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 H\u01af\u1edaNG D\u1eaaN GI\u1ea2I C\u00e2u 1. (1,5 \u0111i\u1ec3m). Cho parabol (P) : y = 1 x2 v\u00e0 \u0111\u01b0\u1eddng th\u1eb3ng (d) : y = x + 4 . 2 a) V\u1ebd (P) v\u00e0 (d) tr\u00ean c\u00f9ng m\u1ed9t m\u1eb7t ph\u1eb3ng t\u1ecda \u0111\u1ed9 Oxy. b) X\u00e1c \u0111\u1ecbnh t\u1ecda \u0111\u1ed9 giao \u0111i\u1ec3m c\u1ee7a (P) v\u00e0 (d) b\u1eb1ng ph\u00e9p t\u00ednh. L\u1eddi gi\u1ea3i y (P) 9 a) V\u1ebd \u0111\u1ed3 th\u1ecb (P) v\u00e0 (d) tr\u00ean c\u00f9ng h\u1ec7 tr\u1ee5c t\u1ecda \u0111\u1ed9. 8 BGT: 7 (d) 6 x \u22124 \u22122 0 2 4 5 4 y = 1 x2 8 2 0 2 8 3 2 2 1 x 0 \u22122 y=x+4 4 2 -4 -3 -2 -1 O -1 1 2 3 4x -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) : 1 x2 = x + 4 2 \uf0db 1 x2 \u2212 x \u2212 4 = 0 2 \uf0db \uf0e9x = 4 \uf0ea\uf0ebx = \u22122 Thay x = 4 v\u00e0o y = 1 x2 , ta \u0111\u01b0\u1ee3c: y = 1 .42 = 8 . 22 Thay x = \u22122 v\u00e0o y = 1 x2 , ta \u0111\u01b0\u1ee3c: y = 1 (\u22122)2 = 2 . 22 V\u1eady (4;8) , (\u22122; 2) l\u00e0 hai giao \u0111i\u1ec3m c\u1ea7n t\u00ecm. C\u00e2u 2. (1 \u0111i\u1ec3m). Cho ph\u01b0\u01a1ng tr\u00ecnh 6x2 + 6x \u2212 13 = 0 c\u00f3 2 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= x1 \u2212 x2 \u22121 + x2 \u2212 x1 \u22121 . 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 L\u1eddi gi\u1ea3i V\u00ec \uf044 = b2 \u2212 4ac = 62 \u2212 4.6.(\u221213) = 348 \uf03e 0 N\u00ean ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m ph\u00e2n bi\u1ec7t x1 ,x2 . \uf0ef\uf0ef\uf0ecS = x1 + x2 = \u2212b = \u22121 \uf0ed = x1 a Theo \u0111\u1ecbnh l\u00ed Vi-et, ta c\u00f3: \uf0ef\uf0ee\uf0efP c \u221213 a = 6 .x2 = Ta c\u00f3: A = x1 \u2212 x2 \u2212 1 + x2 \u2212 x1 \u2212 1 x2 x1 ( ) ( )A = x1 \u2212 x2 \u2212 1 x1 + x2 \u2212 x1 \u2212 1 x2 x1x2 A = x12 \u2212 x1x2 \u2212 x1 + x22 \u2212 x1x2 \u2212 x2 x1x2 A = x12 + x22 \u2212 2x1x2 \u2212 x1 \u2212 x2 x1x2 A = S2 \u2212 2P \u2212 2P \u2212 S P ( \u22121)2 \u2212 4 \uf0e6 \u221213 \uf0f6 \u2212 (\u22121) \u221264 \uf0e7\uf0e8 6 \uf0f7\uf0f8 A = = \u221213 13 6 C\u00e2u 3. (1 \u0111i\u1ec3m). C\u00f3 hai h\u00e3ng \u0111i\u1ec7n tho\u1ea1i c\u1ed1 \u0111\u1ecbnh t\u00ednh ph\u00ed g\u1ecdi cho c\u00e1c thu\u00ea bao nh\u01b0 sau: H\u00e3ng Thu\u00ea bao (ng\u00e0n \u0111\u1ed3ng\/th\u00e1ng) G\u1ecdi n\u1ed9i h\u1ea1t (ng\u00e0n \u0111\u1ed3ng\/ 30 ph\u00fat) H\u00e3ng A 10 6 H\u00e3ng B 15 5 G\u1ecdi y (ng\u00e0n \u0111\u1ed3ng) l\u00e0 gi\u00e1 ti\u1ec1n m\u00e0 kh\u00e1ch h\u00e0ng ph\u1ea3i tr\u1ea3 sau x l\u1ea7n 30 ph\u00fat x N * Bi\u1ebft c\u01b0\u1edbc ph\u00ed h\u00e0ng th\u00e1ng b\u1eb1ng t\u1ed5ng ti\u1ec1n thu\u00ea bao v\u00e0 c\u01b0\u1edbc ph\u00ed g\u1ecdi n\u1ed9i h\u1ea1t. a) H\u00e3y bi\u1ec3u di\u1ec5n y theo x c\u1ee7a t\u1eebng h\u00e3ng, bi\u1ebft r\u1eb1ng y ax b (a,b l\u00e0 s\u1ed1 x\u00e1c \u0111\u1ecbnh). b) H\u00e3y cho bi\u1ebft v\u1edbi c\u00e1ch t\u00ednh ph\u00ed nh\u01b0 tr\u00ean th\u00ec m\u1ed9t kh\u00e1ch h\u00e0ng m\u1ed7i th\u00e1ng g\u1ecdi b\u00ecnh qu\u00e2n 6 gi\u1edd n\u00ean s\u1eed d\u1ee5ng d\u1ecbch v\u1ee5 c\u1ee7a h\u00e3ng n\u00e0o s\u1ebd r\u1ebb h\u01a1n? L\u1eddi gi\u1ea3i a) H\u00e3ng A : y 6x 10 . H\u00e3ng B : y 5x 15 . b) 6 gi\u1edd = 12 l\u1ea7n 30 ph\u00fat Thay x 12 v\u00e0o y 6x 10 , ta c\u00f3: y 6.12 10 82 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 Thay x 12 v\u00e0o y 5x 15 , ta c\u00f3: y 5.12 15 75 V\u1eady kh\u00e1ch h\u00e0ng n\u00ean s\u1eed d\u1ee5ng d\u1ecbch v\u1ee5 c\u1ee7a h\u00e3ng B s\u1ebd r\u1ebb h\u01a1n v\u00ec 75 ng\u00e0n \u0111\u1ed3ng \uf03c 82 ng\u00e0n \u0111\u1ed3ng. C\u00e2u 4. (1 \u0111i\u1ec3m). M\u1ed9t chi\u1ebfc n\u00f3n l\u00e1 c\u00f3 d\u1ea1ng h\u00ecnh n\u00f3n nh\u01b0 h\u00ecnh b\u00ean: \u0111\u1ed9 d\u00e0i \u0111\u01b0\u1eddng sinh l\u00e0 25 cm, b\u00e1n k\u00ednh \u0111\u01b0\u1eddng tr\u00f2n \u0111\u00e1y l\u00e0 15 cm. T\u00ednh th\u1ec3 t\u00edch c\u1ee7a chi\u1ebfc n\u00f3n; bi\u1ebft V 1 S.h , v\u1edbi V 3 l\u00e0 th\u1ec3 t\u00edch, S l\u00e0 di\u1ec7n t\u00edch \u0111\u00e1y, h l\u00e0 chi\u1ec1u cao c\u1ee7a h\u00ecnh n\u00f3n. L\u1eddi gi\u1ea3i Theo \u0111\u1ec1 b\u00e0i ta c\u00f3 h\u00ecnh v\u1ebd: A 25cm B 15cm H X\u00e9t ABH vu\u00f4ng t\u1ea1i H c\u00f3: AH 2 AB2 BH 2 (\u0111\u1ecbnh l\u00ed Pytago) AH 252 152 20 cm Th\u1ec3 t\u00edch chi\u1ebfc n\u00f3n: V 1 S .h 1 . .152.20 1500 cm3 . 3 3 C\u00e2u 5. (1 \u0111i\u1ec3m). Th\u00e1ng tr\u01b0\u1edbc, hai t\u1ed5 c\u00f4ng nh\u00e2n s\u1ea3n xu\u1ea5t \u0111\u01b0\u1ee3c t\u1ed5ng c\u1ed9ng 750 chi ti\u1ebft m\u00e1y. Do k\u0129 thu\u1eadt \u0111\u01b0\u1ee3c c\u1ea3i ti\u1ebfn, th\u00e1ng n\u00e0y s\u1ed1 l\u01b0\u1ee3ng chi ti\u1ebft m\u00e1y t\u1ed5 1 v\u00e0 t\u1ed5 2 s\u1ea3n xu\u1ea5t l\u1ea7n l\u01b0\u1ee3t t\u0103ng 7% v\u00e0 8% so v\u1edbi th\u00e1ng tr\u01b0\u1edbc, \u0111\u1ea1t t\u1ed5ng c\u1ed9ng 806 chi ti\u1ebft m\u00e1y. H\u1ecfi th\u00e1ng tr\u01b0\u1edbc 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 l\u00e0 s\u1ed1 chi ti\u1ebft m\u00e1y t\u1ed5 1 s\u1ea3n xu\u1ea5t th\u00e1ng tr\u01b0\u1edbc x * y l\u00e0 s\u1ed1 chi ti\u1ebft m\u00e1y t\u1ed5 2 s\u1ea3n xu\u1ea5t th\u00e1ng tr\u01b0\u1edbc y * 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 Th\u00e1ng tr\u01b0\u1edbc, hai t\u1ed5 c\u00f4ng nh\u00e2n s\u1ea3n xu\u1ea5t \u0111\u01b0\u1ee3c t\u1ed5ng c\u1ed9ng 750 chi ti\u1ebft m\u00e1y n\u00ean ta c\u00f3: x y 750 1 Th\u00e1ng n\u00e0y s\u1ed1 l\u01b0\u1ee3ng chi ti\u1ebft m\u00e1y t\u1ed5 1 v\u00e0 t\u1ed5 2 s\u1ea3n xu\u1ea5t l\u1ea7n l\u01b0\u1ee3t t\u0103ng 7% v\u00e0 8% so v\u1edbi th\u00e1ng tr\u01b0\u1edbc, \u0111\u1ea1t t\u1ed5ng c\u1ed9ng 806 chi ti\u1ebft m\u00e1y n\u00ean ta c\u00f3: x 1 7% y 1 8% 806 1, 07x 1, 08y 806 2 x y 750 B\u1ed5 sung: T\u1eeb 1 , 2 ta c\u00f3 h\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh: 1, 07x 1, 08y 806 x 400 y 350 (nh\u1eadn) V\u1eady th\u00e1ng tr\u01b0\u1edbc t\u1ed5 1 s\u1ea3n xu\u1ea5t \u0111\u01b0\u1ee3c 400 chi ti\u1ebft m\u00e1y, t\u1ed5 2 s\u1ea3n xu\u1ea5t \u0111\u01b0\u1ee3c 350 chi ti\u1ebft m\u00e1y. C\u00e2u 6. (0,75 \u0111i\u1ec3m). Ng\u01b0\u1eddi h\u00fat thu\u1ed1c l\u00e1 th\u01b0\u1eddng xuy\u00ean s\u1ebd b\u1ecb gi\u1ea3m tu\u1ed5i th\u1ecd, d\u1ec5 m\u1eafc c\u00e1c lo\u1ea1i b\u1ec7nh nguy hi\u1ec3m nh\u01b0: vi\u00eam ph\u1ed5i, vi\u00eam \u0111\u01b0\u1eddng h\u00f4 h\u1ea5p, ung th\u01b0, \u2026 Gi\u1ea3 s\u1eed khi h\u00fat m\u1ed9t \u0111i\u1ebfu thu\u1ed1c th\u00ec ng\u01b0\u1eddi h\u00fat b\u1ecb gi\u1ea3m 15 ph\u00fat tu\u1ed5i th\u1ecd. T\u00ednh xem m\u1ed9t ng\u01b0\u1eddi h\u00fat thu\u1ed1c trung b\u00ecnh m\u1ed7i ng\u00e0y 2 g\u00f3i trong c\u1ea3 n\u0103m 2023 th\u00ec ng\u01b0\u1eddi \u0111\u00f3 s\u1ebd b\u1ecb gi\u1ea3m bao nhi\u00eau tu\u1ed5i th\u1ecd? Bi\u1ebft r\u1eb1ng m\u1ed7i g\u00f3i thu\u1ed1c c\u00f3 20 \u0111i\u1ebfu thu\u1ed1c l\u00e1. L\u1eddi gi\u1ea3i S\u1ed1 ph\u00fat tu\u1ed5i th\u1ecd ng\u01b0\u1eddi \u0111\u00f3 s\u1ebd b\u1ecb gi\u1ea3m l\u00e0: 15.20.2.365 219000 (ph\u00fat). C\u00e2u 7. (0,75 \u0111i\u1ec3m). K\u00ednh \u0111eo m\u1eaft c\u1ee7a ng\u01b0\u1eddi gi\u00e0 th\u01b0\u1eddng l\u00e0 lo\u1ea1i th\u1ea5u k\u00ednh h\u1ed9i t\u1ee5. B\u1ea1n An \u0111\u00e3 d\u00f9ng m\u1ed9t chi\u1ebfc k\u00ednh c\u1ee7a \u00f4ng ngo\u1ea1i (lo\u1ea1i th\u1ea5u k\u00ednh h\u1ed9i t\u1ee5) \u0111\u1ec3 t\u1ea1o ra h\u00ecnh \u1ea3nh c\u1ee7a m\u1ed9t c\u00e2y n\u1ebfn tr\u00ean m\u1ed9t t\u1ea5m m\u00e0n. X\u00e9t c\u00e2y n\u1ebfn l\u00e0 m\u1ed9t v\u1eadt s\u00e1ng c\u00f3 h\u00ecnh d\u1ea1ng l\u00e0 \u0111o\u1ea1n AB \u0111\u1eb7t vu\u00f4ng g\u00f3c v\u1edbi tr\u1ee5c ch\u00ednh c\u1ee7a m\u1ed9t th\u1ea5u k\u00ednh h\u1ed9i t\u1ee5, c\u00e1ch th\u1ea5u k\u00ednh m\u1ed9t \u0111o\u1ea1n OA 4m . Th\u1ea5u k\u00ednh c\u00f3 quang t\u00e2m O v\u00e0 ti\u00eau \u0111i\u1ec3m F, F ' . V\u1eadt AB cho \u1ea3nh th\u1eadt A ' B ' g\u1ea5p 3 l\u1ea7n AB . T\u00ednh ti\u00eau c\u1ef1 c\u1ee7a th\u1ea5u k\u00ednh. Bi\u1ebft r\u1eb1ng \u0111\u01b0\u1eddng \u0111i c\u1ee7a c\u00e1c tia s\u00e1ng \u0111\u01b0\u1ee3c m\u00f4 t\u1ea3 nh\u01b0 trong h\u00ecnh v\u1ebd tr\u00ean. L\u1eddi gi\u1ea3i 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 BD F' A' AO B' OAB \u0111\u1ed3ng d\u1ea1ng OA ' B ' OA AB OA AB 4 1 OA ' 3.4 12 m OA' A' B ' OA' 3AB OA ' 3 1 OD AB ( ABDO l\u00e0 h\u00ecnh ch\u1eef nh\u1eadt) F 'OD \u0111\u1ed3ng d\u1ea1ng F ' A ' B ' F 'O OD F 'O AB F 'O 1 F 'O F 'A ' F 'A' A'B' F ' A' 3AB F 'A' 3 13 \u00c1p d\u1ee5ng t\u00ednh ch\u1ea5t d\u00e3y t\u1ec9 s\u1ed1 b\u1eb1ng nhau, ta c\u00f3: F 'O F ' A ' F 'O ' F ' A ' OA ' 12 3 13 13 44 F 'O 3 1 F 'O 3 m V\u1eady ti\u00eau c\u1ef1 c\u1ee7a th\u1ea5u k\u00ednh l\u00e0 3 m . C\u00e2u 8. (3 \u0111i\u1ec3m) Cho \uf044ABC nh\u1ecdn ( AB \uf03c AC) . \u0110\u01b0\u1eddng tr\u00f2n t\u00e2m O \u0111\u01b0\u1eddng k\u00ednh BC c\u1eaft AC v\u00e0 AB l\u1ea7n l\u01b0\u1ee3t t\u1ea1i E v\u00e0 F , BE v\u00e0 CF c\u1eaft nhau t\u1ea1i H , AH c\u1eaft BC t\u1ea1i D , EF c\u1eaft CB t\u1ea1i M . a) Ch\u1ee9ng minh AD BC v\u00e0 MF.ME MB.MC . b) Tia FD c\u1eaft \u0111\u01b0\u1eddng tr\u00f2n O t\u1ea1i N ( N kh\u00e1c F ). Ch\u1ee9ng minh t\u1ee9 gi\u00e1c OFMN n\u1ed9i ti\u1ebfp. c) G\u1ecdi I v\u00e0 K l\u1ea7n l\u01b0\u1ee3t l\u00e0 h\u00ecnh chi\u1ebfu c\u1ee7a B v\u00e0 C tr\u00ean EF . Cho BC 8cm ; ABC 75 . T\u00ednh SEIB SCKE . 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 A K E F IH M B DO C N a) Ch\u1ee9ng minh AD BC v\u00e0 MF.ME MB.MC . BEC BFC 90 (g\u00f3c n\u1ed9i ti\u1ebfp ch\u1eafn n\u1eeda \u0111\u01b0\u1eddng tr\u00f2n) \uf0de BE AC;CF AB \uf0de BE,CF l\u00e0 \u0111\u01b0\u1eddng cao c\u1ee7a ABC M\u00e0 H l\u00e0 giao \u0111i\u1ec3m c\u1ee7a BE,CF (gt) \uf0de H l\u00e0 tr\u1ef1c t\u00e2m c\u1ee7a ABC \uf0de AH l\u00e0 \u0111\u01b0\u1eddng cao c\u1ee7a ABC \uf0de AD BC H AD . Ta c\u00f3: 4 \u0111i\u1ec3m B,F,E,C c\u00f9ng thu\u1ed9c O \uf0de BFEC n\u1ed9i ti\u1ebfp. X\u00e9t MBF v\u00e0 MEC c\u00f3: FMB EMC (g\u00f3c chung) MFB MCE ( BFEC n\u1ed9i ti\u1ebfp) \uf0de MBF \u0111\u1ed3ng d\u1ea1ng MEC (g-g) MB MF \uf0de ME MC \uf0de MB.MC ME.MF . 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 b) Ch\u1ee9ng minh t\u1ee9 gi\u00e1c OFMN n\u1ed9i ti\u1ebfp. AFC 90 (CF AB ) ADC 90 ( AD BC ) \uf0de 4 \u0111i\u1ec3m A,F,D,C c\u00f9ng thu\u1ed9c \u0111\u01b0\u1eddng tr\u00f2n \u0111\u01b0\u1eddng k\u00ednh AC \uf0de t\u1ee9 gi\u00e1c AFDC n\u1ed9i ti\u1ebfp \uf0de BFD ACB M\u00e0 MFB ACB (cmt) \uf0de MFB BFD \uf0de FB l\u00e0 tia ph\u00e2n gi\u00e1c c\u1ee7a MFN \uf0de MFN 2BFN M\u00e0 BON 2BFN (g\u00f3c \u1edf t\u00e2m v\u00e0 g\u00f3c n\u1ed9i ti\u1ebfp c\u00f9ng ch\u1eafn BN ) \uf0de MFN MON \uf0de t\u1ee9 gi\u00e1c MFON n\u1ed9i ti\u1ebfp. c) T\u00ednh SEIB SCKE . CFB vu\u00f4ng t\u1ea1i F c\u00f3: X\u00e9t EIB vu\u00f4ng t\u1ea1i I v\u00e0 IEB FCB (c\u00f9ng ch\u1eafn BF ) \uf0de EIB \u0111\u1ed3ng d\u1ea1ng CFB (g-g) \uf0de SEIB EB 2 1 SCFB BC 2 X\u00e9t CKE vu\u00f4ng t\u1ea1i K v\u00e0 CFB vu\u00f4ng t\u1ea1i F c\u00f3: KEC FBC ( BFEC n\u1ed9i ti\u1ebfp) \uf0de CKE \u0111\u1ed3ng d\u1ea1ng CFB (g-g) \uf0de SCKE CE 2 2 SCFB BC 2 X\u00e9t BEC vu\u00f4ng t\u1ea1i E c\u00f3: BE 2 EC 2 BC 2 (\u0111\u1ecbnh l\u00ed Pytago) 3 1,2, 3 SEIB SCKE EB 2 CE 2 BC 2 1 SCFB SCFB BC 2 BC 2 BC 2 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 \uf0de SEIB SCKE SCFB 1 BF.FC 2 \uf0de SEIB SCKE 1 BC.cos FBC BC.sin FBC 1 8.cos 75 8.sin 75 8 cm2 . 2 2 ----H\u1ebeT--- 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 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 4 N\u0102M H\u1eccC: 2023 - 2024 M\u00d4N: TO\u00c1N 9 \u0110\u1ec0 THAM KH\u1ea2O \u0110\u1ec1 thi g\u1ed3m 8 c\u00e2u ho\u0309i t\u01b0\u0323 lu\u00e2\u0323n. M\u00c3 \u0110\u1ec0: Qu\u1eadn 4-1 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 parabol (P) : y x2 v\u00e0 \u0111\u01b0\u1eddng th\u1eb3ng (d) : y x 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: 2x2 6x 3 0 c\u00f3 hai nghi\u1ec7m 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 2 2 6029 . x12 x22 x1 x2 C\u00e2u 3. (0.75 \u0111i\u1ec3m). Anh An l\u00e0m vi\u1ec7c cho m\u1ed9t c\u00f4ng ty s\u1ea3n xu\u1ea5t h\u00e0ng cao c\u1ea5p, anh \u0111\u01b0\u1ee3c tr\u1ea3 n\u0103m tri\u1ec7u b\u1ea3y tr\u0103m s\u00e1u m\u01b0\u01a1i ng\u00e0n \u0111\u1ed3ng cho 48 ti\u1ebfng l\u00e0m vi\u1ec7c trong m\u1ed9t tu\u1ea7n. Sau \u0111\u00f3 \u0111\u1ec3 t\u0103ng th\u00eam thu nh\u1eadp, anh An \u0111\u00e3 \u0111\u0103ng k\u00fd l\u00e0m th\u00eam m\u1ed9t s\u1ed1 gi\u1edd n\u1eefa trong tu\u1ea7n, m\u1ed7i gi\u1edd l\u00e0m th\u00eam n\u00e0y anh An \u0111\u01b0\u1ee3c tr\u1ea3 b\u1eb1ng 150% s\u1ed1 ti\u1ec1n m\u00e0 m\u1ed7i gi\u1edd anh An \u0111\u01b0\u1ee3c tr\u1ea3 trong 48 gi\u1edd \u0111\u1ea7u. Cu\u1ed1i tu\u1ea7n sau khi xong vi\u1ec7c, anh An \u0111\u01b0\u1ee3c l\u00e3nh s\u1ed1 ti\u1ec1n l\u00e0 b\u1ea3y tri\u1ec7u hai tr\u0103m ng\u00e0n \u0111\u1ed3ng. H\u1ecfi anh An \u0111\u00e3 l\u00e0m th\u00eam bao nhi\u00eau gi\u1edd trong tu\u1ea7n \u0111\u00f3? C\u00e2u 4. (0,75 \u0111i\u1ec3m). C\u00f4ng ty FPA cung c\u1ea5p d\u1ecbch v\u1ee5 Internet v\u1edbi m\u1ee9c chi ph\u00ed ban \u0111\u1ea7u l\u00e0 400000 \u0111\u1ed3ng v\u00e0 chi ph\u00ed tr\u1ea3 h\u00e0ng th\u00e1ng l\u00e0 272000 \u0111\u1ed3ng. C\u00f4ng ty VNPB cung c\u1ea5p d\u1ecbch v\u1ee5 Internet kh\u00f4ng t\u00ednh chi ph\u00ed ban \u0111\u1ea7u, nh\u01b0ng chi ph\u00ed tr\u1ea3 h\u00e0ng th\u00e1ng l\u00e0 320000 \u0111\u1ed3ng. Anh Minh \u0111\u00e3 \u0111\u0103ng k\u00fd d\u1ecbch v\u1ee5 Internet c\u1ee7a c\u00f4ng ty FPA. H\u1ecfi anh Minh ph\u1ea3i s\u1eed d\u1ee5ng d\u1ecbch v\u1ee5 Internet c\u1ee7a c\u00f4ng ty FPA \u00edt nh\u1ea5t trong bao nhi\u00eau th\u00e1ng th\u00ec t\u1ed5ng chi ph\u00ed s\u1eed d\u1ee5ng s\u1ebd r\u1ebb\u0309 h\u01a1n s\u1eed d\u1ee5ng c\u1ee7a c\u00f4ng ty VNPB? C\u00e2u 5. (1 \u0111i\u1ec3m). Hai ng\u01b0\u1eddi A v\u00e0 B c\u00f9ng \u1edf m\u1ed9t ph\u00eda v\u00e0 c\u00e1ch th\u00e0nh ph\u1ed1 H\u1ed3 Ch\u00ed Minh 50 km . C\u1ea3 hai ng\u01b0\u1eddi c\u00f9ng nhau \u0111i tr\u00ean m\u1ed9t con \u0111\u01b0\u1eddng v\u1ec1 ph\u00eda ng\u01b0\u1ee3c h\u01b0\u1edbng v\u1edbi th\u00e0nh ph\u1ed1, ng\u01b0\u1eddi A \u0111i v\u1edbi v\u1eadn t\u1ed1c trung b\u00ecnh l\u00e0 60 km \/ h v\u00e0 ng\u01b0\u1eddi B \u0111i v\u1edbi v\u1eadn t\u1ed1c trung b\u00ecnh l\u00e0 50 km \/ h . G\u1ecdi d km l\u00e0 kho\u1ea3ng c\u00e1ch t\u1eeb th\u00e0nh ph\u1ed1 H\u1ed3 Ch\u00ed Minh \u0111\u1ebfn hai ng\u01b0\u1eddi A , B sau khi \u0111i \u0111\u01b0\u1ee3c t (gi\u1edd). a) L\u1eadp h\u00e0m s\u1ed1 c\u1ee7a d theo t \u0111\u1ed1i v\u1edbi m\u1ed7i ng\u01b0\u1eddi. b) H\u1ecfi n\u1ebfu hai ng\u01b0\u1eddi xu\u1ea5t ph\u00e1t c\u00f9ng m\u1ed9t l\u00fac th\u00ec v\u00e0o th\u1eddi \u0111i\u1ec3m n\u00e0o k\u1ec3 t\u1eeb l\u00fac xu\u1ea5t ph\u00e1t, kho\u1ea3ng c\u00e1ch gi\u1eefa hai ng\u01b0\u1eddi l\u00e0 20 km . 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). Ch\u00fa H\u00f2a mu\u1ed1n x\u00e2y m\u1ed9t b\u1ec3 n\u01b0\u1edbc b\u00ea t\u00f4ng h\u00ecnh tr\u1ee5 c\u00f3 chi\u1ec1u cao l\u00e0 1,6 m ; b\u00e1n k\u00ednh l\u00f2ng b\u1ec3 (t\u00ednh t\u1eeb t\u00e2m b\u1ec3 \u0111\u1ebfn m\u00e9p trong c\u1ee7a b\u1ebf) l\u00e0 r 1m , b\u1ec1 d\u00e0y c\u1ee7a th\u00e0nh b\u1ec3 l\u00e0 10cm v\u00e0 b\u1ec1 d\u00e0y c\u1ee7a \u0111\u00e1y b\u1ec3 l\u00e0 5cm . H\u1ecfi: a) B\u1ec3 c\u00f3 th\u1ec3 ch\u1ee9a \u0111\u01b0\u1ee3c nhi\u1ec1u nh\u1ea5t bao nhi\u00eau l\u00edt n\u01b0\u1edbc (bi\u1ebft th\u1ec3 t\u00edch h\u00ecnh tr\u1ee5 b\u1eb1ng V r2h v\u1edbi r l\u00e0 b\u00e1n k\u00ednh \u0111\u00e1y; h l\u00e0 chi\u1ec1u cao h\u00ecnh tr\u1ee5; 3,14 ) . b) \u0110\u01b0\u1ee3c bi\u1ebft m\u1ed9t kh\u1ed1i b\u00ea t\u00f4ng c\u1ea7n: 5 bao xi m\u0103ng lo\u1ea1i 50 kg\/bao, 0,5m3 c\u00e1t, 0,9 m3 \u0111\u00e1, 185 l\u00edt n\u01b0\u1edbc. H\u1ecfi ch\u00fa H\u00f2a c\u1ea7n bao nhi\u00eau kg xi m\u0103ng? Bao nhi\u00eau m3 c\u00e1t v\u00e0 bao nhi\u00eau l\u00edt n\u01b0\u1edbc \u0111\u1ec3 x\u00e2y b\u1ec3? (c\u00e1c k\u1ebft qu\u1ea3 \u0111\u01b0\u1ee3c l\u00e0m tr\u00f2n \u0111\u1ebfn ch\u1eef s\u1ed1 th\u1eadp ph\u00e2n th\u1ee9 ba). C\u00e2u 7. (1 \u0111i\u1ec3m). H\u1ea1t ti\u00eau \u0111en th\u01b0\u1eddng \u0111\u01b0\u1ee3c d\u00f9ng l\u00e0m gia v\u1ecb trong n\u1ea5u \u0103n v\u00ec ngo\u00e0i t\u0103ng v\u1ecb ngon c\u1ee7a th\u1ee9c \u0103n, ti\u00eau c\u00f2n c\u00f3 nhi\u1ec1u t\u00e1c d\u1ee5ng t\u1ed1t cho s\u1ee9c kh\u1ecfe nh\u01b0 t\u1ed1t cho d\u1ea1 d\u00e0y, gi\u1ea3m c\u00e2n, s\u1ee9c kh\u1ecfe da, ch\u1ed1ng oxy h\u00f3a v\u00e0 c\u00e1c t\u00e1c d\u1ee5ng kh\u00e1c. \u0110\u01b0\u1ee3c bi\u1ebft t\u1ec9 l\u1ec7 n\u01b0\u1edbc trong h\u1ea1t ti\u00eau xanh c\u00f2n t\u01b0\u01a1i l\u00e0 68% v\u00e0 h\u1ea1t ti\u00eau kh\u00f4 l\u00e0 2% . a) V\u1eady n\u1ebfu \u0111em \u0111i ph\u01a1i kh\u00f4 m\u1ed9t t\u1ea1 ti\u00eau xanh c\u00f2n t\u01b0\u01a1i th\u00ec thu \u0111\u01b0\u1ee3c kh\u1ed1i l\u01b0\u1ee3ng h\u1ea1t ti\u00eau kh\u00f4 l\u00e0 bao nhi\u00eau? (l\u00e0m tr\u00f2n \u0111\u1ebfn ch\u1eef s\u1ed1 th\u1eadp ph\u00e2n th\u1ee9 hai) Gi\u1ea3 s\u1eed l\u01b0\u1ee3ng ti\u00eau hao h\u1ee5t trong qu\u00e1 tr\u00ecnh l\u00e0 5% . b) Bi\u1ebft gi\u00e1 h\u1ed3 ti\u00eau th\u1eddi \u0111i\u1ec3m 11 \/ 4 \/ 2022 nh\u01b0 sau: Ti\u00eau kh\u00f4 c\u00f3 gi\u00e1 55000 Vn\u0111\/kg v\u00e0 ti\u00eau xanh c\u00f2n t\u01b0\u01a1i gi\u00e1 13750 Vn\u0111\/kg. B\u00e1c An c\u00f3 10 t\u1ea5n ti\u00eau t\u01b0\u01a1i v\u00e0 d\u1ef1 \u0111\u1ecbnh thu\u00ea 2 c\u00f4ng nh\u00e2n ph\u01a1i kh\u00f4 trong 10 ng\u00e0y v\u1edbi ti\u1ec1n c\u00f4ng 400000 Vn\u0111\/ng\u00e0y. H\u1ecfi b\u00e1c An l\u00e0m nh\u01b0 v\u1eady s\u1ebd l\u1eddi hay l\u1ed7 so v\u1edbi b\u00e1n ti\u00eau t\u01b0\u01a1i bao nhi\u00eau ti\u1ec1n? C\u00e2u 8. (3 \u0111i\u1ec3m) Cho \u0111\u01b0\u1eddng tr\u00f2n O v\u00e0 \u0111i\u1ec3m M n\u1eb1m ngo\u00e0i \u0111\u01b0\u1eddng tr\u00f2n O . T\u1eeb M v\u1ebd 2 ti\u1ebfp tuy\u1ebfn MA , MB c\u1ee7a \u0111\u01b0\u1eddng tr\u00f2n O ( A v\u00e0 B l\u00e0 2 ti\u1ebfp \u0111i\u1ec3m). G\u1ecdi H l\u00e0 giao \u0111i\u1ec3m c\u1ee7a MO v\u00e0 AB . Qua M v\u1ebd c\u00e1t tuy\u1ebfn MCD c\u1ee7a \u0111\u01b0\u1eddng tr\u00f2n O ( C v\u00e0 D thu\u1ed9c \u0111\u01b0\u1eddng tr\u00f2n O ) sao cho \u0111\u01b0\u1eddng th\u1eb3ng MD c\u1eaft \u0111o\u1ea1n th\u1eb3ng HB . G\u1ecdi I l\u00e0 trung \u0111i\u1ec3m d\u00e2y cung CD . a) Ch\u1ee9ng minh OI CD t\u1ea1i v\u00e0 t\u1ee9 gi\u00e1c MAOI n\u1ed9i ti\u1ebfp. b) Ch\u1ee9ng minh MA2 MC.MD v\u00e0 t\u1ee9 gi\u00e1c OHCD n\u1ed9i ti\u1ebfp. c) Tr\u00ean cung nh\u1ecf AD l\u1ea5y \u0111i\u1ec3m N sao cho DN DB . Qua C v\u1ebd \u0111\u01b0\u1eddng th\u1eb3ng song song v\u1edbi DN c\u1eaft \u0111\u01b0\u1eddng th\u1eb3ng MN t\u1ea1i E v\u00e0 c\u0169ng qua C v\u1ebd \u0111\u01b0\u1eddng th\u1eb3ng song song v\u1edbi BD c\u1eaft c\u1ea1nh AB t\u1ea1i F . Ch\u1ee9ng minh: Tam gi\u00e1c CEF c\u00e2n. ----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 x 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 B\u1ea3ng gi\u00e1 tr\u1ecb h\u00e0m s\u1ed1 (P) : y x2 B\u1ea3ng gi\u00e1 tr\u1ecb h\u00e0m s\u1ed1 (d) : y x 2 \u0110\u1ed3 th\u1ecb: b) Phuong tr\u00ecnh ho\u00e0nh \u0111\u1ed9 giao \u0111i\u1ec3m c\u1ee7a P v\u00e0 d : x2 x 2 x2 x 2 0 x1 1 y1 1 x2 2 y2 4 V\u1eady t\u1ecda \u0111\u1ed9 giao \u0111i\u1ec3m c\u1ee7a P v\u00e0 d l\u00e0 1;1 , 2; 4 . C\u00e2u 2. (1 \u0111i\u1ec3m). Cho ph\u01b0\u01a1ng tr\u00ecnh: 2x2 6x 3 0 c\u00f3 hai nghi\u1ec7m 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 2 2 6029 . x12 x22 x1 x2 L\u1eddi gi\u1ea3i Ta th\u1ea5y a 2 , c 3 , do \u0111\u00f3 a , c tr\u00e1i d\u1ea5u n\u00ean ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m ph\u00e2n bi\u1ec7t. 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 S x1 x2 b3 \u00c1p d\u1ee5ng \u0111\u1ecbnh l\u00fd Vi-\u00e9t ta c\u00f3: a x1x2 c P a 3 2 2 2 6029 2 x12 x22 6029 2 x12 x22 2 2x1x2 6029 x12 x22 x1 x2 x12 x22 x1 x2 x1x2 2 x1 x2 B 2 3 2 2. 3 6029 6061. 2 33 32 2 C\u00e2u 3. (0.75 \u0111i\u1ec3m). Anh An l\u00e0m vi\u1ec7c cho m\u1ed9t c\u00f4ng ty s\u1ea3n xu\u1ea5t h\u00e0ng cao c\u1ea5p, anh \u0111\u01b0\u1ee3c tr\u1ea3 n\u0103m tri\u1ec7u b\u1ea3y tr\u0103m s\u00e1u m\u01b0\u01a1i ng\u00e0n \u0111\u1ed3ng cho 48 ti\u1ebfng l\u00e0m vi\u1ec7c trong m\u1ed9t tu\u1ea7n. Sau \u0111\u00f3 \u0111\u1ec3 t\u0103ng th\u00eam thu nh\u1eadp, anh An \u0111\u00e3 \u0111\u0103ng k\u00fd l\u00e0m th\u00eam m\u1ed9t s\u1ed1 gi\u1edd n\u1eefa trong tu\u1ea7n, m\u1ed7i gi\u1edd l\u00e0m th\u00eam n\u00e0y anh An \u0111\u01b0\u1ee3c tr\u1ea3 b\u1eb1ng 150% s\u1ed1 ti\u1ec1n m\u00e0 m\u1ed7i gi\u1edd anh An \u0111\u01b0\u1ee3c tr\u1ea3 trong 48 gi\u1edd \u0111\u1ea7u. Cu\u1ed1i tu\u1ea7n sau khi xong vi\u1ec7c, anh An \u0111\u01b0\u1ee3c l\u00e3nh s\u1ed1 ti\u1ec1n l\u00e0 b\u1ea3y tri\u1ec7u hai tr\u0103m ng\u00e0n \u0111\u1ed3ng. H\u1ecfi anh An \u0111\u00e3 l\u00e0m th\u00eam bao nhi\u00eau gi\u1edd trong tu\u1ea7n \u0111\u00f3? L\u1eddi gi\u1ea3i G\u1ecdi x (gi\u1edd) l\u00e0 s\u1ed1 gi\u1edd anh An l\u00e0m th\u00eam ( x 0 ). Ta c\u00f3 5760000 x. 5760000 .150% 7 200000 x 8 (th\u1ecfa m\u00e3n). 48 V\u1eady trong tu\u1ea7n \u0111\u00f3 anh An \u0111\u00e3 l\u00e0m th\u00eam 8 gi\u1edd. C\u00e2u 4. (0,75 \u0111i\u1ec3m). C\u00f4ng ty FPA cung c\u1ea5p d\u1ecbch v\u1ee5 Internet v\u1edbi m\u1ee9c chi ph\u00ed ban \u0111\u1ea7u l\u00e0 400000 \u0111\u1ed3ng v\u00e0 chi ph\u00ed tr\u1ea3 h\u00e0ng th\u00e1ng l\u00e0 272000 \u0111\u1ed3ng. C\u00f4ng ty VNPB cung c\u1ea5p d\u1ecbch v\u1ee5 Internet kh\u00f4ng t\u00ednh chi ph\u00ed ban \u0111\u1ea7u, nh\u01b0ng chi ph\u00ed tr\u1ea3 h\u00e0ng th\u00e1ng l\u00e0 320000 \u0111\u1ed3ng. Anh Minh \u0111\u00e3 \u0111\u0103ng k\u00fd d\u1ecbch v\u1ee5 Internet c\u1ee7a c\u00f4ng ty FPA. H\u1ecfi anh Minh ph\u1ea3i s\u1eed d\u1ee5ng d\u1ecbch v\u1ee5 Internet c\u1ee7a c\u00f4ng ty FPA \u00edt nh\u1ea5t trong bao nhi\u00eau th\u00e1ng th\u00ec t\u1ed5ng chi ph\u00ed s\u1eed d\u1ee5ng s\u1ebd r\u1ebb\u0309 h\u01a1n s\u1eed d\u1ee5ng c\u1ee7a c\u00f4ng ty VNPB? L\u1eddi gi\u1ea3i G\u1ecdi x (th\u00e1ng) l\u00e0 th\u1eddi gian s\u1eed d\u1ee5ng d\u1ecbch v\u1ee5 ( x 0 ). S\u1ed1 ti\u1ec1n ph\u1ea3i tr\u1ea3 khi s\u1eed d\u1ee5ng d\u1ecbch v\u1ee5 FPA l\u00e0 272000x 400000 . S\u1ed1 ti\u1ec1n ph\u1ea3i tr\u1ea3 khi s\u1eed d\u1ee5ng d\u1ecbch v\u1ee5 VNPB l\u00e0 322000x . Khi \u0111\u00f3 ta c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh 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 272000x 400000 320000x x 25 8,3 . 3 V\u1eady k\u1ec3 t\u1eeb th\u00e1ng th\u1ee9 9 tr\u1edf \u0111i s\u1eed d\u1ee5ng d\u1ecbch v\u1ee5 FPA r\u1ebb h\u01a1n s\u1eed d\u1ee5ng d\u1ecbch v\u1ee5 VNPB. C\u00e2u 5. (1 \u0111i\u1ec3m). Hai ng\u01b0\u1eddi A v\u00e0 B c\u00f9ng \u1edf m\u1ed9t ph\u00eda v\u00e0 c\u00e1ch th\u00e0nh ph\u1ed1 H\u1ed3 Ch\u00ed Minh 50 km . C\u1ea3 hai ng\u01b0\u1eddi c\u00f9ng nhau \u0111i tr\u00ean m\u1ed9t con \u0111\u01b0\u1eddng v\u1ec1 ph\u00eda ng\u01b0\u1ee3c h\u01b0\u1edbng v\u1edbi th\u00e0nh ph\u1ed1, ng\u01b0\u1eddi A \u0111i v\u1edbi v\u1eadn t\u1ed1c trung b\u00ecnh l\u00e0 60 km \/ h v\u00e0 ng\u01b0\u1eddi B \u0111i v\u1edbi v\u1eadn t\u1ed1c trung b\u00ecnh l\u00e0 50 km \/ h . G\u1ecdi d km l\u00e0 kho\u1ea3ng c\u00e1ch t\u1eeb th\u00e0nh ph\u1ed1 H\u1ed3 Ch\u00ed Minh \u0111\u1ebfn hai ng\u01b0\u1eddi A , B sau khi \u0111i \u0111\u01b0\u1ee3c t (gi\u1edd). a) L\u1eadp h\u00e0m s\u1ed1 c\u1ee7a d theo t \u0111\u1ed1i v\u1edbi m\u1ed7i ng\u01b0\u1eddi. b) H\u1ecfi n\u1ebfu hai ng\u01b0\u1eddi xu\u1ea5t ph\u00e1t c\u00f9ng m\u1ed9t l\u00fac th\u00ec v\u00e0o th\u1eddi \u0111i\u1ec3m n\u00e0o k\u1ec3 t\u1eeb l\u00fac xu\u1ea5t ph\u00e1t, kho\u1ea3ng c\u00e1ch gi\u1eefa hai ng\u01b0\u1eddi l\u00e0 20 km . L\u1eddi gi\u1ea3i a) H\u00e0m s\u1ed1 c\u1ee7a d theo t c\u1ee7a ng\u01b0\u1eddi th\u1ee9 nh\u1ea5t l\u00e0 d1 : d 60t . H\u00e0m s\u1ed1 c\u1ee7a d theo t c\u1ee7a ng\u01b0\u1eddi th\u1ee9 hai l\u00e0 d2 : d 50t 50 . b) Ta c\u00f3 60t 50t 50 20 10t 50 20 t7 10t 50 20 t 3. V\u1eady k\u1ec3 t\u1eeb l\u00fac xu\u1ea5t ph\u00e1t \u0111\u1ebfn khi hai xe c\u00e1ch nhau 20 km l\u00e0 3 gi\u1edd v\u00e0 7 gi\u1edd. C\u00e2u 6. (1 \u0111i\u1ec3m). Ch\u00fa H\u00f2a mu\u1ed1n x\u00e2y m\u1ed9t b\u1ec3 n\u01b0\u1edbc b\u00ea t\u00f4ng h\u00ecnh tr\u1ee5 c\u00f3 chi\u1ec1u cao l\u00e0 1,6 m ; b\u00e1n k\u00ednh l\u00f2ng b\u1ec3 (t\u00ednh t\u1eeb t\u00e2m b\u1ec3 \u0111\u1ebfn m\u00e9p trong c\u1ee7a b\u1ebf) l\u00e0 r 1m , b\u1ec1 d\u00e0y c\u1ee7a th\u00e0nh b\u1ec3 l\u00e0 10cm v\u00e0 b\u1ec1 d\u00e0y c\u1ee7a \u0111\u00e1y b\u1ec3 l\u00e0 5cm . H\u1ecfi: a) B\u1ec3 c\u00f3 th\u1ec3 ch\u1ee9a \u0111\u01b0\u1ee3c nhi\u1ec1u nh\u1ea5t bao nhi\u00eau l\u00edt n\u01b0\u1edbc (bi\u1ebft th\u1ec3 t\u00edch h\u00ecnh tr\u1ee5 b\u1eb1ng V r2h v\u1edbi r l\u00e0 b\u00e1n k\u00ednh \u0111\u00e1y; h l\u00e0 chi\u1ec1u cao h\u00ecnh tr\u1ee5; 3,14 ) . b) \u0110\u01b0\u1ee3c bi\u1ebft m\u1ed9t kh\u1ed1i b\u00ea t\u00f4ng c\u1ea7n: 5 bao xi m\u0103ng lo\u1ea1i 50 kg\/bao, 0,5m3 c\u00e1t, 0,9 m3 \u0111\u00e1, 185 l\u00edt n\u01b0\u1edbc. H\u1ecfi ch\u00fa H\u00f2a c\u1ea7n bao nhi\u00eau kg xi m\u0103ng? Bao nhi\u00eau m3 c\u00e1t v\u00e0 bao nhi\u00eau l\u00edt n\u01b0\u1edbc \u0111\u1ec3 x\u00e2y b\u1ec3? (c\u00e1c k\u1ebft qu\u1ea3 \u0111\u01b0\u1ee3c l\u00e0m tr\u00f2n \u0111\u1ebfn ch\u1eef s\u1ed1 th\u1eadp ph\u00e2n th\u1ee9 ba). L\u1eddi gi\u1ea3i 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 a) Th\u1ec3 t\u00edch n\u01b0\u1edbc m\u00e0 b\u1ec3 c\u00f3 th\u1ec3 ch\u1ee9a l\u00e0 V r2h .12. 1,6 0,05 3,14.1,55 4,867 m3 . b) G\u1ecdi V1 , V2 , V3 l\u1ea7n l\u01b0\u1ee3t l\u00e0 th\u1ec3 t\u00edch c\u1ee7a kh\u1ed1i tr\u1ee5 l\u1edbn, th\u1ec3 t\u00edch ph\u1ea7n ch\u1ee9a n\u01b0\u1edbc, th\u1ec3 t\u00edch kh\u1ed1i b\u00ea t\u00f4ng. V3 V1 V2 . 1 0,1 2 .1,6 .12. 1,6 0,05 1,212 m3 . Kh\u1ed1i l\u01b0\u1ee3ng xi m\u0103ng c\u1ea7n l\u00e0 2,212.50.5 303 kg . Th\u1ec3 t\u00edch c\u00e1t l\u00e0 1.212.0,9 1,091m3 . Th\u1ec3 t\u00edch n\u01b0\u1edbc s\u1eed d\u1ee5ng l\u00e0 1.212.185 224,220 l\u00edt. C\u00e2u 7. (1 \u0111i\u1ec3m). H\u1ea1t ti\u00eau \u0111en th\u01b0\u1eddng \u0111\u01b0\u1ee3c d\u00f9ng l\u00e0m gia v\u1ecb trong n\u1ea5u \u0103n v\u00ec ngo\u00e0i t\u0103ng v\u1ecb ngon c\u1ee7a th\u1ee9c \u0103n, ti\u00eau c\u00f2n c\u00f3 nhi\u1ec1u t\u00e1c d\u1ee5ng t\u1ed1t cho s\u1ee9c kh\u1ecfe nh\u01b0 t\u1ed1t cho d\u1ea1 d\u00e0y, gi\u1ea3m c\u00e2n, s\u1ee9c kh\u1ecfe da, ch\u1ed1ng oxy h\u00f3a v\u00e0 c\u00e1c t\u00e1c d\u1ee5ng kh\u00e1c. \u0110\u01b0\u1ee3c bi\u1ebft t\u1ec9 l\u1ec7 n\u01b0\u1edbc trong h\u1ea1t ti\u00eau xanh c\u00f2n t\u01b0\u01a1i l\u00e0 68% v\u00e0 h\u1ea1t ti\u00eau kh\u00f4 l\u00e0 2% . a) V\u1eady n\u1ebfu \u0111em \u0111i ph\u01a1i kh\u00f4 m\u1ed9t t\u1ea1 ti\u00eau xanh c\u00f2n t\u01b0\u01a1i th\u00ec thu \u0111\u01b0\u1ee3c kh\u1ed1i l\u01b0\u1ee3ng h\u1ea1t ti\u00eau kh\u00f4 l\u00e0 bao nhi\u00eau? (l\u00e0m tr\u00f2n \u0111\u1ebfn ch\u1eef s\u1ed1 th\u1eadp ph\u00e2n th\u1ee9 hai) Gi\u1ea3 s\u1eed l\u01b0\u1ee3ng ti\u00eau hao h\u1ee5t trong qu\u00e1 tr\u00ecnh l\u00e0 5% . b) Bi\u1ebft gi\u00e1 h\u1ed3 ti\u00eau th\u1eddi \u0111i\u1ec3m 11 \/ 4 \/ 2022 nh\u01b0 sau: Ti\u00eau kh\u00f4 c\u00f3 gi\u00e1 55000 Vn\u0111\/kg v\u00e0 ti\u00eau xanh c\u00f2n t\u01b0\u01a1i gi\u00e1 13750 Vn\u0111\/kg. B\u00e1c An c\u00f3 10 t\u1ea5n ti\u00eau t\u01b0\u01a1i v\u00e0 d\u1ef1 \u0111\u1ecbnh thu\u00ea 2 c\u00f4ng nh\u00e2n ph\u01a1i kh\u00f4 trong 10 ng\u00e0y v\u1edbi ti\u1ec1n c\u00f4ng 400000 Vn\u0111\/ng\u00e0y. H\u1ecfi b\u00e1c An l\u00e0m nh\u01b0 v\u1eady s\u1ebd l\u1eddi hay l\u1ed7 so v\u1edbi b\u00e1n ti\u00eau t\u01b0\u01a1i bao nhi\u00eau ti\u1ec1n? L\u1eddi gi\u1ea3i a) Kh\u1ed1i l\u01b0\u1ee3ng m\u1ed9t t\u1ea1 ti\u00eau nguy\u00ean ch\u1ea5t l\u00e0 100.68% 68 kg . Kh\u1ed1i l\u01b0\u1ee3ng m\u1ed9t t\u1ea1 ti\u00eau ch\u1ee9a 2% n\u01b0\u1edbc v\u00e0 hao h\u1ee5t 5% l\u00e0 68. 100 . 95 65,92 kg . 98 100 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 b) S\u1ed1 ti\u1ec1n b\u00e1n 10 t\u1ea5n ti\u00eau t\u01b0\u01a1i l\u00e0 13750.10000 137 500000 Vn\u0111. S\u1ed1 ti\u1ec1n b\u00e1n 10 t\u1ea5n ti\u00eau khi ph\u01a1i kh\u00f4 l\u00e0 55000.10000. 65,92 362 560000 Vn\u0111. 100 S\u1ed1 ti\u1ec1n thu\u00ea nh\u00e2n c\u00f4ng ph\u01a1i l\u00e0 40000.10.2 8000000 Vn\u0111. S\u1ed1 ti\u1ec1n b\u00e1c An thu v\u1ec1 khi b\u00e1n ti\u00eau kh\u00f4 v\u00e0 tr\u1eeb nh\u00e2n c\u00f4ng l\u00e0 362 560000 8000000 354 560000 Vn\u0111. Nh\u01b0 v\u1eady b\u00e1c An b\u00e1n ti\u00eau kh\u00f4 l\u1eddi h\u01a1n b\u00e1n ti\u00eau t\u01b0\u01a1i l\u00e0 354 560000 137 500000 217 060000 Vn\u0111. C\u00e2u 8. (3 \u0111i\u1ec3m) Cho \u0111\u01b0\u1eddng tr\u00f2n O v\u00e0 \u0111i\u1ec3m M n\u1eb1m ngo\u00e0i \u0111\u01b0\u1eddng tr\u00f2n O . T\u1eeb M v\u1ebd 2 ti\u1ebfp tuy\u1ebfn MA , MB c\u1ee7a \u0111\u01b0\u1eddng tr\u00f2n O ( A v\u00e0 B l\u00e0 2 ti\u1ebfp \u0111i\u1ec3m). G\u1ecdi H l\u00e0 giao \u0111i\u1ec3m c\u1ee7a MO v\u00e0 AB . Qua M v\u1ebd c\u00e1t tuy\u1ebfn MCD c\u1ee7a \u0111\u01b0\u1eddng tr\u00f2n O ( C v\u00e0 D thu\u1ed9c \u0111\u01b0\u1eddng tr\u00f2n O ) sao cho \u0111\u01b0\u1eddng th\u1eb3ng MD c\u1eaft \u0111o\u1ea1n th\u1eb3ng HB . G\u1ecdi I l\u00e0 trung \u0111i\u1ec3m d\u00e2y cung CD . a) Ch\u1ee9ng minh OI CD t\u1ea1i v\u00e0 t\u1ee9 gi\u00e1c MAOI n\u1ed9i ti\u1ebfp. b) Ch\u1ee9ng minh MA2 MC.MD v\u00e0 t\u1ee9 gi\u00e1c OHCD n\u1ed9i ti\u1ebfp. c) Tr\u00ean cung nh\u1ecf AD l\u1ea5y \u0111i\u1ec3m N sao cho DN DB . Qua C v\u1ebd \u0111\u01b0\u1eddng th\u1eb3ng song song v\u1edbi DN c\u1eaft \u0111\u01b0\u1eddng th\u1eb3ng MN t\u1ea1i E v\u00e0 c\u0169ng qua C v\u1ebd \u0111\u01b0\u1eddng th\u1eb3ng song song v\u1edbi BD c\u1eaft c\u1ea1nh AB t\u1ea1i F . Ch\u1ee9ng minh: Tam gi\u00e1c CEF c\u00e2n. L\u1eddi gi\u1ea3i a) Do I l\u00e0 trung \u0111i\u1ec3m c\u1ee7a CD n\u00ean OI CD (t\u00ednh ch\u1ea5t \u0111\u01b0\u1eddng k\u00ednh v\u00e0 d\u00e2y cung) 7 X\u00e9t t\u1ee9 gi\u00e1c MAOI c\u00f3 OAM OIM 90 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 OAM OIM 90 90 180 . Suy ra t\u1ee9 gi\u00e1c MAOI n\u1ed9i ti\u1ebfp (t\u1ed5ng hai g\u00f3c d\u1ed1i b\u1eb1ng 180 ). b) X\u00e9t MCB v\u00e0 MBD c\u00f3 BMC DMC MCB MDB (li\u00ean h\u1ec7 gi\u1eefa g\u00f3c n\u1ed9i ti\u1ebfp v\u00e0 g\u00f3c gi\u1eefa tia ti\u1ebfp tuy\u1ebfn v\u00e0 d\u00e2y cung) Suy ra MCB\\\" MBD (g-g) MC MB MC.MD MB2 MC.MD MA2 . 1 MB MD Ta c\u00f3 OA OB R MO l\u00e0 trung tr\u1ef1c c\u1ee7a AB AH l\u00e0 \u0111\u01b0\u1eddng cao trong tam MA MB(tc2tt) gi\u00e1c MAO vu\u00f4ng t\u1ea1i A n\u00ean MA2 MH.MO . 2 T\u1eeb 1 v\u00e0 2 MH.MO MC.MD MH MC . MD MO X\u00e9t MHC v\u00e0 MDO c\u00f3 HMC DMO v\u00e0 MH MC n\u00ean MD MO HMC\\\" DMO (c-g-c). Suy ra MHC ODC n\u00ean t\u1ee9 gi\u00e1c OHCD n\u1ed9i ti\u1ebfp (g\u00f3c ngo\u00e0i b\u1eb1ng g\u00f3c \u0111\u1ed1i trong kh\u00f4ng k\u1ec1) c) G\u1ecdi P l\u00e0 giao \u0111i\u1ec3m c\u1ee7a AB v\u00e0 MD . Ta c\u00f3 CE\/\/DN n\u00ean MC CE . 1 MD DN Ta c\u00f3 CF\/\/BD n\u00ean PC CF . 2 PD BD Do t\u1ee9 gi\u00e1c OHCD n\u1ed9i ti\u1ebfp, suy ra MHC ODC OCD . M\u00e0 OCD OHD . N\u00ean MHC OHD . M\u1eb7t kh\u00e1c HP OM . L\u00fac n\u00e0y ta c\u00f3 HP l\u00e0 ph\u00e2n gi\u00e1c trong, HM l\u00e0 ph\u00e2n gi\u00e1c ngo\u00e0i c\u1ee7a tam gi\u00e1c HCD n\u00ean \u00e1p d\u1ee5ng t\u00ednh ch\u1ea5t ph\u00e2n gi\u00e1c ta c\u00f3 MC PC 3 MD PD T\u1eeb 1 , 2 v\u00e0 3 suy ra CE CF m\u00e0 DB DN CE CF . DN BD V\u1eady tam gi\u00e1c CEF c\u00e2n t\u1ea1i C . ----H\u1ebeT--- 8 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 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 4 NA\u00caM HO\u00cfC: 2023 - 2024 M\u00d4N: TO\u00c1N 9 \u0110\u1ec0 THAM KH\u1ea2O M\u00c3 \u0110\u1ec0: Qu\u1eadn 4 - 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\u00ea\u0309m): Cho Parabol P : y 1 x2 v\u00e0 \u0111\u01b0\u1eddng th\u1eb3ng D :y 3x 1. 4 4 a) V\u1ebd P v\u00e0 D tr\u00ean c\u00f9ng m\u1ed9t h\u1ec7 tr\u1ee5c t\u1ecda \u0111\u1ed9 0xy . b) B\u1eb1ng ph\u00e9p to\u00e1n x\u00e1c \u0111\u1ecbnh t\u1ecda \u0111\u1ed9 giao \u0111i\u1ec3m c\u1ee7a P v\u00e0 D . C\u00e2u 2. (1 \u0111i\u00ea\u0309m) Cho ph\u01b0\u01a1ng tr\u00ecnh: x2 \u221211x + 5 = 0 a) Ch\u1ee9ng minh ph\u01b0\u01a1ng tr\u00ecnh lu\u00f4n c\u00f3 2 nghi\u1ec7m ph\u00e2n bi\u1ec7t x1, x2 r\u1ed3i t\u00ednh t\u1ed5ng v\u00e0 t\u00edch hai ngh\u1ec7m x1, x2 c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh. b) Kh\u00f4ng gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh h\u00e3y t\u00ednh gi\u00e1 tr\u1ecb c\u1ee7a bi\u1ec3u th\u1ee9c: A = \uf0e6 2 \u2212 2 \uf0f6 \uf0d7 ( x1 \u2212 x2 ) \uf0e7 x2 x1 \uf0f7 \uf0e8 \uf0f8 C\u00e2u 3. (0,75 \u0111i\u00ea\u0309m) \u0110\u1ec3 \u0111\u1ea1t k\u1ebft qu\u1ea3 t\u1ed1t nh\u1ea5t trong k\u00ec thi tuy\u1ec3n sinh l\u1edbp 10 THPT v\u00e0o ng\u00e0y 02 \/ 6 \/ 2021 , sau khi t\u1ed5 ch\u1ee9c H\u1ed9i tr\u1ea1i truy\u1ec1n th\u1ed1ng v\u00e0o th\u1ee9 S\u00e1u ng\u00e0y 26 \/ 3 \/ 2021 , h\u1ecdc sinh kh\u1ed1i 9 \u0111\u00e3 \u0111\u1ec1 ra k\u1ebf ho\u1ea1ch h\u1ecdc t\u1eadp m\u00f4n To\u00e1n c\u1ee5 th\u1ec3 nh\u01b0 sau: \\\"M\u1ed7i h\u1ecdc sinh b\u1eaft \u0111\u1ea7u t\u1eeb ng\u00e0y 27 \/ 3 \/ 2021 \u0111\u1ebfn h\u1ebft th\u00e1ng ba m\u1ed7i ng\u00e0y l\u00e0m 3 b\u00e0i to\u00e1n, m\u1ed7i ng\u00e0y trong th\u00e1ng t\u01b0 l\u00e0m 4 b\u00e0i to\u00e1n, m\u1ed7i ng\u00e0y trong th\u00e1ng n\u0103m l\u00e0m 5 b\u00e0i to\u00e1n\\\". Bi\u1ebft th\u00e1ng ba v\u00e0 th\u00e1ng n\u0103m l\u00e0 nh\u1eefng th\u00e1ng c\u00f3 31 ng\u00e0y, th\u00e1ng t\u01b0 c\u00f3 30 ng\u00e0y. H\u1ecfi: a) Theo k\u1ebf ho\u1ea1ch, m\u1ed7i h\u1ecdc sinh l\u00e0m \u0111\u01b0\u1ee3c bao nhi\u00eau b\u00e0i to\u00e1n? b) Ng\u00e0y thi 02 \/ 6 \/ 2021 l\u00e0 th\u1ee9 m\u1ea5y? Gi\u1ea3i th\u00edch v\u00ec sao? C\u00e2u 4. (0,75 \u0111i\u00ea\u0309m) M\u1ed9t lon n\u01b0\u1edbc ng\u1ecdt c\u00f3 gi\u00e1 10000 \u0111\u1ed3ng. M\u1ed9t quy\u1ec3n t\u1eadp c\u00f3 gi\u00e1 b\u1eb1ng 2 gi\u00e1 m\u1ed9t 5 lon n\u01b0\u1edbc ng\u1ecdt, m\u1ed9t h\u1ed9p b\u00fat c\u00f3 gi\u00e1 g\u1ea5p 3 l\u1ea7n gi\u00e1 m\u1ed9t lon n\u01b0\u1edbc ng\u1ecdt. B\u1ea1n An c\u1ea7n mua m\u1ed9t s\u1ed1 quy\u1ec3n t\u1eadp v\u00e0 m\u1ed9t h\u1ed9p b\u00fat. a) G\u1ecdi x l\u00e0 s\u1ed1 quy\u1ec3n t\u1eadp An mua v\u00e0 y l\u00e0 s\u1ed1 ti\u1ec1n An ph\u1ea3i tr\u1ea3 (bao g\u1ed3m ti\u1ec1n mua t\u1eadp v\u00e0 m\u1ed9t h\u1ed9p b\u00fat). Vi\u1ebft c\u00f4ng th\u1ee9c bi\u1ec3u di\u1ec5n y theo x . b) N\u1ebfu An b\u00e1n 2 th\u00f9ng n\u01b0\u1edbc ng\u1ecdt, m\u1ed7i th\u00f9ng g\u1ed3m 24 lon v\u1edbi gi\u00e1 \u0111\u00e3 n\u00eau tr\u00ean \u0111\u1ec3 mua t\u1eadp v\u00e0 m\u1ed9t h\u1ed9p b\u00fat th\u00ec t\u1ed1i \u0111a b\u1ea1n An mua \u0111\u01b0\u1ee3c bao nhi\u00eau quy\u1ec3n t\u1eadp? C\u00e2u 5. (1 \u0111i\u00ea\u0309m) M\u1ed9t c\u00f4ng ty giao cho c\u1eeda h\u00e0ng 100 h\u1ed9p b\u00e1nh \u0111\u1ec3 b\u00e1n ra th\u1ecb tr\u01b0\u1eddng. L\u00fac \u0111\u1ea7u c\u1eeda h\u00e0ng b\u00e1n 24 h\u1ed9p b\u00e1nh v\u1edbi gi\u00e1 b\u00e1n m\u1ed9t h\u1ed9p b\u00e1nh l\u00e0 200000 \u0111\u1ed3ng. Do nhu c\u1ea7u c\u1ee7a th\u1ecb tr\u01b0\u1eddng n\u00ean t\u1eeb h\u1ed9p b\u00e1nh th\u1ee9 25 \u0111\u1ebfn h\u1ed9p b\u00e1nh th\u1ee9 80 m\u1ed7i h\u1ed9p b\u00e1nh c\u00f3 gi\u00e1 b\u00e1n t\u0103ng 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 15% so v\u1edbi gi\u00e1 b\u00e1n l\u00fac \u0111\u1ea7u, t\u1eeb h\u1ed9p b\u00e1nh th\u1ee9 81 \u0111\u1ebfn h\u1ed9p b\u00e1nh th\u1ee9 100 m\u1ed7i h\u1ed9p b\u00e1nh c\u00f3 gi\u00e1 b\u00e1n gi\u1ea3m 10% so v\u1edbi gi\u00e1 b\u00e1n l\u00fac \u0111\u1ea7u. a) H\u1ecfi s\u1ed1 ti\u1ec1n thu c\u1eeda h\u00e0ng \u0111\u01b0\u1ee3c khi b\u00e1n 100 h\u1ed9p b\u00e1nh l\u00e0 bao nhi\u00eau? b) Bi\u1ebft r\u1eb1ng: V\u1edbi s\u1ed1 ti\u1ec1n thu \u0111\u01b0\u1ee3c khi b\u00e1n 100 h\u1ed9p b\u00e1nh, sau khi tr\u1eeb \u0111i 10% ti\u1ec1n thu\u1ebf gi\u00e1 tr\u1ecb gia t\u0103ng VAT c\u1eeda h\u00e0ng v\u1eabn l\u00e3i 1152000 \u0111\u1ed3ng. H\u1ecfi m\u1ed7i h\u1ed9p b\u00e1nh c\u00f4ng ty giao cho c\u1eeda h\u00e0ng c\u00f3 gi\u00e1 l\u00e0 bao nhi\u00eau? C\u00e2u 6. (1 \u0111i\u00ea\u0309m) Ba xe m\u00e1y c\u00f9ng xu\u1ea5t ph\u00e1t t\u1eeb O \u0111i theo ba h\u01b0\u1edbng Ox,Oy,Oz trong \u0111\u00f3 Ox v\u00e0 Oz ng\u01b0\u1ee3c h\u01b0\u1edbng nhau nh\u01b0 h\u00ecnh v\u1ebd. y x Oz Xe th\u1ee9 nh\u1ea5t \u0111i theo h\u01b0\u1edbng Ox , xe th\u1ee9 hai \u0111i theo h\u01b0\u1edbng Oy , xe th\u1ee9 ba \u0111i theo h\u01b0\u1edbng Oz , c\u1ea3 ba xe c\u00f9ng ch\u1ea1y v\u1edbi v\u1eadn t\u1ed1c kh\u00f4ng \u0111\u1ed5i l\u00e0 50km \/ gi\u1edd. Sau 2 gi\u1edd xe th\u1ee9 nh\u1ea5t v\u00e0 xe th\u1ee9 hai \u1edf c\u00e1ch nhau 107 km . H\u1ecfi l\u00fac \u0111\u00f3 xe th\u1ee9 hai v\u00e0 xe th\u1ee9 ba \u1edf c\u00e1ch nhau bao nhi\u00eau ki-l\u00f4-m\u00e9t? (l\u00e0m tr\u00f2n k\u1ebft qu\u1ea3 \u0111\u1ebfn ch\u1eef s\u1ed1 h\u00e0ng \u0111\u01a1n v\u1ecb). C\u00e2u 7. (1 \u0111i\u00ea\u0309m)Hai ng\u01b0\u1eddi th\u1ee3 c\u00f9ng l\u00e0m m\u1ed9t c\u00f4ng vi\u1ec7c trong 16 gi\u1edd th\u00ec xong. N\u1ebfu ng\u01b0\u1eddi th\u1ee3 th\u1ee9 nh\u1ea5t l\u00e0m trong 3 gi\u1edd, ng\u01b0\u1eddi th\u1ee3 th\u1ee9 hai l\u00e0m trong 6 gi\u1edd th\u00ec ho\u00e0n th\u00e0nh 25% c\u00f4ng vi\u1ec7c. H\u1ecfi m\u1ed7i ng\u01b0\u1eddi th\u1ee3 ch\u1ec9 l\u00e0m m\u1ed9t m\u00ecnh th\u00ec trong bao l\u00e2u l\u00e0m xong c\u00f4ng vi\u1ec7c? C\u00e2u 8. (3 \u0111i\u00ea\u0309m)Cho \u0111\u01b0\u1eddng tr\u00f2n t\u00e2m O b\u00e1n k\u00ednh OA v\u00e0 d\u00e2y cung MN vu\u00f4ng g\u00f3c OA(A n\u1eb1m tr\u00ean cung nh\u1ecf MN) . V\u1ebd d\u00e2y cung AB v\u00e0 d\u00e2y cung AC sao cho AB c\u1eaft MN t\u1ea1i I, AC c\u1eaft MN t\u1ea1i K theo th\u1ee9 t\u1ef1 M, I, K, N . a) Ch\u1ee9ng minh: T\u1ee9 gi\u00e1c BIKC n\u1ed9i ti\u1ebfp. b) G\u1ecdi R l\u00e0 giao c\u1ee7a AB v\u00e0 MC,S l\u00e0 giao c\u1ee7a AC v\u00e0 BN . Ch\u1ee9ng minh: MN \/\/ RS v\u00e0 AB.IR = AC.KS . c) Ch\u1ee9ng minh: MA l\u00e0 ti\u1ebfp tuy\u1ebfn c\u1ee7a \u0111\u01b0\u1eddng tr\u00f2n ngo\u1ea1i ti\u1ebfp \uf044MBI v\u00e0 \u0111\u01b0\u1eddng tr\u00f2n ngo\u1ea1i ti\u1ebfp \uf044MBI ti\u1ebfp x\u00fac v\u1edbi \u0111\u01b0\u1eddng tr\u00f2n ngo\u1ea1i ti\u1ebfp \uf044MCK . ----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\u00ea\u0309m): Cho Parabol (P) : y = \u2212 1 x2 v\u00e0 \u0111\u01b0\u01a1\u0300ng th\u1eb3ng (D) : y = \u2212 3 x \u22121. 44 a) V\u1ebd (P) v\u00e0 (D) tr\u00ean c\u00f9ng m\u1ed9t h\u1ec7 tr\u1ee5c t\u1ecda \u0111\u1ed9 0xy. b) B\u1eb1ng ph\u00e9p toa\u0301n xa\u0301c \u0111\u1ecbnh t\u1ecda \u0111\u1ed9 giao \u0111i\u00ea\u0309m c\u1ee7a (P) v\u00e0 (D). 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 \u2013 4 \u22122 02 4 y = \u2212 1 x2 \u2013 4 \u2013 1 0 \u20131 \u20134 4 x 04 y = \u2212 3 x \u22121 \u22121 \u22124 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) : \u2212 1 x2 = \u2212 3 x \u22121 44 \uf0db x2 \u2212 3x \u2212 4 = 0 \uf0db \uf0e9x = \u22121 \uf0ea\uf0ebx = 4 Thay x = \u22121 v\u00e0o y = \u2212 1 x2 , ta \u0111\u01b0\u1ee3c: y = \u2212 1 (\u22121)2 = \u2212 1 4 44 Thay x = 4 v\u00e0o y = \u2212 1 x2 , ta \u0111\u01b0\u1ee3c: y = \u2212 1 .(4)2 = \u22124 44 V\u1eady \uf0e6 \u22121; \u2212 1 \uf0f6 , (4; \u2212 4) l\u00e0 hai giao \u0111i\u1ec3m c\u1ea7n t\u00ecm. \uf0e7 4 \uf0f7 \uf0e8 \uf0f8 C\u00e2u 2. Cho ph\u01b0\u01a1ng tr\u00ecnh: x2 \u221211x + 5 = 0 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) Ch\u1ee9ng minh ph\u01b0\u01a1ng tr\u00ecnh lu\u00f4n c\u00f3 2 nghi\u1ec7m ph\u00e2n bi\u1ec7t x1, x2 r\u1ed3i t\u00ednh t\u1ed5ng v\u00e0 t\u00edch hai ngh\u1ec7m x1, x2 c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh. b) Kh\u00f4ng gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh h\u00e3y t\u00ednh gi\u00e1 tr\u1ecb c\u1ee7a bi\u1ec3u th\u1ee9c: A = \uf0e6 2 \u2212 2 \uf0f6 \uf0d7 ( x1 \u2212 x2 ) \uf0e7 x2 x1 \uf0f7 \uf0e8 \uf0f8 L\u1eddi gi\u1ea3i a) Ch\u1ee9ng minh ph\u01b0\u01a1ng tr\u00ecnh lu\u00f4n c\u00f3 2 nghi\u1ec7m ph\u00e2n bi\u1ec7t x1, x2 r\u1ed3i t\u00ednh t\u1ed5ng v\u00e0 t\u00edch hai ngh\u1ec7m x1, x2 c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh. V\u00ec \uf044 = b2 \u2212 4ac = (\u221211)2 \u2212 4.1.5 = 101 \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 = 11 \uf0ed = x1 a Theo \u0111\u1ecbnh l\u00ed Vi-et, ta c\u00f3: \uf0ef\uf0ef\uf0eeP c =5 a .x2 = b) Kh\u00f4ng gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh h\u00e3y t\u00ednh gia\u0301 tr\u1ecb c\u1ee7a bi\u00ea\u0309u th\u1ee9c: A = \uf0e6 2 \u2212 2 \uf0f6 \uf0d7 ( x1 \u2212 x2 ) \uf0e7 x2 x1 \uf0f7 \uf0e8 \uf0f8 Ta c\u00f3: A = \uf0e6 2 \u2212 2 \uf0f6 \uf0d7( x1 \u2212 x2 ) \uf0e7 x2 x1 \uf0f7 \uf0e8 \uf0f8 A = \uf0e6 2x1 \u2212 2x2 \uf0f6 \uf0d7( x1 \u2212 x2 ) \uf0e7 x1 x2 \uf0f7 \uf0e8 \uf0f8 A = 2 \uf0d7 ( x1 \u2212 x2 )2 x1x2 A = 2 \uf0d7 x12 \u2212 2x1x2 + x22 x1x2 A = 2 \uf0d7 ( x1 + x2 )2 \u2212 2x1x2 \u2212 2x1x2 x1x2 A = 2 \uf0d7 ( x1 + x2 )2 \u2212 4x1x2 x1x2 A = 2 \uf0d7 112 \u2212 4.5 = 202 55 C\u00e2u 3. (0,75 \u0111i\u00ea\u0309m) \u0110\u1ec3 \u0111\u1ea1t k\u1ebft qu\u1ea3 t\u1ed1t nh\u1ea5t trong k\u00ec thi tuy\u1ec3n sinh l\u1edbp 10 THPT v\u00e0o ng\u00e0y 02\/6\/2021, sau khi t\u1ed5 ch\u1ee9c H\u1ed9i tr\u1ea1i truy\u1ec1n th\u1ed1ng v\u00e0o th\u1ee9 S\u00e1u ng\u00e0y 26\/3\/2021, h\u1ecdc 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 sinh kh\u1ed1i 9 \u0111\u00e3 \u0111\u1ec1 ra k\u1ebf ho\u1ea1ch h\u1ecdc t\u1eadp m\u00f4n To\u00e1n c\u1ee5 th\u1ec3 nh\u01b0 sau: \\\"M\u1ed7i h\u1ecdc sinh b\u1eaft \u0111\u1ea7u t\u1eeb ng\u00e0y 27 \/ 3 \/ 2021 \u0111\u1ebfn h\u1ebft th\u00e1ng ba m\u1ed7i ng\u00e0y l\u00e0m 3 b\u00e0i to\u00e1n, m\u1ed7i ng\u00e0y trong th\u00e1ng t\u01b0 l\u00e0m 4 b\u00e0i to\u00e1n, m\u1ed7i ng\u00e0y trong th\u00e1ng n\u0103m l\u00e0m 5 b\u00e0i to\u00e1n\\\". Bi\u1ebft th\u00e1ng ba v\u00e0 th\u00e1ng n\u0103m l\u00e0 nh\u1eefng th\u00e1ng c\u00f3 31 ng\u00e0y, th\u00e1ng t\u01b0 c\u00f3 30 ng\u00e0y. H\u1ecfi: a) Theo k\u1ebf ho\u1ea1ch, m\u1ed7i h\u1ecdc sinh l\u00e0m \u0111\u01b0\u1ee3c bao nhi\u00eau b\u00e0i to\u00e1n? b) Ng\u00e0y thi 02\/6\/2021 l\u00e0 th\u1ee9 m\u1ea5y? Gi\u1ea3i th\u00edch v\u00ec sao? L\u1eddi gi\u1ea3i a) S\u1ed1 ng\u00e0y t\u1eeb 27 \/ 3 \/ 2021 \u0111\u1ebfn 31\/ 3 \/ 2021 S\u1ed1 b\u00e0i to\u00e1n m\u1ed7i h\u1ecdc sinh l\u00e0m \u0111\u01b0\u1ee3c theo k\u1ebf ho\u1ea1ch l\u00e0: 4.(3) + 30.(4) + 51.(5) = 287 (b\u00e0i to\u00e1n) b) S\u1ed1 ng\u00e0y t\u1eeb ng\u00e0y 26 \/ 3 \/ 2021 \u0111\u1ebfn ng\u00e0y 02 \/ 6 \/ 2021 l\u00e0: 5 30 31 2 68 (ng\u00e0y) Ta c\u00f3: 68 : 7 9 (d\u01b0 5 ) m\u00e0 ng\u00e0y 26 \/ 3 \/ 2021 l\u00e0 th\u1ee9 s\u00e1u n\u00ean ng\u00e0y 02 \/ 6 \/ 2021 l\u00e0 ng\u00e0y th\u1ee9 t\u01b0. C\u00e2u 4. (0,75 \u0111i\u00ea\u0309m) M\u1ed9t lon n\u01b0\u1edbc ng\u1ecdt c\u00f3 gi\u00e1 10000 \u0111\u1ed3ng. M\u1ed9t quy\u1ec3n t\u1eadp c\u00f3 gi\u00e1 b\u1eb1ng 2 5 gi\u00e1 m\u1ed9t lon n\u01b0\u1edbc ng\u1ecdt, m\u1ed9t h\u1ed9p b\u00fat c\u00f3 gi\u00e1 g\u1ea5p 3 l\u1ea7n gi\u00e1 m\u1ed9t lon n\u01b0\u1edbc ng\u1ecdt. B\u1ea1n An c\u1ea7n mua m\u1ed9t s\u1ed1 quy\u1ec3n t\u1eadp v\u00e0 m\u1ed9t h\u1ed9p b\u00fat. a) G\u1ecdi x l\u00e0 s\u1ed1 quy\u1ec3n t\u1eadp An mua v\u00e0 y l\u00e0 s\u1ed1 ti\u1ec1n An ph\u1ea3i tr\u1ea3 (bao g\u1ed3m ti\u1ec1n mua t\u1eadp v\u00e0 m\u1ed9t h\u1ed9p b\u00fat). Vi\u1ebft c\u00f4ng th\u1ee9c bi\u1ec3u di\u1ec5n y theo x . b) N\u1ebfu An b\u00e1n 2 th\u00f9ng n\u01b0\u1edbc ng\u1ecdt, m\u1ed7i th\u00f9ng g\u1ed3m 24 lon v\u1edbi gi\u00e1 \u0111\u00e3 n\u00eau tr\u00ean \u0111\u1ec3 mua t\u1eadp v\u00e0 m\u1ed9t h\u1ed9p b\u00fat th\u00ec t\u1ed1i \u0111a b\u1ea1n An mua \u0111\u01b0\u1ee3c bao nhi\u00eau quy\u1ec3n t\u1eadp? L\u1eddi gi\u1ea3i a) Gi\u00e1 ti\u00ea\u0300n m\u1ed9t quy\u00ea\u0309n t\u00e2\u0323p l\u00e0: 2 \uf0d7(10000) = 4000 (\u0111\u1ed3ng) 5 Gi\u00e1 ti\u00ea\u0300n m\u1ed9t h\u1ed9p b\u00fat l\u00e0: 3\uf0d7(10000) = 30000 (\u0111\u1ed3ng) C\u00f4ng th\u1ee9c bi\u1ec3u di\u1ec5n y theo x l\u00e0: y = 4000x + 30000 b) S\u1ed1 ti\u1ec1n b\u00e1n 2 th\u00f9ng n\u01b0\u1edbc ng\u1ecdt l\u00e0: 2.24.10000 = 480000 (\u0111\u1ed3ng) Th\u1ebf y 480000 v\u00e0o y = 4000x + 30000 , ta \u0111\u01b0\u1ee3c: 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 480000 = 4000x + 30000 \uf0db 4000x = 450000 \uf0db x = 112,5 V\u1eady b\u1ea1n An mua \u0111\u01b0\u1ee3c t\u1ed1i \u0111a 112 quy\u1ec3n t\u1eadp C\u00e2u 5. (1 \u0111i\u00ea\u0309m) M\u1ed9t c\u00f4ng ty giao cho c\u1eeda h\u00e0ng 100 h\u1ed9p b\u00e1nh \u0111\u1ec3 b\u00e1n ra th\u1ecb tr\u01b0\u1eddng. L\u00fac \u0111\u1ea7u c\u1eeda h\u00e0ng b\u00e1n 24 h\u1ed9p b\u00e1nh v\u1edbi gi\u00e1 b\u00e1n m\u1ed9t h\u1ed9p b\u00e1nh l\u00e0 200000 \u0111\u1ed3ng. Do nhu c\u1ea7u c\u1ee7a th\u1ecb tr\u01b0\u1eddng n\u00ean t\u1eeb h\u1ed9p b\u00e1nh th\u1ee9 25 \u0111\u1ebfn h\u1ed9p b\u00e1nh th\u1ee9 80 m\u1ed7i h\u1ed9p b\u00e1nh c\u00f3 gi\u00e1 b\u00e1n t\u0103ng 15% so v\u1edbi gi\u00e1 b\u00e1n l\u00fac \u0111\u1ea7u, t\u1eeb h\u1ed9p b\u00e1nh th\u1ee9 81 \u0111\u1ebfn h\u1ed9p b\u00e1nh th\u1ee9 100 m\u1ed7i h\u1ed9p b\u00e1nh c\u00f3 gi\u00e1 b\u00e1n gi\u1ea3m 10% so v\u1edbi gi\u00e1 b\u00e1n l\u00fac \u0111\u1ea7u. a) H\u1ecfi s\u1ed1 ti\u1ec1n thu c\u1eeda h\u00e0ng \u0111\u01b0\u1ee3c khi b\u00e1n 100 h\u1ed9p b\u00e1nh l\u00e0 bao nhi\u00eau? b) Bi\u1ebft r\u1eb1ng: V\u1edbi s\u1ed1 ti\u1ec1n thu \u0111\u01b0\u1ee3c khi b\u00e1n 100 h\u1ed9p b\u00e1nh, sau khi tr\u1eeb \u0111i 10% ti\u1ec1n thu\u1ebf gi\u00e1 tr\u1ecb gia t\u0103ng VAT c\u1eeda h\u00e0ng v\u1eabn l\u00e3i 1152000 \u0111\u1ed3ng. H\u1ecfi m\u1ed7i h\u1ed9p b\u00e1nh c\u00f4ng ty giao cho c\u1eeda h\u00e0ng c\u00f3 gi\u00e1 l\u00e0 bao nhi\u00eau? L\u1eddi gi\u1ea3i a) S\u1ed1 ti\u1ec1n thu \u0111\u01b0\u1ee3c khi b\u00e1n 24 h\u1ed9p b\u00e1nh \u0111\u1ea7u l\u00e0: 24.200000 = 4800000 (\u0111\u1ed3ng) S\u1ed1 ti\u1ec1n thu \u0111\u01b0\u1ee3c khi b\u00e1n \u0111\u01b0\u1ee3c t\u1eeb h\u1ed9p b\u00e1nh th\u1ee9 25 \u0111\u1ebfn h\u1ed9p b\u00e1nh th\u1ee9 80 l\u00e0: 56.200000(1+15%) = 12880000 (\u0111\u1ed3ng) S\u1ed1 ti\u1ec1n thu \u0111\u01b0\u1ee3c khi b\u00e1n \u0111\u01b0\u1ee3c t\u1eeb h\u1ed9p b\u00e1nh th\u1ee9 81 \u0111\u1ebfn h\u1ed9p b\u00e1nh th\u1ee9 100 l\u00e0: 20.200000(1\u221210%) = 3600000 (\u0111\u1ed3ng) S\u1ed1 ti\u1ec1n thu c\u1eeda h\u00e0ng \u0111\u01b0\u1ee3c khi b\u00e1n 100 h\u1ed9p b\u00e1nh l\u00e0: 4800000 +12880000 + 3600000 = 21280000 (\u0111\u1ed3ng) b) S\u1ed1 ti\u1ec1n thu\u1ebf gi\u00e1 tr\u1ecb gia t\u0103ng VAT l\u00e0 10%.21280000 = 2128000 (\u0111\u1ed3ng) S\u1ed1 ti\u1ec1n m\u1ed7i h\u1ed9p b\u00e1nh c\u00f4ng ty giao cho c\u1eeda h\u00e0ng c\u00f3 gi\u00e1 l\u00e0: \uf0e9\uf0eb(21280000 \u2212 2128000) \u22121152000\uf0f9\uf0fb :100 = 180000 (\u0111\u1ed3ng) C\u00e2u 6. (1 \u0111i\u00ea\u0309m) Ba xe m\u00e1y c\u00f9ng xu\u1ea5t ph\u00e1t t\u1eeb O \u0111i theo ba h\u01b0\u1edbng Ox,Oy,Oz trong \u0111\u00f3 Ox v\u00e0 Oz ng\u01b0\u1ee3c h\u01b0\u1edbng nhau nh\u01b0 h\u00ecnh v\u1ebd. y x Oz 6 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 Xe th\u1ee9 nh\u1ea5t \u0111i theo h\u01b0\u1edbng Ox , xe th\u1ee9 hai \u0111i theo h\u01b0\u1edbng Oy , xe th\u1ee9 ba \u0111i theo h\u01b0\u1edbng Oz , c\u1ea3 ba xe c\u00f9ng ch\u1ea1y v\u1edbi v\u1eadn t\u1ed1c kh\u00f4ng \u0111\u1ed5i l\u00e0 50km \/ gi\u1edd. Sau 2 gi\u1edd xe th\u1ee9 nh\u1ea5t v\u00e0 xe th\u1ee9 hai \u1edf c\u00e1ch nhau 107km . H\u1ecfi l\u00fac \u0111\u00f3 xe th\u1ee9 hai v\u00e0 xe th\u1ee9 ba \u1edf c\u00e1ch nhau bao nhi\u00eau ki-l\u00f4-m\u00e9t? (l\u00e0m tr\u00f2n k\u1ebft qu\u1ea3 \u0111\u1ebfn ch\u1eef s\u1ed1 h\u00e0ng \u0111\u01a1n v\u1ecb). L\u1eddi gi\u1ea3i y K B H x A O Cz G\u1ecdi A , B , C l\u1ea7n l\u01b0\u1ee3t l\u00e0 c\u00e1c v\u1ecb tr\u00ed m\u00e0 xe th\u1ee9 nh\u1ea5t, xe th\u1ee9 hai, xe th\u1ee9 ba sau khi xu\u1ea5t ph\u00e1t \u0111\u01b0\u1ee3c 2 gi\u1edd. V\u1ebd OH vu\u00f4ng g\u00f3c v\u1edbi AB t\u1ea1i H v\u00e0 OK vu\u00f4ng g\u00f3c v\u1edbi BC t\u1ea1i K . Ta ch\u1ee9ng \u0111\u01b0\u1ee3c H l\u00e0 trung \u0111i\u1ec3m c\u1ee7a AB v\u00e0 K l\u00e0 trung \u0111i\u1ec3m c\u1ee7a BC . Ta d\u1ec5 t\u00ednh \u0111\u01b0\u1ee3c OA = OB = OC = 100km ; AC = 200km ; BH = 53,5km X\u00e9t \uf044HOB vu\u00f4ng t\u1ea1i H , ta c\u00f3: OH 2 + BH 2 = OB2 (\u0111\u1ecbnh l\u00ed Pitago) \uf0de OH 2 + 53, 52 = 1002 \uf0de OH 2 = 7137, 75 \uf0de OH = 7137, 75 M\u00e0 OH = BK (t\u1ee9 gi\u00e1c OHBK l\u00e0 h\u00ecnh ch\u1eef nh\u1eadt) N\u00ean BK = 7137, 75 M\u1eb7t kh\u00e1c: BC = 2BK (v\u00ec K l\u00e0 trung \u0111i\u1ec3m c\u1ee7a BC ) N\u00ean BC = 2 7137, 75 \uf0bb169 km C\u00e2u 7. (1 \u0111i\u00ea\u0309m)Hai ng\u01b0\u1eddi th\u1ee3 c\u00f9ng l\u00e0m m\u1ed9t c\u00f4ng vi\u1ec7c trong 16 gi\u1edd th\u00ec xong. N\u1ebfu ng\u01b0\u1eddi th\u1ee3 th\u1ee9 nh\u1ea5t l\u00e0m trong 3 gi\u1edd, ng\u01b0\u1eddi th\u1ee3 th\u1ee9 hai l\u00e0m trong 6 gi\u1edd th\u00ec ho\u00e0n th\u00e0nh 25% c\u00f4ng vi\u1ec7c. H\u1ecfi m\u1ed7i ng\u01b0\u1eddi th\u1ee3 ch\u1ec9 l\u00e0m m\u1ed9t m\u00ecnh th\u00ec trong bao l\u00e2u l\u00e0m xong c\u00f4ng vi\u1ec7c? 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 G\u1ecdi x (gi\u1edd) l\u00e0 th\u1eddi gian ng\u01b0\u1eddi th\u1ee9 nh\u1ea5t l\u00e0m xong c\u00f4ng vi\u1ec7c m\u1ed9t m\u00ecnh x \uf03e 16 y (gi\u1edd) l\u00e0 th\u1eddi gian ng\u01b0\u1eddi th\u1ee9 hai l\u00e0m xong c\u00f4ng vi\u1ec7c m\u1ed9t m\u00ecnh y \uf03e 16 Trong m\u1ed7i gi\u1edd ng\u01b0\u1eddi th\u1ee9 nh\u1ea5t l\u00e0m \u0111\u01b0\u1ee3c: 1 (c\u00f4ng vi\u1ec7c) x Trong m\u1ed7i gi\u1edd ng\u01b0\u1eddi th\u1ee9 hai l\u00e0m \u0111\u01b0\u1ee3c: 1 (c\u00f4ng vi\u1ec7c) y V\u00ec hai ng\u01b0\u1eddi th\u1ee3 c\u00f9ng l\u00e0m m\u1ed9t c\u00f4ng vi\u1ec7c trong 16 gi\u1edd th\u00ec xong n\u00ean m\u1ed7i gi\u1edd c\u1ea3 hai ng\u01b0\u1eddi l\u00e0m \u0111\u01b0\u1ee3c 1 (c\u00f4ng vi\u1ec7c) 16 \uf0ec 1 + 1 =1 \uf0ef\uf0ef x y 16 Theo \u0111\u1ec1 b\u00e0i, ta c\u00f3 h\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh: \uf0ed \uf0ef\uf0ef\uf0ee3 \uf0d7 1 + 6\uf0d7 1 = x y 25% \uf0ec\uf0ef\uf0ef\uf0ed\uf0ef\uf0ef\uf0ee3aa++b6=b1=1614 \uf0ef\uf0ec\uf0efa = 1 \uf0ec1 = 1 \uf0ed = 24 \uf0ef\uf0ef x = 24 \u0110\u1eb7t a = 1 ;b = 1 . H\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh tr\u1edf th\u00e0nh: \uf0db \uf0ef\uf0ef\uf0eeb 1 \uf0de \uf0ed 1 \uf0db \uf0ecx = 24 x y 48 \uf0ef 48 \uf0ed = 48 1 \uf0ee y \uf0ef\uf0ee y KL: V\u1eady ng\u01b0\u1eddi th\u1ee3 th\u1ee9 nh\u1ea5t ch\u1ec9 l\u00e0m m\u1ed9t m\u00ecnh trong 48 gi\u1edd th\u00ec xong c\u00f4ng vi\u1ec7c ng\u01b0\u1eddi th\u1ee3 th\u1ee9 hai ch\u1ec9 l\u00e0m m\u1ed9t m\u00ecnh trong 24 gi\u1edd th\u00ec xong c\u00f4ng vi\u1ec7c. C\u00e2u 8. (3 \u0111i\u00ea\u0309m)Cho \u0111\u01b0\u1eddng tr\u00f2n t\u00e2m O b\u00e1n k\u00ednh OA v\u00e0 d\u00e2y cung MN vu\u00f4ng g\u00f3c OA(A n\u1eb1m tr\u00ean cung nh\u1ecf MN) . V\u1ebd d\u00e2y cung AB v\u00e0 d\u00e2y cung AC sao cho AB c\u1eaft MN t\u1ea1i I, AC c\u1eaft MN t\u1ea1i K theo th\u1ee9 t\u1ef1 M, I, K, N . a) Ch\u1ee9ng minh: T\u1ee9 gi\u00e1c BIKC n\u1ed9i ti\u1ebfp. b) G\u1ecdi R l\u00e0 giao c\u1ee7a AB v\u00e0 MC,S l\u00e0 giao c\u1ee7a AC v\u00e0 BN . Ch\u1ee9ng minh: MN \/\/ RS v\u00e0 AB.IR = AC.KS . c) Ch\u1ee9ng minh: MA l\u00e0 ti\u1ebfp tuy\u1ebfn c\u1ee7a \u0111\u01b0\u1eddng tr\u00f2n ngo\u1ea1i ti\u1ebfp \uf044MBI v\u00e0 \u0111\u01b0\u1eddng tr\u00f2n ngo\u1ea1i ti\u1ebfp \uf044MBI ti\u1ebfp x\u00fac v\u1edbi \u0111\u01b0\u1eddng tr\u00f2n ngo\u1ea1i ti\u1ebfp \uf044MCK . L\u1eddi gi\u1ea3i 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 MB R I A K S C N a) Ch\u1ee9ng minh: T\u1ee9 gi\u00e1c BIKC n\u1ed9i ti\u1ebfp. ( )X\u00e9t O , ta c\u00f3: \uf0ecOA la\u00f8 ba\u00f9n k\u00ednh (gt ) ( )\uf0ef\uf0ef\uf0edMN la\u00f8 da\u00e2y cung gt \uf0de A la\u00f8 \u00f1ie\u00e5m ch\u00ednh gi\u00f6\u00f5a MN \uf0ef\uf0ef\uf0eeOA \u22a5 NM (gt) Ta c\u00f3: ( )( )\uf0ec\uf0efAIN=1 2 \uf0ed s\u00f1MB + s\u00f1AN go\u00f9c co\u00f9 \u00f1\u00e6nh be\u00e2n trong \u00f1\u00f6\u00f4\u00f8ng tro\u00f8n cha\u00e9n MB va\u00f8 AN ( )\uf0efs\u00f1AN = s\u00f1AM A la\u00f8 \u00f1ie\u00e5m ch\u00ednh gi\u00f6\u00f5a MN \uf0ee ( )\uf0de AIN = 1 s\u00f1MB + s\u00f1AM 2 \uf0de AIN = 1 s\u00f1AB 2 ( )( )M\u00e0 ACB = 1 s\u00f1AB go\u00f9c no\u00e4i tie\u00e1p cha\u00e9n AB cu\u00fba O 2 N\u00ean ACB = AIN \uf0de T\u1ee9 gi\u00e1c BIKC n\u1ed9i ti\u1ebfp (T\u1ee9 gi\u00e1c c\u00f3 g\u00f3c ngo\u00e0i b\u1eb1ng g\u00f3c \u0111\u1ed1i trong) b) G\u1ecdi R l\u00e0 giao c\u1ee7a AB v\u00e0 MC,S l\u00e0 giao c\u1ee7a AC v\u00e0 BN . Ch\u1ee9ng minh: MN \/\/ RS v\u00e0 AB.IR = AC.KS. Ta c\u00f3: 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 ( )( )\uf0ef\uf0ecRBC = \uf0ef 1 s\u00f1AN go\u00f9c no\u00e4i tie\u00e1p cha\u00e9n AN cu\u00fba O 2 1 s\u00f1AM 2 ( )( )\uf0ef\uf0edRCS = cu\u00fba \uf0ef go\u00f9c no\u00e4i tie\u00e1p cha\u00e9n AM O ( )\uf0ef\uf0ef\uf0ees\u00f1AN = s\u00f1AM A la\u00f8 \u00f1ie\u00e5m ch\u00ednh gi\u00f6\u00f5a MN \uf0de RBC = RCS \uf0de T\u1ee9 gi\u00e1c BCSR n\u1ed9i ti\u1ebfp (T\u1ee9 gi\u00e1c c\u00f3 2 \u0111\u1ec9nh li\u00ean ti\u1ebfp c\u00f9ng nh\u00ecn 1 c\u1ea1nh d\u01b0\u1edbi 2 g\u00f3c b\u1eb1ng nhau) \uf0de IRS = BCS Ma\u00f8 BCS = AIK (T\u00f6\u00f9 gia\u00f9c BCKI no\u00e4i tie\u00e1p) Ne\u00e2n IRS = AIK \uf0de MN \/\/ RS(2 go\u00f9c na\u00f8y \u00f4\u00fb v\u00f2 tr\u00ed \u00f1o\u00e0ng v\u00f2) X\u00e9t \uf044AIK v\u00e0 \uf044ACB, ta c\u00f3: \uf0ec\uf0efAIK = ACB(t\u00f6\u00f9 gia\u00f9c BCKI no\u00e4i tie\u00e1p) \uf0ed \uf0ef\uf0eeIAK = CAB(go\u00f9c chung ) \uf0de \uf044AIK \u223d \uf044ACB(g \u2212 g) \uf0de AI = AK (TS\u00d1D) AC AB \uf0de AI.AB = AK.AC Ch\u1ee9ng minh t\u01b0\u01a1ng t\u1ef1 ta c\u00f3: AR.AB = AS.AC Ta c\u00f3: \uf0ecAR.AB = AS.AC ( cmt ) \uf0ed\uf0eeAI.AB = AK.AC \uf0de AR.AB \u2212 AI.AB = AS.AC \u2212 AK.AC \uf0de AB(AR \u2212 AI) = AC(AS \u2212 AK) \uf0de AB.IR = AC.KS c) Ch\u1ee9ng minh: MA l\u00e0 ti\u1ebfp tuy\u1ebfn c\u1ee7a \u0111\u01b0\u1eddng tr\u00f2n ngo\u1ea1i ti\u1ebfp \uf044MBI v\u00e0 \u0111\u01b0\u1eddng tr\u00f2n ngo\u1ea1i ti\u1ebfp \uf044MBI ti\u1ebfp x\u00fac v\u1edbi \u0111\u01b0\u1eddng tr\u00f2n ngo\u1ea1i ti\u1ebfp \uf044MCK . ( ) ( )G\u1ecdi E,F l\u1ea7n l\u01b0\u1ee3t l\u00e0 t\u00e2m MBI ; MCK Ta c\u00f3: ( )( )AMI = MBI 2 go\u00f9c no\u00e4i tie\u00e1p cha\u00e9n 2 cung AM va\u00f8 AN ba\u00e8ng nhau cu\u00fba O 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 MA l\u00e0 ti\u1ebfp tuy\u1ebfn c\u1ee7a MBI t\u1ea1i M \uf0de AM \u22a5 ME t\u1ea1i M 2 Ch\u1ee9ng minh t\u01b0\u01a1ng t\u1ef1 ta c\u00f3: AM \u22a5 MF t\u1ea1i M 2 T\u1eeb 1 v\u00e0 2 \uf0de ME \uf0ba MF \uf0de M,E,F th\u1eb3ng h\u00e0ng \uf0de E n\u1eb1m gi\u1eefa F v\u00e0 M \uf0de FE = MF \u2212 ME ( ) ( ) ( ) ( )M\u00e0 FE l\u00e0 kho\u1ea3ng c\u00e1ch 2 t\u00e2m c\u1ee7a MBI ; MCK , MF, ME l\u00e0 b\u00e1n k\u00ednh c\u1ee7a MBI ; MCK \uf0de \u0110\u01b0\u1eddng tr\u00f2n ngo\u1ea1i ti\u1ebfp \uf044MBI ti\u1ebfp x\u00fac v\u1edbi \u0111\u01b0\u1eddng tr\u00f2n ngo\u1ea1i ti\u1ebfp \uf044MCK ----H\u1ebeT--- 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 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 4 NA\u00caM HO\u00cfC: 2023 - 2024 M\u00d4N: TO\u00c1N 9 \u0110\u1ec0 THAM KH\u1ea2O M\u00c3 \u0110\u1ec0: Qu\u1eadn 4 \u2013 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 (P) : y = \u22121 x2 v\u00e0 \u0111\u01b0\u1eddng th\u1eb3ng (d) : y = x \u2212 3 . 22 a) V\u1ebd (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 2x2 + 5x \u2212 9 = 0 a) Ch\u1ee9ng t\u1ecf ph\u01b0\u01a1ng tr\u00ecnh lu\u00f4n c\u00f3 nghi\u1ec7m. b) Kh\u00f4ng gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh, t\u00ednh N 11 x1 1 x2 1 . C\u00e2u 3. (0,75 \u0111i\u1ec3m). Quy t\u1eafc sau \u0111\u00e2y cho bi\u1ebft CAN, CHI c\u1ee7a n\u0103m X n\u00e0o \u0111\u00f3. \u2022 \u0110\u1ec3 x\u00e1c \u0111\u1ecbnh CAN, ta t\u00ecm s\u1ed1 d\u01b0 r trong ph\u00e9p chia X cho 10 v\u00e0 tra v\u00e0o b\u1ea3ng 1 \u2022 \u0110\u1ec3 x\u00e1c \u0111\u1ecbnh CHI, ta t\u00ecm s\u1ed1 d\u01b0 s trong ph\u00e9p chia X cho 12 v\u00e0 tra v\u00e0o b\u1ea3ng 2 V\u00ed d\u1ee5: N\u0103m 2020 c\u00f3 CAN l\u00e0 Canh, c\u00f3 CHI l\u00e0 T\u00ed. B\u1ea3ng 1 r 0 1 2 3 456 7 89 CAN Canh T\u00e2n Nh\u00e2m Qu\u00fd Gi\u00e1p \u1ea4t B\u00ednh \u0110inh M\u1eadu K\u1ef7 B\u1ea3ng 2 s 0 1 2 3 4 5 6 7 8 9 10 11 CHI Th\u00e2n D\u1eadu Tu\u1ea5t H\u1ee3i T\u00ed S\u1eedu D\u1ea7n M\u1eb9o Th\u00ecn T\u1ef5 Ng\u1ecd M\u00f9i a) Em h\u00e3y s\u1eed d\u1ee5ng quy t\u1eafc tr\u00ean \u0111\u1ec3 x\u00e1c \u0111\u1ecbnh CAN CHI c\u1ee7a n\u0103m 2023 ? b) L\u00fd Th\u00e1i T\u1ed5 (L\u00fd C\u00f4ng U\u1ea9n) l\u00e0 v\u1ecb vua \u0111\u1ea7u ti\u00ean \u0111\u00e3 m\u1edf n\u00ean tri\u1ec1u \u0111\u1ea1i L\u00fd ph\u1ed3n th\u1ecbnh trong su\u1ed1t trong 200 n\u0103m. \u00d4ng l\u00ean ng\u00f4i v\u00e0o n\u0103m K\u1ef7 D\u1eadu \u0111\u1ea7u th\u1ebf k\u1ef7 XI . Em h\u00e3y cho bi\u1ebft \u00f4ng l\u00ean ng\u00f4i v\u00e0o n\u0103m n\u00e0o? C\u00e2u 4. (0,75 \u0111i\u1ec3m). V\u1edbi s\u1ef1 ph\u00e1t tri\u1ec3n c\u1ee7a khoa h\u1ecdc k\u1ef9 thu\u1eadt, 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\u1ea7u t\u01b0 v\u00e0 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 450 000 000 \u0111\u1ed3ng, chi ph\u00ed \u0111\u1ec3 s\u1ea3n xu\u1ea5t ra 1 chi\u1ebfc xe l\u0103n l\u00e0 2 000 000 \u0111\u1ed3ng, gi\u00e1 b\u00e1n ra m\u1ed7i chi\u1ebfc l\u00e0 3 500 000 \u0111\u1ed3ng. a) G\u1ecdi x l\u00e0 s\u1ed1 xe \u0111\u01b0\u1ee3c s\u1ea3n xu\u1ea5t ra v\u00e0 y l\u00e0 t\u1ed5ng s\u1ed1 ti\u1ec1n \u0111\u00e3 \u0111\u1ea7u t\u01b0 (g\u1ed3m v\u1ed1n ban \u0111\u1ea7u v\u00e0 chi ph\u00ed s\u1ea3n xu\u1ea5t). H\u00e3y l\u1eadp c\u00f4ng th\u1ee9c y theo x . b) C\u00f4ng ty A ph\u1ea3i b\u00e1n bao nhi\u00eau chi\u1ebfc xe m\u1edbi thu h\u1ed3i \u0111\u01b0\u1ee3c v\u1ed1n? 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\u1ed9t c\u1eeda h\u00e0ng b\u00e1nh pizza c\u00f3 ch\u01b0\u01a1ng tr\u00ecnh khuy\u1ebfn m\u00e3i gi\u1ea3m 70% cho b\u00e1nh pizza th\u1ee9 2 c\u00f9ng size c\u00f3 gi\u00e1 b\u1eb1ng ho\u1eb7c th\u1ea5p h\u01a1n pizza th\u1ee9 nh\u1ea5t. Bi\u1ebft b\u00e1nh pizza c\u00f3 gi\u00e1 ban \u0111\u1ea7u l\u00e0 210 000 \u0111\u1ed3ng. a) H\u1ecfi n\u1ebfu kh\u00e1ch h\u00e0ng mua 10 b\u00e1nh pizza th\u00ec ph\u1ea3i tr\u1ea3 bao nhi\u00eau ti\u1ec1n? b) C\u1eeda h\u00e0ng c\u00f3 ch\u01b0\u01a1ng tr\u00ecnh khuy\u1ebfn m\u00e3i th\u00eam, n\u1ebfu h\u00f3a \u0111\u01a1n tr\u00ean 2 000 000 \u0111\u1ed3ng th\u00ec \u0111\u01b0\u1ee3c gi\u1ea3m th\u00eam 5% tr\u00ean t\u1ed5ng s\u1ed1 ti\u1ec1n ph\u1ea3i tr\u1ea3. H\u1ecfi n\u1ebfu kh\u00e1ch h\u00e0ng mua 15 b\u00e1nh pizza th\u00ec ph\u1ea3i tr\u1ea3 bao nhi\u00eau ti\u1ec1n? C\u00e2u 6. (1 \u0111i\u1ec3m). M\u1ed9t b\u1ed3n n\u01b0\u1edbc h\u00ecnh tr\u1ee5 c\u00f3 b\u00e1n k\u00ednh \u0111\u00e1y l\u00e0 3m , chi\u1ec1u cao l\u00e0 4m . Ng\u01b0\u1eddi ta \u0111\u1ed5 n\u01b0\u1edbc v\u00e0o trong b\u1ed3n sao cho chi\u1ec1u cao c\u1ee7a n\u01b0\u1edbc b\u1eb1ng \u0111\u00fang m\u1ed9t n\u1eeda chi\u1ec1u cao c\u1ee7a b\u1ed3n v\u00e0 ti\u1ebfp t\u1ee5c \u0111\u1eb7t v\u00e0o trong b\u1ed3n m\u1ed9t phao n\u01b0\u1edbc c\u00f3 d\u1ea1ng h\u00ecnh c\u1ea7u b\u1eb1ng kim lo\u1ea1i kh\u00f4ng th\u1ea5m n\u01b0\u1edbc c\u00f3 b\u00e1n k\u00ednh l\u00e0 50cm v\u00e0 ch\u00ecm ho\u00e0n to\u00e0n trong n\u01b0\u1edbc. a) H\u1ecfi khi \u0111\u00f3 m\u1ef1c n\u01b0\u1edbc trong b\u1ed3n cao bao nhi\u00eau m\u00e9t (l\u00e0m tr\u00f2n \u0111\u1ebfn ch\u1eef s\u1ed1 th\u1eadp ph\u00e2n th\u1ee9 ba)? b) Sau \u0111\u00f3, ng\u01b0\u1eddi ta l\u1ea1i b\u01a1m th\u00eam n\u01b0\u1edbc v\u00e0o b\u1ed3n b\u1eb1ng m\u1ed9t v\u00f2i c\u00f3 c\u00f4ng su\u1ea5t ch\u1ea3y l\u00e0 30, 0024 l\u00edt cho m\u1ed7i gi\u00e2y. H\u1ecfi sau bao nhi\u00eau ph\u00fat th\u00ec b\u1ed3n \u0111\u1ea7y n\u01b0\u1edbc (l\u00e0m tr\u00f2n \u0111\u1ebfn h\u00e0ng \u0111\u01a1n v\u1ecb)? C\u00e2u 7. (1 \u0111i\u1ec3m). M\u1ed9t c\u00f4ng ty c\u00f3 100 xe ch\u1edf kh\u00e1ch g\u1ed3m hai lo\u1ea1i, lo\u1ea1i xe ch\u1edf \u0111\u01b0\u1ee3c 30 kh\u00e1ch v\u00e0 lo\u1ea1i xe ch\u1edf \u0111\u01b0\u1ee3c 50 kh\u00e1ch. N\u1ebfu d\u00f9ng t\u1ea5t c\u1ea3 s\u1ed1 xe \u0111\u00f3 th\u00ec t\u1ed1i \u0111a c\u00f4ng ty ch\u1edf m\u1ed9t l\u1ea7n \u0111\u01b0\u1ee3c 4300 kh\u00e1ch. H\u1ecfi m\u1ed7i lo\u1ea1i c\u00f4ng ty \u0111\u00f3 c\u00f3 m\u1ea5y xe? C\u00e2u 8. (3 \u0111i\u1ec3m) Cho ABC ( AB \uf03c AC) n\u1ed9i ti\u1ebfp O;R \u0111\u01b0\u1eddng k\u00ednh BC , tr\u00ean cung nh\u1ecf AC l\u1ea5y \u0111i\u1ec3m D , BD c\u1eaft AC t\u1ea1i E , t\u1eeb E v\u1ebd EF BC t\u1ea1i F . a) Ch\u1ee9ng minh t\u1ee9 gi\u00e1c BAEF n\u1ed9i ti\u1ebfp \u0111\u01b0\u1eddng tr\u00f2n. b) Ch\u1ee9ng minh DB l\u00e0 ph\u00e2n gi\u00e1c g\u00f3c ADF . c) G\u1ecdi M l\u00e0 trung \u0111i\u1ec3m EC . Ch\u1ee9ng minh DM.CA CF.CO . ----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 (P) : y = \u22121 x2 v\u00e0 \u0111\u01b0\u1eddng th\u1eb3ng (d) : y = x \u2212 3 . 22 a) V\u1ebd (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 y a) V\u1ebd (P) v\u00e0 (d) tr\u00ean c\u00f9ng h\u1ec7 tr\u1ee5c t\u1ecda \u0111\u1ed9. 1 BGT: -4 -2 2 (d) 0 x 4 20 2 4 24 -3 x 2 y 1x2 8 20 28 2 -2 x 02 yx3 31 2 22 (P) -8 b) T\u00ecm t\u1ecda \u0111\u1ed9 giao \u0111i\u1ec3m c\u1ee7a (P) v\u00e0 (d) b\u1eb1ng ph\u00e9p to\u00e1n. Ph\u01b0\u01a1ng tr\u00ecnh ho\u00e0nh \u0111\u1ed9 giao \u0111i\u1ec3m c\u1ee7a (P) v\u00e0 (d) : 1x2 x 3 0 22 1x2 x 3 22 x1 x3 Thay x = 1 v\u00e0o y = \u22121 x2 , ta \u0111\u01b0\u1ee3c: y 1 .12 1 2 2 2. Thay x 3 v\u00e0o y 1x2 ta \u0111\u01b0\u1ee3c: y 1. 3 2 9 2 2 2. V\u1eady 1; 1 ; 3; 9 l\u00e0 hai giao \u0111i\u1ec3m c\u1ea7n t\u00ecm. 2 2 C\u00e2u 2. (1 \u0111i\u1ec3m) Cho ph\u01b0\u01a1ng tr\u00ecnh 2x 2 5x 9 0 . 3 a) Ch\u1ee9ng t\u1ecf ph\u01b0\u01a1ng tr\u00ecnh lu\u00f4n c\u00f3 nghi\u1ec7m. 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 b) Kh\u00f4ng gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh, t\u00ednh N 11 x1 1 x2 1 . L\u1eddi gi\u1ea3i V\u00ec \uf044 = b2 \u2212 4ac = 52 \u2212 4.2.(\u22129) = 97 \uf03e 0 N\u00ean ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m ph\u00e2n bi\u1ec7t x1 ,x2 . \uf0ef\uf0ef\uf0ecS = x1 + x2 = \u2212b = \u22125 \uf0ed = x1 a 2 Theo \u0111\u1ecbnh l\u00ed Vi-et, ta c\u00f3: \uf0ef\uf0ee\uf0efP \u22129 c = 2 .x2 = a Ta c\u00f3: N 11 x1 1 x2 1 N x2 1 x1 1 x1 x2 2 S2 5 2 9 x1 1 x2 1 x1x2 x1 x2 PS1 2 2. 1 9 5 1 2 2 C\u00e2u 3. (0,75 \u0111i\u1ec3m) Quy t\u1eafc sau \u0111\u00e2y cho bi\u1ebft CAN, CHI c\u1ee7a n\u0103m X n\u00e0o \u0111\u00f3. \u2022 \u0110\u1ec3 x\u00e1c \u0111\u1ecbnh CAN, ta t\u00ecm s\u1ed1 d\u01b0 r trong ph\u00e9p chia X cho 10 v\u00e0 tra v\u00e0o b\u1ea3ng 1 \u2022 \u0110\u1ec3 x\u00e1c \u0111\u1ecbnh CHI, ta t\u00ecm s\u1ed1 d\u01b0 s trong ph\u00e9p chia X cho 12 v\u00e0 tra v\u00e0o b\u1ea3ng 2 V\u00ed d\u1ee5: N\u0103m 2020 c\u00f3 CAN l\u00e0 Canh, c\u00f3 CHI l\u00e0 T\u00ed. B\u1ea3ng 1 r 0 1 2 3 456 7 89 CAN Canh T\u00e2n Nh\u00e2m Qu\u00fd Gi\u00e1p \u1ea4t B\u00ednh \u0110inh M\u1eadu K\u1ef7 B\u1ea3ng 2 s 0 1 2 3 4 5 6 7 8 9 10 11 CHI Th\u00e2n D\u1eadu Tu\u1ea5t H\u1ee3i T\u00ed S\u1eedu D\u1ea7n M\u1eb9o Th\u00ecn T\u1ef5 Ng\u1ecd M\u00f9i a) Em h\u00e3y s\u1eed d\u1ee5ng quy t\u1eafc tr\u00ean \u0111\u1ec3 x\u00e1c \u0111\u1ecbnh CAN CHI c\u1ee7a n\u0103m 2023 ? b) L\u00fd Th\u00e1i T\u1ed5 (L\u00fd C\u00f4ng U\u1ea9n) l\u00e0 v\u1ecb vua \u0111\u1ea7u ti\u00ean \u0111\u00e3 m\u1edf n\u00ean tri\u1ec1u \u0111\u1ea1i L\u00fd ph\u1ed3n th\u1ecbnh trong su\u1ed1t trong 200 n\u0103m. \u00d4ng l\u00ean ng\u00f4i v\u00e0o n\u0103m K\u1ef7 D\u1eadu \u0111\u1ea7u th\u1ebf k\u1ef7 11 . Em h\u00e3y cho bi\u1ebft \u00f4ng l\u00ean ng\u00f4i v\u00e0o n\u0103m n\u00e0o? L\u1eddi gi\u1ea3i a) Em h\u00e3y s\u1eed d\u1ee5ng quy t\u1eafc tr\u00ean \u0111\u1ec3 x\u00e1c \u0111\u1ecbnh CAN CHI c\u1ee7a n\u0103m 2023 ? 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 2023 : 10 202 d\u01b0 3 n\u00ean sau khi tra v\u00e0o b\u1ea3ng 1 , ta \u0111\u01b0\u1ee3c CAN l\u00e0 Qu\u00fd. V\u00ec 2023 : 12 168 d\u01b0 7 n\u00ean sau khi tra v\u00e0o b\u1ea3ng 2 , ta \u0111\u01b0\u1ee3c CHI l\u00e0 M\u1eb9o. V\u1eady n\u0103m 2023 c\u00f3 CAN CHI l\u00e0 Qu\u00fd M\u1eb9o. b) Em h\u00e3y cho bi\u1ebft \u00f4ng l\u00ean ng\u00f4i v\u00e0o n\u0103m n\u00e0o? G\u1ecdi X l\u00e0 n\u0103m L\u00fd Th\u00e1i T\u1ed5 l\u00ean ng\u00f4i. V\u00ec n\u0103m X \u1edf th\u1ebf k\u1ef7 11 n\u00ean 1001 X 1100 Tra v\u00e0o b\u1ea3ng 1 ta \u0111\u01b0\u1ee3c r 9 ; tra v\u00e0o b\u1ea3ng 2 ta \u0111\u01b0\u1ee3c s 1 V\u00ec n\u0103m X l\u00e0 n\u0103m K\u1ef7 D\u1eadu n\u00ean: X : 10 d\u01b0 9 X 1019;1029;1039;1049;1059;1069;1079;1089;1099 X : 12 d\u01b0 1 X 1069 V\u1eady L\u00fd Th\u00e1i T\u1ed5 l\u00ean ng\u00f4i n\u00e0o n\u0103m 1069 . C\u00e2u 4. (0,75 \u0111i\u1ec3m). V\u1edbi s\u1ef1 ph\u00e1t tri\u1ec3n c\u1ee7a khoa h\u1ecdc k\u1ef9 thu\u1eadt, 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\u1ea7u t\u01b0 v\u00e0 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 450 000 000 \u0111\u1ed3ng, chi ph\u00ed \u0111\u1ec3 s\u1ea3n xu\u1ea5t ra 1 chi\u1ebfc xe l\u0103n l\u00e0 2 000 000 \u0111\u1ed3ng, gi\u00e1 b\u00e1n ra m\u1ed7i chi\u1ebfc l\u00e0 3 500 000 \u0111\u1ed3ng. a) G\u1ecdi x l\u00e0 s\u1ed1 xe \u0111\u01b0\u1ee3c s\u1ea3n xu\u1ea5t ra v\u00e0 y l\u00e0 t\u1ed5ng s\u1ed1 ti\u1ec1n \u0111\u00e3 \u0111\u1ea7u t\u01b0 (g\u1ed3m v\u1ed1n ban \u0111\u1ea7u v\u00e0 chi ph\u00ed s\u1ea3n xu\u1ea5t). H\u00e3y l\u1eadp c\u00f4ng th\u1ee9c y theo x . b) C\u00f4ng ty A ph\u1ea3i b\u00e1n bao nhi\u00eau chi\u1ebfc xe m\u1edbi thu h\u1ed3i \u0111\u01b0\u1ee3c v\u1ed1n? L\u1eddi gi\u1ea3i a) H\u00e3y l\u1eadp c\u00f4ng th\u1ee9c y theo x . C\u00f4ng th\u1ee9c y theo x l\u00e0: y 2 000 000x 450 000 000 (\u0111\u1ed3ng) b) C\u00f4ng ty A ph\u1ea3i b\u00e1n bao nhi\u00eau chi\u1ebfc xe m\u1edbi thu h\u1ed3i \u0111\u01b0\u1ee3c v\u1ed1n? S\u1ed1 ti\u1ec1n thu \u0111\u01b0\u1ee3c khi b\u00e1n x chi\u1ebfc xe l\u00e0: 3 500 000x (\u0111\u1ed3ng) \u0110\u1ec3 c\u00f4ng ty A thu h\u1ed3i v\u1ed1n th\u00ec 3 500 000x 2 000 000x 450 000 000 1500 000x 450 000 000 x 300 V\u1eady c\u00f4ng ty A ph\u1ea3i b\u00e1n 300 chi\u1ebfc xe m\u1edbi thu h\u1ed3i \u0111\u01b0\u1ee3c v\u1ed1n. 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 C\u00e2u 5. (1 \u0111i\u1ec3m) M\u1ed9t c\u1eeda h\u00e0ng b\u00e1nh pizza c\u00f3 ch\u01b0\u01a1ng tr\u00ecnh khuy\u1ebfn m\u00e3i gi\u1ea3m 70% cho b\u00e1nh pizza th\u1ee9 2 c\u00f9ng size c\u00f3 gi\u00e1 b\u1eb1ng ho\u1eb7c th\u1ea5p h\u01a1n pizza th\u1ee9 nh\u1ea5t. Bi\u1ebft b\u00e1nh pizza c\u00f3 gi\u00e1 ban \u0111\u1ea7u l\u00e0 210 000 \u0111\u1ed3ng. a) H\u1ecfi n\u1ebfu kh\u00e1ch h\u00e0ng mua 10 b\u00e1nh pizza th\u00ec ph\u1ea3i tr\u1ea3 bao nhi\u00eau ti\u1ec1n? b) C\u1eeda h\u00e0ng c\u00f3 ch\u01b0\u01a1ng tr\u00ecnh khuy\u1ebfn m\u00e3i th\u00eam, n\u1ebfu h\u00f3a \u0111\u01a1n tr\u00ean 2 000 000 \u0111\u1ed3ng th\u00ec \u0111\u01b0\u1ee3c gi\u1ea3m th\u00eam 5% tr\u00ean t\u1ed5ng s\u1ed1 ti\u1ec1n ph\u1ea3i tr\u1ea3. H\u1ecfi n\u1ebfu kh\u00e1ch h\u00e0ng mua 15 b\u00e1nh pizza th\u00ec ph\u1ea3i tr\u1ea3 bao nhi\u00eau ti\u1ec1n? L\u1eddi gi\u1ea3i a) H\u1ecfi n\u1ebfu kh\u00e1ch h\u00e0ng mua 10 b\u00e1nh pizza th\u00ec ph\u1ea3i tr\u1ea3 bao nhi\u00eau ti\u1ec1n? Gi\u00e1 m\u1ed9t b\u00e1nh pizza sau khi \u0111\u01b0\u1ee3c khuy\u1ebfn m\u00e3i 70% l\u00e0: 210 000. 1 70% 63 000 (\u0111\u1ed3ng). N\u1ebfu kh\u00e1ch h\u00e0ng mua 10 b\u00e1nh pizza th\u00ec ph\u1ea3i tr\u1ea3 ti\u1ec1n mua 5 b\u00e1nh pizza v\u1edbi gi\u00e1 210 000 \u0111\u1ed3ng\/b\u00e1nh v\u00e0 tr\u1ea3 ti\u1ec1n mua 5 b\u00e1nh pizza c\u00f2n l\u1ea1i v\u1edbi gi\u00e1 63 000 \u0111\u1ed3ng\/b\u00e1nh. S\u1ed1 ti\u1ec1n kh\u00e1ch h\u00e0ng ph\u1ea3i tr\u1ea3 n\u1ebfu mua 10 b\u00e1nh pizza l\u00e0: 5.210 000 5.63 000 2 730 000 (\u0111\u1ed3ng). b) H\u1ecfi n\u1ebfu kh\u00e1ch h\u00e0ng mua 15 b\u00e1nh pizza th\u00ec ph\u1ea3i tr\u1ea3 bao nhi\u00eau ti\u1ec1n? N\u1ebfu kh\u00e1ch h\u00e0ng mua 15 b\u00e1nh pizza th\u00ec ph\u1ea3i tr\u1ea3 ti\u1ec1n mua 8 b\u00e1nh pizza v\u1edbi gi\u00e1 210 000 \u0111\u1ed3ng\/b\u00e1nh v\u00e0 tr\u1ea3 ti\u1ec1n mua 7 b\u00e1nh pizza v\u1edbi gi\u00e1 63 000 \u0111\u1ed3ng\/b\u00e1nh. S\u1ed1 ti\u1ec1n kh\u00e1ch h\u00e0ng ph\u1ea3i tr\u1ea3 n\u1ebfu mua 15 b\u00e1nh pizza m\u00e0 ch\u01b0a \u00e1p d\u1ee5ng ch\u01b0\u01a1ng tr\u00ecnh khuy\u1ebfn m\u00e3i th\u00eam l\u00e0: 8.210 000 7.63 000 2121000 (\u0111\u1ed3ng). V\u00ec 2121000 2 000 000 n\u00ean kh\u00e1ch h\u00e0ng tr\u00ean \u0111\u01b0\u1ee3c \u00e1p d\u1ee5ng ch\u01b0\u01a1ng tr\u00ecnh khuy\u1ebfn m\u00e3i th\u00eam l\u00e0 gi\u1ea3m 5% tr\u00ean t\u1ed5ng s\u1ed1 ti\u1ec1n ph\u1ea3i tr\u1ea3. V\u1eady s\u1ed1 ti\u1ec1n kh\u00e1ch h\u00e0ng ph\u1ea3i tr\u1ea3 khi mua 15 b\u00e1nh pizza l\u00e0: 2121000 1 5% 2 014 950 (\u0111\u1ed3ng). C\u00e2u 6. (1 \u0111i\u1ec3m) M\u1ed9t b\u1ed3n n\u01b0\u1edbc h\u00ecnh tr\u1ee5 c\u00f3 b\u00e1n k\u00ednh \u0111\u00e1y l\u00e0 3m , chi\u1ec1u cao l\u00e0 4m . Ng\u01b0\u1eddi ta \u0111\u1ed5 n\u01b0\u1edbc v\u00e0o trong b\u1ed3n sao cho chi\u1ec1u cao c\u1ee7a n\u01b0\u1edbc b\u1eb1ng \u0111\u00fang m\u1ed9t n\u1eeda chi\u1ec1u cao c\u1ee7a b\u1ed3n v\u00e0 ti\u1ebfp t\u1ee5c 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 \u0111\u1eb7t v\u00e0o trong b\u1ed3n m\u1ed9t phao n\u01b0\u1edbc c\u00f3 d\u1ea1ng h\u00ecnh c\u1ea7u b\u1eb1ng kim lo\u1ea1i kh\u00f4ng th\u1ea5m n\u01b0\u1edbc c\u00f3 b\u00e1n k\u00ednh l\u00e0 50cm v\u00e0 ch\u00ecm ho\u00e0n to\u00e0n trong n\u01b0\u1edbc. a) H\u1ecfi khi \u0111\u00f3 m\u1ef1c n\u01b0\u1edbc trong b\u1ed3n cao bao nhi\u00eau m\u00e9t (l\u00e0m tr\u00f2n \u0111\u1ebfn ch\u1eef s\u1ed1 th\u1eadp ph\u00e2n th\u1ee9 ba)? b) Sau \u0111\u00f3, ng\u01b0\u1eddi ta l\u1ea1i b\u01a1m th\u00eam n\u01b0\u1edbc v\u00e0o b\u1ed3n b\u1eb1ng m\u1ed9t v\u00f2i c\u00f3 l\u01b0u l\u01b0\u1ee3ng n\u01b0\u1edbc ch\u1ea3y l\u00e0 30, 0024 l\u00edt cho m\u1ed7i gi\u00e2y. H\u1ecfi sau bao nhi\u00eau ph\u00fat th\u00ec b\u1ed3n \u0111\u1ea7y n\u01b0\u1edbc (l\u00e0m tr\u00f2n \u0111\u1ebfn h\u00e0ng \u0111\u01a1n v\u1ecb)? L\u1eddi gi\u1ea3i a) H\u1ecfi khi \u0111\u00f3 m\u1ef1c n\u01b0\u1edbc trong b\u1ed3n cao bao nhi\u00eau m\u00e9t (l\u00e0m tr\u00f2n \u0111\u1ebfn ch\u1eef s\u1ed1 th\u1eadp ph\u00e2n th\u1ee9 ba)? M\u1ed9t n\u1eeda chi\u1ec1u cao c\u1ee7a b\u1ed3n b\u1eb1ng: 4 : 2 2 m . \u0110\u1ed5i: 50cm 0, 5m . m3 . Th\u1ec3 t\u00edch c\u1ee7a phao n\u01b0\u1edbc l\u00e0: 4 .0, 52 Chi\u1ec1u cao c\u1ee7a m\u1ef1c n\u01b0\u1edbc trong b\u1ed3n cao l\u00e0: .32 2 19 2,111 m . 9 b) Sau \u0111\u00f3, ng\u01b0\u1eddi ta l\u1ea1i b\u01a1m th\u00eam n\u01b0\u1edbc v\u00e0o b\u1ed3n b\u1eb1ng m\u1ed9t v\u00f2i c\u00f3 c\u00f4ng su\u1ea5t ch\u1ea3y l\u00e0 30, 0024 l\u00edt cho m\u1ed7i gi\u00e2y. H\u1ecfi sau bao nhi\u00eau ph\u00fat th\u00ec b\u1ed3n \u0111\u1ea7y n\u01b0\u1edbc (l\u00e0m tr\u00f2n \u0111\u1ebfn h\u00e0ng \u0111\u01a1n v\u1ecb)? Th\u1ec3 t\u00edch ph\u1ea7n kh\u00f4ng ch\u1ee9a n\u01b0\u1edbc trong b\u1ed3n l\u00e0: .32. 4 19 17 m3 . 9 \u0110\u1ed5i: 30, 0024 l\u00edt\/gi\u00e2y = 1, 800144 m3 \/ph\u00fat. B\u1ed3n \u0111\u1ea7y n\u01b0\u1edbc sau: 17 : 1, 800144 30 (ph\u00fat). C\u00e2u 7. (1 \u0111i\u1ec3m) M\u1ed9t c\u00f4ng ty c\u00f3 100 xe ch\u1edf kh\u00e1ch g\u1ed3m hai lo\u1ea1i, lo\u1ea1i xe ch\u1edf \u0111\u01b0\u1ee3c 30 kh\u00e1ch v\u00e0 lo\u1ea1i xe ch\u1edf \u0111\u01b0\u1ee3c 50 kh\u00e1ch. N\u1ebfu d\u00f9ng t\u1ea5t c\u1ea3 s\u1ed1 xe \u0111\u00f3 th\u00ec t\u1ed1i \u0111a c\u00f4ng ty ch\u1edf m\u1ed9t l\u1ea7n \u0111\u01b0\u1ee3c 4300 kh\u00e1ch. H\u1ecfi m\u1ed7i lo\u1ea1i c\u00f4ng ty \u0111\u00f3 c\u00f3 m\u1ea5y xe? L\u1eddi gi\u1ea3i G\u1ecdi x , y (xe) l\u00e0 s\u1ed1 xe lo\u1ea1i ch\u1edf \u0111\u01b0\u1ee3c 30 kh\u00e1ch v\u00e0 lo\u1ea1i ch\u1edf \u0111\u01b0\u1ee3c 50 kh\u00e1ch. x,y V\u00ec c\u00f4ng ty c\u00f3 100 xe ch\u1edf kh\u00e1ch n\u00ean ta c\u00f3: x y 100 . 1 V\u00ec n\u1ebfu d\u00f9ng t\u1ea5t c\u1ea3 s\u1ed1 xe th\u00ec t\u1ed1i \u0111a c\u00f4ng ty ch\u1edf m\u1ed9t l\u1ea7n \u0111\u01b0\u1ee3c 4300 kh\u00e1ch n\u00ean ta c\u00f3: 30x 50y 4300 . 2 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 x y 100 x 35 T\u1eeb 1 v\u00e0 2 , ta c\u00f3 h\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh: 30x 50y 4300 y 65 (nh\u1eadn) V\u1eady c\u00f4ng tay c\u00f3 35 xe lo\u1ea1i ch\u1edf \u0111\u01b0\u1ee3c 30 kh\u00e1ch v\u00e0 65 xe lo\u1ea1i ch\u1edf \u0111\u01b0\u1ee3c 50 kh\u00e1ch. C\u00e2u 8. (3 \u0111i\u1ec3m) Cho ABC ( AB \uf03c AC) n\u1ed9i ti\u1ebfp O;R \u0111\u01b0\u1eddng k\u00ednh BC , tr\u00ean cung nh\u1ecf AC l\u1ea5y \u0111i\u1ec3m D , BD c\u1eaft AC t\u1ea1i E , t\u1eeb E v\u1ebd EF BC t\u1ea1i F . a) Ch\u1ee9ng minh t\u1ee9 gi\u00e1c BAEF n\u1ed9i ti\u1ebfp \u0111\u01b0\u1eddng tr\u00f2n. b) Ch\u1ee9ng minh DB l\u00e0 ph\u00e2n gi\u00e1c g\u00f3c ADF . c) G\u1ecdi M l\u00e0 trung \u0111i\u1ec3m EC . Ch\u1ee9ng minh DM.CA CF.CO . L\u1eddi gi\u1ea3i D A E M B C FO a) Ch\u1ee9ng minh t\u1ee9 gi\u00e1c BAEF n\u1ed9i ti\u1ebfp \u0111\u01b0\u1eddng tr\u00f2n. 8 X\u00e9t t\u1ee9 gi\u00e1c BAEF , ta c\u00f3: BFE 90 ( EF BC t\u1ea1i F ) BAE 90 (g\u00f3c n\u1ed9i ti\u1ebfp ch\u1eafn n\u1eeda O ) BFE BAE 180 T\u1ee9 gi\u00e1c BAEF n\u1ed9i ti\u1ebfp (T\u1ee9 gi\u00e1c c\u00f3 hai g\u00f3c \u0111\u1ed1i b\u00f9 nhau) b) Ch\u1ee9ng minh DB l\u00e0 ph\u00e2n gi\u00e1c g\u00f3c ADF . X\u00e9t t\u1ee9 gi\u00e1c CDEF , ta c\u00f3: CDE 90 (g\u00f3c n\u1ed9i ti\u1ebfp ch\u1eafn n\u1eeda O ) BFE 90 (cm c\u00e2u a) 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 BFE CDE T\u1ee9 gi\u00e1c CDEF n\u1ed9i ti\u1ebfp (T\u1ee9 gi\u00e1c c\u00f3 g\u00f3c ngo\u00e0i b\u1eb1ng g\u00f3c trong c\u1ee7a \u0111\u1ec9nh \u0111\u1ed1i di\u1ec7n) EDF BCA M\u00e0 BCA ADB (hai g\u00f3c n\u1ed9i ti\u1ebfp c\u00f9ng ch\u1eafn cung AB c\u1ee7a O ) N\u00ean EDF ADB DB l\u00e0 tia ph\u00e2n gi\u00e1c c\u1ee7a g\u00f3c ADF . c) Ch\u1ee9ng minh DM.CA CF.CO . X\u00e9t CDE vu\u00f4ng t\u1ea1i D , ta c\u00f3: DM l\u00e0 \u0111\u01b0\u1eddng trung tuy\u1ebfn (M l\u00e0 trung \u0111i\u1ec3m c\u1ee7a CE ) DM 1 CE CE 2DM . 2 X\u00e9t CEF v\u00e0 CBA , ta c\u00f3: C l\u00e0 g\u00f3c chung CFE CAB 90 CEF \u223d CBA g g CE CF CB CA (t\u1ec9 s\u1ed1 \u0111\u1ed3ng d\u1ea1ng) CE.CA CF.CB M\u00e0 CE 2DM v\u00e0 CB 2CO ( BC l\u00e0 \u0111\u01b0\u1eddng k\u00ednh c\u1ee7a O ) N\u00ean 2DM.CA CF.2CO DM.CA CF.CO . ----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 5 NA\u00caM HO\u00cfC: 2023 - 2024 \u0110\u1ec0 THAM KH\u1ea2O M\u00d4N: TO\u00c1N 9 M\u00c3 \u0110\u1ec0: Qu\u1eadn 5 - 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 (P) : y = x2 v\u00e0 \u0111\u01b0\u1eddng th\u1eb3ng (d) : y = \u2212x + 2 . 42 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). G\u1ecdi x1 , x2 l\u00e0 hai nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh x2 \u2212 5x \u2212 6 = 0 . Kh\u00f4ng gi\u1ea3i ph\u01b0\u01a1ng C\u00e2u 3. tr\u00ecnh, t\u00ednh gi\u00e1 tr\u1ecb c\u1ee7a c\u00e1c bi\u1ec3u th\u1ee9c: A = (1 \u2212 x1 ) x2 + (1 \u2212 x2 ) x1 . x1 x2 (1 \u0111i\u1ec3m). \u00c1p Cho tam gi\u00e1c ABC vu\u00f4ng t\u1ea1i A c\u00f3 c\u1ea1nh AC = 40cm . \u0110\u01b0\u1eddng tr\u00f2n (O; 4 cm) n\u1ed9i ti\u1ebfp tam gi\u00e1c ABC . T\u00ednh \u0111\u1ed9 d\u00e0i c\u1ea1nh AB c\u1ee7a tam gi\u00e1c ABC . B N 4 cm O C A 40 cm C\u00e2u 4. (1 \u0111i\u1ec3m). Anh B\u00ecnh l\u00e0 c\u00f4ng nh\u00e2n trong m\u1ed9t c\u00f4ng ty may c\u00f3 v\u1ed1n \u0111\u1ea7u t\u01b0 n\u01b0\u1edbc ngo\u00e0i. L\u01b0\u01a1ng c\u01a1 b\u1ea3n kh\u1edfi \u0111i\u1ec3m khi v\u00e0o l\u00e0m l\u00e0 3,5 tri\u1ec7u \u0111\u1ed3ng. C\u00f4ng ty c\u00f3 ch\u1ebf \u0111\u1ed9 t\u00ednh th\u00e2m ni\u00ean cho c\u00f4ng nh\u00e2n l\u00e0m l\u00e2u n\u0103m, c\u1ee9 m\u1ed7i n\u0103m \u0111\u01b0\u1ee3c t\u0103ng m\u1ed9t kho\u1ea3n nh\u1ea5t \u0111\u1ecbnh. V\u00ec th\u1ebf khi l\u00e0m \u0111\u01b0\u1ee3c 5 n\u0103m th\u00ec l\u01b0\u01a1ng c\u01a1 b\u1ea3n c\u1ee7a anh B\u00ecnh l\u00e0 6 tri\u1ec7u \u0111\u1ed3ng. Kh\u00f4ng t\u00ednh c\u00e1c kho\u1ea3n ph\u1ee5 c\u1ea5p, th\u01b0\u1edfng v\u00e0 c\u00e1c kh\u1ea5u tr\u1eeb kh\u00e1c th\u00ec ta th\u1ea5y m\u1ed1i li\u00ean h\u1ec7 gi\u1eefa l\u01b0\u01a1ng c\u01a1 b\u1ea3n v\u00e0 s\u1ed1 n\u0103m l\u00e0m vi\u1ec7c l\u00e0 m\u1ed9t h\u00e0m s\u1ed1 b\u1eadc nh\u1ea5t y = ax + b ( a kh\u00e1c 0 ) c\u00f3 \u0111\u1ed3 th\u1ecb nh\u01b0 h\u00ecnh b\u00ean. 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 a) X\u00e1c \u0111\u1ecbnh h\u1ec7 s\u1ed1 a, b . b) N\u1ebfu th\u00e2m ni\u00ean l\u00e0 7 n\u0103m l\u00e0m vi\u1ec7c th\u00ec l\u01b0\u01a1ng c\u01a1 b\u1ea3n c\u1ee7a anh B\u00ecnh l\u00e0 bao nhi\u00eau? C\u00e2u 5. (1 \u0111i\u1ec3m). C\u00f3 ba th\u00f9ng d\u1ea7u \u0111\u1ef1ng t\u1ed5ng c\u1ed9ng 123 l\u00edt d\u1ea7u. N\u1ebfu \u0111\u1ed5 t\u1eeb th\u00f9ng th\u1ee9 nh\u1ea5t sang th\u00f9ng th\u1ee9 hai 5 l\u00edt, r\u1ed3i \u0111\u1ed5 t\u1eeb th\u00f9ng th\u1ee9 hai sang th\u00f9ng th\u1ee9 ba 7 l\u00edt, ti\u1ebfp t\u1ee5c \u0111\u1ed5 t\u1eeb th\u00f9ng th\u1ee9 ba sang th\u00f9ng th\u1ee9 nh\u1ea5t 9 l\u00edt th\u00ec s\u1ed1 d\u1ea7u \u1edf th\u00f9ng th\u1ee9 nh\u1ea5t s\u1ebd \u00edt h\u01a1n s\u1ed1 d\u1ea7u \u1edf th\u00f9ng th\u1ee9 hai l\u00e0 4 l\u00edt v\u00e0 b\u1eb1ng 2 s\u1ed1 d\u1ea7u \u1edf th\u00f9ng th\u1ee9 ba. T\u00ednh s\u1ed1 l\u00edt d\u1ea7u \u1edf m\u1ed7i th\u00f9ng l\u00fac \u0111\u1ea7u. 3 C\u00e2u 6. (1 \u0111i\u1ec3m). B\u1ea1n Kh\u00f4i d\u00f9ng nh\u1eefng que t\u00ednh c\u00f3 \u0111\u1ed9 d\u00e0i 4 cm \u0111\u1ec3 gh\u00e9p l\u1ea1i th\u00e0nh c\u00e1c h\u00ecnh vu\u00f4ng nh\u01b0 h\u00ecnh v\u1ebd d\u01b0\u1edbi \u0111\u00e2y: AB a) N\u1ebfu c\u1ea1nh AB d\u00e0i 48 cm th\u00ec b\u1ea1n Kh\u00f4i \u0111\u00e3 d\u00f9ng t\u1ea5t c\u1ea3 bao nhi\u00eau que t\u00ednh \u0111\u1ec3 gh\u00e9p \u0111\u01b0\u1ee3c h\u00ecnh tr\u00ean? b) N\u1ebfu b\u1ea1n Kh\u00f4i d\u00f9ng t\u1ea5t c\u1ea3 61 que t\u00ednh th\u00ec c\u1ea1nh AB c\u00f3 \u0111\u1ed9 d\u00e0i l\u00e0 bao nhi\u00eau cm ? C\u00e2u 7. (1 \u0111i\u1ec3m). M\u1ed9t c\u1eeda h\u00e0ng \u0111i\u1ec7n m\u00e1y th\u1ef1c hi\u1ec7n ch\u01b0\u01a1ng tr\u00ecnh khuy\u1ebfn m\u00e3i gi\u1ea3m gi\u00e1 t\u1ea5t c\u1ea3 c\u00e1c m\u1eb7t h\u00e0ng 10% theo gi\u00e1 ni\u00eam y\u1ebft, v\u00e0 n\u1ebfu h\u00f3a \u0111\u01a1n kh\u00e1ch h\u00e0ng tr\u00ean 10 tri\u1ec7u s\u1ebd \u0111\u01b0\u1ee3c gi\u1ea3m th\u00eam 2% s\u1ed1 ti\u1ec1n tr\u00ean h\u00f3a \u0111\u01a1n, h\u00f3a \u0111\u01a1n tr\u00ean 15 tri\u1ec7u s\u1ebd \u0111\u01b0\u1ee3c gi\u1ea3m th\u00eam 3% s\u1ed1 ti\u1ec1n tr\u00ean h\u00f3a \u0111\u01a1n, h\u00f3a \u0111\u01a1n tr\u00ean 40 tri\u1ec7u s\u1ebd \u0111\u01b0\u1ee3c gi\u1ea3m th\u00eam 6% s\u1ed1 ti\u1ec1n tr\u00ean h\u00f3a \u0111\u01a1n. \u00d4ng Nam mu\u1ed1n mua m\u1ed9t ti vi v\u1edbi gi\u00e1 ni\u00eam y\u1ebft l\u00e0 9 200000 \u0111\u1ed3ng v\u00e0 m\u1ed9t t\u1ee7 l\u1ea1nh v\u1edbi gi\u00e1 ni\u00eam y\u1ebft l\u00e0 8100000 \u0111\u1ed3ng. H\u1ecfi v\u1edbi ch\u01b0\u01a1ng tr\u00ecnh khuy\u1ebfn m\u00e3i c\u1ee7a c\u1eeda h\u00e0ng, \u00f4ng Nam ph\u1ea3i tr\u1ea3 bao nhi\u00eau ti\u1ec1n? C\u00e2u 8. (2,5 \u0111i\u1ec3m) Cho h\u00ecnh thang ABCD \u0111\u00e1y l\u1edbn AD , n\u1ed9i ti\u1ebfp trong m\u1ed9t \u0111\u01b0\u1eddng tr\u00f2n t\u00e2m O . C\u00e1c c\u1ea1nh b\u00ean AB v\u00e0 CD c\u1eaft nhau t\u1ea1i I . Ti\u1ebfp tuy\u1ebfn v\u1edbi \u0111\u01b0\u1eddng tr\u00f2n t\u00e2m O t\u1ea1i B v\u00e0 D c\u1eaft nhau t\u1ea1i K . a) Ch\u1ee9ng minh tam gi\u00e1c IAD c\u00e2n v\u00e0 BID = 180\uf0b0 \u2212 BOD . b) Ch\u1ee9ng minh n\u0103m \u0111i\u1ec3m O, B, I , K, D c\u00f9ng thu\u1ed9c m\u1ed9t \u0111\u01b0\u1eddng tr\u00f2n v\u00e0 IK \/\/ AD . c) V\u1ebd h\u00ecnh b\u00ecnh h\u00e0nh BDKM . \u0110\u01b0\u1eddng tr\u00f2n t\u00e2m O c\u1eaft \u0111\u01b0\u1eddng tr\u00f2n ngo\u1ea1i ti\u1ebfp tam gi\u00e1c BKM t\u1ea1i N (N kh\u00e1c B) . Ch\u1ee9ng minh r\u1eb1ng ba \u0111i\u1ec3m M, N, D 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 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 = \u2212x + 2 . 42 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. B\u1ea3ng gi\u00e1 tr\u1ecb: x \u22124 \u22122 0 2 4 y = x2 4 10 14 4 x0 2 1 y = \u2212x + 2 2 2 \u0110\u1ed3 th\u1ecb: a) 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) : \u2212x + 2 = x2 24 \uf0db x2 + x \u2212 2 = 0 42 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 \uf0db \uf0e9x =2 \uf0ea \u22124 \uf0eb = Thay x = 2 v\u00e0o y = x2 , ta \u0111\u01b0\u1ee3c: y = 22 = 1. 44 ( )Thay x = \u22124 v\u00e0o y = x2 , ta \u0111\u01b0\u1ee3c: y = \u22124 2 =4. 44 V\u1eady (2;1) v\u00e0 (\u22124; 4) l\u00e0 hai giao \u0111i\u1ec3m c\u1ea7n t\u00ecm. C\u00e2u 2. (1 \u0111i\u1ec3m) G\u1ecdi x1 , x2 l\u00e0 hai nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh x2 \u2212 5x \u2212 6 = 0 . Kh\u00f4ng gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh, t\u00ednh gi\u00e1 tr\u1ecb c\u1ee7a c\u00e1c bi\u1ec3u th\u1ee9c: A = (1 \u2212 x1 ) x2 + (1 \u2212 x2 ) x1 . x1 x2 L\u1eddi gi\u1ea3i V\u00ec \uf044 = b2 \u2212 4ac = (\u22125)2 \u2212 4.1.(\u22126) = 49 \uf03e 0 N\u00ean ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 2 nghi\u1ec7m ph\u00e2n bi\u1ec7t x1, ,x2 . \uf0ec = x1 + x2 = \u2212b = \u2212(\u22125) = 5 \uf0ef\uf0efS a Theo \u0111\u1ecbnh l\u00fd Vi-et, ta c\u00f3: \uf0ed 1 \uf0ef P = x1 .x2 = c = \u22126 = \u22126 \uf0ef\uf0ee a 1 Ta c\u00f3: A = (1 \u2212 )x1 x2 + (1 \u2212 x2 ) x1 x1 x2 ( ) ( )A = 1 \u2212 x1 x22 + 1 \u2212 x2 x12 x1x2 x1x2 A = x22 \u2212 x1x22 + x12 \u2212 x12x2 x1x2 ( ) ( )A = x1 + x2 2 \u2212 2x1x2 \u2212 x1x2 x1 + x2 x1x2 A = 52 \u2212 2.(\u22126) \u2212 (\u22126).5 = \u221267 . (\u22126) 6 ( )C\u00e2u 3. (1 \u0111i\u1ec3m). Cho tam gi\u00e1c ABC vu\u00f4ng t\u1ea1i A c\u00f3 c\u1ea1nh AC = 40cm . \u0110\u01b0\u1eddng tr\u00f2n O; 4 cm n\u1ed9i ti\u1ebfp tam gi\u00e1c ABC . T\u00ednh \u0111\u1ed9 d\u00e0i c\u1ea1nh AB c\u1ee7a tam gi\u00e1c ABC . L\u1eddi gi\u1ea3i 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 4 cm M O AP C 40 cm G\u1ecdi M, N, P l\u1ea7n l\u01b0\u1ee3t l\u00e0 h\u00ecnh chi\u1ebfu c\u1ee7a O l\u00ean c\u00e1c c\u1ea1nh AB, BC, AC . Ta c\u00f3: + OM = ON = OP = 4cm , + AMOP l\u00e0 h\u00ecnh vu\u00f4ng, suy ra AM = AP = 4cm . + BM = BN , CN = CP (t\u00ednh ch\u1ea5t hai ti\u1ebfp tuy\u1ebfn c\u1eaft nhau). * AB + AC = AM + MB + AP + PC = ( AM + AP) + (MB + PC) = 2.4 + (BN + CN) = 8 + BC . \uf0db AB + 40 = 8 + BC \uf0db BC = AB + 32 M\u1eb7c kh\u00e1c: tam gi\u00e1c ABC vu\u00f4ng t\u1ea1i A suy ra BC2 = AB2 + AC2 ( )Thay BC = AB + 32 v\u00e0o, ta \u0111\u01b0\u1ee3c AB + 32 2 = AB2 + 402 \uf0db 64.AB = 576 \uf0db AB = 9 . V\u1eady \u0111\u1ed9 d\u00e0i c\u1ea1nh AB c\u1ee7a tam gi\u00e1c ABC l\u00e0 9 cm . C\u00e2u 4. (1 \u0111i\u1ec3m). Anh B\u00ecnh l\u00e0 c\u00f4ng nh\u00e2n trong m\u1ed9t c\u00f4ng ty may c\u00f3 v\u1ed1n \u0111\u1ea7u t\u01b0 n\u01b0\u1edbc ngo\u00e0i. L\u01b0\u01a1ng c\u01a1 b\u1ea3n kh\u1edfi \u0111i\u1ec3m khi v\u00e0o l\u00e0m l\u00e0 3,5 tri\u1ec7u \u0111\u1ed3ng. C\u00f4ng ty c\u00f3 ch\u1ebf \u0111\u1ed9 t\u00ednh th\u00e2m ni\u00ean cho c\u00f4ng nh\u00e2n l\u00e0m l\u00e2u n\u0103m, c\u1ee9 m\u1ed7i n\u0103m \u0111\u01b0\u1ee3c t\u0103ng m\u1ed9t kho\u1ea3n nh\u1ea5t \u0111\u1ecbnh. V\u00ec th\u1ebf khi l\u00e0m \u0111\u01b0\u1ee3c 5 n\u0103m th\u00ec l\u01b0\u01a1ng c\u01a1 b\u1ea3n c\u1ee7a anh B\u00ecnh l\u00e0 6 tri\u1ec7u \u0111\u1ed3ng. Kh\u00f4ng t\u00ednh c\u00e1c kho\u1ea3n ph\u1ee5 c\u1ea5p, th\u01b0\u1edfng v\u00e0 c\u00e1c kh\u1ea5u tr\u1eeb kh\u00e1c th\u00ec ta th\u1ea5y m\u1ed1i li\u00ean h\u1ec7 gi\u1eefa l\u01b0\u01a1ng c\u01a1 b\u1ea3n v\u00e0 s\u1ed1 n\u0103m l\u00e0m vi\u1ec7c l\u00e0 m\u1ed9t h\u00e0m s\u1ed1 b\u1eadc nh\u1ea5t y = ax + b ( a kh\u00e1c 0 ) c\u00f3 \u0111\u1ed3 th\u1ecb nh\u01b0 h\u00ecnh b\u00ean. 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 a) X\u00e1c \u0111\u1ecbnh h\u1ec7 s\u1ed1 a, b . b) N\u1ebfu th\u00e2m ni\u00ean l\u00e0 7 n\u0103m l\u00e0m vi\u1ec7c th\u00ec l\u01b0\u01a1ng c\u01a1 b\u1ea3n c\u1ee7a anh B\u00ecnh l\u00e0 bao nhi\u00eau? L\u1eddi gi\u1ea3i a) X\u00e1c \u0111\u1ecbnh h\u1ec7 s\u1ed1 a, b . \u0110\u1ed3 th\u1ecb h\u00e0m s\u1ed1 qua 2 \u0111i\u1ec3m (0; 3,5), (5;6) n\u00ean ta c\u00f3 h\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh: \uf0ec0.a + b = 3,5 \uf0ed \uf0ee 5.a + b =6 \uf0db \uf0ef\uf0ef\uf0eca = 1 \uf0ed = \uf0ee\uf0ef\uf0efb 2 7 2 V\u1eady a = 1 , b = 7 . 22 H\u00e0m s\u1ed1 \u0111\u00e3 cho l\u00e0: y = 1 x + 7 . 22 b) Th\u00e2m ni\u00ean l\u00e0 7 n\u0103m th\u00ec x = 7 , thay v\u00e0o h\u00e0m s\u1ed1 y = 1 x + 7 , ta c\u00f3: 22 y = 1 .7 + 7 22 \uf0dey =7 V\u1eady n\u1ebfu th\u00e2m ni\u00ean l\u00e0 7 n\u0103m l\u00e0 vi\u1ec7c th\u00ec l\u01b0\u01a1ng c\u01a1 b\u1ea3n c\u1ee7a anh B\u00ecnh l\u00e0 7 tri\u1ec7u \u0111\u1ed3ng. C\u00e2u 5. (1 \u0111i\u1ec3m). C\u00f3 ba th\u00f9ng d\u1ea7u \u0111\u1ef1ng t\u1ed5ng c\u1ed9ng 123 l\u00edt d\u1ea7u. N\u1ebfu \u0111\u1ed5 t\u1eeb th\u00f9ng th\u1ee9 nh\u1ea5t sang th\u00f9ng th\u1ee9 hai 5 l\u00edt, r\u1ed3i \u0111\u1ed5 t\u1eeb th\u00f9ng th\u1ee9 hai sang th\u00f9ng th\u1ee9 ba 7 l\u00edt, ti\u1ebfp t\u1ee5c \u0111\u1ed5 t\u1eeb th\u00f9ng th\u1ee9 ba sang 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 th\u00f9ng th\u1ee9 nh\u1ea5t 9 l\u00edt th\u00ec s\u1ed1 d\u1ea7u \u1edf th\u00f9ng th\u1ee9 nh\u1ea5t s\u1ebd \u00edt h\u01a1n s\u1ed1 d\u1ea7u \u1edf th\u00f9ng th\u1ee9 hai l\u00e0 4 l\u00edt v\u00e0 b\u1eb1ng 2 s\u1ed1 d\u1ea7u \u1edf th\u00f9ng th\u1ee9 ba. T\u00ednh s\u1ed1 l\u00edt d\u1ea7u \u1edf m\u1ed7i th\u00f9ng l\u00fac \u0111\u1ea7u. 3 L\u1eddi gi\u1ea3i G\u1ecdi x (l\u00edt), y (l\u00edt) l\u1ea7n l\u01b0\u1ee3t l\u00e0 s\u1ed1 l\u00edt d\u1ea7u \u1edf th\u00f9ng th\u1ee9 nh\u1ea5t, th\u00f9ng th\u1ee9 hai l\u00fac \u0111\u1ea7u. (\u0110K : x, y \uf03e 0; x \uf03c 123; y \uf03c 123) Do \u0111\u00f3, s\u1ed1 l\u00edt d\u1ea7u \u1edf th\u00f9ng th\u1ee9 ba l\u00fac \u0111\u1ea7u l\u00e0: 123 \u2212 x \u2212 y (l\u00edt). N\u1ebfu \u0111\u1ed5 t\u1eeb th\u00f9ng th\u1ee9 nh\u1ea5t sang th\u00f9ng th\u1ee9 hai 5 l\u00edt, r\u1ed3i \u0111\u1ed5 t\u1eeb th\u00f9ng th\u1ee9 hai sang th\u00f9ng th\u1ee9 ba 7 l\u00edt, ti\u1ebfp t\u1ee5c \u0111\u1ed5 t\u1eeb th\u00f9ng th\u1ee9 ba sang th\u00f9ng th\u1ee9 nh\u1ea5t 9 l\u00edt. S\u1ed1 l\u00edt d\u1ea7u \u1edf: + Th\u00f9ng th\u1ee9 nh\u1ea5t l\u00e0: x \u2212 5 + 9 = x + 4 (l\u00edt) + Th\u00f9ng th\u1ee9 hai l\u00e0: y + 5 \u2212 7 = y \u2212 2 (l\u00edt) + Th\u00f9ng th\u1ee9 ba l\u00e0: 123 \u2212 x \u2212 y + 7 \u2212 9 = 121 \u2212 x \u2212 y (l\u00edt) Theo \u0111\u1ec1, ta c\u00f3 h\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh: \uf0ef\uf0ecx + 4 = y \u2212 2 \u2212 4 \uf0ed\uf0ee\uf0efx 2 + 4 = 3 (121 \u2212 x \u2212 y) \uf0db \uf0ecx \u2212 y = \u221210 \uf0ee\uf0ed5x + 2y = 230 \uf0db \uf0ecx = 30 (nh\u1eadn) \uf0ee\uf0edy = 40 V\u1eady l\u00fac \u0111\u1ea7u, th\u00f9ng th\u1ee9 nh\u1ea5t ch\u1ee9a l\u00e0 30 l\u00edt d\u1ea7u, th\u00f9ng th\u1ee9 hai ch\u1ee9a 40 l\u00edt d\u1ea7u, th\u00f9ng th\u1ee9 ba ch\u1ee9a 123 \u2212 30 \u2212 40 = 53 l\u00edt d\u1ea7u. C\u00e2u 6. (1 \u0111i\u1ec3m). B\u1ea1n Kh\u00f4i d\u00f9ng nh\u1eefng que t\u00ednh c\u00f3 \u0111\u1ed9 d\u00e0i 4 cm \u0111\u1ec3 gh\u00e9p l\u1ea1i th\u00e0nh c\u00e1c h\u00ecnh vu\u00f4ng nh\u01b0 h\u00ecnh v\u1ebd d\u01b0\u1edbi \u0111\u00e2y: AB a) N\u1ebfu c\u1ea1nh AB d\u00e0i 48 cm th\u00ec b\u1ea1n Kh\u00f4i \u0111\u00e3 d\u00f9ng t\u1ea5t c\u1ea3 bao nhi\u00eau que t\u00ednh \u0111\u1ec3 gh\u00e9p \u0111\u01b0\u1ee3c h\u00ecnh tr\u00ean? b) N\u1ebfu b\u1ea1n Kh\u00f4i d\u00f9ng t\u1ea5t c\u1ea3 61 que t\u00ednh th\u00ec c\u1ea1nh AB c\u00f3 \u0111\u1ed9 d\u00e0i l\u00e0 bao nhi\u00eau cm ? L\u1eddi gi\u1ea3i a) S\u1ed1 que t\u00ednh c\u1ea7n \u0111\u1ec3 gh\u00e9p c\u1ea1nh AB d\u00e0i 48 cm l\u00e0: 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 48 : 4 = 12 (que t\u00ednh) S\u1ed1 que t\u00ednh c\u1ea7n \u0111\u1ec3 gh\u00e9p h\u00ecnh c\u00f3 c\u1ea1nh AB d\u00e0i 48 cm l\u00e0: (12\uf0b4 2) + (12 + 1) = 37 (que t\u00ednh) b) S\u1ed1 que t\u00ednh tr\u00ean m\u1ed9t c\u1ea1nh AB l\u00e0: (61 \u2212 1) : 3 = 20 (que t\u00ednh) \u0110\u1ed9 d\u00e0i c\u1ea1nh AB l\u00e0: 20 \uf0b4 4 = 80 cm V\u1eady c\u1ea7n 37 que t\u00ednh v\u00e0 chi\u1ec1u d\u00e0i c\u1ea1nh AB l\u00e0 80cm . C\u00e2u 7. (1 \u0111i\u1ec3m). M\u1ed9t c\u1eeda h\u00e0ng \u0111i\u1ec7n m\u00e1y th\u1ef1c hi\u1ec7n ch\u01b0\u01a1ng tr\u00ecnh khuy\u1ebfn m\u00e3i gi\u1ea3m gi\u00e1 t\u1ea5t c\u1ea3 c\u00e1c m\u1eb7t h\u00e0ng 10% theo gi\u00e1 ni\u00eam y\u1ebft, v\u00e0 n\u1ebfu h\u00f3a \u0111\u01a1n kh\u00e1ch h\u00e0ng tr\u00ean 10 tri\u1ec7u s\u1ebd \u0111\u01b0\u1ee3c gi\u1ea3m th\u00eam 2% s\u1ed1 ti\u1ec1n tr\u00ean h\u00f3a \u0111\u01a1n, h\u00f3a \u0111\u01a1n tr\u00ean 15 tri\u1ec7u s\u1ebd \u0111\u01b0\u1ee3c gi\u1ea3m th\u00eam 3% s\u1ed1 ti\u1ec1n tr\u00ean h\u00f3a \u0111\u01a1n, h\u00f3a \u0111\u01a1n tr\u00ean 40 tri\u1ec7u s\u1ebd \u0111\u01b0\u1ee3c gi\u1ea3m th\u00eam 6% s\u1ed1 ti\u1ec1n tr\u00ean h\u00f3a \u0111\u01a1n. \u00d4ng Nam mu\u1ed1n mua m\u1ed9t ti vi v\u1edbi gi\u00e1 ni\u00eam y\u1ebft l\u00e0 9 200000 \u0111\u1ed3ng v\u00e0 m\u1ed9t t\u1ee7 l\u1ea1nh v\u1edbi gi\u00e1 ni\u00eam y\u1ebft l\u00e0 8100000 \u0111\u1ed3ng. H\u1ecfi v\u1edbi ch\u01b0\u01a1ng tr\u00ecnh khuy\u1ebfn m\u00e3i c\u1ee7a c\u1eeda h\u00e0ng, \u00f4ng Nam ph\u1ea3i tr\u1ea3 bao nhi\u00eau ti\u1ec1n? L\u1eddi gi\u1ea3i Gi\u00e1 ti\u1ec1n c\u1ee7a m\u1ed9t ti vi sau khi gi\u1ea3m gi\u00e1 10% l\u00e0: 9200000.(100% \u2212 10%) = 8280000 (\u0111\u1ed3ng) Gi\u00e1 ti\u1ec1n c\u1ee7a m\u1ed9t t\u1ee7 l\u1ea1nh sau khi gi\u1ea3m gi\u00e1 10% l\u00e0: 8100000.(100% \u2212 10%) = 7 290000 (\u0111\u1ed3ng) T\u1ed5ng s\u1ed1 ti\u1ec1n \u00f4ng Nam ph\u1ea3i tr\u1ea3 tr\u01b0\u1edbc khuy\u1ebfn m\u00e3i theo h\u00f3a \u0111\u01a1n l\u00e0: 8 280000 + 7 290000 = 15570000 (\u0111\u1ed3ng) V\u00ec h\u00f3a \u0111\u01a1n tr\u00ean 15000000 \u0111\u1ed3ng n\u00ean \u00f4ng Nam \u0111\u01b0\u1ee3c \u00e1p d\u1ee5ng khuy\u1ebfn m\u00e3i gi\u1ea3m th\u00eam 3% tr\u00ean t\u1ed5ng h\u00f3a \u0111\u01a1n. S\u1ed1 ti\u1ec1n \u00f4ng Nam th\u1ef1c t\u1ebf ph\u1ea3i tr\u1ea3 l\u00e0: 15570000.(100% \u2212 3%) = 15102900 (\u0111\u1ed3ng) V\u1eady, v\u1edbi ch\u01b0\u01a1ng tr\u00ecnh khuy\u1ebfn m\u00e3i c\u1ee7a c\u1eeda h\u00e0ng, \u00f4ng Nam ph\u1ea3i tr\u1ea3 cho c\u1eeda h\u00e0ng s\u1ed1 ti\u1ec1n 15102900 (\u0111\u1ed3ng). 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 C\u00e2u 8. (2,5 \u0111i\u1ec3m) Cho h\u00ecnh thang ABCD \u0111\u00e1y l\u1edbn AD , n\u1ed9i ti\u1ebfp trong m\u1ed9t \u0111\u01b0\u1eddng tr\u00f2n t\u00e2m O . C\u00e1c c\u1ea1nh b\u00ean AB v\u00e0 CD c\u1eaft nhau t\u1ea1i I . Ti\u1ebfp tuy\u1ebfn v\u1edbi \u0111\u01b0\u1eddng tr\u00f2n t\u00e2m O t\u1ea1i B v\u00e0 D c\u1eaft nhau t\u1ea1i K . a) Ch\u1ee9ng minh tam gi\u00e1c IAD c\u00e2n v\u00e0 BID = 180\uf0b0 \u2212 BOD . b) Ch\u1ee9ng minh n\u0103m \u0111i\u1ec3m O, B, I , K, D c\u00f9ng thu\u1ed9c m\u1ed9t \u0111\u01b0\u1eddng tr\u00f2n v\u00e0 IK \/\/ AD . c) V\u1ebd h\u00ecnh b\u00ecnh h\u00e0nh BDKM . \u0110\u01b0\u1eddng tr\u00f2n t\u00e2m O c\u1eaft \u0111\u01b0\u1eddng tr\u00f2n ngo\u1ea1i ti\u1ebfp tam gi\u00e1c BKM t\u1ea1i N (N kh\u00e1c B) . Ch\u1ee9ng minh r\u1eb1ng ba \u0111i\u1ec3m M, N, D th\u1eb3ng h\u00e0ng. L\u1eddi gi\u1ea3i M IK C B N O D A a) Ch\u1ee9ng minh tam gi\u00e1c IAD c\u00e2n v\u00e0 BID = 180\uf0b0 \u2212 BOD . 9 * Ch\u1ee9ng minh tam gi\u00e1c IAD c\u00e2n: T\u1ee9 gi\u00e1c ABCD n\u1ed9i ti\u1ebfp \u0111\u01b0\u1eddng tr\u00f2n t\u00e2m O , suy ra ABC + ADC = 180\uf0b0 M\u00e0 ABC + DAB = 180\uf0b0 (hai g\u00f3c trong c\u00f9ng ph\u00eda do BC \/\/ AD ) 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 N\u00ean ADC = DAB Suy ra tam gi\u00e1c IAD c\u00e2n t\u1ea1i I . * Ch\u1ee9ng minh BID = 180\uf0b0 \u2212 BOD . + BOD = 2BAD = s\u0111 BD (g\u00f3c n\u1ed9i ti\u1ebfp v\u00e0 g\u00f3c \u1edf t\u00e2m c\u00f9ng ch\u1eafn cung BD . + Tam gi\u00e1c IAD c\u00e2n t\u1ea1i I \uf0de BID + 2BAD = 180\uf0b0 \uf0de BID + BOD = 180\uf0b0 (do BOD = 2BAD ) \uf0de BID = 180\uf0b0 \u2212 BOD . b) Ch\u1ee9ng minh n\u0103m \u0111i\u1ec3m O, B, I , K, D c\u00f9ng thu\u1ed9c m\u1ed9t \u0111\u01b0\u1eddng tr\u00f2n v\u00e0 IK \/\/ AD . * Ch\u1ee9ng minh n\u0103m \u0111i\u1ec3m O, B, I , K, D c\u00f9ng thu\u1ed9c m\u1ed9t \u0111\u01b0\u1eddng tr\u00f2n: + BK, BD l\u00e0 2 ti\u1ebfp tuy\u1ebfn c\u1ee7a (O) n\u00ean OBK = ODK = 90\uf0b0 , suy ra t\u1ee9 gi\u00e1c OBKD n\u1ed9i ti\u1ebfp \u0111\u01b0\u1ee3c \u0111\u01b0\u1eddng tr\u00f2n (t\u1ed5ng hai g\u00f3c \u0111\u1ed1i b\u1eb1ng 180\uf0b0 ) (1) + t\u1ee9 gi\u00e1c OBKD n\u1ed9i ti\u1ebfp \u0111\u01b0\u1ee3c \u0111\u01b0\u1eddng tr\u00f2n \uf0de BKA + BOD = 180\uf0b0 hay BKD = 180\uf0b0 \u2212 BOD \uf0de BKD = BID (do BID = 180\uf0b0 \u2212 BOD ) T\u1ee9 gi\u00e1c BIKD c\u00f3 hai \u0111\u1ec9nh I, K c\u00f9ng nh\u00ecn c\u1ea1nh BD d\u01b0\u1edbi m\u1ed9t g\u00f3c b\u1eb1ng nhau n\u00ean BIKD n\u1ed9i ti\u1ebfp \u0111\u01b0\u1ee3c \u0111\u01b0\u1eddng tr\u00f2n. (2) T\u1eeb (1) v\u00e0 (2) suy ra n\u0103m \u0111i\u1ec3m O, B, I , K, D c\u00f9ng thu\u1ed9c m\u1ed9t \u0111\u01b0\u1eddng tr\u00f2n. * Ch\u1ee9ng minh IK \/\/ AD : + T\u1ee9 gi\u00e1c BIKD n\u1ed9i ti\u1ebfp \u0111\u01b0\u1ee3c \u0111\u01b0\u1eddng tr\u00f2n (ch\u1ee9ng minh tr\u00ean) \uf0de KID = KBD = 1 s\u0111 KD (hai g\u00f3c n\u1ed9i ti\u1ebfp c\u00f9ng ch\u1eafn cung KD ) (3) 2 + KBD = BAD = 1 s\u0111 BD (g\u00f3c t\u1ea1o b\u1edfi ti\u1ebfp tuy\u1ebfn KB v\u00e0 d\u00e2y BD v\u00e0 g\u00f3c n\u1ed9i ti\u1ebfp ch\u1eafn cung 2 BD ) (4) + BAD \uf0ba IAD = ADI (tam gi\u00e1c IAD c\u00e2n t\u1ea1i I ) (5) T\u1eeb (3) , (4) , (5) suy ra KID = ADI , m\u00e0 hai g\u00f3c n\u00e0y \u1edf v\u1ecb tr\u00ed so le trong n\u00ean IK \/\/ AD . c) V\u1ebd h\u00ecnh b\u00ecnh h\u00e0nh BDKM . \u0110\u01b0\u1eddng tr\u00f2n t\u00e2m O c\u1eaft \u0111\u01b0\u1eddng tr\u00f2n ngo\u1ea1i ti\u1ebfp tam gi\u00e1c BKM t\u1ea1i N (N kh\u00e1c B) . Ch\u1ee9ng minh r\u1eb1ng ba \u0111i\u1ec3m M, N, D th\u1eb3ng h\u00e0ng. 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 Ta ch\u1ee9ng minh BND + BNM = 180\uf0b0 . Ta c\u00f3: BAD = KBD (ch\u1ee9ng minh tr\u00ean) KBD = BKM (so le trong do MK \/\/ BD ) BKM = BNM (hai g\u00f3c n\u1ed9i ti\u1ebfp c\u00f9ng ch\u1eafn cung BM c\u1ee7a (BKM) ) \uf0de BAD = BNM * T\u1ee9 gi\u00e1c BNDA n\u1ed9i ti\u1ebfp (O) \uf0de BND + BAD = 180\uf0b0 \uf0db BND + BNM = 180\uf0b0 Suy ra ba \u0111i\u1ec3m M, N, D th\u1eb3ng h\u00e0ng. ----H\u1ebeT--- 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 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 5 NA\u00caM HO\u00cfC: 2021 - 2022 \u0110\u1ec0 THAM KH\u1ea2O M\u00d4N: TO\u00c1N 9 M\u00c3 \u0110\u1ec0: Qu\u1eadn 5 - 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 h\u00e0m s\u1ed1 (P) : y = x2 v\u00e0 \u0111\u01b0\u1eddng th\u1eb3ng (d) : y = 2x \u2212 2 . 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 x2 \u2212 5x + 4 = 0 . Kh\u00f4ng gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh, h\u00e3y t\u00ednh gi\u00e1 tr\u1ecb bi\u1ec3u th\u1ee9c A = 5x1 \u2212 x2 \u2212 x1 \u2212 5x2 . x1 x2 C\u00e2u 3. (1 \u0111i\u1ec3m). L\u00fac 6 gi\u1edd s\u00e1ng, m\u1ed9t xe \u00f4 t\u00f4 \u1edf v\u1ecb tr\u00ed c\u00e1ch th\u00e0nh ph\u1ed1 H\u1ed3 Ch\u00ed Minh 50km v\u00e0 kh\u1edfi h\u00e0nh \u0111i H\u00e0 N\u1ed9i (\u1edf ng\u01b0\u1ee3c chi\u1ec1u v\u1edbi TPHCM). G\u1ecdi y = ax + b l\u00e0 h\u00e0m s\u1ed1 bi\u1ec3u di\u1ec5n \u0111\u1ed9 d\u00e0i qu\u00e3ng \u0111\u01b0\u1eddng t\u1eeb TPHCM \u0111\u1ebfn v\u1ecb tr\u00ed c\u1ee7a xe \u00f4 t\u00f4 sau x gi\u1edd theo \u0111\u1ed3 th\u1ecb \u1edf h\u00ecnh sau. y (km) 230 50 50 km H\u00e0 N\u1ed9i 0 3 x (gi\u1edd) TPHCM a) T\u00ecm a v\u00e0 b . b) V\u00e0o l\u00fac m\u1ea5y gi\u1edd th\u00ec xe \u00f4 t\u00f4 c\u00e1ch TPHCM 410km ? C\u00e2u 4. (1 \u0111i\u1ec3m). Hai x\u00ed nghi\u1ec7p theo k\u1ebf ho\u1ea1ch ph\u1ea3i l\u00e0m t\u1ed5ng c\u1ed9ng 360 d\u1ee5ng c\u1ee5. Tr\u00ean th\u1ef1c t\u1ebf, x\u00ed nghi\u1ec7p A v\u01b0\u1ee3t m\u1ee9c 12% , x\u00ed nghi\u1ec7p B v\u01b0\u1ee3t m\u1ee9c 10% do \u0111\u00f3 c\u1ea3 hai x\u00ed nghi\u1ec7p l\u00e0m t\u1ed5ng c\u1ed9ng 400 d\u1ee5ng c\u1ee5. T\u00ednh s\u1ed1 d\u1ee5ng c\u1ee5 m\u1ed7i x\u00ed nghi\u1ec7p ph\u1ea3i l\u00e0m. C\u00e2u 5. (0,75 \u0111i\u1ec3m). M\u1ed9t gi\u1ea3i b\u00f3ng \u0111\u00e1 g\u1ed3m 6 \u0111\u1ed9i b\u00f3ng thi \u0111\u1ea5u theo th\u1ec3 th\u1ee9c vo\u0300ng tro\u0300n 1 l\u01b0\u1ee3t. \u0110\u1ed9i th\u1eafng \u0111\u01b0\u1ee3c 3 \u0111i\u1ec3m, ho\u00e0 \u0111\u01b0\u1ee3c 1 \u0111i\u1ec3m, thua 0 \u0111i\u1ec3m. K\u1ebft thu\u0301c gi\u1ea3i \u0111\u1ea5u, t\u1ed5ng s\u1ed1 \u0111i\u1ec3m c\u1ee7a c\u1ea3 6 \u0111\u1ed9i l\u00e0 41 \u0111i\u1ec3m. a) Ho\u0309i gi\u1ea3i \u0111\u1ea5u c\u00f3 bao nhi\u00eau tr\u1eadn? b) T\u00ednh s\u1ed1 tr\u1eadn h\u00f2a c\u1ee7a gi\u1ea3i \u0111\u1ea5u? C\u00e2u 6. (1 \u0111i\u1ec3m). Ba b\u1ea1n D\u0169ng, T\u00e0i v\u00e0 Tr\u00ed \u0111\u1ee9ng \u1edf ba v\u1ecb tr\u00ed A , B , C tr\u00ean m\u1ed9t \u0111\u01b0\u1eddng tr\u00f2n t\u00e2m O \u0111\u1ec3 ch\u01a1i tro\u0300 truy\u1ec1n c\u1ea7u. Bi\u1ebft kho\u1ea3ng c\u00e1ch t\u1eeb D\u0169ng \u0111\u1ebfn T\u00e0i b\u1eb1ng kho\u1ea3ng c\u00e1ch t\u1eeb D\u0169ng \u0111\u1ebfn Tr\u00ed l\u00e0 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 16m ( AB = AC = 16m ), kho\u1ea3ng c\u00e1ch t\u1eeb T\u00e0i \u0111\u1ebfn Tr\u00ed l\u00e0 19,2m ( BC = 19,2m) (H\u00ecnh b\u00ean). Em h\u00e3y t\u00ednh b\u00e1n k\u00ednh c\u1ee7a \u0111\u01b0\u1eddng tr\u00f2n (O) . C\u00e2u 7. (0,75 \u0111i\u1ec3m). B\u1ea1n An \u0111i mua giu\u0301p b\u1ed1 c\u00e2y l\u0103n s\u01a1n \u1edf c\u1eeda h\u00e0ng nh\u00e0 b\u00e1c To\u00e0n. M\u1ed9t c\u00e2y l\u0103n s\u01a1n t\u01b0\u1eddng c\u00f3 d\u1ea1ng m\u1ed9t kh\u1ed1i tr\u1ee5 v\u1edbi b\u00e1n k\u00ednh \u0111\u00e1y l\u00e0 5 cm v\u00e0 chi\u1ec1u cao l\u00e0 23 cm (h\u00ecnh v\u1ebd b\u00ean). Nh\u00e0 s\u1ea3n xu\u1ea5t cho bi\u1ebft sau khi l\u0103n 1000 vo\u0300ng th\u00ec c\u00e2y s\u01a1n t\u01b0\u1eddng c\u00f3 th\u1ec3 b\u1ecb ho\u0309ng. Ho\u0309i b\u1ea1n An c\u1ea7n mua \u00edt nh\u1ea5t m\u1ea5y c\u00e2y l\u0103n s\u01a1n t\u01b0\u1eddng bi\u1ebft di\u1ec7n t\u00edch t\u01b0\u1eddng m\u00e0 b\u1ed1 b\u1ea1n An c\u1ea7n s\u01a1n l\u00e0 100 m2 ? C\u00e2u 8. (3 \u0111i\u1ec3m) Cho \uf044ABC nh\u1ecdn ( AB \uf03c AC ) n\u1ed9i ti\u1ebfp \u0111\u01b0\u1eddng tr\u00f2n (O) . C\u00e1c \u0111\u01b0\u1eddng cao AD , BE , CF c\u1eaft nhau t\u1ea1i H . Tia EF c\u1eaft tia CB t\u1ea1i K . a) Ch\u1ee9ng minh t\u1ee9 gi\u00e1c BFEC n\u1ed9i ti\u1ebfp v\u00e0 KE.KF = KB.KC . b) \u0110\u01b0\u1eddng th\u1eb3ng KA c\u1eaft (O) t\u1ea1i M . Ch\u1ee9ng minh t\u1ee9 gi\u00e1c AEFM n\u1ed9i ti\u1ebfp. c) G\u1ecdi N l\u00e0 trung \u0111i\u1ec3m c\u1ee7a BC . Ch\u1ee9ng minh M,H,N 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"]