00f2ng t\u1eeb b\u1ebfn A \u0111\u1ebfn b\u1ebfn B m\u1ea5t 2 gi\u1edd; \u0111i ng\u01a3\u1ee3c d\u00f2ng t\u1eeb b\u1ebfn B \u0111\u1ebfn b\u1ebfn A m\u1ea5t 3 gi\u1edd. Bi\u1ebft v\u1eadn t\u1ed1c d\u00f2ng n\u01a3\u1edbc l\u00e0 10km\/gi\u1edd. T\u00ednh chi\u1ec1u d\u00e0i qu\u00e3ng \u0111\u01a3\u1eddng AB? G\u1ecdi VX l\u00e0 v\u1eadn t\u1ed1c xu\u00f4i d\u00f2ng v\u00e0 VN l\u00e0 v\u1eadn t\u1ed1c ng\u01b0\u1ee3c d\u00f2ng. Hi\u1ec7u v\u1eadn t\u1ed1c: VX \u2013 VN = 10 x 2 = 20 (km\/gi\u1edd) C\u00f9ng qu\u00e3ng \u0111\u01b0\u1eddng th\u00ec th\u1eddi gian t\u1ec9 l\u1ec7 ngh\u1ecbch v\u1edbi v\u1eadn t\u1ed1c. T\u1ec9 s\u1ed1 v\u1eadn t\u1ed1c gi\u1eefa xu\u00f4i d\u00f2ng v\u00e0 ng\u01b0\u1ee3c d\u00f2ng l\u00e0: 3\/2. V\u1eadn t\u1ed1c xu\u00f4i d\u00f2ng l\u00e0: 20 : (3-2) x 3 = 60 (km\/gi\u1edd) Qu\u00e3ng \u0111\u01b0\u1eddng AB l\u00e0: 60 x 2 = 120 (km) B\u00e0i 12: M\u1ed9t can\u00f4 \u0111i t\u1eeb A v\u1ec1 B h\u00eat 3 gi\u1edd v\u00e0 \u0111i t\u1eeb B v\u1ec1 A h\u1ebft 4 gi\u1edd .Bi\u1ebft v\u1eadn t\u1ed1c dong n\u01a3\u1edbc l\u00e0 4km\/gi\u1edd .T\u00ednh qu\u00e3ng d\u01a3\u01a1ng AB? C\u00f9ng qu\u00e3ng \u0111\u01b0\u1eddng th\u00ec v\u1eadn t\u1ed1c t\u1ec9 l\u1ec7 ng\u1ecbch v\u1edbi th\u1eddi gian. G\u1ecdi Vx l\u00e0 v\u1eadn t\u1ed1c xu\u00f4i d\u00f2ng v\u00e0 Vn l\u00e0 v\u1eadn t\u1ed1c ng\u01b0\u1ee3c d\u00f2ng. Ta c\u00f3 Vx\/ Vn = 4\/3 V\u1eadn t\u1ed1c xu\u00f4i d\u00f2ng h\u01a1n v\u1eadn t\u1ed1c ng\u01b0\u1ee3c d\u00f2ng: 4 x 2 = 8 (km\/gi\u1edd) (v\u1eadn t\u1ed1c xu\u00f4i d\u00f2ng h\u01a1n v\u1eadn t\u1ed1c ng\u01b0\u1ee3c d\u00f2ng 2 l\u1ea7n v\u1eadn t\u1ed1c d\u00f2ng n\u01b0\u1edbc). Vx: |___|___|___|___| Vn: |___|___|___| ..8.. V\u1eadn t\u1ed1c xu\u00f4i d\u00f2ng: 8 x 4 = 32 (km\/gi\u1edd) Qu\u00e3ng \u0111\u01b0\u1eddng AB: 32 x 3 = 96 (km) B\u00e0i 13: M\u1ed9t \u00f4 t\u00f4 \u0111i t\u1eeb A \u0111\u1ebfn B v\u1edbi v\u1eadn t\u1ed1c 60 km\/gi\u1edd, l\u00fac t\u1eeb B v\u1ec1 A \u00f4 t\u00f4 \u0111i v\u1edbi v\u1eadn t\u1ed1c 40km\/gi\u1edd. T\u00ednh v\u1eadn t\u1ed1c trung b\u00ecnh \u00f4 t\u00f4 \u0111\u00e3 \u0111i tr\u00ean c\u1ea3 qu\u00e3ng \u0111\u01a3\u1eddng? Mu\u1ed1n t\u00ednh v\u1eadn t\u1ed1c trung b\u00ecnh ph\u1ea3i c\u1ea7n c\u00f3 qu\u00e3ng \u0111\u01b0\u1eddng v\u00e0 th\u1eddi gian t\u01b0\u01a1ng \u1ee9ng. Gi\u1ea3 s\u1eed qu\u00e3ng \u0111\u01b0\u1eddng d\u00e0i 120 km (v\u00ec 120 chia h\u1ebft cho 60 v\u00e0 40 \u0111\u1ec3 d\u1ec5 t\u00ednh) Th\u1eddi gian \u0111i t\u1eeb A \u0111\u1ebfn B: 120 : 60 = 2 (gi\u1edd) Th\u1eddi gian \u0111i t\u1eeb B \u0111\u1ebfn A: 120 : 40 = 3 (gi\u1edd) T\u1ed5ng th\u1eddi gian c\u1ea3 \u0111i l\u1eabn v\u1ec1: 2 + 3 = 5 (gi\u1edd) V\u1eadn t\u1ed1c trung b\u00ecnh c\u1ea3 \u0111i l\u1ea7n v\u1ec1: 120 x 2 : (3 + 2) = 48 (km\/gi\u1edd) B\u00e0i 14: L\u00fac 5gi\u1edd 15 ph\u00fat, m\u1ed9t \u00f4 t\u00f4 ch\u1edf h\u00e0ng t\u1eeb t\u1ec9nh A \u0111\u1ebfn t\u1ec9nh B v\u1edbi v\u1eadn t\u1ed1c 55km\/gi\u1edd. \u0110\u1ebfn 8 gi\u1edd 51 ph\u00fat th\u00ec \u00f4 t\u00f4 \u0111\u1ebfn t\u1ec9nh B. Sau khi tr\u1ea3 h\u00e0ng cho t\u1ec9nh B h\u1ebft 45 ph\u00fat, \u00f4 t\u00f4 quay v\u1ec1 A v\u1edbi v\u1eadn t\u1ed1 60km\/gi\u1edd. H\u1ecfi \u00f4t\u00f4 quay v\u1ec1 \u0111\u1ebfn A l\u00fac m\u1ea5y gi\u1edd? Th\u1eddi gian \u00f4 t\u00f4 \u0111i t\u1eeb A \u0111\u1ebfn B l\u00e0: 8 gi\u1edd 51 ph\u00fat - 5 gi\u1edd 15 ph\u00fat = 3 gi\u1edd 36 ph\u00fat = 3,6 (gi\u1edd) Qu\u00e3ng \u0111\u01b0\u1eddng AB l\u00e0: 55 x 3,6 = 198 km Th\u1eddi gian \u00d4 t\u00f4 quay v\u1ec1 t\u1eeb B v\u1ec1 A l\u00e0: 198 : 60 = 3,3 (gi\u1edd) = 3 gi\u1edd 18 ph\u00fat \u00d4 t\u00f4 quay v\u1ec1 \u0111\u1ebfn A l\u00fac: 8 gi\u1edd 51 + 45 ph\u00fat + 3 gi\u1edd 18 ph\u00fat = 12 gi\u1edd 54 ph\u00fat B\u00e0i 15: M\u1ed9t ca n\u00f4 \u0111i xu\u00f4i d\u00f2ng t\u1eeb A \u0111\u1ebfn B v\u1edbi v\u1eadn t\u1ed1c 50km\/gi\u1edd. \u0110i ng\u01a3\u1ee3c d\u00f2ng t\u1eeb B \u0111\u1ebfn A v\u1edbi v\u1eadn t\u1ed1c 40km\/gi\u1edd. Bi\u1ebft t\u1ed5ng th\u1eddi gian c\u1ea3 \u0111i l\u1eabn v\u1ec1 l\u00e0 3,6 gi\u1edd? T\u00ednh \u0111\u1ed9 d\u00e0i qu\u00e3ng \u0111\u01a3\u1eddng AB? C\u00f9ng qu\u00e3ng \u0111\u01b0\u1eddng th\u00ec v\u1eadn t\u1ed1c t\u1ec9 l\u1ec7 ngh\u1ecbc v\u1edbi th\u1eddi gian: T\u1ec9 l\u1ec7 th\u1eddi gian c\u1ee7a xu\u00f4i v\u00e0 ng\u01b0\u1ee3c d\u00f2ng: 40\/50 = 4\/5 Th\u1eddi gian xu\u00f4i d\u00f2ng: 3,6: (4+5)x4= 1,6 (gi\u1edd) Qu\u00e3ng \u0111\u01b0\u1eddng AB: 50 x 1,6 = 80 (km) 60","61"]
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