TS 10 CO DAP AN 23-24

["TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH a) Ng\u00e0y 02 \/ 10 \/ 2020 nh\u00f3m c\u1ee7a Lan g\u1ed3m 5 ng\u01b0\u1eddi \u0111\u1ebfn c\u1eeda h\u00e0ng mua m\u1ed7i ng\u01b0\u1eddi 1 \u00e1o thun. H\u1ecfi nh\u00f3m c\u1ee7a Lan ph\u1ea3i tr\u1ea3 bao nhi\u00eau ti\u1ec1n? b) Ng\u00e0y 10 \/ 10 \/ 2020 Lan quay l\u1ea1i c\u1eeda h\u00e0ng mua th\u00eam 1 c\u00e1i n\u00f3n bi\u1ebft tr\u01b0\u1edbc \u0111\u00f3 ng\u00e0y 02 \/ 10 \/ 2020 , ngo\u00e0i mua 1 \u00e1o thun, Lan \u0111\u00e3 mua th\u00eam 1 \u00e1o kho\u00e1c. Khi v\u1ec1 \u0111\u1ebfn nh\u00e0, Lan t\u00ednh t\u1ed5ng s\u1ed1 ti\u1ec1n mua 1 \u00e1o thun, 1 \u00e1o kho\u00e1c, 1 n\u00f3n t\u1ed5ng c\u1ed9ng l\u00e0 872000 \u0111\u1ed3ng. H\u1ecfi gi\u00e1 ni\u00eam y\u1ebft c\u1ee7a 1 c\u00e1i n\u00f3n l\u00e0 bao nhi\u00eau? C\u00e2u 5. (1 \u0111i\u1ec3m). C\u00f4ng ty A th\u1ef1c hi\u1ec7n m\u1ed9t cu\u1ed9c kh\u1ea3o s\u00e1t \u0111\u1ec3 t\u00ecm hi\u1ec3u v\u1ec1 m\u1ed1i li\u00ean h\u1ec7 gi\u1eefa y (s\u1ea3n ph\u1ea9m) v\u00e0 s\u1ed1 l\u01b0\u1ee3ng s\u1ea3n ph\u1ea9m T b\u00e1n ra v\u1edbi x (ngh\u00ecn \u0111\u1ed3ng) l\u00e0 gi\u00e1 b\u00e1n ra c\u1ee7a m\u1ed7i s\u1ea3n ph\u1ea9m T v\u00e0 nh\u1eadn th\u1ea5y r\u1eb1ng y = ax + b ( a,b l\u00e0 h\u1eb1ng s\u1ed1). Bi\u1ebft v\u1edbi gi\u00e1 b\u00e1n l\u00e0 400000 (\u0111\u1ed3ng)\/ s\u1ea3n ph\u1ea9m th\u00ec s\u1ed1 l\u01b0\u1ee3ng s\u1ea3n ph\u1ea9m b\u00e1n ra l\u00e0 1200 (s\u1ea3n ph\u1ea9m); v\u1edbi gi\u00e1 b\u00e1n l\u00e0 460000 (\u0111\u1ed3ng)\/ s\u1ea3n ph\u1ea9m th\u00ec s\u1ed1 l\u01b0\u1ee3ng s\u1ea3n ph\u1ea9m b\u00e1n ra l\u00e0 1800 (s\u1ea3n ph\u1ea9m) a) X\u00e1c \u0111\u1ecbnh a,b. b) B\u1eb1ng ph\u00e9p t\u00ednh, h\u00e3y t\u00ednh s\u1ed1 l\u01b0\u1ee3ng s\u1ea3n ph\u1ea9m b\u00e1n ra v\u1edbi gi\u00e1 b\u00e1n l\u00e0 440000 (\u0111\u1ed3ng)\/ s\u1ea3n ph\u1ea9m? C\u00e2u 6. (1 \u0111i\u1ec3m). M\u1ed9t \u0111\u1ecba ph\u01b0\u01a1ng c\u1ea5y 10ha gi\u1ed1ng l\u00faa lo\u1ea1i I v\u00e0 8ha gi\u1ed1ng l\u00faa lo\u1ea1i II . Sau m\u1ed9t m\u00f9a v\u1ee5, \u0111\u1ecba ph\u01b0\u01a1ng \u0111\u00f3 thu ho\u1ea1ch v\u00e0 t\u00ednh to\u00e1n s\u1ea3n l\u01b0\u1ee3ng th\u1ea5y: + T\u1ed5ng s\u1ea3n l\u01b0\u1ee3ng c\u1ee7a hai gi\u1ed1ng l\u00faa thu v\u1ec1 l\u00e0 139 t\u1ea5n; +S\u1ea3n l\u01b0\u1ee3ng thu v\u1ec1 t\u1eeb 4ha gi\u1ed1ng l\u00faa lo\u1ea1i I nhi\u1ec1u h\u01a1n s\u1ea3n l\u01b0\u1ee3ng thu v\u1ec1 t\u1eeb 3ha gi\u1ed1ng l\u00faa lo\u1ea1i II l\u00e0 6 t\u1ea5n. H\u00e3y t\u00ednh n\u0103ng su\u1ea5t l\u00faa trung b\u00ecnh (\u0111\u01a1n v\u1ecb: t\u1ea5n\/ha) c\u1ee7a m\u1ed7i lo\u1ea1i gi\u1ed1ng l\u00faa. C\u00e2u 7. (1 \u0111i\u1ec3m). Cho c\u1ed1c r\u01b0\u1ee3u (nh\u01b0 h\u00ecnh v\u1ebd), ph\u1ea7n ph\u00eda tr\u00ean l\u00e0 m\u1ed9t h\u00ecnh n\u00f3n c\u00f3 chi\u1ec1u cao 6cm v\u00e0 \u0111\u00e1y l\u00e0 \u0111\u01b0\u1eddng tr\u00f2n b\u00e1n k\u00ednh 3cm . T\u00ednh th\u1ec3 t\u00edch r\u01b0\u1ee3u trong ly. (K\u1ebft qu\u1ea3 l\u00e0m tr\u00f2n \u0111\u1ebfn ch\u1eef s\u1ed1 th\u1eadp ph\u00e2n th\u1ee9 nh\u1ea5t) . C\u00e2u 8. (3 \u0111i\u1ec3m) Cho n\u1eeda \u0111\u01b0\u1eddng tr\u00f2n (O) , \u0111\u01b0\u1eddng k\u00ednh BC . Tr\u00ean n\u1eeda \u0111\u01b0\u1eddng tr\u00f2n (O) , l\u1ea5y hai \u0111i\u1ec3m A v\u00e0 D (theo th\u1ee9 t\u1ef1 B, A,D,C ). Tia BA v\u00e0 CD c\u1eaft nhau t\u1ea1i S , \u0111o\u1ea1n th\u1eb3ng AC c\u1eaft BD t\u1ea1i H. a) Ch\u1ee9ng minh SH \u22a5 BC t\u1ea1i E v\u00e0 t\u1ee9 gi\u00e1c HECD n\u1ed9i ti\u1ebfp. b) G\u1ecdi T l\u00e0 trung \u0111i\u1ec3m SH , tia AT c\u1eaft SC t\u1ea1i I , DE c\u1eaft HC t\u1ea1i K . Ch\u1ee9ng minh: TAH = KDC . T\u1eeb \u0111\u00f3 suy ra CK.CA = CD.CI c) D\u01b0\u1eddng trung tr\u1ef1c c\u1ea3 \u0111o\u1ea1n th\u1eb3ng AK c\u1eaft BH t\u1ea1i Q . Ch\u1ee9ng minh \uf044IAK c\u00e2n v\u00e0 ba \u0111i\u1ec3m A,O,Q th\u1eb3ng h\u00e0ng. ----H\u1ebeT--- T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N 2","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N 3","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH H\u01af\u1edaNG D\u1eaaN GI\u1ea2I C\u00e2u 1. (1,5 \u0111i\u1ec3m). Cho (P) : y = x2 v\u00e0 \u0111\u01b0\u1eddng th\u1eb3ng (d) : y = 2x \u2212 7 . 44 a) V\u1ebd \u0111\u1ed3 th\u1ecb (P) v\u00e0 (d) tr\u00ean c\u00f9ng h\u1ec7 tr\u1ee5c t\u1ecda \u0111\u1ed9. b) T\u00ecm t\u1ecda \u0111\u1ed9 giao \u0111i\u1ec3m c\u1ee7a (P) v\u00e0 (d) b\u1eb1ng ph\u00e9p t\u00ednh. L\u1eddi gi\u1ea3i a) V\u1ebd \u0111\u1ed3 th\u1ecb (P) v\u00e0 (d) tr\u00ean c\u00f9ng h\u1ec7 tr\u1ee5c t\u1ecda \u0111\u1ed9. a) BGT: x \u22122 \u22121 0 1 2 y = x2 11 4 14 0 4 1 x 01 y = 2x \u2212 7 \u22127 1 44 4 b) T\u00ecm t\u1ecda \u0111\u1ed9 giao \u0111i\u1ec3m c\u1ee7a (P) v\u00e0 (d) b\u1eb1ng ph\u00e9p t\u00ednh. Ph\u01b0\u01a1ng tr\u00ecnh ho\u00e0nh \u0111\u1ed9 giao \u0111i\u1ec3m c\u1ee7a (P) v\u00e0 (d) : x2 = 2x \u2212 7 44 \uf0db x2 \u2212 2x + 7 = 0 44 \uf0db \uf0e9x = 1 \uf0ea\uf0ebx = 7 Thay x = 1 v\u00e0o y = x2 , ta \u0111\u01b0\u1ee3c: y = 12 = 1 . 4 44 Thay x = 7 v\u00e0o y = x2 , ta \u0111\u01b0\u1ee3c: y = 72 = 49 . 4 44 V\u1eady \uf0e6\uf0e7 1; 1\uf0f6 , \uf0e6 7; 49 \uf0f6 l\u00e0 hai giao \u0111i\u1ec3m c\u1ea7n t\u00ecm. \uf0e8 \uf0f7 \uf0e7 \uf0f7 4 \uf0f8 \uf0e8 4 \uf0f8 T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N 4","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH C\u00e2u 2. (1 \u0111i\u1ec3m). Cho ph\u01b0\u01a1ng tr\u00ecnh 4x2 \u2212 2x \u2212 5 = 0 c\u00f3 2 nghi\u1ec7m l\u00e0 x1 , x2 . Kh\u00f4ng gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh, h\u00e3y t\u00ednh gi\u00e1 tr\u1ecb c\u1ee7a bi\u1ec3u th\u1ee9c B = x1 + x2 x2 \u2212 1 x1 \u2212 1 L\u1eddi gi\u1ea3i V\u00ec \uf044 = b2 \u2212 4ac = (\u22122)2 \u2212 4.4.(\u22125) = 64 \uf03e 0 N\u00ean ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m ph\u00e2n bi\u1ec7t x1 ,x2 . \uf0ec\uf0ef\uf0efS = x1 + x2 = \u2212b = 1 \uf0ed = x1 a 2 Theo \u0111\u1ecbnh l\u00ed Vi-et, ta c\u00f3: \uf0ef\uf0ef\uf0eeP \u22125 c = 4 .x2 = a Ta c\u00f3: B = x1 + x2 x2 \u2212 1 x1 \u2212 1 B = x1 + x2 x2 \u2212 1 x1 \u2212 1 B = x1 (x1 \u2212 1) + x2 (x2 \u2212 1) (x1 \u2212 1)(x2 \u2212 1) B = x12 \u2212 x1 + x22 \u2212 x2 x1x2 \u2212 x2 \u2212 x1 + 1 ( ) ( )B = x1 + x2 2 \u2212 2x1x2 \u2212 x1 + x2 ( )x1x2 \u2212 x1 + x2 + 1 \uf0e6 1 \uf0f62 \u2212 2.\uf0e6\uf0e7\uf0e8 \u22125 \uf0f6 \u2212 1 \uf0e7\uf0e8 2 \uf0f7\uf0f8 4 \uf0f7 2 B = \uf0f8 = \u22123 \u22125 \u2212 1 + 1 42 C\u00e2u 3. (1 \u0111i\u1ec3m). M\u1ed7i n\u01a1i tr\u00ean th\u1ebf gi\u1edbi c\u00f3 m\u1ed9t m\u00fai gi\u1edd. Gi\u1edd m\u1ed7i ng\u00e0y t\u1ea1i m\u1ed7i n\u01a1i \u0111\u01b0\u1ee3c t\u00ednh theo c\u00f4ng th\u1ee9c T GMT H , trong \u0111\u00f3 T l\u00e0 gi\u1edd t\u1ea1i n\u01a1i \u0111\u00f3, GMT l\u00e0 gi\u1edd g\u1ed1c, gi\u1edd \u1edf m\u00fai gi\u1edd l\u00e0 0 , H \u0111\u01b0\u1ee3c x\u00e1c \u0111\u1ecbnh b\u1edfi b\u1ea3ng sau: a M\u00fai gi\u1edd 0 1 2 3 4 5 6 7 H 0 123 4 567 M\u00fai gi\u1edd 8 9 10 11 12 13 14 15 HH 8 9 10 11 12 11 10 9 M\u00fai M\u00fai gi\u1edd 16 17 18 19 20 21 22 23 H 87 654 321 Nh\u01b0 v\u1eady khi bi\u1ebft gi\u1edd \u1edf m\u1ed9t n\u01a1i c\u00f3 m\u00fai gi\u1edd n\u00e0y, ta c\u00f3 th\u1ec3 t\u00ednh gi\u1edd \u1edf n\u01a1i c\u00f3 m\u00fai gi\u1edd kh\u00e1c. T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N 5","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH M\u00fai gi\u1edd c\u1ee7a m\u1ed9t s\u1ed1 th\u00e0nh ph\u1ed1 \u0111\u01b0\u1ee3c cho b\u1edfi b\u1ea3ng sau: Th\u00e0nh ph\u1ed1 H\u1ed3 Ch\u00ed Minh New York Moscow Los Angeles M\u00fai gi\u1edd 7 19 3 16 D\u1ef1a v\u00e0o c\u00e1ch t\u00ednh tr\u00ean em h\u00e3y t\u00ednh xem: d) L\u00fac 11 gi\u1edd ng\u00e0y 03 \/ 06 \u1edf NewYork th\u00ec \u1edf Moscow l\u00e0 m\u1ea5y gi\u1edd ng\u00e0y n\u00e0o? e) Qu\u1ef3nh \u0111i chuy\u1ebfn bay t\u1eeb Tp.HCM \u0111\u1ebfn Moscow c\u1ee7a h\u00e3ng h\u00e0ng kh\u00f4ng Aeroflot. Chuy\u1ebfn bay xu\u1ea5t ph\u00e1t l\u00fac 14 gi\u1edd 30 ph\u00fat ng\u00e0y 01 \/ 09 theo gi\u1edd t\u1ea1i Tp.HCM. Em h\u00e3y t\u00ednh xem chuy\u1ebfn bay k\u00e9o d\u00e0i bao l\u00e2u bi\u1ebft Qu\u1ef3nh \u0111\u1ebfn s\u00e2n bay qu\u1ed1c t\u1ebf Sheremetyevo c\u1ee7a Moscow l\u00fac 21 gi\u1edd ng\u00e0y 01 \/ 09 ? L\u1eddi gi\u1ea3i a) 11h ng\u00e0y 03 \/ 06 \u1edf NewYork TNY GMT HNY 11 GMT ( 5) GMT 16 V\u1eady GMT l\u00e0 16h ng\u00e0y 03 \/ 06 TMC GMT H MC TMC 16 3 19 V\u1eady \u1edf Moscow l\u00e0 19h ng\u00e0y 03 \/ 06 b) HCM l\u00e0 14h30p ng\u00e0y 01 \/ 09 THCM GMT H HCM 14, 5 GMT 7 GMT 7, 5 Moscow l\u00e0 21h ng\u00e0y 01 \/ 09 TMC GMT HMC 21 GMT 3 GMT 18 V\u1eady chuy\u1ebfn bay d\u00e0i: 18 7, 5 10, 5h . C\u00e2u 4. (0,75 \u0111i\u1ec3m). Nh\u00e2n d\u1ecbp k\u1ec9 ni\u1ec7m 1 n\u0103m th\u00e0nh l\u1eadp, m\u1ed9t c\u1eeda h\u00e0ng th\u1eddi trang \u0111\u01b0a ra ch\u01b0\u01a1ng tr\u00ecnh khuy\u1ebfn m\u00e3i: \u0111\u1ee3t 1 : Gi\u1ea3m 20% tr\u00ean t\u1ea5t c\u1ea3 c\u00e1c m\u1eb7t h\u00e0ng t\u1eeb ng\u00e0y 01 \/ 10 \/ 2020 \u0111\u1ebfn 05 \/ 10 \/ 2020 ; \u0111\u1ee3t 2 : gi\u1ea3m 5% t\u1ea5t c\u1ea3 c\u00e1c m\u1eb7t h\u00e0ng tr\u00ean gi\u00e1 \u0111\u00e3 gi\u1ea3m c\u1ee7a \u0111\u1ee3t 1 t\u1eeb ng\u00e0y T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N 6","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH 06 \/ 10 \/ 2020 \u0111\u1ebfn 10 \/ 10 \/ 2020 bi\u00eat gi\u00e1 ni\u00eam y\u1ebft c\u1ee7a \u00e1o thun l\u00e0 400 000 \u0111\u1ed3ng\/ \u00e1o, \u00e1o kho\u00e1c 500 000 \u0111\u1ed3ng\/ \u00e1o. f) Ng\u00e0y 02 \/ 10 \/ 2020 nh\u00f3m c\u1ee7a Lan g\u1ed3m 5 ng\u01b0\u1eddi \u0111\u1ebfn c\u1eeda h\u00e0ng mua m\u1ed7i ng\u01b0\u1eddi 1 \u00e1o thun. H\u1ecfi nh\u00f3m c\u1ee7a Lan ph\u1ea3i tr\u1ea3 bao nhi\u00eau ti\u1ec1n? g) Ng\u00e0y 10 \/ 10 \/ 2020 Lan quay l\u1ea1i c\u1eeda h\u00e0ng mua th\u00eam 1 c\u00e1i n\u00f3n bi\u1ebft tr\u01b0\u1edbc \u0111\u00f3 ng\u00e0y 02 \/ 10 \/ 2020 , ngo\u00e0i mua 1 \u00e1o thun, Lan \u0111\u00e3 mua th\u00eam 1 \u00e1o kho\u00e1c. Khi v\u1ec1 \u0111\u1ebfn nh\u00e0, Lan t\u00ednh t\u1ed5ng s\u1ed1 ti\u1ec1n mua 1 \u00e1o thun, 1 \u00e1o kho\u00e1c, 1 n\u00f3n t\u1ed5ng c\u1ed9ng l\u00e0 872000 \u0111\u1ed3ng. H\u1ecfi gi\u00e1 ni\u00eam y\u1ebft c\u1ee7a 1 c\u00e1i n\u00f3n l\u00e0 bao nhi\u00eau? L\u1eddi gi\u1ea3i a) Nh\u00f3m c\u1ee7a Lan ph\u1ea3i tr\u1ea3: 5.400000. 1 20% 1 5% 1600000(d) d b) G\u1ecdi x l\u00e0 gi\u00e1 ni\u00eam y\u1ebft c\u1ee7a n\u00f3n (x 0) S\u1ed1 ti\u1ec1n Lan chi cho 3 m\u00f3n l\u00e0: 872000 400000 500000 1 20% x 1 20% 1 5% x 200000(N ) x 200000 (nh\u1eadn) V\u1eady gi\u00e1 ni\u00eam y\u1ebft c\u1ee7a n\u00f3n l\u00e0 200 000 \u0111\u1ed3ng. C\u00e2u 5. (1 \u0111i\u1ec3m). C\u00f4ng ty ty A th\u1ef1c hi\u1ec7n m\u1ed9t cu\u1ed9c kh\u1ea3o s\u00e1t \u0111\u1ec3 t\u00ecm hi\u1ec3u v\u1ec1 m\u1ed1i li\u00ean h\u1ec7 gi\u1eefa y (s\u1ea3n ph\u1ea9m) v\u00e0 s\u1ed1 l\u01b0\u1ee3ng s\u1ea3n ph\u1ea9m T b\u00e1n ra v\u1edbi x (ngh\u00ecn \u0111\u1ed3ng) l\u00e0 gi\u00e1 b\u00e1n ra c\u1ee7a m\u1ed7i s\u1ea3n ph\u1ea9m T v\u00e0 nh\u1eadn th\u1ea5y r\u1eb1ng y ax b (a,b l\u00e0 h\u1eb1ng s\u1ed1). Bi\u1ebft v\u1edbi gi\u00e1 b\u00e1n l\u00e0 400 000 (\u0111\u1ed3ng)\/ s\u1ea3n ph\u1ea9m th\u00ec s\u1ed1 l\u01b0\u1ee3ng s\u1ea3n ph\u1ea9m b\u00e1n ra l\u00e0 1200 (s\u1ea3n ph\u1ea9m); v\u1edbi gi\u00e1 b\u00e1n l\u00e0 460 000 (\u0111\u1ed3ng)\/ s\u1ea3n ph\u1ea9m th\u00ec s\u1ed1 l\u01b0\u1ee3ng s\u1ea3n ph\u1ea9m b\u00e1n ra l\u00e0 1800 (s\u1ea3n ph\u1ea9m) h) X\u00e1c \u0111\u1ecbnh a,b. i) B\u1eb1ng ph\u00e9p t\u00ednh, h\u00e3y t\u00ednh s\u1ed1 l\u01b0\u1ee3ng s\u1ea3n ph\u1ea9m b\u00e1n ra v\u1edbi gi\u00e1 b\u00e1n l\u00e0 440 000 (\u0111\u1ed3ng)\/ s\u1ea3n ph\u1ea9m? L\u1eddi gi\u1ea3i a) Gi\u00e1 b\u00e1n l\u00e0 400 000 (\u0111\u1ed3ng)\/ s\u1ea3n ph\u1ea9m th\u00ec s\u1ed1 l\u01b0\u1ee3ng s\u1ea3n ph\u1ea9m b\u00e1n ra l\u00e0 1200 (s\u1ea3n ph\u1ea9m): Thay x 400;y 1200 v\u00e0o ph\u01b0\u01a1ng tr\u00ecnhy ax b 1200 a.400 b T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N 7","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH Gi\u00e1 b\u00e1n l\u00e0 460 000 (\u0111\u1ed3ng)\/ s\u1ea3n ph\u1ea9m th\u00ec s\u1ed1 l\u01b0\u1ee3ng s\u1ea3n ph\u1ea9m b\u00e1n ra l\u00e0 1800 (s\u1ea3n ph\u1ea9m) Thay x 460;y 1800 v\u00e0o ph\u01b0\u01a1ng tr\u00ecnhy ax b 1800 a.460 b Ta c\u00f3 h\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh: 1200 a.400 b 400a b 1200 a 10 1800 a.460 b 460a b 1800 b 2800 V\u1eady y 10x 2800 b) Thay x 440 v\u00e0o ph\u01b0\u01a1ng tr\u00ecnh y 10x 2800 y 10.440 2800 1600 sp C\u00e2u 6. (1 \u0111i\u1ec3m). M\u1ed9t \u0111\u1ecba ph\u01b0\u01a1ng c\u1ea5y 10ha gi\u1ed1ng l\u00faa lo\u1ea1i I v\u00e0 8ha gi\u1ed1ng l\u00faa lo\u1ea1i II . Sau m\u1ed9t m\u00f9a v\u1ee5, \u0111\u1ecba ph\u01b0\u01a1ng \u0111\u00f3 thu ho\u1ea1ch v\u00e0 t\u00ednh to\u00e1n s\u1ea3n l\u01b0\u1ee3ng th\u1ea5y: + T\u1ed5ng s\u1ea3n l\u01b0\u1ee3ng c\u1ee7a hai gi\u1ed1ng l\u00faa thu v\u1ec1 l\u00e0 139 t\u1ea5n; +S\u1ea3n l\u01b0\u1ee3ng thu v\u1ec1 t\u1eeb 4ha gi\u1ed1ng l\u00faa lo\u1ea1i I nhi\u1ec1u h\u01a1n s\u1ea3n l\u01b0\u1ee3ng thu v\u1ec1 t\u1eeb 3ha gi\u1ed1ng l\u00faa lo\u1ea1i II l\u00e0 6 t\u1ea5n. H\u00e3y t\u00ednh n\u0103ng su\u1ea5t l\u00faa trung b\u00ecnh (\u0111\u01a1n v\u1ecb: t\u1ea5n\/ha) c\u1ee7a m\u1ed7i lo\u1ea1i gi\u1ed1ng l\u00faa. L\u1eddi gi\u1ea3i G\u1ecdi x,y l\u1ea7n l\u01b0\u1ee3t l\u00e0 n\u0103ng su\u1ea5t lua trung b\u00ecnh c\u1ee7a l\u00faa lo\u1ea1i I v\u00e0 l\u00faa lo\u1ea1i II x,y 0 T\u1ed5ng s\u1ea3n l\u01b0\u1ee3ng c\u1ee7a hai gi\u1ed1ng l\u00faa thu v\u1ec1 l\u00e0 139 t\u1ea5n: 10x 8y 139 S\u1ea3n l\u01b0\u1ee3ng thu v\u1ec1 t\u1eeb 4ha gi\u1ed1ng l\u00faa lo\u1ea1i I nhi\u1ec1u h\u01a1n s\u1ea3n l\u01b0\u1ee3ng thu v\u1ec1 t\u1eeb 3ha gi\u1ed1ng l\u00faa lo\u1ea1i II l\u00e0 6 t\u1ea5n: 4x 3y 6 Ta c\u00f3 h\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh: 10x 8y 139 x 7, 5 4x 3y 6 y 8 (N ) N\u0103ng su\u1ea5t l\u00faa lo\u1ea1i I l\u00e0 7, 5 t\u1ea5n\/ha N\u0103ng su\u1ea5t l\u00faa lo\u1ea1i II l\u00e0 8 t\u1ea5n\/ha C\u00e2u 7. (1 \u0111i\u1ec3m). Cho c\u1ed1c r\u01b0\u1ee3u (nh\u01b0 h\u00ecnh v\u1ebd), ph\u1ea7n ph\u00eda tr\u00ean l\u00e0 m\u1ed9t h\u00ecnh n\u00f3n c\u00f3 chi\u1ec1u cao 6cm v\u00e0 \u0111\u00e1y l\u00e0 \u0111\u01b0\u1eddng tr\u00f2n b\u00e1n k\u00ednh 3cm . T\u00ednh th\u1ec3 t\u00edch r\u01b0\u1ee3u trong ly. (K\u1ebft qu\u1ea3 l\u00e0m tr\u00f2n \u0111\u1ebfn ch\u1eef s\u1ed1 th\u1eadp ph\u00e2n th\u1ee9 nh\u1ea5t) . L\u1eddi gi\u1ea3i G\u1ecdi c\u00e1c \u0111i\u1ec3m nh\u01b0 h\u00ecnh v\u1ebd T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N 8","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH OE O ' E OO ' 6 2 4cm Ta c\u00f3: CO \/\/AO ' CO ' EO (H\u1ec7 qu\u1ea3 Talet) AO ' EO ' CO 4 CO 2cm 36 Vruou 1 . .22.4 16, 8cm3 3 C\u00e2u 9. (3 \u0111i\u1ec3m) Cho n\u1eeda \u0111\u01b0\u1eddng tr\u00f2n (O) , \u0111\u01b0\u1eddng k\u00ednh BC . Tr\u00ean n\u1eeda \u0111\u01b0\u1eddng tr\u00f2n (O) , l\u1ea5y hai \u0111i\u1ec3m A v\u00e0 D (theo th\u1ee9 t\u1ef1 B,A,D,C ). Tia BA v\u00e0 CD c\u1eaft nhau t\u1ea1i S , \u0111o\u1ea1n th\u1eb3ng AC c\u1eaft BD t\u1ea1i H. a) Ch\u1ee9ng minh SH BC t\u1ea1i E v\u00e0 t\u1ee9 gi\u00e1c HECD n\u1ed9i ti\u1ebfp. b) G\u1ecdi T l\u00e0 trung \u0111i\u1ec3m SH , tia AT c\u1eaft SC t\u1ea1i I , DE c\u1eaft HC t\u1ea1i K . Ch\u1ee9ng minh: TAH KDC . T\u1eeb \u0111\u00f3 suy ra CK.CA CD.CI c) D\u01b0\u1eddng trung tr\u1ef1c c\u1ea3 \u0111o\u1ea1n th\u1eb3ngAK c\u1eaft BH t\u1ea1iQ . Ch\u1ee9ng minh IAK c\u00e2n v\u00e0 ba \u0111i\u1ec3m A,O,Q th\u1eb3ng h\u00e0ng. L\u1eddi gi\u1ea3i a) Ch\u1ee9ng minh SH BC t\u1ea1i E v\u00e0 t\u1ee9 gi\u00e1c HECD n\u1ed9i ti\u1ebfp. 9 T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH X\u00e9t O , c\u00f3: \uf0ec\uf0ef\uf0efCDB = 1 s\u00f1 BC = 90\uf0b0 \uf0ed = 2 s\u00f1 BC = 90\uf0b0 \uf0ef\uf0ef\uf0eeBAC 1 2 X\u00e9t SBC c\u00f3: BD,CAl\u00e0 \u0111\u01b0\u1eddng cao c\u1eaft nhau t\u1ea1i H H l\u00e0 tr\u1ef1c t\u00e2m c\u1ee7a ABC SH BC t\u1ea1i E X\u00e9t t\u1ee9 gi\u00e1c HECD c\u00f3: HDC 90 HEC 90 HDC HEC 180 T\u1ee9 gi\u00e1c HECD n\u1ed9i ti\u1ebfp. b) G\u1ecdi T l\u00e0 trung \u0111i\u1ec3m SH , tia AT c\u1eaft SC t\u1ea1i I , DE c\u1eaft HC t\u1ea1i K . Ch\u1ee9ng minh: TAH KDC . T\u1eeb \u0111\u00f3 suy ra CK.CA CD.CI X\u00e9t SHA vu\u00f4ng t\u1ea1i A , c\u00f3 AI l\u00e0 \u0111\u01b0\u1eddng trung tuy\u1ebfn. TA TH TS TAH c\u00e2n t\u1ea1i T TAH AHT M\u00e0 AHT EHC (\u0111\u1ed1i \u0111\u1ec9nh) V\u00e0 EHC EDC (tg EHDC nt) TAH KDC X\u00e9t CKD v\u00e0 CIA c\u00f3: C chung TAH KDC (cmt) T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N 10","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH CKD \u223d CIA(g g) CK CI (cctl) CD CA CK.CA CD.CI c) \u0110\u01b0\u1eddng trung tr\u1ef1c c\u1ea3 \u0111o\u1ea1n th\u1eb3ng AK c\u1eaft BH t\u1ea1iQ . Ch\u1ee9ng minh IAK c\u00e2n v\u00e0 ba \u0111i\u1ec3m A,O,Q th\u1eb3ng h\u00e0ng. D\u1ec5 d\u00e0ng Cm: T\u1ee9 gi\u00e1c SAHD n\u1ed9i ti\u1ebfp. Suy ra: HAD = HSD(1) . Ta c\u00f3: KDC HAI (cmt) Tg AIDK nt KAD KID 2 1 v\u00e0 2 HSD KID M\u00e0 2 g\u00f3c n\u00e0y \u1edf v\u1ecb tr\u00ed \u0111\u1ed3ng v\u1ecb IK \/ \/SH AHS AKI M\u00e0 AHS HAT IKA IAK IAK c\u00e2n t\u1ea1i I Ta c\u00f3: QA QK IA IK QI l\u00e0 \u0111\u01b0\u1eddng trung tr\u1ef1c c\u1ee7a AK AIQ SAI (c\u00f9ng ph\u1ee5 IAK ) M\u00e0 SAI TSA ADH AIQ ADQ T\u1ee9 gi\u00e1c AIDQ n\u1ed9i ti\u1ebfp M\u00e0 t\u1ee9 gi\u00e1c AIDK n\u1ed9i ti\u1ebfp 5 \u0111i\u1ec3m A,I,D,K,Q c\u00f9ng thu\u1ed9c \u0111\u01b0\u1eddng tr\u00f2n. T\u1ee9 gi\u00e1c AIKQ n\u1ed9i ti\u1ebfp. T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N 11","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH IAQ IKQ 180 M\u00e0 IAQ IKQ IAQ 90 IA AQ 3 Ta c\u00f3: TAH THA EHC OAC OCA OCA EHC 90 TAH OAC 90 TAO 90 TA AO 4 3 v\u00e0 4 A,Q,O th\u1eb3ng h\u00e0ng. ----H\u1ebeT--- T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N 12","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH S\u1ede GD&\u0110T TP H\u1ed2 CH\u00cd MINH \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 PH\u00d2NG GD&\u0110T QU\u1eacN T\u00c2N PH\u00da N\u0102M H\u1eccC: 2023 - 2024 \u0110\u1ec0 THAM KH\u1ea2O M\u00d4N: TO\u00c1N 9 M\u00c3 \u0110\u1ec0: Qu\u1eadn TP - 2 \u0110\u1ec1 thi g\u1ed3m 8 c\u00e2u ho\u0309i t\u01b0\u0323 lu\u00e2\u0323n. Th\u01a1\u0300i gian: 120 phu\u0301t (kh\u00f4ng k\u00ea\u0309 th\u01a1\u0300i gian pha\u0301t \u0111\u1ec1) C\u00e2u 1. (1,5 \u0111i\u1ec3m). Cho (P) : y = \u2212x2 v\u00e0 \u0111\u01b0\u1eddng th\u1eb3ng (d) : y = x \u2212 1,5 . 2 a) V\u1ebd \u0111\u1ed3 th\u1ecb (P) v\u00e0 (d) tr\u00ean c\u00f9ng h\u1ec7 tr\u1ee5c t\u1ecda \u0111\u1ed9. b) T\u00ecm t\u1ecda \u0111\u1ed9 giao \u0111i\u1ec3m c\u1ee7a (P) v\u00e0 (d) b\u1eb1ng ph\u00e9p t\u00ednh. C\u00e2u 2. (1 \u0111i\u1ec3m). Cho ph\u01b0\u01a1ng tr\u00ecnh 3x2 \u2212 2x \u2212 6 = 0 c\u00f3 2 nghi\u1ec7m l\u00e0 x1 , x2 . Kh\u00f4ng gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh, h\u00e3y t\u00ednh gi\u00e1 tr\u1ecb c\u1ee7a bi\u1ec3u th\u1ee9c M \uf0e6 x1 \uf0f6\uf0e6 x2 \uf0f6 = \uf0e71\u2212 2x2 \uf0f7.\uf0e7 1 \u2212 2x1 \uf0f7 \uf0f8\uf0e8 \uf0f8 \uf0e8 C\u00e2u 3. (0,75 \u0111i\u1ec3m). Ng\u00e0y n trong th\u00e1ng t c\u1ee7a n\u0103m 2021 l\u00e0 th\u1ee9 m\u1ea5y trong tu\u1ea7n? C\u00f3 ph\u1ea3i ch\u1ee7 nh\u1eadt kh\u00f4ng? \u0110\u1ec3 tr\u1ea3 l\u1eddi c\u00e2u h\u1ecfi n\u00e0y, ta \u00e1p d\u1ee5ng quy t\u1eafc Newday nh\u01b0 sau: B\u01b0\u1edbc 1: T\u00ednh M nh\u01b0 sau: Trong tr\u01b0\u1eddng h\u1ee3p t \uf0b3 3 th\u00ec M = (t \u2212 3).30 + (n + 1) + N . Trong \u0111\u00f3 N l\u00e0 s\u1ed1 l\u01b0\u1ee3ng th\u00e1ng c\u00f3 31 ng\u00e0y t\u00ednh t\u1eeb th\u00e1ng 3 \u0111\u1ebfn th\u00e1ng th\u1ee9 (t \u2212 1) Trong tr\u01b0\u1eddng h\u1ee3p th\u00e1ng t l\u00e0 th\u00e1ng 1 th\u00ec M = n + 26 . Trong tr\u01b0\u1eddng h\u1ee3p th\u00e1ng t l\u00e0 th\u00e1ng 2 th\u00ec M = n + 1 . B\u01b0\u1edbc 2: T\u00ecm r l\u00e0 d\u01b0 c\u1ee7a ph\u00e9p chia M cho 7 . N\u1ebfu r = 0 th\u00ec ng\u00e0y n trong th\u00e1ng t l\u00e0 th\u1ee9 b\u1ea3y. N\u1ebfu r = 1 th\u00ec ng\u00e0y n trong th\u00e1ng t l\u00e0 ch\u1ee7 nh\u1eadt. N\u1ebfu r = 2 th\u00ec ng\u00e0y n trong th\u00e1ng t l\u00e0 th\u1ee9 hai. \u2026 N\u1ebfu r = 6 th\u00ec ng\u00e0y n trong th\u00e1ng t l\u00e0 th\u1ee9 s\u00e1u. Bi\u1ebft s\u1ed1 ng\u00e0y trong th\u00e1ng c\u1ee7a n\u0103m 2021 th\u1ec3 hi\u1ec7n trong b\u1ea3ng sau: Th\u00e1ng 1, 3, 5, 7 , 8,10,12 2 4,6,9,11 S\u1ed1 ng\u00e0y 31 28 30 a) Ng\u00e0y Qu\u1ed1c t\u1ebf Gia \u0110\u00ecnh 28 \/ 06 trong n\u0103m 2021 l\u00e0 th\u1ee9 m\u1ea5y? K\u1ef7 ni\u1ec7m ng\u00e0y th\u00e0nh l\u1eadp \u0110\u1ea3ng C\u1ed9ng S\u1ea3n 03 \/ 02 trong n\u0103m 2021 l\u00e0 th\u1ee9 m\u1ea5y? b) Tr\u00ean Th\u1ebf Gi\u1edbi ch\u1ecdn ng\u00e0y Ch\u1ee7 Nh\u1eadt trong tu\u1ea7n l\u1ec5 th\u1ee9 hai c\u1ee7a th\u00e1ng 5 \u0111\u1ec3 l\u00e0m \u201cNg\u00e0y c\u1ee7a M\u1eb9\u201d. Ng\u00e0y c\u1ee7a m\u1eb9 trong n\u0103m 2021 l\u00e0 ng\u00e0y m\u1ea5y? C\u00e2u 4. (0,75 \u0111i\u1ec3m). C\u00e2n n\u1eb7ng trung b\u00ecnh c\u1ee7a tr\u1ebb s\u01a1 sinh \u0111\u1ee7 th\u00e1ng l\u00e0 kho\u1ea3ng 3000g . Tr\u1ebb l\u00fac 6 th\u00e1ng c\u00f3 c\u00e2n n\u1eb7ng g\u1ea5p \u0111\u00f4i l\u00fac s\u01a1 sinh, 6 th\u00e1ng ti\u1ebfp theo m\u1ed7i th\u00e1ng t\u0103ng 500g . T\u1eeb n\u0103m th\u1ee9 hai tr\u1edf \u0111i, trung b\u00ecnh m\u1ed7i n\u0103m t\u0103ng th\u00eam 1,5kg . G\u1ecdi P(kg) l\u00e0 c\u00e2n n\u1eb7ng c\u1ee7a tr\u1ebb em d\u01b0\u1edbi 14 tu\u1ed5i; N (tu\u1ed5i) l\u00e0 s\u1ed1 tu\u1ed5i (d\u1ef1a v\u00e0o h\u1eb1ng s\u1ed1 sinh h\u1ecdc ng\u01b0\u1eddi Vi\u1ec7t Nam n\u0103m 1975 ). T\u00ecm c\u00e2n n\u1eb7ng trung b\u00ecnh c\u1ee7a tr\u1ebb tr\u00f2n 1 tu\u1ed5i v\u00e0 x\u00e1c \u0111\u1ecbnh h\u00e0m s\u1ed1 P theo N . T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N 1","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH C\u00e2u 5. (1 \u0111i\u1ec3m). M\u1ed7i ng\u00e0y, b\u1ea1n An \u0111\u1ec1u s\u1eed d\u1ee5ng \u0111i\u1ec7n tho\u1ea1i smartphone \u0111\u1ec3 ch\u01a1i game 90 ph\u00fat, l\u01b0\u1edbt facebook 30 ph\u00fat, nh\u1eafn tin \u201cchat\u201d c\u00f9ng b\u1ea1n b\u00e8 h\u1ebft 20 ph\u00fat, xem c\u00e1c chuy\u00ean m\u1ee5c gi\u1ea3i tr\u00ed v\u00e0 c\u00e1c th\u00f4ng tin tr\u00ean m\u1ea1ng h\u1ebft 100 ph\u00fat. a) Th\u1eddi gian b\u1ea1n An s\u1eed d\u1ee5ng \u0111i\u1ec7n tho\u1ea1i smartphone chi\u1ebfm bao nhi\u00eau ph\u1ea7n tr\u0103m th\u1eddi gian c\u1ee7a m\u1ed9t ng\u00e0y ( 24 gi\u1edd)? (l\u00e0m tr\u00f2n k\u1ebft qu\u1ea3 \u0111\u1ebfn h\u00e0ng \u0111\u01a1n v\u1ecb). b) V\u00ec s\u1eed d\u1ee5ng \u0111i\u1ec7n tho\u1ea1i smartphone nhi\u1ec1u r\u1ea5t c\u00f3 h\u1ea1i cho s\u1ee9c kh\u1ecfe, \u0111\u1eb7c bi\u1ec7t v\u1ec1 tr\u00ed nh\u1edb n\u00ean m\u1eb9 b\u1ea1n An \u0111\u00e3 y\u00eau c\u1ea7u m\u1ed7i ng\u00e0y b\u1ea1n ch\u1ec9 \u0111\u01b0\u1ee3c ph\u00e9p d\u00f9ng 2,5% th\u1eddi gian c\u1ee7a m\u1ed9t ng\u00e0y \u0111\u1ec3 s\u1eed d\u1ee5ng \u0111i\u1ec7n tho\u1ea1i smartphone v\u00e0 10% th\u1eddi gian c\u00f2n l\u1ea1i c\u1ee7a ng\u00e0y ph\u1ea3i d\u00f9ng v\u00e0o vi\u1ec7c h\u1ecdc \u1edf nh\u00e0. V\u1eady m\u1ed7i ng\u00e0y b\u1ea1n An ph\u1ea3i h\u1ecdc \u1edf nh\u00e0 trong bao nhi\u00eau ph\u00fat? C\u00e2u 6. (1 \u0111i\u1ec3m). H\u00ecnh b\u00ean l\u00e0 m\u1ed9t m\u1eabu pho m\u00e1t \u0111\u01b0\u1ee3c c\u1eaft ra t\u1eeb m\u1ed9t kh\u1ed1i pho m\u00e1t d\u1ea1ng h\u00ecnh tr\u1ee5 (c\u00f3 c\u00e1c k\u00edch th\u01b0\u1edbc nh\u01b0 h\u00ecnh v\u1ebd). Bi\u1ebft kh\u1ed1i l\u01b0\u1ee3ng ri\u00eang c\u1ee7a pho m\u00e1t l\u00e0 3g \/ cm3 v\u00e0 c\u00f4ng th\u1ee9c kh\u1ed1i l\u01b0\u1ee3ng ri\u00eang l\u00e0 D = m V ( )(Trong \u0111\u00f3 D g \/ cm3 l\u00e0 kh\u1ed1i l\u01b0\u1ee3ng ri\u00eang, m( g) l\u00e0 kh\u1ed1i ( )l\u01b0\u1ee3ng, V cm3 l\u00e0 th\u00ea\u0309 t\u00edch) a) H\u00e3y t\u00ednh di\u1ec7n t\u00edch 1 m\u1eb7t \u0111\u00e1y v\u00e0 kh\u1ed1i l\u01b0\u1ee3ng c\u1ee7a m\u1eabu pho m\u00e1t tr\u00ean. Bi\u1ebft Th\u00ea\u0309 t\u00edch h\u00ecnh tr\u1ee5 Vtru = S.h . Trong \u0111\u00f3 S l\u00e0 di\u1ec7n t\u00edch 1 \u0111a\u0301y v\u00e0 h l\u00e0 chi\u1ec1u cao c\u1ee7a h\u00ecnh tr\u1ee5. b) Chi\u1ebfc h\u1ed9p th\u1ef1c ph\u1ea9m h\u00ecnh h\u1ed9p ch\u1eef nh\u1eadt c\u00f3 k\u00edch th\u01b0\u1edbc l\u1ea7n l\u01b0\u1ee3t l\u00e0 189mm , 103mm , 101mm c\u00f3 th\u1ec3 ch\u1ee9a h\u1ebft ph\u1ea7n c\u00f2n l\u1ea1i c\u1ee7a kh\u1ed1i pho m\u00e1t kh\u00f4ng? C\u00e2u 7. (1 \u0111i\u1ec3m). Khi th\u00eam 200g axit v\u00e0o dung d\u1ecbch axit A v\u00e0 thu \u0111\u01b0\u1ee3c dung d\u1ecbch B c\u00f3 n\u1ed3ng \u0111\u1ed9 axit l\u00e0 50% . Sau \u0111\u00f3, ta l\u1ea1i th\u00eam 300g n\u01b0\u1edbc v\u00e0o dung d\u1ecbch B v\u00e0 thu \u0111\u01b0\u1ee3c dung d\u1ecbch C c\u00f3 n\u1ed3ng \u0111\u1ed9 l\u00e0 40% . T\u00ednh n\u1ed3ng \u0111\u1ed9 axit trong dung d\u1ecbch A . Bi\u1ebft C% = mct .100% , trong \u0111\u00f3 C% mdd l\u00e0 n\u1ed3ng \u0111\u1ed9 ph\u1ea7n tr\u0103m, mct l\u00e0 kh\u1ed1i l\u01b0\u1ee3ng ch\u1ea5t tan, mdd l\u00e0 kh\u1ed1i l\u01b0\u1ee3ng dung d\u1ecbch. C\u00e2u 8. (3 \u0111i\u1ec3m) Cho \u0111i\u1ec3m S n\u1eb1m ngo\u00e0i \u0111\u01b0\u1eddng tr\u00f2n (O) . K\u1ebb SB v\u00e0 SC l\u00e0 2 ti\u1ebfp tuy\u1ebfn c\u1ee7a \u0111\u01b0\u1eddng tr\u00f2n ( B , C l\u00e0 2 ti\u1ebfp \u0111i\u1ec3m). L\u1ea5y \u0111i\u1ec3m A n\u1eb1m tr\u00ean cung l\u1edbn BC . K\u1ebb CF \u22a5 AB t\u1ea1i F , k\u1ebb BE \u22a5 AC t\u1ea1i E . L\u1ea5y M l\u00e0 giao \u0111i\u1ec3m c\u1ee7a OS v\u00e0 BC . G\u1ecdi K l\u00e0 giao \u0111i\u1ec3m c\u1ee7a EF v\u1edbi SB v\u00e0 H l\u00e0 giao \u0111i\u1ec3m c\u1ee7a BE v\u1edbi CF . a) Ch\u1ee9ng minh: KFB = ACB v\u00e0 BC = 2.FM b) Ch\u1ee9ng minh: KM \u22a5 AB v\u00e0 AH \u22a5 BC \u1edf D . c) Ch\u1ee9ng minh: \uf044FHD \u0111\u1ed3ng d\u1ea1ng \uf044KMS . ----H\u1ebeT--- T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N 2","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N 3","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH H\u01af\u1edaNG D\u1eaaN GI\u1ea2I C\u00e2u 1. (1,5 \u0111i\u1ec3m). Cho ( P) : y = \u2212x2 v\u00e0 \u0111\u01b0\u1eddng th\u1eb3ng (d ) : y = x \u22121,5 2 a) V\u1ebd \u0111\u1ed3 th\u1ecb ( P) v\u00e0 (d ) tr\u00ean c\u00f9ng h\u1ec7 tr\u1ee5c t\u1ecda \u0111\u1ed9. b) T\u00ecm t\u1ecda \u0111\u1ed9 giao \u0111i\u1ec3m c\u1ee7a ( P) v\u00e0 (d ) b\u1eb1ng ph\u00e9p t\u00ednh.. L\u1eddi gi\u1ea3i a) V\u1ebd \u0111\u1ed3 th\u1ecb ( P) v\u00e0 (d ) tr\u00ean c\u00f9ng h\u1ec7 tr\u1ee5c t\u1ecda \u0111\u1ed9. BGT: x \u22122 \u22121 0 1 2 y = \u2212x2 \u22124 \u22121 0 \u22121 \u22124 x5 3 y = x \u22121,5 1 0 2 b) T\u00ecm t\u1ecda \u0111\u1ed9 giao \u0111i\u1ec3m c\u1ee7a ( P) v\u00e0 (d ) b\u1eb1ng ph\u00e9p t\u00ednh. Ph\u01b0\u01a1ng tr\u00ecnh ho\u00e0nh \u0111\u1ed9 giao \u0111i\u1ec3m c\u1ee7a ( P) v\u00e0 (d ) : \u2212x2 = x \u22121,5 2 \uf0db \u2212x2 \u2212 x +1,5 = 0 2 \uf0e9 x =1 \uf0ea \uf0db \uf0eax = \u2212 3 \uf0eb2 Thay x = 1 v\u00e0o y = \u2212x2 , ta \u0111\u01b0\u1ee3c: y = \u221212 = 1. Thay x=\u22123 v\u00e0o y = \u2212x2 , ta \u0111\u01b0\u1ee3c: y = \u2212 \uf0e6 \u22123 \uf0f62 =\u22129. 2 \uf0e7\uf0e8 2 \uf0f7\uf0f8 4 V\u1eady (1;1) , \uf0e6 \u2212 3;\u2212 9\uf0f6 l\u00e0 hai giao \u0111i\u1ec3m c\u1ea7n t\u00ecm. \uf0e7\uf0e8 2 4 \uf0f7\uf0f8 C\u00e2u 2. (1 \u0111i\u1ec3m). Cho ph\u01b0\u01a1ng tr\u00ecnh 3x2 \u2212 2x \u2212 6 = 0 c\u00f3 2 nghi\u1ec7m l\u00e0 x1, x2 . T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N 4","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH Kh\u00f4ng gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh, h\u00e3y t\u00ednh gi\u00e1 tr\u1ecb c\u1ee7a bi\u1ec3u th\u1ee9c M = \uf0e6 x1 \uf0f6\uf0e6 x2 \uf0f6 \uf0e71\u2212 2x2 \uf0f7.\uf0e71\u2212 2x1 \uf0f7 \uf0e8 \uf0f8\uf0e8 \uf0f8 L\u1eddi gi\u1ea3i V\u00ec \uf044 = b2 \u2212 4ac = (\u22122)2 \u2212 4.3.(\u22126) = 76 \uf03e 0 N\u00ean ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m ph\u00e2n bi\u1ec7t x1, x2 . \uf0ec\uf0ef\uf0efS = x1 + x2 = \u2212b = 2 \uf0ed = a 3 Theo \u0111\u1ecbnh l\u00ed Vi-et, ta c\u00f3: \uf0ef\uf0ef\uf0eeP c x1.x2 = a = \u22122 Ta c\u00f3: M = \uf0e6 x1 \uf0f6\uf0e6 x2 \uf0f6 \uf0e71\u2212 2x2 \uf0f7.\uf0e71\u2212 2x1 \uf0f7 \uf0e8 \uf0f8\uf0e8 \uf0f8 \uf0db M = \uf0e6 2x2 \u2212 x1 \uf0f6\uf0e6 2x1 \u2212 x2 \uf0f6 \uf0e7 2 x2 \uf0f7.\uf0e7 2 x1 \uf0f7 \uf0e8 \uf0f8\uf0e8 \uf0f8 \uf0db M = 4x1x2 \u2212 2x12 \u2212 2x22 + x1x2 4x1. x2 ( )\u22122 \uf0dbM = x12 + x22 + 4x1x2 + x1x2 4x1. x2 \uf0db M = \u22122( x1 + )x2 2 \u2212 2x1x2 + 5x1x2 4 x1 x2 \uf0db M = \u22122( x1 + )x2 2 + 3x1x2 4 x1 x2 \uf0db M = 31 36 C\u00e2u 3. (0,75 \u0111i\u1ec3m) Ng\u00e0y n trong th\u00e1ng t c\u1ee7a n\u0103m 2021 l\u00e0 th\u1ee9 m\u1ea5y trong tu\u1ea7n? C\u00f3 ph\u1ea3i ch\u1ee7 nh\u1eadt kh\u00f4ng? \u0110\u1ec3 tr\u1ea3 l\u1eddi c\u00e2u h\u1ecfi n\u00e0y, ta \u00e1p d\u1ee5ng quy t\u1eafc Newday nh\u01b0 sau: B\u01b0\u1edbc 1: T\u00ednh M nh\u01b0 sau: Trong tr\u01b0\u1eddng h\u1ee3p t \uf0b3 3 th\u00ec M = (t \u2212 3).30 + (n +1) + N . Trong \u0111\u00f3 N l\u00e0 s\u1ed1 l\u01b0\u1ee3ng th\u00e1ng c\u00f3 31 ng\u00e0y t\u00ednh t\u1eeb th\u00e1ng 3 \u0111\u1ebfn th\u00e1ng th\u1ee9 (t \u22121) Trong tr\u01b0\u1eddng h\u1ee3p th\u00e1ng t l\u00e0 th\u00e1ng 1 th\u00ec M = n + 26 . Trong tr\u01b0\u1eddng h\u1ee3p th\u00e1ng t l\u00e0 th\u00e1ng 2 th\u00ec M = n +1 . B\u01b0\u1edbc 2: T\u00ecm r l\u00e0 d\u01b0 c\u1ee7a ph\u00e9p chia M cho 7. N\u1ebfu r = 0 th\u00ec ng\u00e0y n trong th\u00e1ng t l\u00e0 th\u1ee9 b\u1ea3y. T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N 5","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH N\u1ebfu r = 1 th\u00ec ng\u00e0y n trong th\u00e1ng t l\u00e0 ch\u1ee7 nh\u1eadt. N\u1ebfu r = 2 th\u00ec ng\u00e0y n trong th\u00e1ng t l\u00e0 th\u1ee9 hai. \u2026 N\u1ebfu r = 6 th\u00ec ng\u00e0y n trong th\u00e1ng t l\u00e0 th\u1ee9 s\u00e1u. Bi\u1ebft s\u1ed1 ng\u00e0y trong th\u00e1ng c\u1ee7a n\u0103m 2021 th\u1ec3 hi\u1ec7n trong b\u1ea3ng sau: Th\u00e1ng 1,3,5,7,8,10,12 2 4,6,9,11 S\u1ed1 ng\u00e0y 31 28 30 a) Ng\u00e0y Qu\u1ed1c t\u1ebf Gia \u0110\u00ecnh 28\/06 trong n\u0103m 2021 l\u00e0 th\u1ee9 m\u1ea5y? K\u1ef7 ni\u1ec7m ng\u00e0y th\u00e0nh l\u1eadp \u0110\u1ea3ng C\u1ed9ng S\u1ea3n 03\/02 trong n\u0103m 2021 l\u00e0 th\u1ee9 m\u1ea5y? b) Tr\u00ean Th\u1ebf Gi\u1edbi ch\u1ecdn ng\u00e0y Ch\u1ee7 Nh\u1eadt trong tu\u1ea7n l\u1ec5 th\u1ee9 hai c\u1ee7a th\u00e1ng 5 \u0111\u1ec3 l\u00e0m \u201cNg\u00e0y c\u1ee7a M\u1eb9\u201d. Ng\u00e0y c\u1ee7a m\u1eb9 trong n\u0103m 2021 l\u00e0 ng\u00e0y m\u1ea5y? L\u1eddi gi\u1ea3i a) Ta c\u00f3 B\u01b0\u1edbc 1: Th\u00e1ng 6 : t \uf0b3 3 M = (t \u2212 3).30 + (n +1) + N \uf0de M = (6 \u2212 3).30 + (28 +1) + 2 \uf0de M = 121 B\u01b0\u1edbc 2: M chia 7 d\u01b0 2 v\u1eady n l\u00e0 ng\u00e0y th\u1ee9 2 b) Ng\u00e0y C\u1ee7a M\u1eb9\\\" \u0111\u01b0\u1ee3c ch\u1ecdn v\u00e0o m\u1ed9t ng\u00e0y c\u1ee7a th\u00e1ng 5 n\u00ean ta c\u00f3 t = 5 \uf0de t \u22121 = 4 S\u1ed1 l\u01b0\u1ee3ng th\u00e1ng c\u00f3 31 ng\u00e0y t\u1eeb th\u00e1ng 03 \u0111\u1ebfn th\u00e1ng 04 l\u00e0 1 th\u00e1ng N\u00ean ta c\u00f3 N = 1 Ng\u00e0y 30 \/ 4 M = (t \u2212 3).30 + (n +1) + N \uf0de M = (4 \u2212 3).30 + (30 +1) +1 \uf0de M = 62 theo quy t\u1eafc Newday ta c\u00f3 M = 62 chia 7 d\u01b0 6 \uf0de Ng\u00e0y 30 \/ 4 l\u00e0 th\u1ee9 6 Ch\u1ee7 nh\u1eadt \u0111\u1ea7u ti\u00ean c\u1ee7a th\u00e1ng 5 l\u00e0 2 \/ 5 . V\u1eady ch\u1ee7 nh\u1eadt th\u1ee9 hai c\u1ee7a th\u00e1ng 5 l\u00e0 ng\u00e0y 9 \/ 5 . C\u00e2u 4. (0,75 \u0111i\u1ec3m). C\u00e2n n\u1eb7ng trung b\u00ecnh c\u1ee7a tr\u1ebb s\u01a1 sinh \u0111\u1ee7 th\u00e1ng l\u00e0 kho\u1ea3ng 3000g. Tr\u1ebb l\u00fac 6 th\u00e1ng c\u00f3 c\u00e2n n\u1eb7ng g\u1ea5p \u0111\u00f4i l\u00fac s\u01a1 sinh, 6 th\u00e1ng ti\u1ebfp theo m\u1ed7i th\u00e1ng t\u0103ng 500g. T\u1eeb n\u0103m th\u1ee9 hai tr\u1edf \u0111i, trung b\u00ecnh m\u1ed7i n\u0103m t\u0103ng th\u00eam 1,5kg. G\u1ecdi P (kg) l\u00e0 c\u00e2n n\u1eb7ng c\u1ee7a tr\u1ebb em d\u01b0\u1edbi 14 T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N 6","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH tu\u1ed5i; N (tu\u1ed5i) l\u00e0 s\u1ed1 tu\u1ed5i (d\u1ef1a v\u00e0o h\u1eb1ng s\u1ed1 sinh h\u1ecdc ng\u01b0\u1eddi Vi\u1ec7t Nam n\u0103m 1975). T\u00ecm c\u00e2n n\u1eb7ng trung b\u00ecnh c\u1ee7a tr\u1ebb tr\u00f2n 1 tu\u1ed5i v\u00e0 x\u00e1c \u0111\u1ecbnh h\u00e0m s\u1ed1 P theo N.? L\u1eddi gi\u1ea3i C\u00e2n n\u1eb7ng trung b\u00ecnh c\u1ee7a tr\u1ebb tr\u00f2n 1 tu\u1ed5i l\u00e0: 3000.2 + 500.6 = 9000( g ) = 9(kg ) Ta c\u00f3: P (kg ) c\u00e2n n\u1eb7ng tr\u1ebb em d\u01b0\u1edbi 14 tu\u1ed5i N (s\u1ed1 tu\u1ed5i) T\u1eeb \u0111\u00f3 ta c\u00f3 h\u00e0m s\u1ed1 P theo N : P = 9 +1,5.N C\u00e2u 5. (1 \u0111i\u1ec3m) M\u1ed7i ng\u00e0y, b\u1ea1n An \u0111\u1ec1u s\u1eed d\u1ee5ng \u0111i\u1ec7n tho\u1ea1i smartphone \u0111\u1ec3 ch\u01a1i game 90 ph\u00fat, l\u01b0\u1edbt facebook 30 ph\u00fat, nh\u1eafn tin \u201cchat\u201d c\u00f9ng b\u1ea1n b\u00e8 h\u1ebft 20 ph\u00fat, xem c\u00e1c chuy\u00ean m\u1ee5c gi\u1ea3i tr\u00ed v\u00e0 c\u00e1c th\u00f4ng tin tr\u00ean m\u1ea1ng h\u1ebft 100 ph\u00fat. a) Th\u1eddi gian b\u1ea1n An s\u1eed d\u1ee5ng \u0111i\u1ec7n tho\u1ea1i smartphone chi\u1ebfm bao nhi\u00eau ph\u1ea7n tr\u0103m th\u1eddi gian c\u1ee7a m\u1ed9t ng\u00e0y (24 gi\u1edd)? (l\u00e0m tr\u00f2n k\u1ebft qu\u1ea3 \u0111\u1ebfn h\u00e0ng \u0111\u01a1n v\u1ecb). b) V\u00ec s\u1eed d\u1ee5ng \u0111i\u1ec7n tho\u1ea1i smartphone nhi\u1ec1u r\u1ea5t c\u00f3 h\u1ea1i cho s\u1ee9c kh\u1ecfe, \u0111\u1eb7c bi\u1ec7t v\u1ec1 tr\u00ed nh\u1edb n\u00ean m\u1eb9 b\u1ea1n An \u0111\u00e3 y\u00eau c\u1ea7u m\u1ed7i ng\u00e0y b\u1ea1n ch\u1ec9 \u0111\u01b0\u1ee3c ph\u00e9p d\u00f9ng 2,5% th\u1eddi gian c\u1ee7a m\u1ed9t ng\u00e0y \u0111\u1ec3 s\u1eed d\u1ee5ng \u0111i\u1ec7n tho\u1ea1i smartphone v\u00e0 10% th\u1eddi gian c\u00f2n l\u1ea1i c\u1ee7a ng\u00e0y ph\u1ea3i d\u00f9ng v\u00e0o vi\u1ec7c h\u1ecdc \u1edf nh\u00e0. V\u1eady m\u1ed7i ng\u00e0y b\u1ea1n An ph\u1ea3i h\u1ecdc \u1edf nh\u00e0 trong bao nhi\u00eau ph\u00fat? L\u1eddi gi\u1ea3i a) Th\u1eddi gian b\u1ea1n An s\u1eed d\u1ee5ng \u0111i\u1ec7n tho\u1ea1i smartphone chi\u1ebfm s\u1ed1 ph\u1ea7n tr\u0103m th\u1eddi gian c\u1ee7a m\u1ed9t ng\u00e0y ( 24 gi\u1edd) l\u00e0: (90 + 30 + 20 +100) : 60 .100% \uf0bb 17% 24 b) M\u1ed7i ng\u00e0y b\u1ea1n An ph\u1ea3i h\u1ecdc \u1edf nh\u00e0 s\u1ed1 ph\u00fat l\u00e0: 24.(1\u2212 2,5%).10%.60 =140, 4 (ph\u00fat) C\u00e2u 6. (1 \u0111i\u1ec3m) H\u00ecnh b\u00ean l\u00e0 m\u1ed9t m\u1eabu pho m\u00e1t \u0111\u01b0\u1ee3c c\u1eaft ra t\u1eeb m\u1ed9t kh\u1ed1i pho m\u00e1t d\u1ea1ng h\u00ecnh tr\u1ee5 (c\u00f3 c\u00e1c k\u00edch th\u01b0\u1edbc nh\u01b0 h\u00ecnh v\u1ebd). Bi\u1ebft kh\u1ed1i l\u01b0\u1ee3ng ri\u00eang c\u1ee7a pho m\u00e1t l\u00e0 3g \/ cm3 v\u00e0 c\u00f4ng th\u1ee9c kh\u1ed1i l\u01b0\u1ee3ng ri\u00eang l\u00e0 D = m V (Trong \u0111\u00f3 D ( g \/ cm3 ) l\u00e0 kh\u1ed1i l\u01b0\u1ee3ng ri\u00eang, m (g) l\u00e0 kh\u1ed1i l\u01b0\u1ee3ng, V ( cm3 ) l\u00e0 th\u00ea\u0309 t\u00edch) T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N 7","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH a) H\u00e3y t\u00ednh di\u1ec7n t\u00edch 1 m\u1eb7t \u0111\u00e1y v\u00e0 kh\u1ed1i l\u01b0\u1ee3ng c\u1ee7a m\u1eabu pho m\u00e1t tr\u00ean. Bi\u1ebft Th\u00ea\u0309 t\u00edch h\u00ecnh tr\u1ee5 Vtru = S.h . Trong \u0111\u00f3 S l\u00e0 di\u1ec7n t\u00edch 1 \u0111a\u0301y v\u00e0 h l\u00e0 chi\u1ec1u cao c\u1ee7a h\u00ecnh tr\u1ee5 b) Chi\u1ebfc h\u1ed9p th\u1ef1c ph\u1ea9m h\u00ecnh h\u1ed9p ch\u1eef nh\u1eadt c\u00f3 k\u00edch th\u01b0\u1edbc l\u1ea7n l\u01b0\u1ee3t l\u00e0 189mm, 103mm, 101mm c\u00f3 th\u1ec3 ch\u1ee9a h\u1ebft ph\u1ea7n c\u00f2n l\u1ea1i c\u1ee7a kh\u1ed1i pho m\u00e1t kh\u00f4ng? L\u1eddi gi\u1ea3i n0\uf070 R2 = 15o.\uf070 .102 25 \uf070 3600 3600 6 ( )a) Di\u1ec7n t\u00edch 1 m\u1eb7t \u0111\u00e1y l\u00e0: Squat = = cm2 Kh\u1ed1i l\u01b0\u1ee3ng c\u1ee7a m\u1eabu pho m\u00e1t tr\u00ean l\u00e0: 25 \uf070.8.3 = 100\uf070 \uf0bb 314, 2( g ) 6 189 .103 .101 10 10 10 ( )b) Th\u1ec3 t\u00edch h\u00ecnh h\u1ed9p l\u00e0: Vhop = \uf0bb 1966, 2 cm3 Th\u1ec3 t\u00edch ph\u1ea7n c\u00f2n l\u1ea1i c\u1ee7a pho m\u00e1t l\u00e0: = \uf070 .R2.h \u2212 25 \uf070.8 = \uf070.102.8 \u2212 25 \uf070.8 = 2300 \uf070 6 63 ( )Vpho\u2212mat \uf0bb 2408, 6 cm3 V\u00ec Vhop \uf03c Vpho\u2212mat n\u00ean chi\u1ebfc h\u1ed9p th\u1ef1c ph\u1ea9m h\u00ecnh h\u1ed9p ch\u1eef nh\u1eadt kh\u00f4ng th\u1ec3 ch\u1ee9a h\u1ebft ph\u1ea7n c\u00f2n l\u1ea1i c\u1ee7a kh\u1ed1i pho m\u00e1t. C\u00e2u 7. (1 \u0111i\u1ec3m) Khi th\u00eam 200g axit v\u00e0o dung d\u1ecbch axit A v\u00e0 thu \u0111\u01b0\u1ee3c dung d\u1ecbch B c\u00f3 n\u1ed3ng \u0111\u1ed9 axit l\u00e0 50%. Sau \u0111\u00f3, ta l\u1ea1i th\u00eam 300g n\u01b0\u1edbc v\u00e0o dung d\u1ecbch B v\u00e0 thu \u0111\u01b0\u1ee3c dung d\u1ecbch C c\u00f3 n\u1ed3ng \u0111\u1ed9 l\u00e0 40%. T\u00ednh n\u1ed3ng \u0111\u1ed9 axit trong dung d\u1ecbch A. Bi\u1ebft C% = mct .100% , trong \u0111\u00f3 C% mdd l\u00e0 n\u1ed3ng \u0111\u1ed9 ph\u1ea7n tr\u0103m, mct l\u00e0 kh\u1ed1i l\u01b0\u1ee3ng ch\u1ea5t tan, mdd l\u00e0 kh\u1ed1i l\u01b0\u1ee3ng dung d\u1ecbch. L\u1eddi gi\u1ea3i G\u1ecdi x ( g ) , y ( g ) l\u1ea7n l\u01b0\u1ee3t l\u00e0 kh\u1ed1i l\u01b0\u1ee3ng ch\u1ea5t tan v\u00e0 kh\u1ed1i l\u01b0\u1ee3ng dung d\u1ecbch trong dung d\u1ecbch axit A ( x; y \uf03e 0) V\u00ec khi th\u00eam 200g axit v\u00e0o dung d\u1ecbch axit A th\u00ec thu \u0111\u01b0\u1ee3c dung d\u1ecbch axit B c\u00f3 n\u1ed3ng \u0111\u1ed9 axit l\u00e0 50% n\u00ean ta c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh: x + 200 .100% = 50% \uf0db x + 200 = 1 \uf0db 2.( x + 200) = y + 200 \uf0db 2x \u2212 y = \u2212200 (1) y + 200 y + 200 2 V\u00ec sau \u0111\u00f3 th\u00eam 300g n\u01b0\u1edbc v\u00e0o dung d\u1ecbch B th\u00ec thu \u0111\u01b0\u1ee3c dung d\u1ecbch C c\u00f3 n\u1ed3ng \u0111\u1ed9 l\u00e0 40% n\u00ean ta c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh: x + 200 .100% = 40% \uf0db x + 200 = 2 \uf0db 5.( x + 200) = 2.( y + 500) \uf0db 5x \u2212 2 y = 0(2) y + 500 y + 500 5 T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N 8","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH T\u1eeb (1) v\u00e0 (2) ta c\u00f3 h\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh: \uf0ec2x \u2212 y = \u2212200 \uf0ed \uf0ee 5x \u2212 2y = 0 \uf0db \uf0ec\uf0ef x = 400(n) \uf0ed y = 1000(n) \uf0ef\uf0ee N\u1ed3ng \u0111\u1ed9 axit trong dung d\u1ecbch A l\u00e0: 400 .100% = 40% . 1000 C\u00e2u 8. (3 \u0111i\u1ec3m) Cho \u0111i\u1ec3m S n\u1eb1m ngo\u00e0i \u0111\u01b0\u1eddng tr\u00f2n (O) . K\u1ebb SB v\u00e0 SC l\u00e0 2 ti\u1ebfp tuy\u1ebfn c\u1ee7a \u0111\u01b0\u1eddng tr\u00f2n (B, C l\u00e0 2 ti\u1ebfp \u0111i\u1ec3m). L\u1ea5y \u0111i\u1ec3m A n\u1eb1m tr\u00ean cung l\u1edbn BC. K\u1ebb CF \u22a5 AB t\u1ea1i F, k\u1ebb BE \u22a5 AC t\u1ea1i E. L\u1ea5y M l\u00e0 giao \u0111i\u1ec3m c\u1ee7a OS v\u00e0 BC. G\u1ecdi K l\u00e0 giao \u0111i\u1ec3m c\u1ee7a EF v\u1edbi SB v\u00e0 H l\u00e0 giao \u0111i\u1ec3m c\u1ee7a BE v\u1edbi CF. a) Ch\u1ee9ng minh: KFB = ACB v\u00e0 BC = 2.FM b) Ch\u1ee9ng minh: KM \u22a5 AB v\u00e0 AH \u22a5 BC \u1edf D. c) Ch\u1ee9ng minh: \uf044FHD \u0111\u1ed3ng d\u1ea1ng \uf044KMS . L\u1eddi gi\u1ea3i a) Ch\u1ee9ng minh: KFB = ACB v\u00e0 BC = 2.FM 9 X\u00e9t t\u1ee9 gi\u00e1c BFEC, ta c\u00f3: T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH \uf0ec\uf0efBFC = 90\uf0b0 \uf0ed \uf0ef\uf0eeBEC = 90\uf0b0 \uf0de BFC = BEC \uf0de T\u1ee9 gi\u00e1c BFEC n\u1ed9i ti\u1ebfp \u0111\u01b0\u1eddng tr\u00f2n \uf0de KFB = ACB X\u00e9t (O) , ta c\u00f3: \uf0ec SB = SC \uf0ed\uf0eeOB = OC \uf0de OS l\u00e0 \u0111\u01b0\u1eddng trung tr\u1ef1c c\u1ee7a \u0111o\u1ea1n th\u1eb3ng BC . \uf0de OS \u22a5 BC t\u1ea1i M X\u00e9t \uf044OBC , ta c\u00f3: OB = OC (= R) \uf0de \uf044OBC c\u00e2n t\u1ea1i O X\u00e9t \uf044OBC c\u00e2n t\u1ea1i O , ta c\u00f3: OM l\u00e0 \u0111\u01b0\u1eddng cao ( OM \u22a5 BC ) \uf0de OM l\u00e0 \u0111\u01b0\u1eddng trung tuy\u1ebfn. \uf0de M l\u00e0 trung \u0111i\u1ec3m BC . X\u00e9t \uf044BFC vu\u00f4ng t\u1ea1i F , ta c\u00f3: FM l\u00e0 \u0111\u01b0\u1eddng trung tuy\u1ebfn ( M l\u00e0 trung \u0111i\u1ec3m BC ) \uf0de FM = 1 BC (H\u1ec7 qu\u1ea3 h\u00ecnh ch\u1eef nh\u1eadt) 2 \uf0de BC = 2.FM b) Ch\u1ee9ng minh: KM \u22a5 AB v\u00e0 AH \u22a5 BC \u1edf D . Ta c\u00f3: \uf0ef\uf0ef\uf0ecFM = BC (cmt) \uf0ed = 2 \uf0ef\uf0ef\uf0eeBM MC = BC 2 \uf0de FM = BM = MC Ta c\u00f3: \uf0ec\uf0efKBF = ACB(gtbtt & dc = gnt) \uf0ed \uf0ef\uf0ee KFB = ACB(cmt) \uf0de KBF = KFB \uf0de \uf044KFB c\u00e2n t\u1ea1i K . T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N 10","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH \uf0de KF = KB Ta c\u00f3: \uf0ecMF = MB(cmt) \uf0ed = KB(cmt) \uf0ee KF \uf0de KM l\u00e0 \u0111\u01b0\u1eddng trung tr\u1ef1c c\u1ee7a \u0111o\u1ea1n th\u1eb3ng FB \uf0de KM \u22a5 AB X\u00e9t \uf044ABC , ta c\u00f3: CF l\u00e0 \u0111\u01b0\u1eddng cao ( CF \u22a5 AB ) BE l\u00e0 \u0111\u01b0\u1eddng cao ( BE \u22a5 AC ) CF v\u00e0 BE c\u1eaft nhau t\u1ea1i H (gt) \uf0de H l\u00e0 tr\u1ef1c t\u00e2m c\u1ee7a \uf044ABC \uf0de AH l\u00e0 \u0111\u01b0\u1eddng cao th\u1ee9 3 \uf0de AH \u22a5 BC \u1edf D . c) Ch\u1ee9ng minh: \uf044FHD \u0111\u1ed3ng d\u1ea1ng \uf044KMS X\u00e9t \uf044KFB c\u00e2n t\u1ea1i K , ta c\u00f3: KM l\u00e0 \u0111\u01b0\u1eddng cao ( KM \u22a5 AB ) \uf0de KM l\u00e0 \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c \uf0de FKM = MKS X\u00e9t \uf044SBC c\u00e2n t\u1ea1i S , ta c\u00f3: SM l\u00e0 \u0111\u01b0\u1eddng cao ( SM \u22a5 BC ) \uf0de SM l\u00e0 \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c \uf0de KSM = CSM X\u00e9t t\u1ee9 gi\u00e1c AFHE , ta c\u00f3: AFH + AEH = 90\uf0b0 + 90\uf0b0 = 180\uf0b0 \uf0de T\u1ee9 gi\u00e1c AFHE n\u1ed9i ti\u1ebfp \uf0de HAE = HFE (1) X\u00e9t t\u1ee9 gi\u00e1c AEDB , ta c\u00f3: AEB = ADB = 90\uf0b0 \uf0de T\u1ee9 gi\u00e1c AEDB n\u1ed9i ti\u1ebfp 11 \uf0de HAE = DBE (2) X\u00e9t t\u1ee9 gi\u00e1c BFHD , ta c\u00f3: T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH HFB + HDB = 90\uf0b0 + 90\uf0b0 = 180\uf0b0 \uf0de HFD = HBD (3) T\u1eeb (2) v\u00e0 (3) \uf0de HFD = HAE (4) T\u1eeb (1) v\u00e0 (4) \uf0de HFD = HFE Ta c\u00f3: \uf0ecKM \u22a5 AB \uf0ed \uf0ee CF \u22a5 AB \uf0de KM \/ \/CF Ta c\u00f3: \uf0ecHFE = FKM (KM \/ \/CF, dongvi) \uf0ef\uf0ef \uf0ed MKS = FKM (cmt) \uf0ef HFD = HFE(cmt) \uf0ef\uf0ee \uf0de HFD = MKS X\u00e9t t\u1ee9 gi\u00e1c AFDC , ta c\u00f3: AFC = ADC = 90\uf0b0 \uf0de T\u1ee9 gi\u00e1c AFDC n\u1ed9i ti\u1ebfp \uf0de ACF = HDF Ta c\u00f3: \uf0ec ACF + BAC = 90\uf0b0 \uf0ef\uf0ef \uf0ed KSM + SBC = 90\uf0b0 \uf0ef = SBC ( gtbtt & dc = gnt) \uf0ef\uf0eeBAC \uf0de ACF = KSM M\u00e0 ACF = HDF(cmt) N\u00ean KSM = HDF X\u00e9t \uf044FHD v\u00e0 \uf044KMS .ta c\u00f3: \uf0ec\uf0efHDF = KSM (cmt) \uf0ed \uf0ef\uf0ee HFD = MKS(cmt) \uf0de \uf044FHD \u0111\u1ed3ng d\u1ea1ng \uf044KMS (g-g) T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N 12","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH ----H\u1ebeT--- T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N 13","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH S\u00d4\u00db GD&\u00d1T TP HO\u00c0 CH\u00cd MINH \u00d1E\u00c0 THAM KHA\u00dbO TUYE\u00c5N SINH 10 PHO\u00d8NG G\u00d1&\u00d1T QUA\u00c4N PHU\u00d9 NHUA\u00c4N NA\u00caM HO\u00cfC: 2022 - 2023 \u0110\u1ec0 THAM KH\u1ea2O M\u00d4N: TO\u00c1N 9 M\u00c3 \u0110\u1ec0: Qu\u1eadn Ph\u00fa Nhu\u1eadn - 1 \u0110\u00ea thi g\u1ed3m 8 c\u00e2u ho\u0309i t\u01b0\u0323 lu\u00e2\u0323n. Th\u01a1\u0300i gian: 120 phu\u0301t (kh\u00f4ng k\u00ea\u0309 th\u01a1\u0300i gian pha\u0301t \u0111\u00ea\u0300) C\u00e2u 1. (1,5 \u0111i\u1ec3m). Cho h\u00e0m s\u1ed1 y = \u2212x2 c\u00f3 \u0111\u1ed3 th\u1ecb (P) v\u00e0 h\u00e0m s\u1ed1 y = \u22122x + 1 c\u00f3 \u0111\u1ed3 th\u1ecb (D) . a) V\u1ebd (P) v\u00e0 (D) tr\u00ean c\u00f9ng m\u1eb7t ph\u1eb3ng t\u1ecda \u0111\u1ed9. b) T\u00ecm t\u1ecda \u0111\u1ed9 giao \u0111i\u1ec3m c\u1ee7a (P) v\u00e0 (D) b\u1eb1ng ph\u00e9p to\u00e1n. C\u00e2u 2. (1 \u0111i\u1ec3m). Cho ph\u01b0\u01a1ng tr\u00ecnh: x2 \u2212 6x + 8 = 0 c\u00f3 2 nghi\u1ec7m x1 , x2 .Kh\u00f4ng gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh, h\u00e3y t\u00ecnh gi\u00e1 tr\u1ecb c\u1ee7a bi\u1ec3u th\u1ee9c bi\u1ec3u th\u1ee9c A= x12 \u2212 x1 + x22 \u2212 x2 . x2 x1 C\u00e2u 3. (0,75 \u0111i\u1ec3m). V\u1edbi s\u1ef1 ph\u00e1t tri\u1ec3n c\u1ee7a khoa h\u1ecdc k\u0129 thu\u1eadt hi\u1ec7n nay, ng\u01b0\u1eddi ta t\u1ea1o ra nhi\u1ec1u m\u1eabu xe l\u0103n \u0111\u1eb9p v\u00e0 ti\u1ec7n d\u1ee5ng cho ng\u01b0\u1eddi khuy\u1ebft t\u1eadt. C\u00f4ng ty A \u0111\u00e3 s\u1ea3n xu\u1ea5t ra nh\u1eefng chi\u1ebfc xe l\u0103n cho ng\u01b0\u1eddi khuy\u1ebft t\u1eadt v\u1edbi s\u1ed1 v\u1ed1n ban \u0111\u1ea7u l\u00e0 500 000 000 \u0111\u1ed3ng v\u00e0 th\u00eam chi ph\u00ed \u0111\u1ec3 s\u1ea3n xu\u1ea5t ra m\u1ed9t chi\u1ebfc xe l\u0103n l\u00e0 2 500 000 \u0111\u1ed3ng. Bi\u1ebft r\u1eb1ng gi\u00e1 b\u00e1n ra m\u1ed7i chi\u1ebfc l\u00e0 3 500 000 \u0111\u1ed3ng. a) Vi\u1ebft h\u00e0m s\u1ed1 bi\u1ec3u di\u1ec5n s\u1ed1 ti\u1ec1n l\u1eddi ho\u1eb7c l\u1ed7 y (\u0111\u1ed3ng) khi b\u00e1n ra x (chi\u1ebfc xe l\u0103n). b) H\u1ecfi c\u1ea7n ph\u1ea3i b\u00e1n \u00edt nh\u1ea5t bao nhi\u00eau chi\u1ebfc xe l\u0103n th\u00ec c\u00f4ng ty A kh\u00f4ng b\u1ecb l\u1ed7? C\u00e2u 4. (1 \u0111i\u1ec3m). Hai l\u1edbp 9A v\u00e0 9B c\u00f3 t\u1ed5ng c\u1ed9ng 86 h\u1ecdc sinh. Trong \u0111\u1ee3t thu nh\u1eb7t gi\u1ea5y b\u00e1o c\u0169 th\u1ef1c hi\u1ec7n k\u1ebf ho\u1ea1ch nh\u1ecf, c\u00f3 1 h\u1ecdc sinh l\u1edbp 9A g\u00f3p \u0111\u01b0\u1ee3c 5kg , c\u00e1c b\u1ea1n c\u00f2n l\u1ea1i trong l\u1edbp, m\u1ed7i b\u1ea1n g\u00f3p \u0111\u01b0\u1ee3c 4kg . L\u1edbp 9B c\u00f3 1 h\u1ecdc sinh g\u00f3p \u0111\u01b0\u1ee3c 7kg , c\u00e1c b\u1ea1n c\u00f2n l\u1ea1i trong l\u1edbp, m\u1ed7i b\u1ea1n g\u00f3p \u0111\u01b0\u1ee3c 8kg . T\u00ednh s\u1ed1 h\u1ecdc sinh c\u1ee7a m\u1ed7i l\u1edbp, bi\u1ebft c\u1ea3 hai l\u1edbp g\u00f3p \u0111\u01b0\u1ee3c 520kg gi\u1ea5y b\u00e1o c\u0169? C\u00e2u 5. (1 \u0111i\u1ec3m). M\u1ed9t nh\u00f3m h\u1ecdc sinh tham gia th\u1ef1c h\u00e0nh m\u00f4n Sinh h\u1ecdc v\u1edbi nhi\u1ec7m v\u1ee5 \u0111\u01b0\u1ee3c giao l\u00e0 ch\u0103m s\u00f3c v\u00e0 ghi nh\u1eadn s\u1ef1 ph\u00e1t tri\u1ec3n v\u1ec1 chi\u1ec1u cao c\u1ee7a c\u00e2y. Nh\u00f3m \u0111\u01b0\u1ee3c gi\u00e1o vi\u00ean giao ch\u0103m s\u00f3c m\u1ed9t c\u00e2y non c\u00f3 chi\u1ec1u cao ban \u0111\u1ea7u l\u00e0 2,56cm . Sau hai tu\u1ea7n ch\u0103m s\u00f3c, nh\u00f3m ghi nh\u1eadn chi\u1ec1u cao c\u1ee7a c\u00e2y \u0111\u00e3 t\u0103ng th\u00eam 1,28cm . G\u1ecdi h(cm) l\u00e0 chi\u1ec1u cao c\u1ee7a c\u00e2y sau . t . (tu\u1ea7n) ch\u0103m s\u00f3c, h v\u00e0 t li\u00ean h\u1ec7 v\u1edbi nhau b\u1eb1ng h\u00e0m s\u1ed1 h = at + b (gi\u1ea3 s\u1eed, m\u1ee9c t\u0103ng chi\u1ec1u cao trung b\u00ecnh c\u1ee7a c\u00e2y \u1edf m\u1ed7i tu\u1ea7n ch\u00eanh l\u1ec7ch kh\u00f4ng \u0111\u00e1ng k\u1ec3) a) X\u00e1c \u0111\u1ecbnh h\u1ec7 s\u1ed1 a,b c\u1ee7a h\u00e0m s\u1ed1 h = at + b b) H\u1ecfi sau bao nhi\u00eau ng\u00e0y th\u00ec c\u00e2y s\u1ebd \u0111\u1ea1t chi\u1ec1u cao 5,76cm , t\u00ednh t\u1eeb khi c\u00e2y \u0111\u01b0\u1ee3c giao cho nh\u00f3m ch\u0103m s\u00f3c. C\u00e2u 6. (1 \u0111i\u1ec3m). M\u1ed9t b\u1ed3n h\u00ecnh tr\u1ee5 \u0111ang ch\u1ee9a d\u1ea7u, \u0111\u01b0\u1ee3c \u0111\u1eb7t n\u1eb1m ngang, c\u00f3 chi\u1ec1u d\u00e0i b\u1ed3n l\u00e0 5m , b\u00e1n k\u00ednh \u0111\u00e1y 1m , v\u1edbi n\u1eafp b\u1ed3n \u0111\u1eb7t tr\u00ean m\u1eb7t n\u1eb1m ngang c\u1ee7a h\u00ecnh tr\u1ee5. Ng\u01b0\u1eddi ta \u0111\u00e3 r\u00fat d\u1ea7u trong b\u1ed3n t\u01b0\u01a1ng \u1ee9ng v\u1edbi 0,5m c\u1ee7a \u0111\u01b0\u1eddng k\u00ednh \u0111\u00e1y (nh\u01b0 h\u00ecnh v\u1ebd). T\u00ednh l\u01b0\u1ee3ng d\u1ea7u c\u00f2n l\u1ea1i trong b\u1ed3n (gi\u1ea3 s\u1eed \u0111\u1ed9 d\u00e0y c\u1ee7a b\u1ed3n l\u00e0 kh\u00f4ng \u0111\u00e1ng k\u1ec3 v\u00e0 k\u1ebft qu\u1ea3 l\u00e0m tr\u00f2n \u0111\u1ebfn ch\u1eef s\u1ed1 th\u1eadp ph\u00e2n th\u1ee9 2 ). Bi\u1ebft: Vhinh tru\u00ef = \uf070 R2.h , R : b\u00e1n k\u00ednh \u0111\u00e1y, h : chi\u1ec1u cao h\u00ecnh tr\u1ee5. T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N 1","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH 0.5m B A C H 1m O 5m C\u00e2u 7. (0,75 \u0111i\u1ec3m) \u0110\u1ec3 t\u00ednh ti\u1ec1n \u0111i\u1ec7n c\u1ee7a m\u1ed9t thi\u1ebft b\u1ecb \u0111i\u1ec7n ta l\u1ea5y l\u01b0\u1ee3ng \u0111i\u1ec7n thi\u1ebft b\u1ecb ti\u00eau th\u1ee5 nh\u00e2n v\u1edbi gi\u00e1 \u0111i\u1ec7n t\u1ea1i th\u1eddi \u0111i\u1ec3m \u0111\u00f3.C\u00f4ng th\u1ee9c t\u00ednh l\u01b0\u1ee3ng \u0111i\u1ec7n ti\u00eau ti\u00eau th\u1ee5 c\u1ee7a thi\u1ebft b\u1ecb \u0111i\u1ec7n nh\u01b0 sau: T = P.t (trong \u0111\u00f3: T l\u00e0 l\u01b0\u1ee3ng \u0111i\u1ec7n ti\u00eau th\u1ee5 trong kho\u1ea3ng th\u1eddi gian t (gi\u1edd); P c\u00f4ng su\u1ea5t c\u1ee7a thi\u1ebft b\u1ecb). a) M\u1ed9t m\u00e1y l\u1ea1nh c\u00f3 c\u00f4ng su\u1ea5t 800W , m\u1ed9t ng\u00e0y s\u1eed d\u1ee5ng trung b\u00ecnh 4 gi\u1edd. T\u00ednh l\u01b0\u1ee3ng \u0111i\u1ec7n ti\u00eau th\u1ee5 c\u1ee7a m\u00e1y l\u1ea1nh \u0111\u00f3 trong th\u00e1ng 5 \/ 2022 ? b) Nh\u00e0 anh An c\u00f3 c\u00e1c thi\u1ebft b\u1ecb nh\u01b0 sau: C\u00f4ng su\u1ea5t Th\u01a1\u0300i gian B\u1ea2NG GI\u00c1 \u0110I\u1ec6N SINH HO\u1ea0T 1 thi\u1ebft b\u1ecb d\u00f9ng 1 ng\u00e0y S\u1ed1 l\u01b0\u1ee3ng S\u1ed1 kWh s\u1eed d\u1ee5ng Gi\u00e1 (\u0111\u1ed3ng\/kWh) 4 \u0111\u00e8n Led 18W \/gi\u1edd 4 gi\u1edd Cho 50 kWh \u0111\u1ea7u ti\u00ean 1 678 1 m\u00e1y l\u1ea1nh 100W \/gi\u1edd 8 gi\u1edd Cho kWh 51 \u0111\u1ebfn 100 1734 1n\u1ed3i c\u01a1m \u0111i\u1ec7n 900W \/gi\u1edd 3 gi\u1edd Cho kWh 101 \u0111\u1ebfn 200 2 014 1 t\u1ee7 l\u1ea1nh 1040W \/ng\u00e0y 24 gi\u1edd Cho kWh 201 \u0111\u1ebfn 300 2 536 2 qu\u1ea1t m\u00e1y 48W \/gi\u1edd 10 gi\u1edd Cho kWh 301 \u0111\u1ebfn 400 2 834 Cho kWh t\u1eeb 401 tr\u1edf l\u00ean 2927 T\u00ednh ti\u1ec1n \u0111i\u1ec7n gia \u0111\u00ecnh anh An ph\u1ea3i tr\u1ea3 trong th\u00e1ng 5 \/ 2022 ? Bi\u1ebft thu\u1ebf gi\u00e1 tr\u1ecb gia t\u0103ng l\u00e0 10% . (l\u00e0m tr\u00f2n k\u1ebft qu\u1ea3 \u0111\u1ebfn h\u00e0ng ngh\u00ecn) C\u00e2u 8. (3 \u0111i\u1ec3m) T\u1eeb \u0111i\u1ec3m A n\u1eb1m ngo\u00e0i \u0111\u01b0\u1eddng tr\u00f2n (O,R) v\u1edbi OA \uf03e 2R. V\u1ebd ti\u1ebfp tuy\u1ebfn AB v\u00e0 c\u00e1t tuy\u1ebfn ACD v\u1edbi \u0111\u01b0\u1eddng tr\u00f2n (O) , ( B l\u00e0 ti\u1ebfp \u0111i\u1ec3m; AC \uf03c AD , tia AD kh\u00f4ng c\u1eaft \u0111o\u1ea1n th\u1eb3ng OB ). G\u1ecdi CE,DF l\u00e0 c\u00e1c \u0111\u01b0\u1eddng cao c\u1ee7a tam gi\u00e1c BCD . a) Ch\u1ee9ng minh: t\u1ee9 gi\u00e1c DEFC n\u1ed9i ti\u1ebfp v\u00e0 EF \/\/ AB . b) Tia EF c\u1eaft AD t\u1ea1i G , BG c\u1eaft (O) t\u1ea1i H . Ch\u1ee9ng minh: \uf044FHC \u0111\u1ed3ng d\u1ea1ng \uf044GAB . c) G\u1ecdi I l\u00e0 giao \u0111i\u1ec3m c\u1ee7a CE v\u00e0 DF . Tia HI c\u1eaft DC t\u1ea1i M . Ch\u1ee9ng minh: OM \u22a5 CD ----H\u1ebeT--- T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N 2","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH H\u01af\u1edaNG D\u1eaaN GI\u1ea2I C\u00e2u 1. (1,5 \u0111i\u1ec3m). Cho h\u00e0m s\u1ed1 y = \u2212x2 c\u00f3 \u0111\u1ed3 th\u1ecb (P) v\u00e0 h\u00e0m s\u1ed1 y = \u22122x + 1 c\u00f3 \u0111\u1ed3 th\u1ecb (D) . a) V\u1ebd (P) v\u00e0 (D) tr\u00ean c\u00f9ng m\u1eb7t ph\u1eb3ng t\u1ecda \u0111\u1ed9. b) T\u00ecm t\u1ecda \u0111\u1ed9 giao \u0111i\u1ec3m c\u1ee7a (P) v\u00e0 (D) b\u1eb1ng ph\u00e9p to\u00e1n. L\u1eddi gi\u1ea3i ( )a) V\u1ebd \u0111\u1ed3 th\u1ecb (P) v\u00e0 D tr\u00ean c\u00f9ng h\u1ec7 tr\u1ee5c t\u1ecda \u0111\u1ed9. BGT: x \u22122 \u22121 0 1 2 y = \u2212x2 \u22124 \u22121 0 \u22121 \u22124 x 01 y = \u22122x + 1 1 \u22121 ( ) ( )b) T\u00ecm t\u1ecda \u0111\u1ed9 giao \u0111i\u1ec3m c\u1ee7a P v\u00e0 D b\u1eb1ng ph\u00e9p t\u00ednh. ( ) ( )Ph\u01b0\u01a1ng tr\u00ecnh ho\u00e0nh \u0111\u1ed9 giao \u0111i\u1ec3m c\u1ee7a P v\u00e0 D : \u2212x2 = \u22122x + 1 \uf0db x2 \u2212 2x + 1 = 0 \uf0db x =1 Thay x = 1 v\u00e0o y = \u2212x2 , ta \u0111\u01b0\u1ee3c: y = \u221212 = \u22121 . V\u1eady (1; \u2212 1) l\u00e0 hai giao \u0111i\u1ec3m c\u1ea7n t\u00ecm. C\u00e2u 2. (1 \u0111i\u1ec3m). Cho ph\u01b0\u01a1ng tr\u00ecnh: x2 \u2212 6x + 8 = 0 c\u00f3 2 nghi\u1ec7m x1 , x2 .Kh\u00f4ng gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh, h\u00e3y t\u00ecnh gi\u00e1 tr\u1ecb c\u1ee7a bi\u1ec3u th\u1ee9c bi\u1ec3u th\u1ee9c A= x12 \u2212 x1 + x22 \u2212 x2 . x2 x1 L\u1eddi gi\u1ea3i V\u00ec \uf044 = b2 \u2212 4ac = 62 \u2212 4.1.8 = 4 \uf03e 0 N\u00ean ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m ph\u00e2n bi\u1ec7t x1 , x2 . \uf0ec\uf0ef\uf0efS = x1 + x2 = \u2212b = 6 \uf0ed = x1 a Theo \u0111\u1ecbnh l\u00ed Vi-et, ta c\u00f3: \uf0ef\uf0ef\uf0eeP c =8 a .x2 = Ta c\u00f3: A= x12 \u2212 x1+ x22 \u2212 x2 x2 x1 T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N 3","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH A = x12 \u2212 x1+ x22 \u2212 x2 x2 x1 A = \uf0e6 x12 + x22 \uf0f6 \u2212 ( x1 + x2 ) \uf0e7 x2 x1 \uf0f7 \uf0e8 \uf0f8 ( )A = x13 + x23 \u2212 x1x2 x1 + x2 A = ( x1 + x2 )3 \u2212 3x1x2 ( x1 + x2 ) \u2212 ( x1 + x2 ) x1x2 A = S 3 \u2212 3PS \u2212 S P A = 63 \u2212 3.8.6 \u2212 6 = 3 8 C\u00e2u 3. (0,75 \u0111i\u1ec3m). V\u1edbi s\u1ef1 ph\u00e1t tri\u1ec3n c\u1ee7a khoa h\u1ecdc k\u0129 thu\u1eadt hi\u1ec7n nay, ng\u01b0\u1eddi ta t\u1ea1o ra nhi\u1ec1u m\u1eabu xe l\u0103n \u0111\u1eb9p v\u00e0 ti\u1ec7n d\u1ee5ng cho ng\u01b0\u1eddi khuy\u1ebft t\u1eadt. C\u00f4ng ty A \u0111\u00e3 s\u1ea3n xu\u1ea5t ra nh\u1eefng chi\u1ebfc xe l\u0103n cho ng\u01b0\u1eddi khuy\u1ebft t\u1eadt v\u1edbi s\u1ed1 v\u1ed1n ban \u0111\u1ea7u l\u00e0 500 000 000 \u0111\u1ed3ng v\u00e0 th\u00eam chi ph\u00ed \u0111\u1ec3 s\u1ea3n xu\u1ea5t ra m\u1ed9t chi\u1ebfc xe l\u0103n l\u00e0 2 500 000 \u0111\u1ed3ng. Bi\u1ebft r\u1eb1ng gi\u00e1 b\u00e1n ra m\u1ed7i chi\u1ebfc l\u00e0 3 500 000 \u0111\u1ed3ng. a) Vi\u1ebft h\u00e0m s\u1ed1 bi\u1ec3u di\u1ec5n s\u1ed1 ti\u1ec1n l\u1eddi ho\u1eb7c l\u1ed7 y (\u0111\u1ed3ng) khi b\u00e1n ra x (chi\u1ebfc xe l\u0103n). b) H\u1ecfi c\u1ea7n ph\u1ea3i b\u00e1n \u00edt nh\u1ea5t bao nhi\u00eau chi\u1ebfc xe l\u0103n th\u00ec c\u00f4ng ty A kh\u00f4ng b\u1ecb l\u1ed7? L\u1eddi gi\u1ea3i a) Vi\u1ebft h\u00e0m s\u1ed1 bi\u1ec3u di\u1ec5n s\u1ed1 ti\u1ec1n l\u1eddi ho\u1eb7c l\u1ed7 y (\u0111\u1ed3ng) khi b\u00e1n ra x (chi\u1ebfc xe l\u0103n). G\u1ecdi y (\u0111\u1ed3ng) s\u1ed1 ti\u1ec1n l\u1eddi ho\u1eb7c l\u1ed7 khi b\u00e1n ra \u0111\u01b0\u1ee3c x chi\u1ebfc xe l\u0103n: y = (3 500 000 \u2212 2 500 000)x \u2212 500 000 000 = 1 000 000x \u2212 500 000 000 (\u0111\u1ed3ng) b) H\u1ecfi c\u1ea7n ph\u1ea3i b\u00e1n \u00edt nh\u1ea5t bao nhi\u00eau chi\u1ebfc xe l\u0103n th\u00ec c\u00f4ng ty A kh\u00f4ng b\u1ecb l\u1ed7? S\u1ed1 chi\u1ebfc xe \u00edt nh\u1ea5t c\u00f4ng ty A c\u1ea7n b\u00e1n \u0111\u1ec3 kh\u00f4ng b\u1ecb l\u1ed7 l\u00e0: 3,5x = 2,5x + 500 \uf0db x = 500 V\u1eady c\u00f4ng ty A c\u1ea7n b\u00e1n \u0111\u01b0\u1ee3c \u00edt nh\u1ea5t 500 chi\u1ebfc xe l\u0103n s\u1ebd kh\u00f4ng b\u1ecb l\u1ed7. C\u00e2u 4. (1 \u0111i\u1ec3m). Hai l\u1edbp 9A v\u00e0 9B c\u00f3 t\u1ed5ng c\u1ed9ng 86 h\u1ecdc sinh. Trong \u0111\u1ee3t thu nh\u1eb7t gi\u1ea5y b\u00e1o c\u0169 th\u1ef1c hi\u1ec7n k\u1ebf ho\u1ea1ch nh\u1ecf, c\u00f3 1 h\u1ecdc sinh l\u1edbp 9A g\u00f3p \u0111\u01b0\u1ee3c 5kg , c\u00e1c b\u1ea1n c\u00f2n l\u1ea1i trong l\u1edbp, m\u1ed7i b\u1ea1n g\u00f3p \u0111\u01b0\u1ee3c 4kg . L\u1edbp 9B c\u00f3 1 h\u1ecdc sinh g\u00f3p \u0111\u01b0\u1ee3c 7kg , c\u00e1c b\u1ea1n c\u00f2n l\u1ea1i trong l\u1edbp, m\u1ed7i b\u1ea1n g\u00f3p \u0111\u01b0\u1ee3c 8kg . T\u00ednh s\u1ed1 h\u1ecdc sinh c\u1ee7a m\u1ed7i l\u1edbp, bi\u1ebft c\u1ea3 hai l\u1edbp g\u00f3p \u0111\u01b0\u1ee3c 520kg gi\u1ea5y b\u00e1o c\u0169? L\u1eddi gi\u1ea3i G\u1ecdi x (h\u1ecdc sinh), y (h\u1ecdc sinh), l\u1ea7n l\u01b0\u1ee3t l\u00e0 s\u1ed1 h\u1ecdc sinh c\u1ee7a l\u1edbp 9A , 9B (x \uf0ce *, y \uf0ce *, x \uf03c 86, y \uf03c 86) ) V\u00ec hai l\u1edbp 9A v\u00e0 9B c\u00f3 t\u1ed5ng c\u1ed9ng 86 h\u1ecdc sinh n\u00ean ta c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh: x + y = 86 (1) L\u1edbp 9A g\u00f3p \u0111\u01b0\u1ee3c: 5 + ( x \u22121).4 = 4x +1 (kg) L\u1edbp 9B g\u00f3p \u0111\u01b0\u1ee3c: 7 + ( y \u22121).8 = 8y \u22121 (kg) T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N 4","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH V\u00ec c\u1ea3 hai l\u1edbp g\u00f3p \u0111\u01b0\u1ee3c 520kg gi\u1ea5y b\u00e1o c\u0169 n\u00ean ta c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh 4x +1+ 8y \u22121 = 520 \uf0db x + 2y = 130 (2) T\u1eeb (1) v\u00e0 (2) ta c\u00f3 hpt: \uf0ecx + y = 86 \uf0db \uf0ecx = 42 (nh\u1eadn) \uf0ed\uf0eex + 2y = 130 \uf0ed\uf0eey = 44 V\u1eady l\u1edbp 9A c\u00f3 42 h\u1ecdc sinh, l\u1edbp 9B c\u00f3 44 h\u1ecdc sinh. C\u00e2u 5. (1 \u0111i\u1ec3m). M\u1ed9t nh\u00f3m h\u1ecdc sinh tham gia th\u1ef1c h\u00e0nh m\u00f4n Sinh h\u1ecdc v\u1edbi nhi\u1ec7m v\u1ee5 \u0111\u01b0\u1ee3c giao l\u00e0 ch\u0103m s\u00f3c v\u00e0 ghi nh\u1eadn s\u1ef1 ph\u00e1t tri\u1ec3n v\u1ec1 chi\u1ec1u cao c\u1ee7a c\u00e2y. Nh\u00f3m \u0111\u01b0\u1ee3c gi\u00e1o vi\u00ean giao ch\u0103m s\u00f3c m\u1ed9t c\u00e2y non c\u00f3 chi\u1ec1u cao ban \u0111\u1ea7u l\u00e0 2,56cm . Sau hai tu\u1ea7n ch\u0103m s\u00f3c, nh\u00f3m ghi nh\u1eadn chi\u1ec1u cao c\u1ee7a c\u00e2y \u0111\u00e3 t\u0103ng th\u00eam 1,28cm . G\u1ecdi h(cm) l\u00e0 chi\u1ec1u cao c\u1ee7a c\u00e2y sau . t . (tu\u1ea7n) ch\u0103m s\u00f3c, h v\u00e0 t li\u00ean h\u1ec7 v\u1edbi nhau b\u1eb1ng h\u00e0m s\u1ed1 h = at + b (gi\u1ea3 s\u1eed, m\u1ee9c t\u0103ng chi\u1ec1u cao trung b\u00ecnh c\u1ee7a c\u00e2y \u1edf m\u1ed7i tu\u1ea7n ch\u00eanh l\u1ec7ch kh\u00f4ng \u0111\u00e1ng k\u1ec3) a) X\u00e1c \u0111\u1ecbnh h\u1ec7 s\u1ed1 a,b c\u1ee7a h\u00e0m s\u1ed1 h = at + b b) H\u1ecfi sau bao nhi\u00eau ng\u00e0y th\u00ec c\u00e2y s\u1ebd \u0111\u1ea1t chi\u1ec1u cao 5,76cm , t\u00ednh t\u1eeb khi c\u00e2y \u0111\u01b0\u1ee3c giao cho nh\u00f3m ch\u0103m s\u00f3c. L\u1eddi gi\u1ea3i a) X\u00e1c \u0111\u1ecbnh h\u1ec7 s\u1ed1 a,b c\u1ee7a h\u00e0m s\u1ed1 h = at + b V\u00ec c\u00e2y non c\u00f3 chi\u1ec1u cao ban \u0111\u1ea7u l\u00e0 2,56cm n\u00ean t= 0 v\u00e0 h = 2,56 \uf0de 2,56 = a.0 + b \uf0db b = 2,56 (1) V\u00ec sau hai tu\u1ea7n ch\u0103m s\u00f3c, nh\u00f3m ghi nh\u1eadn chi\u1ec1u cao c\u1ee7a c\u00e2y \u0111\u00e3 t\u0103ng th\u00eam 1,28cm \uf0de t= 2 v\u00e0 h = 2,56 + 1,28 = 3,84 \uf0de 3,84 = a.2 + b (2) T\u1eeb (1) v\u00e0 (2) ta c\u00f3 h\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh \uf0ecb = 2,56 3,84 \uf0db \uf0ecb = 2,56 \uf0ed\uf0ee2a + b = \uf0ed\uf0eea = 0,64 b) H\u1ecfi sau bao nhi\u00eau ng\u00e0y th\u00ec c\u00e2y s\u1ebd \u0111\u1ea1t chi\u1ec1u cao 5,76cm , t\u00ednh t\u1eeb khi c\u00e2y \u0111\u01b0\u1ee3c giao cho nh\u00f3m ch\u0103m s\u00f3c. Ta c\u00f3: h = 0,64t + 2,56 Thay h = 5,67v\u00e0o h = 0,64t + 2,56 ta c\u00f3: 5,67 = 0,64t + 2,56 \uf0db 0,64t = 3,2 \uf0db t = 5 V\u1eady sau 5.7 = 35 ng\u00e0y th\u00ec c\u00e2y \u0111\u1ea1t chi\u1ec1u cao 5,76cm . C\u00e2u 6. (1 \u0111i\u1ec3m). M\u1ed9t b\u1ed3n h\u00ecnh tr\u1ee5 \u0111ang ch\u1ee9a d\u1ea7u, \u0111\u01b0\u1ee3c \u0111\u1eb7t n\u1eb1m ngang, c\u00f3 chi\u1ec1u d\u00e0i b\u1ed3n l\u00e0 5m , b\u00e1n k\u00ednh \u0111\u00e1y 1m , v\u1edbi n\u1eafp b\u1ed3n \u0111\u1eb7t tr\u00ean m\u1eb7t n\u1eb1m ngang c\u1ee7a h\u00ecnh tr\u1ee5. Ng\u01b0\u1eddi ta \u0111\u00e3 r\u00fat d\u1ea7u trong b\u1ed3n t\u01b0\u01a1ng \u1ee9ng v\u1edbi 0,5m c\u1ee7a \u0111\u01b0\u1eddng k\u00ednh \u0111\u00e1y (nh\u01b0 h\u00ecnh v\u1ebd). T\u00ednh l\u01b0\u1ee3ng d\u1ea7u c\u00f2n l\u1ea1i trong b\u1ed3n (gi\u1ea3 s\u1eed \u0111\u1ed9 d\u00e0y c\u1ee7a b\u1ed3n l\u00e0 kh\u00f4ng \u0111\u00e1ng k\u1ec3 v\u00e0 k\u1ebft qu\u1ea3 l\u00e0m tr\u00f2n \u0111\u1ebfn ch\u1eef s\u1ed1 th\u1eadp ph\u00e2n th\u1ee9 2 ). Bi\u1ebft: Vhinh tru\u00ef = \uf070 R2.h , R : b\u00e1n k\u00ednh \u0111\u00e1y, h : chi\u1ec1u cao h\u00ecnh tr\u1ee5. T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N 5","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH 0.5m B A C H 1m O 5m L\u1eddi gi\u1ea3i Ta c\u00f3: OH = OB \u2212 BH = 1\u2212 0,5 = 0,5m X\u00e9t \uf044OHC vu\u00f4ng t\u1ea1i H ta c\u00f3: cos HOC = OH \uf0de HOC = 60\uf0b0 \uf0de AOC = 2HOC = 120\uf0b0 OC Ta c\u00f3: AC = 2HC = 2OC.sin HOC = 2.1. 3 = 3 2 Di\u1ec7n t\u00edch m\u1eb7t \u0111\u00e1y ph\u1ea7n d\u1ea7u r\u00fat ra : \uf070 .12.120\uf0b0 0 , 5. 3 =\uf070 \u2212 3 360\uf0b0 2 3 4 ( )Sq(AOC) m2 \u2212 S\uf044AOC = \u2212 ( )Th\u1ec3 \uf0e6 \uf070 3 \uf0f6 m3 t\u00edch d\u1ea7u r\u00fat ra: \uf0e7\uf0e7\uf0e8 3 \u2212 4 \uf0f7\uf0f7\uf0f8.5 ( )Vhinh tru\u00ef = \uf070 R2 .h = \uf070 .12.5 = 5\uf070 m3 ( )L\u01b0\u1ee3ng d\u1ea7u c\u00f2n l\u1ea1i trong b\u1ed3n: 5\uf070 \u2212 \uf0e6 \uf070 \u2212 3 \uf0f6 \uf0bb m3 \uf0e7\uf0e7\uf0e8 3 4 \uf0f7\uf0f7\uf0f8.5 12,64 C\u00e2u 7. (0,75 \u0111i\u1ec3m) \u0110\u1ec3 t\u00ednh ti\u1ec1n \u0111i\u1ec7n c\u1ee7a m\u1ed9t thi\u1ebft b\u1ecb \u0111i\u1ec7n ta l\u1ea5y l\u01b0\u1ee3ng \u0111i\u1ec7n thi\u1ebft b\u1ecb ti\u00eau th\u1ee5 nh\u00e2n v\u1edbi gi\u00e1 \u0111i\u1ec7n t\u1ea1i th\u1eddi \u0111i\u1ec3m \u0111\u00f3.C\u00f4ng th\u1ee9c t\u00ednh l\u01b0\u1ee3ng \u0111i\u1ec7n ti\u00eau ti\u00eau th\u1ee5 c\u1ee7a thi\u1ebft b\u1ecb \u0111i\u1ec7n nh\u01b0 sau: T = P.t (trong \u0111\u00f3: T l\u00e0 l\u01b0\u1ee3ng \u0111i\u1ec7n ti\u00eau th\u1ee5 trong kho\u1ea3ng th\u1eddi gian t (gi\u1edd); P c\u00f4ng su\u1ea5t c\u1ee7a thi\u1ebft b\u1ecb). c) M\u1ed9t m\u00e1y l\u1ea1nh c\u00f3 c\u00f4ng su\u1ea5t 800W , m\u1ed9t ng\u00e0y s\u1eed d\u1ee5ng trung b\u00ecnh 4 gi\u1edd. T\u00ednh l\u01b0\u1ee3ng \u0111i\u1ec7n ti\u00eau th\u1ee5 c\u1ee7a m\u00e1y l\u1ea1nh \u0111\u00f3 trong th\u00e1ng 5 \/ 2022 ? d) Nh\u00e0 anh An c\u00f3 c\u00e1c thi\u1ebft b\u1ecb nh\u01b0 sau: C\u00f4ng su\u1ea5t Th\u01a1\u0300i gian B\u1ea2NG GI\u00c1 \u0110I\u1ec6N SINH HO\u1ea0T 1 thi\u1ebft b\u1ecb d\u00f9ng 1 ng\u00e0y S\u1ed1 l\u01b0\u1ee3ng S\u1ed1 kWh s\u1eed d\u1ee5ng Gi\u00e1 (\u0111\u1ed3ng\/kWh) 4 \u0111\u00e8n Led 18W \/gi\u1edd 4 gi\u1edd Cho 50 kWh \u0111\u1ea7u ti\u00ean 1 678 1 m\u00e1y l\u1ea1nh 100W \/gi\u1edd 8 gi\u1edd Cho kWh 51 \u0111\u1ebfn 100 1734 1n\u1ed3i c\u01a1m \u0111i\u1ec7n 900W \/gi\u1edd 3 gi\u1edd Cho kWh 101 \u0111\u1ebfn 200 2 014 T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N 6","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH 1 t\u1ee7 l\u1ea1nh 1040W \/ng\u00e0y 24 gi\u1edd Cho kWh 201 \u0111\u1ebfn 300 2 536 2 qu\u1ea1t m\u00e1y 48W \/gi\u1edd 10 gi\u1edd Cho kWh 301 \u0111\u1ebfn 400 2 834 Cho kWh t\u1eeb 401 tr\u1edf l\u00ean 2927 T\u00ednh ti\u1ec1n \u0111i\u1ec7n gia \u0111\u00ecnh anh An ph\u1ea3i tr\u1ea3 trong th\u00e1ng 5 \/ 2022 ? Bi\u1ebft thu\u1ebf gi\u00e1 tr\u1ecb gia t\u0103ng l\u00e0 10% . (l\u00e0m tr\u00f2n k\u1ebft qu\u1ea3 \u0111\u1ebfn h\u00e0ng ngh\u00ecn). L\u1eddi gi\u1ea3i a) M\u1ed9t m\u00e1y l\u1ea1nh c\u00f3 c\u00f4ng su\u1ea5t 800W , m\u1ed9t ng\u00e0y s\u1eed d\u1ee5ng trung b\u00ecnh 4 gi\u1edd. T\u00ednh l\u01b0\u1ee3ng \u0111i\u1ec7n ti\u00eau th\u1ee5 c\u1ee7a m\u00e1y l\u1ea1nh \u0111\u00f3 trong th\u00e1ng 5 \/ 2022 ? L\u01b0\u1ee3ng \u0111i\u1ec7n ti\u00eau th\u1ee5 trung b\u00ecnh c\u1ee7a m\u00e1y l\u1ea1nh \u0111\u00f3 trong 1 ng\u00e0y l\u00e0: T = P.t = 8.400 = 3200W V\u00ec th\u00e1ng 5 \/ 2022 c\u00f3 31 ng\u00e0y n\u00ean l\u01b0\u1ee3ng \u0111i\u1ec7n ti\u00eau th\u1ee5 c\u1ee7a m\u00e1y l\u1ea1nh trong th\u00e1ng 5 \/ 2022 l\u00e0: 3200.31 = 99200W b) T\u00ednh ti\u1ec1n \u0111i\u1ec7n gia \u0111\u00ecnh anh An ph\u1ea3i tr\u1ea3 trong th\u00e1ng 5 \/ 2022 ? Bi\u1ebft thu\u1ebf gi\u00e1 tr\u1ecb gia t\u0103ng l\u00e0 10% . (l\u00e0m tr\u00f2n k\u1ebft qu\u1ea3 \u0111\u1ebfn h\u00e0ng ngh\u00ecn) L\u01b0\u1ee3ng \u0111i\u1ec7n ti\u00eau th\u1ee5 c\u1ee7a nh\u00e0 anh An trong th\u00e1ng 5 \/ 2022 l\u00e0: (4.18.4 + 100.8 + 900.3 + 1040.24 + 2.48.10).31 = 920 948W = 920, 948kWh Ti\u1ec1n \u0111i\u1ec7n gia \u0111\u00ecnh anh An ph\u1ea3i tr\u1ea3 trong th\u00e1ng 5 \/ 2022 l\u00e0: (1678.50 + 1734.50 + 2014.100 + 2536.100 + 2834.100 + 2927.520,948).(100% + 10%) = 2 677 000 (\u0111\u1ed3ng) ( )C\u00e2u 8. (3 \u0111i\u1ec3m) T\u1eeb \u0111i\u1ec3m A n\u1eb1m ngo\u00e0i \u0111\u01b0\u1eddng tr\u00f2n O,R v\u1edbi OA \uf03e 2R . V\u1ebd ti\u1ebfp tuy\u1ebfn AB v\u00e0 c\u00e1t ( )tuy\u1ebfn ACD v\u1edbi \u0111\u01b0\u1eddng tr\u00f2n O , ( B l\u00e0 ti\u1ebfp \u0111i\u1ec3m; AC \uf03c AD , tia AD kh\u00f4ng c\u1eaft \u0111o\u1ea1n th\u1eb3ng OB ). G\u1ecdi CE,DF l\u00e0 c\u00e1c \u0111\u01b0\u1eddng cao c\u1ee7a tam gi\u00e1c BCD . d) Ch\u1ee9ng minh: t\u1ee9 gi\u00e1c DEFC n\u1ed9i ti\u1ebfp v\u00e0 EF \/\/ AB . e) Tia EF c\u1eaft AD t\u1ea1i G , BG c\u1eaft (O) t\u1ea1i H . Ch\u1ee9ng minh: \uf044FHC \u0111\u1ed3ng d\u1ea1ng \uf044GAB . f) G\u1ecdi I l\u00e0 giao \u0111i\u1ec3m c\u1ee7a CE v\u00e0 DF . Tia HI c\u1eaft DC t\u1ea1i M . Ch\u1ee9ng minh: OM \u22a5 CD L\u1eddi gi\u1ea3i T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N 7","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH B HE FI AO G C M D a) Ch\u1ee9ng minh: t\u1ee9 gi\u00e1c DEFC n\u1ed9i ti\u1ebfp v\u00e0 EF \/\/ AB . X\u00e9t t\u1ee9 gi\u00e1c DEFC ta c\u00f3 : DEC = DFC = 90\uf0b0 ( CE, DF l\u00e0 \u0111\u01b0\u1eddng cao c\u1ee7a \uf044BCD) \uf0de T\u1ee9 gi\u00e1c DEFC n\u1ed9i ti\u1ebfp ( 2 \u0111\u1ec9nh E,F li\u00ean ti\u1ebfp c\u00f9ng nh\u00ecn c\u1ea1nh DC d\u01b0\u1edbi g\u00f3c kh\u00f4ng \u0111\u1ed5i) Ta c\u00f3: + ABC = EDC (G\u00f3c nt v\u00e0 g\u00f3c t\u1ea1o b\u1edfi tia ti\u1ebfp tuy\u1ebfn v\u00e0 d\u00e2y c\u00f9ng ch\u1eafn BC ) + BFE = EDC (T\u1ee9 gi\u00e1c DEFC n\u1ed9i ti\u1ebfp) \uf0de ABC = BFE M\u00e0 ABC,BFE n\u1eb1m \u1edf v\u1ecb tr\u00ed so le trong \uf0de EF \/\/ AB b) Tia EF c\u1eaft AD t\u1ea1i G , BG c\u1eaft (O) t\u1ea1i H . Ch\u1ee9ng minh: \uf044FHC \u0111\u1ed3ng d\u1ea1ng \uf044GAB Ta c\u00f3 EF \/\/ AB \uf0de ABH = FGH (so le trong) M\u00e0 ABH = FCH (G\u00f3c nt v\u00e0 g\u00f3c t\u1ea1o b\u1edfi tia ti\u1ebfp tuy\u1ebfn v\u00e0 d\u00e2y c\u00f9ng ch\u1eafn BH ) \uf0de FGH = FCH \uf0de T\u1ee9 gi\u00e1c FHGC n\u1ed9i ti\u1ebfp ( 2 \u0111\u1ec9nh C,G li\u00ean ti\u1ebfp c\u00f9ng nh\u00ecn c\u1ea1nh FH d\u01b0\u1edbi 1 g\u00f3c kh\u00f4ng \u0111\u1ed5i) \uf0de BGA = CFH (g\u00f3c ngo\u00e0i b\u1eb1ng g\u00f3c \u0111\u1ed1i trong) X\u00e9t \uf044FHC v\u00e0 \uf044GAB , c\u00f3: + BGA = CFH + ABH = FCH (cmt) \uf0de \uf044FHC \u0111\u1ed3ng d\u1ea1ng \uf044GAB c) G\u1ecdi I l\u00e0 giao \u0111i\u1ec3m c\u1ee7a CE v\u00e0 DF . Tia HI c\u1eaft DC t\u1ea1i M . Ch\u1ee9ng minh: OM \u22a5 CD K\u00e9o d\u00e0i HM c\u1eaft (O) t\u1ea1i K Ta c\u00f3: FHG = FCD (tg FHGC n\u1ed9i ti\u1ebfp) M\u00e0 FCD = BEF ( tg DEFC n\u1ed9i ti\u1ebfp) T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N 8","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH ( )\uf0de FHG = BEF \uf0de tg BEFH n\u1ed9i ti\u1ebfp v\u00ec c\u00f3 g\u00f3c ngo\u00e0i b\u1eb1ng g\u00f3c \u0111\u1ed1i trong 1 X\u00e9t t\u1ee9 gi\u00e1c BEIF ta c\u00f3 : BEI = BFI = 90\uf0b0 ( CE, DF l\u00e0 \u0111\u01b0\u1eddng cao c\u1ee7a \uf044BCD) \uf0de BEI + BFI = 180\uf0b0 \uf0de T\u1ee9 gi\u00e1c BEIF n\u1ed9i ti\u1ebfp \u0111\u01b0\u1eddng tr\u00f2n \u0111\u01b0\u1eddng k\u00ednh BI v\u00ec c\u00f3 t\u1ed5ng s\u1ed1 \u0111o 2 g\u00f3c \u0111\u1ed1i b\u1eb1ng 180\uf0b0 (2) ( ) ( )T\u1eeb 1 v\u00e0 2 \uf0deB,E,F,H,I c\u00f9ng thu\u1ed9c m\u1ed9t \u0111\u01b0\u1eddng tr\u00f2n \u0111\u01b0\u1eddng k\u00ednh BI \uf0de BHK = 90\uf0b0 (G\u00f3c n\u1ed9i ti\u1ebfp ch\u1eafn n\u1eeda \u0111\u01b0\u1eddng tr\u00f2n \u0111\u01b0\u1eddng k\u00ednh BI ) \uf0de BK l\u00e0 \u0111\u01b0\u1eddng k\u00ednh c\u1ee7a \u0111\u01b0\u1eddng tr\u00f2n (O) Ta c\u00f3: + DF \u22a5 BC ( DF l\u00e0 \u0111\u01b0\u1eddng cao c\u1ee7a \uf044BCD) + KC \u22a5 BC ( BCK l\u00e0 gnt ch\u1eafn n\u1eeda \u0111\u01b0\u1eddng tr\u00f2n \u0111\u01b0\u1eddng k\u00ednh BK ) \uf0de DF \/\/ KC (3) Ta c\u00f3: + CI \u22a5 DB ( CE l\u00e0 \u0111\u01b0\u1eddng cao c\u1ee7a \uf044BCD) + DK \u22a5 DB ( BDK l\u00e0 gnt ch\u1eafn n\u1eeda \u0111\u01b0\u1eddng tr\u00f2n \u0111\u01b0\u1eddng k\u00ednh BK ) \uf0de CI \/\/ DK (4) T\u1eeb (3) v\u00e0 (4)\uf0de T\u1ee9 gi\u00e1c DICK l\u00e0 h\u00ecnh b\u00ecnh h\u00e0nh M\u00e0 M l\u00e0 giao \u0111i\u1ec3m c\u1ee7a 2 \u0111\u01b0\u1eddng ch\u00e9o KI v\u00e0 DC \uf0de M l\u00e0 trung \u0111i\u1ec3m c\u1ee7a DC \uf0de OM \u22a5 CD (Quan h\u1ec7 gi\u1eefa \u0111\u01b0\u1eddng k\u00ednh v\u00e0 d\u00e2y) ----H\u1ebeT--- T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N 9","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH S\u00d4\u00db GD&\u00d1T TP HO\u00c0 CH\u00cd MINH \u00d1E\u00c0 THAM KHA\u00dbO TUYE\u00c5N SINH 10 PHO\u00d8NG G\u00d1&\u00d1T QUA\u00c4N TA\u00c2N PHU\u00d9 NA\u00caM HO\u00cfC: 2023 - 2024 M\u00d4N: TO\u00c1N 9 \u0110\u1ec0 THAM KH\u1ea2O M\u00c3 \u0110\u1ec0: Qu\u1eadn T\u00e2n Ph\u00fa - 3 \u0110\u00ea thi g\u1ed3m 8 c\u00e2u ho\u0309i t\u01b0\u0323 lu\u00e2\u0323n. Th\u01a1\u0300i gian: 120 phu\u0301t (kh\u00f4ng k\u00ea\u0309 th\u01a1\u0300i gian pha\u0301t \u0111\u00ea\u0300) C\u00e2u 1. (1,5 \u0111i\u1ec3m). Cho parabol (P) : y = x2 v\u00e0 \u0111\u01b0\u1eddng th\u1eb3ng (d) : y = 2x \u2212 1 a) V\u1ebd \u0111\u1ed3 th\u1ecb (P) v\u00e0 (d) tr\u00ean c\u00f9ng h\u1ec7 tr\u1ee5c t\u1ecda \u0111\u1ed9. b) T\u00ecm t\u1ecda \u0111\u1ed9 giao \u0111i\u1ec3m c\u1ee7a (P) v\u00e0 (d) b\u1eb1ng ph\u00e9p to\u00e1n. C\u00e2u 2. (1 \u0111i\u1ec3m). Cho ph\u01b0\u01a1ng tr\u00ecnh: x2 \u2212 4x \u2212 5 = 0 c\u00f3 hai nghi\u1ec7m l\u00e0 x1 v\u00e0 x2 . Kh\u00f4ng gi\u1ea3i ph\u01b0\u01a1ng ( )tr\u00ecnh, h\u00e3y t\u00ednh gi\u00e1 tr\u1ecb c\u1ee7a bi\u1ec3u th\u1ee9c: A = 2 x1 \u2212 x2 2 + 3x1x2 C\u00e2u 3. (0,75 \u0111i\u1ec3m). \u0110\u1ec3 x\u00e1c \u0111\u1ecbnh ng\u00e0y n , th\u00e1ng t , n\u0103m K r\u01a1i v\u00e0o th\u1ee9 m\u1ea5y, ta t\u00ednh theo quy t\u1eafc sau: M = (K \u2212 1).365 + th\u01b0\u01a1ng c\u1ee7a \uf0e6 K \u2212 1 \uf0f6 + C \uf0e7 4 \uf0f7 \uf0e8 \uf0f8 Trong \u0111\u00f3, K l\u00e0 s\u1ed1 n\u0103m, C l\u00e0 s\u1ed1 ng\u00e0y t\u00ednh t\u1eeb ng\u00e0y \u0111\u1ea7u ti\u00ean c\u1ee7a n\u0103m K t\u1edbi ng\u00e0y n , th\u00e1ng t , n\u0103m K . Sau \u0111\u00f3, l\u1ea5y M chia cho 7 ta \u0111\u01b0\u1ee3c s\u1ed1 d\u01b0 r v\u00e0 tra b\u1ea3ng sau: r 0 1 2 345 6 Th\u1ee9 th\u1ee9 Ch\u1ee7 th\u1ee9 th\u1ee9 th\u1ee9 th\u1ee9 th\u1ee9 B\u1ea3y Nh\u1eadt Hai Ba T\u01b0 N\u0103m S\u00e1u V\u00ed d\u1ee5: Ng\u00e0y 1 \/ 6 \/ 2000 bi\u1ebft n\u0103m 2000 l\u00e0 n\u0103m nhu\u1eadn. C = 31+ 29 + 31+ 30 + 31+ 1 = 153 . Th\u01b0\u01a1ng c\u1ee7a (2000 \u2212 1) chia 4 l\u00e0 499 . M = (2000 \u2212 1).365 + 499 + 153 = 730 287 . V\u00ec 730287 : 7 d\u01b0 5 n\u00ean ng\u00e0y 1 \/ 6 \/ 2000 l\u00e0 th\u1ee9 n\u0103m. a) Em h\u00e3y cho bi\u1ebft ng\u00e0y 15 \/ 4 \/ 2021 l\u00e0 ng\u00e0y th\u1ee9 m\u1ea5y ? Bi\u1ebft n\u0103m 2021 kh\u00f4ng ph\u1ea3i n\u0103m nhu\u1eadn. b) N\u1ebfu ng\u00e0y 12 c\u1ee7a th\u00e1ng t thu\u1ed9c n\u1eeda \u0111\u1ea7u n\u0103m 2021 r\u01a1i v\u00e0o th\u1ee9 hai. Em h\u00e3y cho bi\u1ebft \u0111\u00f3 l\u00e0 th\u00e1ng n\u00e0o? C\u00e2u 4. (0,75 \u0111i\u1ec3m). Do c\u00e1c ho\u1ea1t \u0111\u1ed9ng c\u00f4ng nghi\u1ec7p thi\u1ebfu ki\u1ec3m so\u00e1t c\u1ee7a con ng\u01b0\u1eddi l\u00e0m cho nhi\u1ec7t \u0111\u1ed9 tr\u00e1i \u0111\u1ea5t t\u0103ng d\u1ea7n m\u1ed9t c\u00e1ch r\u1ea5t \u0111\u00e1ng lo ng\u1ea1i. C\u00e1c nh\u00e0 khoa h\u1ecdc \u0111\u00e3 \u0111\u01b0a ra c\u00f4ng th\u1ee9c d\u1ef1 b\u00e1o T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N 1","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH nhi\u1ec7t \u0111\u1ed9 trung b\u00ecnh tr\u00ean b\u1ec1 m\u1eb7t tr\u00e1i \u0111\u1ea5t nh\u01b0 sau: T = at + b . Trong \u0111\u00f3 T l\u00e0 nhi\u1ec7t \u0111\u1ed9 trung b\u00ecnh c\u1ee7a b\u1ec1 m\u1eb7t tr\u00e1i \u0111\u1ea5t t\u00ednh theo \u0111\u1ed9 C ; t l\u00e0 s\u1ed1 n\u0103m k\u1ec3 t\u1eeb n\u0103m 1950 . V\u00e0o n\u0103m 1950 , ng\u01b0\u1eddi ta kh\u1ea3o s\u00e1t th\u1ea5y nhi\u1ec7t \u0111\u1ed9 tr\u00e1i \u0111\u1ea5t l\u00e0 15\uf0b0C v\u00e0 sau 30 n\u0103m kh\u1ea3o s\u00e1t c\u00e1c nh\u00e0 khoa h\u1ecdc \u0111\u00e3 th\u1ea5y nhi\u1ec7t \u0111\u1ed9 tr\u00e1i \u0111\u1ea5t \u0111\u00e3 t\u0103ng 0,6\uf0b0C . X\u00e1c \u0111\u1ecbnh h\u1ec7 s\u1ed1 a, b . C\u00e2u 5. (1 \u0111i\u1ec3m). M\u1ed9t bu\u1ed5i li\u00ean hoan l\u1edbp c\u00f4 gi\u00e1o \u0111\u1ecbnh chia m\u1ed9t s\u1ed1 k\u1eb9o th\u00e0nh c\u00e1c ph\u1ea7n qu\u00e0 cho c\u00e1c em h\u1ecdc sinh. N\u1ebfu m\u1ed7i ph\u1ea7n gi\u1ea3m \u0111i 6 vi\u00ean th\u00ec c\u00e1c em c\u00f3 th\u00eam 5 ph\u1ea7n qu\u00e0, n\u1ebfu gi\u1ea3m \u0111i 10 vi\u00ean th\u00ec c\u00e1c em c\u00f3 th\u00eam 10 ph\u1ea7n qu\u00e0. H\u1ecfi t\u1ed5ng s\u1ed1 k\u1eb9o l\u00e0 bao nhi\u00eau vi\u00ean? C\u00e2u 6. (1 \u0111i\u1ec3m). M\u1ed9t chi ti\u1ebft x\u00e2y d\u1ef1ng b\u1eb1ng b\u00ea t\u00f4ng c\u00f3 k\u00edch th\u01b0\u1edbc nh\u01b0 h\u00ecnh v\u1ebd b\u00ean, g\u1ed3m: \u2022 Ph\u00eda tr\u00ean l\u00e0 m\u1ed9t h\u00ecnh tr\u1ee5 c\u00f3 chi\u1ec1u cao 2 (m) , \u0111\u01b0\u1eddng k\u00ednh \u0111\u00e1y 0,5 (m) . \u2022 Ph\u00eda d\u01b0\u1edbi l\u00e0 n\u1eeda h\u00ecnh c\u1ea7u c\u00f3 \u0111\u01b0\u1eddng k\u00ednh 0,5 (m) . a) T\u00ednh th\u1ec3 t\u00edch c\u1ee7a chi ti\u1ebft tr\u00ean (l\u00e0m tr\u00f2n \u0111\u1ebfn ch\u1eef s\u1ed1 th\u00e2\u0323p ph\u00e2n th\u1ee9 ba). Bi\u1ebft V h\u00ecnh tr\u1ee5 = \uf070 R2h ; V h\u00ecnh c\u1ea7u = 4 \uf070 R3 3 b) M\u1ed7i xe tr\u1ed9n b\u00ea t\u00f4ng cung c\u1ea5p \u0111\u01b0\u1ee3c 6m3 b\u00ea t\u00f4ng. M\u1ed9t c\u00f4ng tr\u00ecnh x\u00e2y d\u1ef1ng c\u1ea7n s\u1eed d\u1ee5ng 40 chi ti\u1ebft nh\u01b0 \u1edf c\u00e2u a th\u00ec c\u1ea7n \u00edt nh\u1ea5t bao nhi\u00eau xe \u0111\u1ec3 \u0111\u00e1p \u1ee9ng \u0111\u01b0\u1ee3c nhu c\u1ea7u? C\u00e2u 7. (1 \u0111i\u1ec3m). M\u1ed9t c\u1eeda h\u00e0ng b\u00e1n hoa ni\u00eam y\u1ebft gi\u00e1 1 b\u00f4ng h\u1ed3ng l\u00e0 15000 \u0111\u1ed3ng. N\u1ebfu kh\u00e1ch h\u00e0ng mua 10 b\u00f4ng tr\u1edf l\u00ean th\u00ec t\u1eeb b\u00f4ng th\u1ee9 10 m\u1ed7i b\u00f4ng gi\u1ea3m 10% tr\u00ean gi\u00e1 ni\u00eam y\u1ebft. N\u1ebfu mua 20 b\u00f4ng tr\u1edf l\u00ean th\u00ec t\u1eeb b\u00f4ng th\u1ee9 20 \u0111\u01b0\u1ee3c gi\u1ea3m th\u00eam 5% tr\u00ean gi\u00e1 \u0111\u00e3 gi\u1ea3m. N\u1ebfu mua nhi\u1ec1u h\u01a1n 50 b\u00f4ng th\u00ec \u0111\u01b0\u1ee3c gi\u1ea3m th\u00eam 2% tr\u00ean t\u1ed5ng h\u00f3a \u0111\u01a1n. a) N\u1ebfu mua 60 b\u00f4ng th\u00ec ph\u1ea3i tr\u1ea3 bao nhi\u00eau ti\u1ec1n? (l\u00e0m tr\u00f2n \u0111\u1ebfn h\u00e0ng ngh\u00ecn). b) \u00d4ng A \u0111\u00e3 mua m\u1ed9t s\u1ed1 b\u00f4ng v\u00e0 tr\u1ea3 334125 \u0111\u1ed3ng. H\u00e3y t\u00ednh s\u1ed1 b\u00f4ng \u00f4ng \u0111\u00e3 mua. C\u00e2u 8. (3 \u0111i\u1ec3m) Cho tam gi\u00e1c ABC nh\u1ecdn ( AB \uf03c AC) n\u1ed9i ti\u1ebfp (O) , \u0111\u01b0\u1eddng cao AD v\u00e0 BE c\u1eaft nhau t\u1ea1i H . K\u1ebb \u0111\u01b0\u1eddng k\u00ednh AK , v\u1ebd BF vu\u00f4ng g\u00f3c v\u1edbi AK t\u1ea1i F . a) Ch\u1ee9ng minh n\u0103m \u0111i\u1ec3m A, B, D, E, F c\u00f9ng thu\u1ed9c m\u1ed9t \u0111\u01b0\u1eddng tr\u00f2n. b) G\u1ecdi M l\u00e0 trung \u0111i\u1ec3m BC . Ch\u1ee9ng minh M, H, K th\u1eb3ng h\u00e0ng. c) G\u1ecdi I l\u00e0 trung \u0111i\u1ec3m AB . Ch\u1ee9ng minh D \u0111\u1ed1i x\u1ee9ng v\u1edbi F qua IM . ----H\u1ebeT--- T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N 2","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH H\u01af\u1edaNG D\u1eaaN GI\u1ea2I C\u00e2u 1. (1,5 \u0111i\u1ec3m). Cho parabol (P) : y = x2 v\u00e0 \u0111\u01b0\u1eddng th\u1eb3ng (d) : y = 2x \u2212 1 a) V\u1ebd \u0111\u1ed3 th\u1ecb (P) v\u00e0 (d) tr\u00ean c\u00f9ng h\u1ec7 tr\u1ee5c t\u1ecda \u0111\u1ed9. b) T\u00ecm t\u1ecda \u0111\u1ed9 giao \u0111i\u1ec3m c\u1ee7a (P) v\u00e0 (d) b\u1eb1ng ph\u00e9p to\u00e1n. L\u1eddi gi\u1ea3i Parabol ( P) : y = x2 v\u00e0 \u0111\u01b0\u1eddng th\u1eb3ng (d) : y = 2x \u2212 1 B\u1ea3ng gi\u00e1 tr\u1ecb x \u22122 \u22121 0 1 2 y = x2 4 1 0 1 4 x 0 2 y = 2x \u22121 1 3 Ph\u01b0\u01a1ng tr\u00ecnh ho\u00e0nh \u0111\u1ed9 giao \u0111i\u1ec3m c\u1ee7a (P) v\u00e0 (d) 3 x2 = 2x \u22121 \uf0db x2 \u2212 2x + 1 = 0 \uf0db x =1 Thay x = 1 v\u00e0o y = x2 , ta \u0111\u01b0\u1ee3c \uf0de y = 1 T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH V\u1eady t\u1ecda \u0111\u1ed9 giao \u0111i\u1ec3m c\u1ee7a (P) v\u00e0 (d) l\u00e0 (1;1) C\u00e2u 2. (1 \u0111i\u1ec3m). Cho ph\u01b0\u01a1ng tr\u00ecnh: x2 \u2212 4x \u2212 5 = 0 c\u00f3 hai nghi\u1ec7m l\u00e0 x1 , x2 . Kh\u00f4ng gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh, h\u00e3y t\u00ednh gi\u00e1 tr\u1ecb c\u1ee7a bi\u1ec3u th\u1ee9c: ( )A = 2 x1 \u2212 x2 2 + 3x1x2 . L\u1eddi gi\u1ea3i Theo h\u1ec7 th\u1ee9c Vi \u2013\u00e9t ta c\u00f3 S = x1 + x2 = 4 P = x1.x2 = \u22125 ( )Ta c\u00f3 A = 2 x1 \u2212 x2 2 + 3x1x2 ( )A = 2 S2 \u2212 4P + 3P A = 2 \uf0eb\uf0e942 \u2212 4.(\u22125)\uf0f9\uf0fb + 3.(\u22125) A = 2 \uf0e9\uf0eb42 \u2212 4.(\u22125)\uf0f9\uf0fb + 3.(\u22125) A = 57 . C\u00e2u 3. (0,75 \u0111i\u1ec3m). \u0110\u1ec3 x\u00e1c \u0111\u1ecbnh ng\u00e0y n , th\u00e1ng t , n\u0103m K r\u01a1i v\u00e0o th\u1ee9 m\u1ea5y, ta t\u00ednh theo quy t\u1eafc sau: M = (K \u2212 1).365 + th\u01b0\u01a1ng c\u1ee7a \uf0e6 K \u2212 1 \uf0f6 + C \uf0e7\uf0e8 4 \uf0f7\uf0f8 Trong \u0111\u00f3, K l\u00e0 s\u1ed1 n\u0103m, C l\u00e0 s\u1ed1 ng\u00e0y t\u00ednh t\u1eeb ng\u00e0y \u0111\u1ea7u ti\u00ean c\u1ee7a n\u0103m K t\u1edbi ng\u00e0y n , th\u00e1ng t , n\u0103m K . Sau \u0111\u00f3, l\u1ea5y M chia cho 7 ta \u0111\u01b0\u1ee3c s\u1ed1 d\u01b0 r v\u00e0 tra b\u1ea3ng sau: r 0 1 2 345 6 Th\u1ee9 th\u1ee9 Ch\u1ee7 th\u1ee9 th\u1ee9 th\u1ee9 th\u1ee9 th\u1ee9 B\u1ea3y Nh\u1eadt Hai Ba T\u01b0 N\u0103m S\u00e1u V\u00ed d\u1ee5: Ng\u00e0y 1 \/ 6 \/ 2000 bi\u1ebft n\u0103m 2000 l\u00e0 n\u0103m nhu\u1eadn. C = 31+ 29 + 31+ 30 + 31+ 1 = 153 . ( )Th\u01b0\u01a1ng c\u1ee7a 2000 \u2212 1 chia 4 l\u00e0 499 . M = (2000 \u2212 1).365 + 499 + 153 = 730 287 . V\u00ec 730287 : 7 d\u01b0 5 n\u00ean ng\u00e0y 1 \/ 6 \/ 2000 l\u00e0 th\u1ee9 n\u0103m. a) Em h\u00e3y cho bi\u1ebft ng\u00e0y 15 \/ 4 \/ 2021 l\u00e0 ng\u00e0y th\u1ee9 m\u1ea5y ? Bi\u1ebft n\u0103m 2021 kh\u00f4ng ph\u1ea3i n\u0103m nhu\u1eadn. T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N 4","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH b) N\u1ebfu ng\u00e0y 12 c\u1ee7a th\u00e1ng t thu\u1ed9c n\u1eeda \u0111\u1ea7u n\u0103m 2021 r\u01a1i v\u00e0o th\u1ee9 hai. Em h\u00e3y cho bi\u1ebft \u0111\u00f3 l\u00e0 th\u00e1ng n\u00e0o? L\u1eddi gi\u1ea3i a) C = 31+ 28 + 31+ 15 = 105 . Th\u01b0\u01a1ng c\u1ee7a (2000 \u2212 1) chia 4 l\u00e0 505 . M 2021 1 .365 505 105 737910 . V\u00ec 737910 chia 7 d\u01b0 5 n\u00ean 15 \/ 4 \/ 2021 l\u00e0 th\u1ee9 n\u0103m. b) M 2021 1 .365 505 C 737805 C V\u00ec ng\u00e0y 12 \/ t \/ 2021 l\u00e0 th\u1ee9 hai n\u00ean M = (737805 + C) chia 7 d\u01b0 2 (v\u1edbi t = 1; 2; 3; 4; 5; 6) N\u1ebfu t = 1 th\u00ec C = 12 \uf0de M = 737805 +12 = 737817 chia 7 d\u01b0 3 (lo\u1ea1i) N\u1ebfu t = 2 th\u00ec C = 31+ 12 = 43 \uf0de M = 737805 + 43 = 737848 chia 7 d\u01b0 6 (lo\u1ea1i) N\u1ebfu t = 3 th\u00ec C = 31+ 28 + 12 = 71 \uf0de M = 737805 + 71 = 737876 chia 7 d\u01b0 6 (lo\u1ea1i) N\u1ebfu t = 4 th\u00ec C = 31+ 28 + 31+ 12 = 102 \uf0de M = 737805 +102 = 737907 chia 7 d\u01b0 2 (nh\u1eadn) N\u1ebfu t = 5 th\u00ec C = 31+ 28 + 31+ 30 + 12 = 132 \uf0de M = 737805 +132 = 737937 chia 7 d\u01b0 4 (lo\u1ea1i) N\u1ebfu t = 6 th\u00ec C = 31+ 28 + 31 + 30 + 31 + 12 = 163 \uf0de M = 737805 +163 = 737968 chia 7 d\u01b0 0 (lo\u1ea1i) V\u1eady t l\u00e0 th\u00e1ng 4 . C\u00e2u 4. (0,75 \u0111i\u1ec3m). Do c\u00e1c ho\u1ea1t \u0111\u1ed9ng c\u00f4ng nghi\u1ec7p thi\u1ebfu ki\u1ec3m so\u00e1t c\u1ee7a con ng\u01b0\u1eddi l\u00e0m cho nhi\u1ec7t \u0111\u1ed9 tr\u00e1i \u0111\u1ea5t t\u0103ng d\u1ea7n m\u1ed9t c\u00e1ch r\u1ea5t \u0111\u00e1ng lo ng\u1ea1i. C\u00e1c nh\u00e0 khoa h\u1ecdc \u0111\u00e3 \u0111\u01b0a ra c\u00f4ng th\u1ee9c d\u1ef1 b\u00e1o nhi\u1ec7t \u0111\u1ed9 trung b\u00ecnh tr\u00ean b\u1ec1 m\u1eb7t tr\u00e1i \u0111\u1ea5t nh\u01b0 sau: T = at + b . Trong \u0111\u00f3 T l\u00e0 nhi\u1ec7t \u0111\u1ed9 trung b\u00ecnh c\u1ee7a b\u1ec1 m\u1eb7t tr\u00e1i \u0111\u1ea5t t\u00ednh theo \u0111\u1ed9 C; t l\u00e0 s\u1ed1 n\u0103m k\u1ec3 t\u1eeb n\u0103m 1950. V\u00e0o n\u0103m 1950, ng\u01b0\u1eddi ta kh\u1ea3o s\u00e1t th\u1ea5y nhi\u1ec7t \u0111\u1ed9 tr\u00e1i \u0111\u1ea5t l\u00e0 150C v\u00e0 sau 30 n\u0103m kh\u1ea3o s\u00e1t c\u00e1c nh\u00e0 khoa h\u1ecdc \u0111\u00e3 th\u1ea5y nhi\u1ec7t \u0111\u1ed9 tr\u00e1i \u0111\u1ea5t \u0111\u00e3 t\u0103ng 0, 60C . X\u00e1c \u0111\u1ecbnh h\u1ec7 s\u1ed1 a,b. L\u1eddi gi\u1ea3i Khi t = 0 th\u00ec y = 79 n\u00ean y = at + b \uf0db a.0 + b = 79 (1) Khi t = 1 th\u00ec y = 79 + 11 = 90 n\u00ean y = at + b \uf0db a.1+ b = 90 (2) T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N 5","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH T\u1eeb (1) v\u00e0 (2) ta c\u00f3 hpt: \uf0eca.0 + b = 79 \uf0db \uf0eca = 11 \uf0ed\uf0eea.1+ b = 90 \uf0ed\uf0eeb . = 79 C\u00e2u 5. (1 \u0111i\u1ec3m). M\u1ed9t bu\u1ed5i li\u00ean hoan l\u1edbp c\u00f4 gi\u00e1o \u0111\u1ecbnh chia m\u1ed9t s\u1ed1 k\u1eb9o th\u00e0nh c\u00e1c ph\u1ea7n qu\u00e0 cho c\u00e1c em h\u1ecdc sinh. N\u1ebfu m\u1ed7i ph\u1ea7n gi\u1ea3m \u0111i 6 vi\u00ean th\u00ec c\u00e1c em c\u00f3 th\u00eam 5 ph\u1ea7n qu\u00e0, n\u1ebfu gi\u1ea3m \u0111i 10 vi\u00ean th\u00ec c\u00e1c em c\u00f3 th\u00eam 10 ph\u1ea7n qu\u00e0. H\u1ecfi t\u1ed5ng s\u1ed1 k\u1eb9o l\u00e0 bao nhi\u00eau vi\u00ean? L\u1eddi gi\u1ea3i ( )G\u1ecdi x (vi\u00ean) l\u00e0 s\u1ed1 k\u1eb9o c\u1ee7a m\u1ed7i ph\u1ea7n qu\u00e0 ban \u0111\u1ea7u, y l\u00e0 s\u1ed1 ph\u1ea7n qu\u00e0 ban \u0111\u1ea7u x, y \uf0ce * . V\u00ec n\u1ebfu m\u1ed7i ph\u1ea7n gi\u1ea3m \u0111i 6 vi\u00ean th\u00ec c\u00e1c em c\u00f3 th\u00eam 5 ph\u1ea7n qu\u00e0 n\u00ean (x \u2212 6)(y + 5) = xy \uf0db 5x \u2212 6y = 30 (1) V\u00ec n\u1ebfu gi\u1ea3m \u0111i 10 vi\u00ean th\u00ec c\u00e1c em c\u00f3 th\u00eam 10 ph\u1ea7n qu\u00e0 n\u00ean ( x \u221210)( y +10) = xy \uf0db 10x \u221210y =100 (2) T\u1eeb (1) v\u00e0 (2) ta c\u00f3 h\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh \uf0ec5x \u2212 6y = 30 \uf0db \uf0ecx = 30 (n) \uf0ee\uf0ed10x \u2212 10y = 100 \uf0ed\uf0eey = 20 (n) V\u1eady t\u1ed5ng s\u1ed1 k\u1eb9o l\u00e0 30.20 = 600 vi\u00ean C\u00e2u 6. (1 \u0111i\u1ec3m). M\u1ed9t chi ti\u1ebft x\u00e2y d\u1ef1ng b\u1eb1ng b\u00ea t\u00f4ng c\u00f3 k\u00edch th\u01b0\u1edbc nh\u01b0 h\u00ecnh v\u1ebd b\u00ean, g\u1ed3m: \u2022 Ph\u00eda tr\u00ean l\u00e0 m\u1ed9t h\u00ecnh tr\u1ee5 c\u00f3 chi\u1ec1u cao 2 (m), \u0111\u01b0\u1eddng k\u00ednh \u0111\u00e1y 0,5 (m). \u2022 Ph\u00eda d\u01b0\u1edbi l\u00e0 n\u1eeda h\u00ecnh c\u1ea7u c\u00f3 \u0111\u01b0\u1eddng k\u00ednh 0,5 (m). a) T\u00ednh th\u1ec3 t\u00edch c\u1ee7a chi ti\u1ebft tr\u00ean (l\u00e0m tr\u00f2n \u0111\u1ebfn ch\u1eef s\u1ed1 th\u00e2\u0323p ph\u00e2n th\u1ee9 ba). Bi\u1ebft V h\u00ecnh tr\u1ee5 = \uf070 R2h ; V h\u00ecnh c\u1ea7u = 4 \uf070 R3 3 b) M\u1ed7i xe tr\u1ed9n b\u00ea t\u00f4ng cung c\u1ea5p \u0111\u01b0\u1ee3c 6m3 b\u00ea t\u00f4ng. M\u1ed9t c\u00f4ng tr\u00ecnh x\u00e2y d\u1ef1ng c\u1ea7n s\u1eed d\u1ee5ng 40 chi ti\u1ebft nh\u01b0 \u1edf c\u00e2u a th\u00ec c\u1ea7n \u00edt nh\u1ea5t bao nhi\u00eau xe \u0111\u1ec3 \u0111\u00e1p \u1ee9ng \u0111\u01b0\u1ee3c nhu c\u1ea7u? L\u1eddi gi\u1ea3i a) B\u00e1n k\u00ednh \u0111\u00e1y: 0, 5 : 2 0, 25 m 4 \uf070 R3 4 \uf070 0,253 Th\u1ec3 t\u00edch c\u1ee7a chi\u1ebfc b\u00ecnh l\u00e0: \uf070 R2h + 3 = \uf070 .0,252.2 + 3 ( )= 0,425 m3 22 b) Th\u1ec3 t\u00edch c\u1ee7a 40 chi ti\u1ebft l\u00e0: 0, 425.40 17 m3 S\u1ed1 xe c\u1ea7n l\u00e0: 17 : 6 2,8 (xe). V\u1eady c\u1ea7n 3 xe \u0111\u1ec3 \u0111\u00e1p \u1ee9ng nhu c\u1ea7u x\u00e2y d\u1ef1ng. T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N 6","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH C\u00e2u 7. (1 \u0111i\u1ec3m). M\u1ed9t c\u1eeda h\u00e0ng b\u00e1n hoa ni\u00eam y\u1ebft gi\u00e1 1 b\u00f4ng h\u1ed3ng l\u00e0 15000 \u0111\u1ed3ng. N\u1ebfu kh\u00e1ch h\u00e0ng mua 10 b\u00f4ng tr\u1edf l\u00ean th\u00ec t\u1eeb b\u00f4ng th\u1ee9 10 m\u1ed7i b\u00f4ng gi\u1ea3m 10% tr\u00ean gi\u00e1 ni\u00eam y\u1ebft. N\u1ebfu mua 20 b\u00f4ng tr\u1edf l\u00ean th\u00ec t\u1eeb b\u00f4ng th\u1ee9 20 \u0111\u01b0\u1ee3c gi\u1ea3m th\u00eam 5% tr\u00ean gi\u00e1 \u0111\u00e3 gi\u1ea3m. N\u1ebfu mua nhi\u1ec1u h\u01a1n 50 b\u00f4ng th\u00ec \u0111\u01b0\u1ee3c gi\u1ea3m th\u00eam 2% tr\u00ean t\u1ed5ng h\u00f3a \u0111\u01a1n. a) N\u1ebfu mua 60 b\u00f4ng th\u00ec ph\u1ea3i tr\u1ea3 bao nhi\u00eau ti\u1ec1n? (l\u00e0m tr\u00f2n \u0111\u1ebfn h\u00e0ng ngh\u00ecn). b) \u00d4ng A \u0111\u00e3 mua m\u1ed9t s\u1ed1 b\u00f4ng v\u00e0 tr\u1ea3 334125 \u0111\u1ed3ng. H\u00e3y t\u00ednh s\u1ed1 b\u00f4ng \u00f4ng \u0111\u00e3 mua. L\u1eddi gi\u1ea3i a) Gi\u00e1 1 b\u00f4ng t\u1eeb b\u00f4ng th\u1ee9 10 \u0111\u1ebfn b\u00f4ng th\u1ee9 19 l\u00e0: 15000 1 10% 13500 (\u0111\u1ed3ng) Gi\u00e1 1 b\u00f4ng t\u1eeb b\u00f4ng th\u1ee9 20 l\u00e0: 13500 1 5% 12825 (\u0111\u1ed3ng) S\u1ed1 ti\u1ec1n ph\u1ea3i tr\u1ea3 n\u1ebfu mua 60 b\u00f4ng l\u00e0: 9.15000 19 9 .13500 60 19 .12825 . 1 2% 780000 (\u0111\u1ed3ng) b) S\u1ed1 ti\u1ec1n mua 9 b\u00f4ng l\u00e0 : 9.15000 135000 (\u0111\u1ed3ng) S\u1ed1 ti\u1ec1n mua 19 b\u00f4ng l\u00e0: 135000 19 9 .13500 270000 (\u0111\u1ed3ng) S\u1ed1 ti\u1ec1n n\u1ebfu mua 50 b\u00f4ng l\u00e0: 270000 50 19 .12825 667575 V\u00ec 270000 \u0111\u1ed3ng 334125 \u0111\u1ed3ng 667 575 \u0111\u1ed3ng n\u00ean \u00f4ng A \u0111\u00e3 mua nhi\u1ec1u h\u01a1n 19 b\u00f4ng v\u00e0 \u00edt h\u01a1n 50 b\u00f4ng. S\u1ed1 ti\u1ec1n c\u00f2n l\u1ea1i: 334125 270000 64125 (\u0111\u1ed3ng) S\u1ed1 b\u00f4ng c\u00f2n l\u1ea1i: 64125 : 12825 5 (b\u00f4ng). S\u1ed1 b\u00f4ng \u0111\u00e3 mua: 5 19 24 (b\u00f4ng) C\u00e2u 8. (3 \u0111i\u1ec3m) Cho tam gi\u00e1c ABC nh\u1ecdn AB AC n\u1ed9i ti\u1ebfp O , \u0111\u01b0\u1eddng cao AD v\u00e0 BE c\u1eaft nhau t\u1ea1i H . K\u1ebb \u0111\u01b0\u1eddng k\u00ednh AK , v\u1ebd BF vu\u00f4ng g\u00f3c v\u1edbi AK t\u1ea1i F . a) Ch\u1ee9ng minh n\u0103m \u0111i\u1ec3m A, B, D, E, F c\u00f9ng thu\u1ed9c m\u1ed9t \u0111\u01b0\u1eddng tr\u00f2n. b) G\u1ecdi M l\u00e0 trung \u0111i\u1ec3m BC . Ch\u1ee9ng minh M,H,K th\u1eb3ng h\u00e0ng. c) G\u1ecdi I l\u00e0 trung \u0111i\u1ec3m AB . Ch\u1ee9ng minh D \u0111\u1ed1i x\u1ee9ng v\u1edbi F qua IM . L\u1eddi gi\u1ea3i A E O F H B DM C K a) Ch\u1ee9ng minh n\u0103m \u0111i\u1ec3m A, B, D, E, F c\u00f9ng thu\u1ed9c m\u1ed9t \u0111\u01b0\u1eddng tr\u00f2n. T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N 7","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH Ta c\u00f3 ADB = AFB = AEB = 90\uf0b0 \uf0de n\u0103m \u0111i\u1ec3m A, B, D, E, F c\u00f9ng thu\u1ed9c m\u1ed9t \u0111\u01b0\u1eddng tr\u00f2n. b) ACK = ABK = 90\uf0b0 (g\u00f3c nt ch\u1eafn n\u1eeda O ) BH \/ \/ CK (c\u00f9ng vu\u00f4ng g\u00f3c AC ) CH \/ \/ BK (c\u00f9ng vu\u00f4ng g\u00f3c AB ) N\u00ean BHCK l\u00e0 h\u00ecnh b\u00ecnh h\u00e0nh M\u00e0 M l\u00e0 trung \u0111i\u1ec3m BC N\u00ean M l\u00e0 trung \u0111i\u1ec3m HK Suy ra M, H, K th\u1eb3ng h\u00e0ng. A I E H F O L BD M C K c) G\u1ecdi L l\u00e0 giao \u0111i\u1ec3m IM v\u00e0 DF MDF = BAK ( ABDF n\u1ed9i ti\u1ebfp) BAK = BCK ( 2 g\u00f3c n\u1ed9i ti\u1ebfp c\u00f9ng ch\u1eafn BK ) \uf0de MDF = BCK (2) Ch\u1ee9ng minh \u0111\u01b0\u1ee3c: IM l\u00e0 \u0111\u01b0\u1eddng trung b\u00ecnh \uf044ABC . \uf0de IM \/ \/ AC \uf0de IMD = ACB (\u0111\u1ed3ng v\u1ecb) (2) M\u00e0 BCK + ACB = ACK = 900 (gnt ch\u1eafn n\u1eeda \u0111\u01b0\u1eddng tr\u00f2n) (3) T\u1eeb (1), (2), (3) \uf0de MDF + IMD = 900 \uf0de \uf044DLM vu\u00f4ng t\u1ea1i L \uf0de IM \u22a5 DF Ta c\u00f3 I l\u00e0 trung \u0111i\u1ec3m AB . \uf0de I l\u00e0 t\u00e2m \u0111\u01b0\u1eddng tr\u00f2n \u0111\u01b0\u1eddng k\u00ednh AB . T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N 8","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH \uf0de ID = IF (b\u00e1n k\u00ednh (I) ) \uf0de \uf044IDF c\u00e2n t\u1ea1i I . M\u00e0 IM l\u00e0 \u0111\u01b0\u1eddng cao \uf044IDF (IM \u22a5 DF) \uf0de IM l\u00e0 trung tr\u1ef1c c\u1ee7a AF . \uf0de D \u0111\u1ed1i x\u1ee9ng v\u1edbi F qua IM . ----H\u1ebeT--- T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N 9","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH S\u00d4\u00db GD&\u00d1T TP HO\u00c0 CH\u00cd MINH \u00d1E\u00c0 THAM KHA\u00dbO TUYE\u00c5N SINH 10 PHO\u00d8NG G\u00d1&\u00d1T QUA\u00c4N PHU\u00d9 NHUA\u00c4N NA\u00caM HO\u00cfC: 2023 - 2024 M\u00d4N: TO\u00c1N 9 \u0110\u1ec0 THAM KH\u1ea2O M\u00c3 \u0110\u1ec0: Qu\u1eadn Ph\u00fa Nhu\u1eadn - 2 \u0110\u00ea thi g\u1ed3m 8 c\u00e2u ho\u0309i t\u01b0\u0323 lu\u00e2\u0323n. Th\u01a1\u0300i gian: 120 phu\u0301t (kh\u00f4ng k\u00ea\u0309 th\u01a1\u0300i gian pha\u0301t \u0111\u00ea\u0300) C\u00e2u 1. (1,5 \u0111i\u1ec3m). Cho Parabol (P) : y = \u2212 1 x2 v\u00e0 \u0111\u01b0\u1eddng th\u1eb3ng (d) : y = \u22122x + 2 . 2 a) V\u1ebd (P) v\u00e0 (d) tr\u00ean c\u00f9ng m\u1ed9t m\u1eb7t ph\u1eb3ng t\u1ecda \u0111\u1ed9. b) T\u00ecm t\u1ecda \u0111\u1ed9 giao \u0111i\u1ec3m c\u1ee7a (P) v\u00e0 (d) b\u1eb1ng ph\u00e9p t\u00ednh. C\u00e2u 2. (1 \u0111i\u1ec3m). Cho ph\u01b0\u01a1ng tr\u00ecnh 2x2 \u2212 3x \u2212 3 = 0 c\u00f3 hai nghi\u1ec7m l\u00e0 x1 v\u00e0 x2 . T\u00ednh gi\u00e1 tr\u1ecb bi\u1ec3u th\u1ee9c sau: B = x12x2 + x22x1 . L\u01b0u \u00fd: T\u1eeb b\u00e0i n\u00e0y, c\u00e1c s\u1ed1 li\u1ec7u t\u00ednh to\u00e1n v\u1ec1 \u0111\u1ed9 d\u00e0i khi l\u00e0m tr\u00f2n (n\u1ebfu c\u00f3) l\u1ea5y \u0111\u1ebfn m\u1ed9t ch\u1eef s\u1ed1 th\u1eadp ph\u00e2n, s\u1ed1 \u0111o g\u00f3c l\u00e0m tr\u00f2n \u0111\u1ebfn ph\u00fat. C\u00e2u 3. (1 \u0111i\u1ec3m). Th\u00e1ng th\u1ee9 nh\u1ea5t hai t\u1ed5 s\u1ea3n xu\u1ea5t \u0111\u01b0\u1ee3c 800 chi ti\u1ebft m\u00e1y. Th\u00e1ng th\u1ee9 hai t\u1ed5 I v\u01b0\u1ee3t m\u1ee9c 10% v\u00e0 t\u1ed5 II v\u01b0\u1ee3t m\u1ee9c 20% so v\u1edbi th\u00e1ng th\u1ee9 nh\u1ea5t, v\u00ec v\u1eady hai t\u1ed5 \u0111\u00e3 s\u1ea3n xu\u1ea5t \u0111\u01b0\u1ee3c 910 chi ti\u1ebft m\u00e1y. H\u1ecfi th\u00e1ng th\u1ee9 nh\u1ea5t m\u1ed7i t\u1ed5 s\u1ea3n xu\u1ea5t \u0111\u01b0\u1ee3c bao nhi\u00eau chi ti\u1ebft m\u00e1y. C\u00e2u 4. (0,75 \u0111i\u1ec3m). M\u1ed9t ng\u01b0\u1eddi g\u1eedi ti\u1ebft ki\u1ec7m 300 tri\u1ec7u \u0111\u1ed3ng v\u00e0o t\u00e0i kho\u1ea3n ng\u00e2n h\u00e0ng Nam \u00c1. C\u00f3 2 s\u1ef1 l\u1ef1a ch\u1ecdn: L\u1ef1a ch\u1ecdn 1: Ng\u01b0\u1eddi g\u1eedi c\u00f3 th\u1ec3 nh\u1eadn \u0111\u01b0\u1ee3c l\u00e3i su\u1ea5t 7% m\u1ed9t n\u0103m. L\u1ef1a ch\u1ecdn 2 : Ng\u01b0\u1eddi g\u1eedi nh\u1eadn ti\u1ec1n th\u01b0\u1edfng ngay l\u00e0 4 tri\u1ec7u v\u1edbi l\u00e3i su\u1ea5t 6% m\u1ed9t n\u0103m. Ng\u01b0\u1eddi g\u1eedi n\u00ean ch\u1ecdn l\u1ef1a ch\u1ecdn n\u00e0o \u0111\u1ec3 nh\u1eadn \u0111\u01b0\u1ee3c ti\u1ec1n l\u00e3i cao h\u01a1n sau th\u1eddi h\u1ea1n 1 n\u0103m? Sau th\u1eddi h\u1ea1n 2 n\u0103m? C\u00e2u 5. (1 \u0111i\u1ec3m). T\u1eeb nh\u00e0 b\u1ea1n Nam \u0111\u1ebfn tr\u01b0\u1eddng ph\u1ea3i qua m\u1ed9t kh\u00fac s\u00f4ng r\u1ed9ng 100 m (T\u1eeb B \u0111\u1ebfn A ). Nh\u01b0ng th\u1ef1c t\u1ebf khi b\u1ea1n Nam \u0111i \u0111\u00f2 qua s\u00f4ng th\u00ec d\u00f2ng n\u01b0\u1edbc \u0111\u1ea9y xi\u00ean chi\u1ebfc \u0111\u00f2 m\u1ed9t g\u00f3c 40\uf0b0 (\u0111\u1ebfn \u0111i\u1ec3m C \u1edf b\u1edd b\u00ean kia). T\u1eeb C b\u1ea1n Nam ph\u1ea3i \u0111i b\u1ed9 \u0111\u1ebfn tr\u01b0\u1eddng (\u0111i\u1ec3m D ) m\u1ea5t th\u1eddi gian g\u1ea5p \u0111\u00f4i khi \u0111i t\u1eeb A . H\u1ecfi qu\u00e3ng \u0111\u01b0\u1eddng m\u00e0 Nam \u0111i b\u1ed9 \u0111\u1ebfn tr\u01b0\u1eddng l\u00e0 bao nhi\u00eau m\u00e9t? Bi\u1ebft r\u1eb1ng v\u1eadn t\u1ed1c Nam \u0111i b\u1ed9 l\u00e0 kh\u00f4ng thay \u0111\u1ed5i. T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N 1","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH C\u00e2u 6. (1 \u0111i\u1ec3m). Ng\u01b0\u1eddi ta tr\u1ed9n 8g ch\u1ea5t l\u1ecfng n\u00e0y v\u1edbi 6g ch\u1ea5t l\u1ecfng kh\u00e1c c\u00f3 kh\u1ed1i l\u01b0\u1ee3ng ri\u00eang l\u1edbn h\u01a1n n\u00f3 l\u00e0 0,2 g \/ cm3 \u0111\u1ec3 \u0111\u01b0\u1ee3c h\u1ed7n h\u1ee3p c\u00f3 kh\u1ed1i l\u01b0\u1ee3ng ri\u00eang 0,7 g \/ cm3 . T\u00ecm kh\u1ed1i l\u01b0\u1ee3ng ri\u00eang c\u1ee7a m\u1ed7i ch\u1ea5t l\u1ecfng. C\u00e2u 7. (1 \u0111i\u1ec3m). B\u1ea1n H\u01b0ng l\u00e0m vi\u1ec7c t\u1ea1i nh\u00e0 h\u00e0ng n\u1ecd, b\u1ea1n \u1ea5y \u0111\u01b0\u1ee3c tr\u1ea3 2 tri\u1ec7u \u0111\u1ed3ng cho 40 gi\u1edd l\u00e0m vi\u1ec7c t\u1ea1i qu\u00e1n trong m\u1ed9t tu\u1ea7n. M\u1ed7i gi\u1edd l\u00e0m th\u00eam trong tu\u1ea7n b\u1ea1n \u0111\u01b0\u1ee3c tr\u1ea3 b\u1eb1ng 1 1 s\u1ed1 ti\u1ec1n m\u00e0 2 m\u1ed7i gi\u1edd b\u1ea1n \u1ea5y ki\u1ebfm \u0111\u01b0\u1ee3c trong 40 gi\u1edd \u0111\u1ea7u. N\u1ebfu trong tu\u1ea7n \u0111\u00f3 b\u1ea1n Hung \u0111\u01b0\u1ee3c tr\u1ea3 2,3 tri\u1ec7u \u0111\u1ed3ng th\u00ec b\u1ea1n \u1ea5y \u0111\u00e3 ph\u1ea3i l\u00e0m th\u00eam bao nhi\u00eau gi\u1edd ? C\u00e2u 8. (3 \u0111i\u1ec3m) Cho \uf044ABC nh\u1ecdn n\u1ed9i ti\u1ebfp trong (O) . L\u1ea5y D \uf0ce BC nh\u1ecf, k\u1ebb d\u00e2y AE \/\/ BC , DE c\u1eaft BC t\u1ea1i F . V\u1ebd DH , DK , DI l\u1ea7n l\u01b0\u1ee3t vu\u00f4ng g\u00f3c v\u1edbi c\u00e1c c\u1ea1nh BC , AC , AB . a) Ch\u1ee9ng minh \uf044BDF \uf044ADC , \uf044DCF \uf044DAB . b) Ch\u00fang minh ba \u0111i\u1ec3m H , I \u00b8 K th\u1eb3ng h\u00e0ng . c) Ch\u1ee9ng minh BC = AB + AC . DH DI DK ----H\u1ebeT--- T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N 2","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH H\u01af\u1edaNG D\u1eaaN GI\u1ea2I C\u00e2u 1. (1,5 \u0111i\u1ec3m). Cho Parabol (P) : y = \u2212 1 x2 v\u00e0 \u0111\u01b0\u1eddng th\u1eb3ng (d) : y = \u22122x + 2 . 2 a) V\u1ebd (P) v\u00e0 (d) tr\u00ean c\u00f9ng m\u1ed9t m\u1eb7t ph\u1eb3ng t\u1ecda \u0111\u1ed9. b) T\u00ecm t\u1ecda \u0111\u1ed9 giao \u0111i\u1ec3m c\u1ee7a (P) v\u00e0 (d) b\u1eb1ng ph\u00e9p t\u00ednh. L\u1eddi gi\u1ea3i a) BGT: x \u22124 \u22122 0 2 4 y = \u2212 1 x2 \u22128 \u22122 0 \u22122 \u22128 2 x 12 y = \u22122x + 2 0 \u22122 b) T\u00ecm t\u1ecda \u0111\u1ed9 giao \u0111i\u1ec3m c\u1ee7a (P) v\u00e0 (d) b\u1eb1ng ph\u00e9p t\u00ednh. 3 Ph\u01b0\u01a1ng tr\u00ecnh ho\u00e0nh \u0111\u1ed9 giao \u0111i\u1ec3m c\u1ee7a (P) v\u00e0 (d) : \u2212 1 x2 = \u22122x + 2 2 \uf0db x2 \u2212 4x + 4 = 0 \uf0dbx=2 Thay x = 2 v\u00e0o y = \u22122x + 2 , ta \u0111\u01b0\u1ee3c: y = \u22122.2 + 2 = \u22122 . V\u1eady (2; \u2212 2) l\u00e0 hai giao \u0111i\u1ec3m c\u1ea7n t\u00ecm. T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH C\u00e2u 2. (1 \u0111i\u1ec3m). Cho ph\u01b0\u01a1ng tr\u00ecnh 2x2 \u2212 3x \u2212 3 = 0 c\u00f3 hai nghi\u1ec7m l\u00e0 x1 v\u00e0 x2 . T\u00ednh gi\u00e1 tr\u1ecb bi\u1ec3u th\u1ee9c sau: B = x12x2 + x22x1 . L\u1eddi gi\u1ea3i V\u00ec \uf044 = b2 \u2212 4ac = (\u22123)2 \u2212 4.2.(\u22123) = 33 \uf03e 0 N\u00ean ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m ph\u00e2n bi\u1ec7t x1 , x2 . \uf0ec\uf0ef\uf0efS = x1 + x2 = \u2212b = 3 \uf0ed = x1 a2 Theo \u0111\u1ecbnh l\u00ed Vi-et, ta c\u00f3: \uf0ef\uf0ef\uf0eeP c =\u22123 a .x2 = 2 ( )Ta c\u00f3: A = x12x2 + x22x1 = x1x2 =\u22123.3 =\u22129 . x1 + x2 22 4 L\u01b0u \u00fd: T\u1eeb b\u00e0i n\u00e0y, c\u00e1c s\u1ed1 li\u1ec7u t\u00ednh to\u00e1n v\u1ec1 \u0111\u1ed9 d\u00e0i khi l\u00e0m tr\u00f2n (n\u1ebfu c\u00f3) l\u1ea5y \u0111\u1ebfn m\u1ed9t ch\u1eef s\u1ed1 th\u1eadp ph\u00e2n, s\u1ed1 \u0111o g\u00f3c l\u00e0m tr\u00f2n \u0111\u1ebfn ph\u00fat. C\u00e2u 3. (1 \u0111i\u1ec3m). Th\u00e1ng th\u1ee9 nh\u1ea5t hai t\u1ed5 s\u1ea3n xu\u1ea5t \u0111\u01b0\u1ee3c 800 chi ti\u1ebft m\u00e1y. Th\u00e1ng th\u1ee9 hai t\u1ed5 I v\u01b0\u1ee3t m\u1ee9c 10% v\u00e0 t\u1ed5 II v\u01b0\u1ee3t m\u1ee9c 20% so v\u1edbi th\u00e1ng th\u1ee9 nh\u1ea5t, v\u00ec v\u1eady hai t\u1ed5 \u0111\u00e3 s\u1ea3n xu\u1ea5t \u0111\u01b0\u1ee3c 910 chi ti\u1ebft m\u00e1y. H\u1ecfi th\u00e1ng th\u1ee9 nh\u1ea5t m\u1ed7i t\u1ed5 s\u1ea3n xu\u1ea5t \u0111\u01b0\u1ee3c bao nhi\u00eau chi ti\u1ebft m\u00e1y. L\u1eddi gi\u1ea3i G\u1ecdi x , y l\u1ea7n l\u01b0\u1ee3t l\u00e0 s\u1ed1 chi ti\u1ebft m\u00e1y t\u1ed5 I v\u00e0 t\u1ed5 II s\u1ea3n xu\u1ea5t \u0111\u01b0\u1ee3c trong th\u00e1ng th\u1ee9 nh\u1ea5t (x, y \uf0ce *) V\u00ec trong th\u00e1ng th\u1ee9 nh\u1ea5t, c\u1ea3 hai t\u1ed5 s\u1ea3n xu\u1ea5t \u0111\u01b0\u1ee3c 800 chi ti\u1ebft m\u00e1y n\u00ean ta c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh: x + y = 800 (1) Th\u00e1ng th\u1ee9 hai, c\u1ea3 hai t\u1ed5 \u0111\u00e3 v\u01b0\u1ee3t m\u1ee9c v\u00e0 s\u1ea3n xu\u1ea5t \u0111\u01b0\u1ee3c 910 chi ti\u1ebft m\u00e1y, n\u00ean ta c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh: x(1+ 10%) + y(1+ 20%) = 910 \uf0db 1,1x + 1,2y = 910 (2) T\u1eeb (1) v\u00e0 (2) ta c\u00f3 h\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh: \uf0ecx + y = 800 = 910 \uf0db \uf0ec\uf0efx = 500(n) . \uf0ed\uf0ee1,1x + 1,2y \uf0ed = 300 ( n) \uf0ef\uf0eey V\u1eady th\u00e1ng th\u1ee9 nh\u1ea5t, t\u1ed5 I s\u1ea3n xu\u1ea5t \u0111\u01b0\u1ee3c 500 chi ti\u1ebft m\u00e1y, t\u1ed5 II s\u1ea3n xu\u1ea5t \u0111\u01b0\u1ee3c 300 chi ti\u1ebft m\u00e1y. C\u00e2u 4. (0,75 \u0111i\u1ec3m). M\u1ed9t ng\u01b0\u1eddi g\u1eedi ti\u1ebft ki\u1ec7m 300 tri\u1ec7u \u0111\u1ed3ng v\u00e0o t\u00e0i kho\u1ea3n ng\u00e2n h\u00e0ng Nam \u00c1. C\u00f3 2 s\u1ef1 l\u1ef1a ch\u1ecdn: L\u1ef1a ch\u1ecdn 1: Ng\u01b0\u1eddi g\u1eedi c\u00f3 th\u1ec3 nh\u1eadn \u0111\u01b0\u1ee3c l\u00e3i su\u1ea5t 7% m\u1ed9t n\u0103m. L\u1ef1a ch\u1ecdn 2 : Ng\u01b0\u1eddi g\u1eedi nh\u1eadn ti\u1ec1n th\u01b0\u1edfng ngay l\u00e0 4 tri\u1ec7u v\u1edbi l\u00e3i su\u1ea5t 6% m\u1ed9t n\u0103m. Ng\u01b0\u1eddi g\u1eedi n\u00ean ch\u1ecdn l\u1ef1a ch\u1ecdn n\u00e0o \u0111\u1ec3 nh\u1eadn \u0111\u01b0\u1ee3c ti\u1ec1n l\u00e3i cao h\u01a1n sau th\u1eddi h\u1ea1n 1 n\u0103m? Sau th\u1eddi h\u1ea1n 2 n\u0103m? T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N 4","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH L\u1eddi gi\u1ea3i Tr\u01b0\u1eddng h\u1ee3p kh\u00e1ch h\u00e0ng g\u1eedi k\u1ef3 h\u1ea1n 1 n\u0103m: T\u1ed5ng s\u1ed1 ti\u1ec1n nh\u1eadn \u0111\u01b0\u1ee3c sau 1 n\u0103m v\u1edbi l\u1ef1a ch\u1ecdn 1: 300.(1 + 7%) = 321 tri\u1ec7u \u0111\u1ed3ng. T\u1ed5ng s\u1ed1 ti\u1ec1n nh\u1eadn \u0111\u01b0\u1ee3c sau 1 n\u0103m v\u1edbi l\u1ef1a ch\u1ecdn 2 : 300.(1 + 6%) + 4 = 322 tri\u1ec7u \u0111\u1ed3ng. V\u1eady kh\u00e1ch h\u00e0ng n\u00ean ch\u1ecdn l\u1ef1a ch\u1ecdn 2 s\u1ebd nh\u1eadn \u0111\u01b0\u1ee3c t\u1ed5ng s\u1ed1 ti\u1ec1n nhi\u1ec1u h\u01a1n. Tr\u01b0\u1eddng h\u1ee3p kh\u00e1ch h\u00e0ng g\u1eedi k\u1ef3 h\u1ea1n 2 n\u0103m: T\u1ed5ng s\u1ed1 ti\u1ec1n nh\u1eadn \u0111\u01b0\u1ee3c sau 2 n\u0103m v\u1edbi l\u1ef1a ch\u1ecdn 1: 321.(1 + 7%) = 343,47 tri\u1ec7u \u0111\u1ed3ng. T\u1ed5ng s\u1ed1 ti\u1ec1n nh\u1eadn \u0111\u01b0\u1ee3c sau 2 n\u0103m v\u1edbi l\u1ef1a ch\u1ecdn 2 : 300.(1+ 6%)(1+ 6%) + 4 = 341,08 tri\u1ec7u \u0111\u1ed3ng. V\u1eady kh\u00e1ch h\u00e0ng n\u00ean ch\u1ecdn l\u1ef1a ch\u1ecdn 1 s\u1ebd nh\u1eadn \u0111\u01b0\u1ee3c t\u1ed5ng s\u1ed1 ti\u1ec1n nhi\u1ec1u h\u01a1n. C\u00e2u 5. (1 \u0111i\u1ec3m). T\u1eeb nh\u00e0 b\u1ea1n Nam \u0111\u1ebfn tr\u01b0\u1eddng ph\u1ea3i qua m\u1ed9t kh\u00fac s\u00f4ng r\u1ed9ng 100 m (T\u1eeb B \u0111\u1ebfn A ). Nh\u01b0ng th\u1ef1c t\u1ebf khi b\u1ea1n Nam \u0111i \u0111\u00f2 qua s\u00f4ng th\u00ec d\u00f2ng n\u01b0\u1edbc \u0111\u1ea9y xi\u00ean chi\u1ebfc \u0111\u00f2 m\u1ed9t g\u00f3c 40\uf0b0 (\u0111\u1ebfn \u0111i\u1ec3m C \u1edf b\u1edd b\u00ean kia). T\u1eeb C b\u1ea1n Nam ph\u1ea3i \u0111i b\u1ed9 \u0111\u1ebfn tr\u01b0\u1eddng (\u0111i\u1ec3m D ) m\u1ea5t th\u1eddi gian g\u1ea5p \u0111\u00f4i khi \u0111i t\u1eeb A . H\u1ecfi qu\u00e3ng \u0111\u01b0\u1eddng m\u00e0 Nam \u0111i b\u1ed9 \u0111\u1ebfn tr\u01b0\u1eddng l\u00e0 bao nhi\u00eau m\u00e9t? Bi\u1ebft r\u1eb1ng v\u1eadn t\u1ed1c Nam \u0111i b\u1ed9 l\u00e0 kh\u00f4ng thay \u0111\u1ed5i. L\u1eddi gi\u1ea3i X\u00e9t \uf044ABC vu\u00f4ng t\u1ea1i A , ta c\u00f3: tan ABC = AC \uf0de AC = AB.tan 40\uf0b0 = 100.tan 40\uf0b0 \uf0bb 83,9 m . AB V\u1edbi v\u1eadn t\u1ed1c \u0111i b\u1ed9 tr\u00ean c\u00e1c tuy\u1ebfn \u0111\u01b0\u1eddng kh\u00e1c nhau kh\u00f4ng \u0111\u1ed5i v\u00e0 th\u1eddi gian \u0111i t\u1eeb C \u2192 D g\u1ea5p \u0111\u00f4i th\u1eddi gian \u0111i t\u1eeb A \u2192 D n\u00ean DC = 2AD . X\u00e9t \uf044ACD vu\u00f4ng t\u1ea1i A , ta c\u00f3: DC2 = AD2 + AC2 (\u0110L Pytago) T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N 5","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH \uf0de DC 2 = \uf0e6 DC \uf0f62 + 83,92 \uf0e7\uf0e8 2 \uf0f7 \uf0f8 \uf0de 3 DC2 = 83,92 4 \uf0de DC = 83,92 : 3 \uf0bb 96,9 m. 4 V\u1eady Nam \u0111\u00e3 \u0111i b\u1ed9 qu\u00e3ng \u0111\u01b0\u1eddng DC \uf0bb 96,9 m . C\u00e2u 6. (1 \u0111i\u1ec3m). Ng\u01b0\u1eddi ta tr\u1ed9n 8g ch\u1ea5t l\u1ecfng n\u00e0y v\u1edbi 6g ch\u1ea5t l\u1ecfng kh\u00e1c c\u00f3 kh\u1ed1i l\u01b0\u1ee3ng ri\u00eang l\u1edbn h\u01a1n n\u00f3 l\u00e0 0,2 g \/ cm3 \u0111\u1ec3 \u0111\u01b0\u1ee3c h\u1ed7n h\u1ee3p c\u00f3 kh\u1ed1i l\u01b0\u1ee3ng ri\u00eang 0,7 g \/ cm3 . T\u00ecm kh\u1ed1i l\u01b0\u1ee3ng ri\u00eang c\u1ee7a m\u1ed7i ch\u1ea5t l\u1ecfng. L\u1eddi gi\u1ea3i ( ) ( )G\u1ecdi V1 , V2 cm3 l\u1ea7n l\u01b0\u1ee3t l\u00e0 th\u1ec3 t\u00edch c\u1ee7a m\u1ed7i ch\u1ea5t l\u1ecfng V1 ,V2 \uf03e 0 Kh\u1ed1i l\u01b0\u1ee3ng ri\u00eang c\u1ee7a m\u1ed7i ch\u1ea5t l\u1ecfng: D1 = m1 = 8 v\u00e0 D2 = m2 =6. V1 V1 V2 V2 V\u00ec kh\u1ed1i l\u01b0\u1ee3ng ri\u00eang D1 l\u1edbn h\u01a1n D2 n\u00ean ta c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh: D1 D2 0,2 86 0, 2 V1 V2 1 V\u00ec h\u1ed7n h\u1ee3p sau khi tr\u1ed9n c\u00f3 kh\u1ed1i l\u01b0\u1ee3ng ri\u00eang l\u00e0 0, 7g \/ cm3 n\u00ean ta c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh: D m1 m2 0, 7 V1 V2 m1 m2 20 V2 20 V1 2 V1 V2 0, 7 Thay 2 v\u00e0o 1 , ta \u0111\u01b0\u1ee3c: 8 6 0, 2 V1 20 V1 8 20 V1 6V1 0, 2V1. 20 V1 160 8V1 6V1 4V1 0, 2V12 0, 2V12 14V1 160 0 V1 80 V2 20 80 60 l V1 10 V2 20 10 10 n V\u1eady kh\u1ed1i l\u01b0\u1ee3ng ri\u00eang c\u1ee7a m\u1ed7i ch\u1ea5t l\u1ecfng: D1 8 0, 8g \/ cm3 , D2 6 0, 6g \/ cm3 . 10 10 T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N 6","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH C\u00e2u 7. (1 \u0111i\u1ec3m). B\u1ea1n H\u01b0ng l\u00e0m vi\u1ec7c t\u1ea1i nh\u00e0 h\u00e0ng n\u1ecd, b\u1ea1n \u1ea5y \u0111\u01b0\u1ee3c tr\u1ea3 2 tri\u1ec7u \u0111\u1ed3ng cho 40 gi\u1edd l\u00e0m vi\u1ec7c t\u1ea1i qu\u00e1n trong m\u1ed9t tu\u1ea7n. M\u1ed7i gi\u1edd l\u00e0m th\u00eam trong tu\u1ea7n b\u1ea1n \u0111\u01b0\u1ee3c tr\u1ea3 b\u1eb1ng 1 1 s\u1ed1 ti\u1ec1n m\u00e0 2 m\u1ed7i gi\u1edd b\u1ea1n \u1ea5y ki\u1ebfm \u0111\u01b0\u1ee3c trong 40 gi\u1edd \u0111\u1ea7u. N\u1ebfu trong tu\u1ea7n \u0111\u00f3 b\u1ea1n H\u01b0ng \u0111\u01b0\u1ee3c tr\u1ea3 2,3 tri\u1ec7u \u0111\u1ed3ng th\u00ec b\u1ea1n \u1ea5y \u0111\u00e3 ph\u1ea3i l\u00e0m th\u00eam bao nhi\u00eau gi\u1edd ? L\u1eddi gi\u1ea3i S\u1ed1 ti\u1ec1n b\u1ea1n H\u01b0ng \u0111\u01b0\u1ee3c tr\u1ea3 trong 1 gi\u1edd l\u00e0m vi\u1ec7c trong tu\u1ea7n: 2 : 40 0, 05 tri\u1ec7u \u0111\u1ed3ng 50 000 \u0111\u1ed3ng. S\u1ed1 ti\u1ec1n m\u1ed9t gi\u1edd l\u00e0m th\u00eam b\u1ea1n H\u01b0ng nh\u1eadn \u0111\u01b0\u1ee3c trong tu\u1ea7n: 50 000.1 1 75 000 \u0111\u1ed3ng. 2 S\u1ed1 gi\u1edd l\u00e0m th\u00eam trong tu\u1ea7n c\u1ee7a b\u1ea1n H\u01b0ng: 2 300 000 2 000 000 : 75 000 4 gi\u1edd. C\u00e2u 8. (3 \u0111i\u1ec3m) Cho \uf044ABC nh\u1ecdn n\u1ed9i ti\u1ebfp trong (O) . L\u1ea5y D \uf0ce BC nh\u1ecf, k\u1ebb d\u00e2y AE \/\/ BC , DE c\u1eaft BC t\u1ea1i F . V\u1ebd DH , DK , DI l\u1ea7n l\u01b0\u1ee3t vu\u00f4ng g\u00f3c v\u1edbi c\u00e1c c\u1ea1nh BC , AC , AB . a) Ch\u1ee9ng minh \uf044BDF \uf044ADC , \uf044DCF \uf044DAB . b) Ch\u00fang minh ba \u0111i\u1ec3m H , I \u00b8 K th\u1eb3ng h\u00e0ng . c) Ch\u1ee9ng minh BC = AB + AC . DH DI DK L\u1eddi gi\u1ea3i a) Ch\u1ee9ng minh \uf044BDF \uf044ADC , \uf044DCF \uf044DAB . 7 Trong (O) , ta c\u00f3: AE \/\/ BC (gt) s\u00f1 AB = s\u00f1CE . ( )M\u00e0: BFD = 1 s\u00f1BD = s\u00f1CE (g\u00f3c c\u00f3 \u0111\u1ec9nh \u1edf trong \u0111\u01b0\u1eddng tr\u00f2n ch\u1eafn BD v\u00e0 CE ) 2 ( )N\u00ean: BFD = 1 s\u00f1BD = s\u00f1 AB = 1 s\u00f1 AD 22 T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH ( )L\u1ea1i c\u00f3: ACD == 1 s\u00f1 AD (gnt O ch\u1eafn AD ) 2 Suy ra: BFD = ACD X\u00e9t \uf044BDF v\u00e0 \uf044ADC , ta c\u00f3: DBF = DAC ( 2 gnt (O) c\u00f9ng ch\u1eafn DC ) BFD = ACD (cmt) \uf0de \uf044BDF \uf044ADC (g.g). X\u00e9t \uf044DCF v\u00e0 \uf044DAB , ta c\u00f3: DCF = DAB ( 2 gnt (O) c\u00f9ng ch\u1eafn BD ) CDF = ADB ( 2 gnt (O) c\u00f9ng ch\u1eafn 2 cung AB = CE ) \uf0de \uf044DCF \uf044DAB (g.g). b) Ch\u00fang minh ba \u0111i\u1ec3m H , I \u00b8 K th\u1eb3ng h\u00e0ng . X\u00e9t t\u1ee9 gi\u00e1c BHDI , c\u00f3: \uf0ec\uf0efBHD = 90\uf0b0(DH \u22a5 BC) \uf0ed \uf0ef\uf0eeBID = 90\uf0b0 ( DI \u22a5 AB) \uf0de BHD + BID = 180\uf0b0 \uf0de T\u1ee9 gi\u00e1c BHDI n\u1ed9i ti\u1ebfp v\u00ec c\u00f3 hai g\u00f3c \u0111\u1ed1i b\u00f9 nhau. X\u00e9t t\u1ee9 gi\u00e1c DHKC , c\u00f3: \uf0ef\uf0ecDHC = 90\uf0b0(DH \u22a5 BC) \uf0ed \uf0ef\uf0eeDKC = 90\uf0b0( DK \u22a5 AC ) \uf0de DHC = DKC (= 90\uf0b0) \uf0de T\u1ee9 gi\u00e1c DHKC n\u1ed9i ti\u1ebfp v\u00ec c\u00f3 hai \u0111\u1ec9nh k\u1ec1 c\u00f9ng nh\u00ecn m\u1ed9t c\u1ea1nh d\u01b0\u1edbi hai g\u00f3c b\u1eb1ng nhau. \uf0de DHK + DCL = 180\uf0b0 M\u00e0: DBI = DCK (tg ABDC n\u1ed9i ti\u1ebfp) V\u00e0 DBI = DHI (tg BIDH n\u1ed9i ti\u1ebfp) N\u00ean: DHK + DHI = 180\uf0b0 \uf0de KHI = 180\uf0b0 . T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N 8"]