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 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 7 NA\u00caM HO\u00cfC: 2021 - 2022 M\u00d4N: TO\u00c1N 9 \u0110\u1ec0 THAM KH\u1ea2O M\u00c3 \u0110\u1ec0: Qu\u1eadn 7 - 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 = \u2212 x2 v\u00e0 \u0111\u01b0\u1eddng th\u1eb3ng (D) : y = \u2212 3 x \u2212 1 . 44 a) V\u1ebd \u0111\u1ed3 th\u1ecb (P) v\u00e0 (D) tr\u00ean c\u00f9ng h\u1ec7 tr\u1ee5c t\u1ecda \u0111\u1ed9. b) Vi\u1ebft ph\u01b0\u01a1ng tr\u00ecnh \u0111\u01b0\u1eddng th\u1eb3ng D1 song song v\u1edbi D v\u00e0 \u0111i qua \u0111i\u1ec3m A thu\u1ed9c P c\u00f3 ho\u00e0nh \u0111\u1ed9 b\u1eb1ng 2 . C\u00e2u 2. (1 \u0111i\u1ec3m). Cho ph\u01b0\u01a1ng tr\u00ecnh x2 3x 5 1 0 . Kh\u00f4ng gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh, h\u00e3y t\u00ednh gi\u00e1 tr\u1ecb c\u1ee7a bi\u1ec3u th\u1ee9c M 2x12x22 x1x2 3x1 3x2 . C\u00e2u 3. (1 \u0111i\u1ec3m). \u0110\u1ea7u n\u0103m h\u1ecdc, m\u1ed9t tr\u01b0\u1eddng THPT tuy\u1ec3n \u0111\u01b0\u1ee3c 75 h\u1ecdc sinh v\u00e0o 2 l\u1edbp chuy\u00ean V\u0103n v\u00e0 chuy\u00ean S\u1eed. N\u1ebfu chuy\u1ec3n 15 h\u1ecdc sinh t\u1eeb l\u1edbp chuy\u00ean V\u0103n sang l\u1edbp chuy\u00ean S\u1eed th\u00ec s\u1ed1 h\u1ecdc sinh l\u1edbp chuy\u00ean S\u1eed b\u1eb1ng 8 s\u1ed1 h\u1ecdc sinh l\u1edbp chuy\u00ean V\u0103n. H\u00e3y t\u00ednh s\u1ed1 h\u1ecdc sinh c\u1ee7a m\u1ed7i l\u1edbp. 7 C\u00e2u 4. (1 \u0111i\u1ec3m). X\u00ed nghi\u1ec7p may Vi\u1ec7t Ti\u1ebfn h\u00e0ng th\u00e1ng ph\u1ea3i chi 410 000 000 \u0111\u1ed3ng \u0111\u1ec3 tr\u1ea3 l\u01b0\u01a1ng cho c\u00f4ng nh\u00e2n, mua v\u1eadt t\u01b0 v\u00e0 c\u00e1c kho\u1ea3n ph\u00ed kh\u00e1c. M\u1ed7i chi\u1ebfc \u00e1o \u0111\u01b0\u1ee3c b\u00e1n v\u1edbi gi\u00e1 350 000 \u0111\u1ed3ng. G\u1ecdi s\u1ed1 ti\u1ec1n l\u1eddi (ho\u1eb7c l\u1ed7) m\u00e0 x\u00ed nghi\u1ec7p thu \u0111\u01b0\u1ee3c sau m\u1ed7i th\u00e1ng l\u00e0 T v\u00e0 m\u1ed7i th\u00e1ng x\u00ed nghi\u1ec7p b\u00e1n \u0111\u01b0\u1ee3c x chi\u1ebfc \u00e1o. a) L\u1eadp h\u00e0m s\u1ed1 c\u1ee7a T theo x . b) C\u1ea7n ph\u1ea3i b\u00e1n trung b\u00ecnh bao nhi\u00eau chi\u1ebfc \u00e1o m\u1ed7i th\u00e1ng \u0111\u1ec3 sau 1 n\u0103m, x\u00ed nghi\u1ec7p thu \u0111\u01b0\u1ee3c ti\u1ec1n l\u1eddi l\u00e0 1380 000 000 \u0111\u1ed3ng. C\u00e2u 5. (0,75 \u0111i\u1ec3m). M\u1ed9t vi\u00ean g\u1ea1ch h\u00ecnh vu\u00f4ng ( 40cm 40cm ) \u0111\u01b0\u1ee3c trang tr\u00ed ho\u1ea1 ti\u1ebfp nh\u01b0 tr\u00ean h\u00ecnh, t\u00ednh di\u1ec7n t\u00edch ph\u1ea7n t\u00f4 m\u00e0u. C\u00e2u 6. (0,75 \u0111i\u1ec3m). M\u1ed7i c\u00f4ng nh\u00e2n c\u1ee7a c\u00f4ng ty C\u1ed5 ph\u1ea7n ABC c\u00f3 s\u1ed1 ti\u1ec1n th\u01b0\u1edfng t\u1ebft n\u0103m 2018 l\u00e0 1 th\u00e1ng l\u01b0\u01a1ng. \u0110\u1ebfn n\u0103m 2019 , s\u1ed1 ti\u1ec1n th\u01b0\u1edfng t\u1ebft c\u1ee7a h\u1ecd \u0111\u01b0\u1ee3c t\u0103ng th\u00eam 6% so v\u1edbi s\u1ed1 ti\u1ec1n th\u01b0\u1edfng t\u1ebft c\u1ee7a n\u0103m 2018 . V\u00e0o n\u0103m 2020 , s\u1ed1 ti\u1ec1n th\u01b0\u1edfng t\u1ebft c\u1ee7a h\u1ecd \u0111\u01b0\u1ee3c t\u0103ng th\u00eam 10% so v\u1edbi s\u1ed1 ti\u1ec1n th\u01b0\u1edfng t\u1ebft c\u1ee7a n\u0103m 2019 , ngo\u00e0i ra n\u1ebfu c\u00f4ng nh\u00e2n n\u00e0o \u0111\u01b0\u1ee3c l\u00e0 c\u00f4ng \u0111o\u00e0n vi\u00ean xu\u1ea5t s\u1eafc s\u1ebd \u0111\u01b0\u1ee3c th\u01b0\u1edfng th\u00eam 500 000 \u0111\u1ed3ng. Anh Ba l\u00e0 c\u00f4ng \u0111o\u00e0n vi\u00ean xu\u1ea5t s\u1eafc c\u1ee7a n\u0103m 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 2019 , n\u00ean anh nh\u1eadn \u0111\u01b0\u1ee3c s\u1ed1 ti\u1ec1n th\u01b0\u1edfng t\u1ebft l\u00e0 6 330 000 \u0111\u1ed3ng. H\u1ecfi n\u0103m 2018 , ti\u1ec1n l\u01b0\u01a1ng 1 th\u00e1ng c\u1ee7a anh Ba l\u00e0 bao nhi\u00eau? C\u00e2u 7. (1 \u0111i\u1ec3m). M\u1ed9t kh\u1ed1i l\u1eadp ph\u01b0\u01a1ng c\u00f3 c\u1ea1nh 1m ch\u1ee9a \u0111\u1ea7y n\u01b0\u1edbc. \u0110\u1eb7t v\u00e0o trong kh\u1ed1i \u0111\u00f3 m\u1ed9t kh\u1ed1i n\u00f3n c\u00f3 \u0111\u1ec9nh tr\u00f9ng v\u1edbi t\u00e2m m\u1ed9t m\u1eb7t c\u1ee7a l\u1eadp ph\u01b0\u01a1ng, \u0111\u00e1y kh\u1ed1i n\u00f3n ti\u1ebfp x\u00fac v\u1edbi c\u00e1c c\u1ea1nh c\u1ee7a m\u1eb7t \u0111\u1ed1i di\u1ec7n (xem h\u00ecnh v\u1ebd b\u00ean). T\u00ednh t\u1ec9 s\u1ed1 th\u1ec3 t\u00edch c\u1ee7a l\u01b0\u1ee3ng n\u01b0\u1edbc tr\u00e0n ra ngo\u00e0i v\u00e0 l\u01b0\u1ee3ng n\u01b0\u1edbc ban \u0111\u1ea7u trong kh\u1ed1i h\u1ed9p. C\u00e2u 8. (3.0 \u0111i\u1ec3m) Cho tam gi\u00e1c ABC ( AB \uf03c AC) c\u00f3 \u0111\u01b0\u1eddng cao AH . V\u1ebd \u0111\u01b0\u1eddng tr\u00f2n t\u00e2m O \u0111\u01b0\u1eddng k\u00ednh AB c\u1eaft AC t\u1ea1i I . G\u1ecdi E l\u00e0 \u0111i\u1ec3m \u0111\u1ed1i x\u1ee9ng c\u1ee7a H qua AC , EI c\u1eaft AB t\u1ea1i K v\u00e0 c\u1eaft O t\u1ea1i \u0111i\u1ec3m th\u1ee9 hai l\u00e0 D . a) Ch\u1ee9ng minh t\u1ee9 gi\u00e1c ADBH n\u1ed9i ti\u1ebfp v\u00e0 AD = AE . b) Ch\u1ee9ng minh DH AB . Suy ra HA l\u00e0 ph\u00e2n gi\u00e1c c\u1ee7a g\u00f3c IHK . c) Ch\u1ee9ng minh n\u0103m \u0111i\u1ec3m A , E , C , H , K c\u00f9ng thu\u1ed9c \u0111\u01b0\u1eddng tr\u00f2n. ----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 = \u2212 x2 v\u00e0 \u0111\u01b0\u1eddng th\u1eb3ng (D) : y = \u2212 3 x \u2212 1 . 44 a) V\u1ebd \u0111\u1ed3 th\u1ecb (P) v\u00e0 (D) tr\u00ean c\u00f9ng h\u1ec7 tr\u1ee5c t\u1ecda \u0111\u1ed9. b) Vi\u1ebft ph\u01b0\u01a1ng tr\u00ecnh \u0111\u01b0\u1eddng th\u1eb3ng D1 song song v\u1edbi D v\u00e0 \u0111i qua \u0111i\u1ec3m A thu\u1ed9c P c\u00f3 ho\u00e0nh \u0111\u1ed9 b\u1eb1ng 2 . 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 \u22124 \u22122 0 2 4 y = \u2212 x2 \u22124 \u22121 0 \u22121 \u22124 4 x 0 \u22124 y = \u22123x\u22121 \u22121 2 4 b) Vi\u1ebft ph\u01b0\u01a1ng tr\u00ecnh \u0111\u01b0\u1eddng th\u1eb3ng D1 song song v\u1edbi D v\u00e0 \u0111i qua \u0111i\u1ec3m A thu\u1ed9c P c\u00f3 ho\u00e0nh \u0111\u1ed9 b\u1eb1ng 2 . G\u1ecdi A 2;a P , n\u00ean ta c\u00f3 a 22 1. 4 Suy ra A 2; 1 . V\u00ec D1 \/\/ D : y 3 x 1 n\u00ean D1 : y 3 x c v\u1edbi c 1. 4 4 V\u00ec A D1 n\u00ean 1 3 .2 c c 1 (nh\u1eadn). 4 2 V\u1eady D1 : y 3 x 1 4 2. C\u00e2u 2. (1 \u0111i\u1ec3m). Cho ph\u01b0\u01a1ng tr\u00ecnh x2 3x 5 1 0 . 2x12x22 x1x2 3x1 3x2 . Kh\u00f4ng gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh, h\u00e3y t\u00ednh gi\u00e1 tr\u1ecb c\u1ee7a bi\u1ec3u th\u1ee9c M L\u1eddi gi\u1ea3i V\u00ec a.c 1. 5 1 0 n\u00ean ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m x1 v\u00e0 x2 . 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 \uf0ef\uf0ef\uf0ecS = x1 + x2 = \u2212b = 3 \uf0ed = x1 a 5 +1 Theo \u0111\u1ecbnh l\u00ed Vi-et, ta c\u00f3: \uf0ee\uf0ef\uf0efP c =\u2212 a .x2 = Ta c\u00f3: M 2x12x22 x1x2 3x1 3x2 ( ) ( )M = 2 x1x2 2 \u2212 x1x2 \u2212 3 x1 + x2 ( ) ( )2 M = 2 \u2212 5 + 1 \u2212 \u2212 5 + 1 \u2212 3.3 ( )M = 2 6 \u2212 2 5 + 5 \u2212 10 = 2 \u2212 3 5 V\u1eady M 2 3 5 . C\u00e2u 3. (1 \u0111i\u1ec3m). \u0110\u1ea7u n\u0103m h\u1ecdc, m\u1ed9t tr\u01b0\u1eddng THPT tuy\u1ec3n \u0111\u01b0\u1ee3c 75 h\u1ecdc sinh v\u00e0o 2 l\u1edbp chuy\u00ean V\u0103n v\u00e0 chuy\u00ean S\u1eed. N\u1ebfu chuy\u1ec3n 15 h\u1ecdc sinh t\u1eeb l\u1edbp chuy\u00ean V\u0103n sang l\u1edbp chuy\u00ean S\u1eed th\u00ec s\u1ed1 h\u1ecdc sinh 8 l\u1edbp chuy\u00ean S\u1eed b\u1eb1ng 7 s\u1ed1 h\u1ecdc sinh l\u1edbp chuy\u00ean V\u0103n. H\u00e3y t\u00ednh s\u1ed1 h\u1ecdc sinh c\u1ee7a m\u1ed7i l\u1edbp. L\u1eddi gi\u1ea3i * ) (h\u1ecdc sinh). G\u1ecdi x l\u00e0 s\u1ed1 h\u1ecdc sinh l\u1edbp chuy\u00ean V\u0103n l\u00fac \u0111\u1ea7u ( 0 x 75 , x * ) (h\u1ecdc sinh). G\u1ecdi y l\u00e0 s\u1ed1 h\u1ecdc sinh l\u1edbp chuy\u00ean S\u1eed l\u00fac \u0111\u1ea7u ( 0 x 75 , x V\u00ec t\u1ed5ng s\u1ed1 h\u1ecdc sinh l\u00fac \u0111\u1ea7u c\u1ee7a hai l\u1edbp l\u00e0 75 h\u1ecdc sinh n\u00ean ta c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh x y 75 1 S\u1ed1 h\u1ecdc sinh l\u1edbp chuy\u00ean V\u0103n l\u00fac sau l\u00e0 x 15 (h\u1ecdc sinh). S\u1ed1 h\u1ecdc sinh l\u1edbp chuy\u00ean S\u1eed l\u00fac sau l\u00e0 y 15 (h\u1ecdc sinh). V\u00ec s\u1ed1 h\u1ecdc sinh l\u1edbp chuy\u00ean S\u1eed l\u00fac sau b\u1eb1ng 8 s\u1ed1 h\u1ecdc sinh l\u1edbp chuy\u00ean V\u0103n l\u00fac sau n\u00ean ta c\u00f3 7 ph\u01b0\u01a1ng tr\u00ecnh y 15 8 x 15 8 x y 225 2 7 77 x y 75 x 50 y 25 (nh\u1eadn). T\u1eeb 1 v\u00e0 2 , ta c\u00f3 h\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh 8x y 225 7 7 V\u1eady s\u1ed1 h\u1ecdc sinh l\u1edbp chuy\u00ean V\u0103n l\u00e0 50 (h\u1ecdc sinh). S\u1ed1 h\u1ecdc sinh l\u1edbp chuy\u00ean S\u1eed l\u00e0 25 (h\u1ecdc sinh). C\u00e2u 4. (1 \u0111i\u1ec3m). X\u00ed nghi\u1ec7p may Vi\u1ec7t Ti\u1ebfn h\u00e0ng th\u00e1ng ph\u1ea3i chi 410 000 000 \u0111\u1ed3ng \u0111\u1ec3 tr\u1ea3 l\u01b0\u01a1ng cho c\u00f4ng nh\u00e2n, mua v\u1eadt t\u01b0 v\u00e0 c\u00e1c kho\u1ea3n ph\u00ed kh\u00e1c. M\u1ed7i chi\u1ebfc \u00e1o \u0111\u01b0\u1ee3c b\u00e1n v\u1edbi gi\u00e1 350 000 \u0111\u1ed3ng. 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 G\u1ecdi s\u1ed1 ti\u1ec1n l\u1eddi (ho\u1eb7c l\u1ed7) m\u00e0 x\u00ed nghi\u1ec7p thu \u0111\u01b0\u1ee3c sau m\u1ed7i th\u00e1ng l\u00e0 T v\u00e0 m\u1ed7i th\u00e1ng x\u00ed nghi\u1ec7p b\u00e1n \u0111\u01b0\u1ee3c x chi\u1ebfc \u00e1o. a) L\u1eadp h\u00e0m s\u1ed1 c\u1ee7a T theo x . b) C\u1ea7n ph\u1ea3i b\u00e1n trung b\u00ecnh bao nhi\u00eau chi\u1ebfc \u00e1o m\u1ed7i th\u00e1ng \u0111\u1ec3 sau 1 n\u0103m, x\u00ed nghi\u1ec7p thu \u0111\u01b0\u1ee3c ti\u1ec1n l\u1eddi l\u00e0 1380 000 000 \u0111\u1ed3ng. L\u1eddi gi\u1ea3i a) H\u00e0m s\u1ed1 T 350 000x 410 000 000 . b) X\u00ed nghi\u1ec7p thu \u0111\u01b0\u1ee3c ti\u1ec1n l\u1eddi l\u00e0 1380 000 000 \u0111\u1ed3ng n\u00ean ta c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh 1380 000 000 12 350 000x 410 000 000 x 1500 (chi\u1ebfc \u00e1o). V\u1eady c\u1ea7n b\u00e1n \u00edt nh\u1ea5t l\u00e0 1500 chi\u1ebfc \u00e1o m\u1ed7i th\u00e1ng th\u00ec x\u00ed nghi\u1ec7p thu \u0111\u01b0\u1ee3c ti\u1ec1n l\u1eddi l\u00e0 1380 000 000 \u0111\u1ed3ng. C\u00e2u 5. (0,75 \u0111i\u1ec3m). M\u1ed9t vi\u00ean g\u1ea1ch h\u00ecnh vu\u00f4ng ( 40 cm x 40 cm) \u0111\u01b0\u1ee3c nh\u01b0 tr\u00ean h\u00ecnh, t\u00ednh di\u1ec7n t\u00edch ph\u1ea7n t\u00f4 m\u00e0u. L\u1eddi gi\u1ea3i Di\u1ec7n t\u00edch ph\u1ea7n t\u00f4 m\u00e0u b\u1eb1ng 8 l\u1ea7n di\u1ec7n t\u00edch h\u00ecnh vi\u00ean ph\u00e2n BKH . 5 Di\u1ec7n t\u00edch h\u00ecnh vi\u00ean ph\u00e2n Squat S BKH 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 = \uf070 KB2 \uf0d7 90\uf0b0 \u2212 1 BK \uf0d7 BN 360\uf0b0 2 202 \uf0d7 90\uf0b0 1 = \uf070 \uf0d7 360\uf0b0 \u2212 2 \uf0d7 20 \uf0d7 20 = 100\uf070 \u2212 200 ( )V\u1eady di\u1ec7n t\u00edch c\u1ea7n t\u00ecm l\u00e0 8 \uf0d7 100\uf070 \u2212 200 = 913cm2 . C\u00e2u 6. (0,75 \u0111i\u1ec3m). M\u1ed7i c\u00f4ng nh\u00e2n c\u1ee7a c\u00f4ng ty C\u1ed5 ph\u1ea7n ABC c\u00f3 s\u1ed1 ti\u1ec1n th\u01b0\u1edfng t\u1ebft n\u0103m 2018 l\u00e0 1 th\u00e1ng l\u01b0\u01a1ng. \u0110\u1ebfn n\u0103m 2019 , s\u1ed1 ti\u1ec1n th\u01b0\u1edfng t\u1ebft c\u1ee7a h\u1ecd \u0111\u01b0\u1ee3c t\u0103ng th\u00eam 6% so v\u1edbi s\u1ed1 ti\u1ec1n th\u01b0\u1edfng t\u1ebft c\u1ee7a n\u0103m 2018 . V\u00e0o n\u0103m 2020 , s\u1ed1 ti\u1ec1n th\u01b0\u1edfng t\u1ebft c\u1ee7a h\u1ecd \u0111\u01b0\u1ee3c t\u0103ng th\u00eam 10% so v\u1edbi s\u1ed1 ti\u1ec1n th\u01b0\u1edfng t\u1ebft c\u1ee7a n\u0103m 2019 , ngo\u00e0i ra n\u1ebfu c\u00f4ng nh\u00e2n n\u00e0o \u0111\u01b0\u1ee3c l\u00e0 c\u00f4ng \u0111o\u00e0n vi\u00ean xu\u1ea5t s\u1eafc s\u1ebd \u0111\u01b0\u1ee3c th\u01b0\u1edfng th\u00eam 500 000 \u0111\u1ed3ng. Anh Ba l\u00e0 c\u00f4ng \u0111o\u00e0n vi\u00ean xu\u1ea5t s\u1eafc c\u1ee7a n\u0103m 2019 , n\u00ean anh nh\u1eadn \u0111\u01b0\u1ee3c s\u1ed1 ti\u1ec1n th\u01b0\u1edfng t\u1ebft l\u00e0 6 330 000 \u0111\u1ed3ng. H\u1ecfi n\u0103m 2018 , ti\u1ec1n l\u01b0\u01a1ng 1 th\u00e1ng c\u1ee7a anh Ba l\u00e0 bao nhi\u00eau? L\u1eddi gi\u1ea3i S\u1ed1 ti\u1ec1n l\u01b0\u01a1ng m\u1ed9t th\u00e1ng v\u00e0o n\u0103m 2018 c\u1ee7a anh Ba l\u00e0 6 330 000 500 000 1 10% 1 6% 5 000 000 (\u0111\u1ed3ng) V\u1eady ti\u1ec1n l\u01b0\u01a1ng m\u1ed9t th\u00e1ng v\u00e0o n\u0103m 2018 c\u1ee7a anh Ba l\u00e0 5 000 000 (\u0111\u1ed3ng) C\u00e2u 7. (1 \u0111i\u1ec3m). M\u1ed9t kh\u1ed1i l\u1eadp ph\u01b0\u01a1ng c\u00f3 c\u1ea1nh 1 m ch\u1ee9a 1 1 2 12 m3 . \u0111\u1ea7y n\u01b0\u1edbc. \u0110\u1eb7t v\u00e0o trong kh\u1ed1i \u0111\u00f3 m\u1ed9t kh\u1ed1i n\u00f3n c\u00f3 3 \u0111\u1ec9nh tr\u00f9ng v\u1edbi t\u00e2m m\u1ed9t m\u1eb7t c\u1ee7a l\u1eadp ph\u01b0\u01a1ng, \u0111\u00e1y 2 .1 kh\u1ed1i n\u00f3n ti\u1ebfp x\u00fac v\u1edbi c\u00e1c c\u1ea1nh c\u1ee7a m\u1eb7t \u0111\u1ed1i di\u1ec7n (xem h\u00ecnh v\u1ebd b\u00ean). T\u00ednh t\u1ec9 s\u1ed1 th\u1ec3 t\u00edch c\u1ee7a l\u01b0\u1ee3ng n\u01b0\u1edbc tr\u00e0n ra ngo\u00e0i v\u00e0 l\u01b0\u1ee3ng n\u01b0\u1edbc ban \u0111\u1ea7u trong kh\u1ed1i h\u1ed9p. L\u1eddi gi\u1ea3i Th\u1ec3 t\u00edch l\u01b0\u1ee3ng n\u01b0\u1edbc tr\u00e0n ra b\u1eb1ng v\u1edbi th\u1ec3 t\u00edch kh\u1ed1i n\u00f3n n\u00ean Vnon Th\u1ec3 t\u00edch kh\u1ed1i l\u1eadp ph\u01b0\u01a1ng l\u00e0 Vlp 13 1 m3 . 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 V\u1eady t\u1ec9 s\u1ed1 c\u1ea7n t\u00ecm l\u00e0 12 12 . 1 12 C\u00e2u 8. (3.0 \u0111i\u1ec3m) Cho tam gi\u00e1c ABC ( AB \uf03c AC) c\u00f3 \u0111\u01b0\u1eddng cao AH . V\u1ebd \u0111\u01b0\u1eddng tr\u00f2n t\u00e2m O \u0111\u01b0\u1eddng k\u00ednh AB c\u1eaft AC t\u1ea1i I . G\u1ecdi E l\u00e0 \u0111i\u1ec3m \u0111\u1ed1i x\u1ee9ng c\u1ee7a H qua AC , EI c\u1eaft AB t\u1ea1i K v\u00e0 c\u1eaft O t\u1ea1i \u0111i\u1ec3m th\u1ee9 hai l\u00e0 D . a) Ch\u1ee9ng minh t\u1ee9 gi\u00e1c ADBH n\u1ed9i ti\u1ebfp v\u00e0 AD = AE . b) Ch\u1ee9ng minh DH AB . Suy ra HA l\u00e0 ph\u00e2n gi\u00e1c c\u1ee7a g\u00f3c IHK . c) Ch\u1ee9ng minh n\u0103m \u0111i\u1ec3m A , E , C , H , K c\u00f9ng thu\u1ed9c \u0111\u01b0\u1eddng tr\u00f2n. L\u1eddi gi\u1ea3i a) Ta c\u00f3: AHB vu\u00f4ng t\u1ea1i H 7 H thu\u1ed9c \u0111\u01b0\u1eddng tr\u00f2n \u0111\u01b0\u1eddng k\u00ednh AB hay thu\u1ed9c \u0111\u01b0\u1eddng tr\u00f2n O . ADBH n\u1ed9i ti\u1ebfp \u0111\u01b0\u1eddng tr\u00f2n O . Ta c\u00f3: ADI AHI (c\u00f9ng ch\u1eafn cung AI ) (1) V\u00ec E v\u00e0 H \u0111\u1ed1i x\u1ee9ng qua AC n\u00ean AC l\u00e0 \u0111\u01b0\u1eddng trung tr\u1ef1c c\u1ee7a EH AE AH v\u00e0 IE IH . X\u00e9t AIE v\u00e0 AIH c\u00f3: T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH \u2022 AI chung. \u2022 AE AH \u2022 IE IH AIE AIH (c \u2013 c \u2013 c) AEI AHI (2) T\u1eeb 1 v\u00e0 2 AEI ADI AED c\u00e2n t\u1ea1i A AD AE . b) Ta c\u00f3: ADB 90 (g\u00f3c n\u1ed9i ti\u1ebfp ch\u1eafn n\u1eeda \u0111\u01b0\u1eddng tr\u00f2n) X\u00e9t AHB v\u00e0 ADB c\u00f3: \u2022 AH AD AE \u2022 AB l\u00e0 c\u1ea1nh chung \u2022 AHB ADB 90 AHB ADB (c\u1ea1nh huy\u1ec1n \u2013 c\u1ea1nh g\u00f3c vu\u00f4ng) AB HD BH BD m\u00e0 AH AD AB l\u00e0 \u0111\u01b0\u1eddng trung tr\u1ef1c c\u1ee7a HD V\u00ec AB l\u00e0 \u0111\u01b0\u1eddng trung tr\u1ef1c c\u1ee7a HD n\u00ean KH KD . X\u00e9t AKH v\u00e0 AKD c\u00f3: \u2022 AH AD \u2022 AK l\u00e0 c\u1ea1nh chung \u2022 KH KD AKH AKD (c \u2013 c \u2013 c) AHK ADK m\u00e0 ADI AHI AHK AHI HA l\u00e0 tia ph\u00e2n gi\u00e1c c\u1ee7a IHK . c) X\u00e9t AEC v\u00e0 AHC c\u00f3: \u2022 AE AH ( AC l\u00e0 \u0111\u01b0\u1eddng trung tr\u1ef1c c\u1ee7a EH ) \u2022 CE CH ( AC l\u00e0 \u0111\u01b0\u1eddng trung tr\u1ef1c c\u1ee7a EH ) \u2022 AC l\u00e0 c\u1ea1nh chung AEC AHC (c \u2013 c \u2013 c) AEC AHC 90 AEC AHC 180 t\u1ee9 gi\u00e1c AECH n\u1ed9i ti\u1ebfp \u0111\u01b0\u1eddng tr\u00f2n (3) 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 (4) Ta c\u00f3: AHK AHI m\u00e0 AHI AEI AHK AEI T\u1ee9 gi\u00e1c AEHK n\u1ed9i ti\u1ebfp \u0111\u01b0\u1eddng tr\u00f2n T\u1eeb (3) v\u00e0 (4) n\u0103m \u0111i\u1ec3m A , E , C , H , K c\u00f9ng thu\u1ed9c \u0111\u01b0\u1eddng tr\u00f2n. ----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 7 NA\u00caM HO\u00cfC: 2023 - 2024 \u0110\u1ec0 THAM KH\u1ea2O M\u00d4N: TO\u00c1N 9 \u0110\u00ea thi g\u1ed3m 8 c\u00e2u ho\u0309i t\u01b0\u0323 lu\u00e2\u0323n. M\u00c3 \u0110\u1ec0: Qu\u1eadn 7 - 2 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). Trong m\u1eb7t ph\u1eb3ng t\u1ecda \u0111\u1ed9 Oxy , cho parabol (P) : y = x2 v\u00e0 \u0111\u01b0\u1eddng th\u1eb3ng (d) : y = mx + 3 ( m l\u00e0 tham s\u1ed1). a) V\u1ebd parabol (P) . b) Khi m = 2 , t\u00ecm t\u1ecda \u0111\u1ed9 giao \u0111i\u1ec3m c\u1ee7a (P) v\u00e0 (d) b\u1eb1ng ph\u00e9p to\u00e1n. C\u00e2u 2. (1 \u0111i\u1ec3m). Cho ph\u01b0\u01a1ng tr\u00ecnh x2 \u2212 x \u2212 13 = 0 c\u00f3 2 nghi\u1ec7m l\u00e0 x1 , x2 . Kh\u00f4ng gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh, h\u00e3y t\u00ednh gi\u00e1 tr\u1ecb c\u1ee7a bi\u1ec3u th\u1ee9c A = (x1 \u2212 2x2 )(2x1 \u2212 x2 ) C\u00e2u 3. (1 \u0111i\u1ec3m). T\u1ed5ng s\u1ed1 h\u1ecdc sinh c\u1ee7a hai l\u1edbp 9A v\u00e0 9B \u1edf m\u1ed9t tr\u01b0\u1eddng THCS l\u00e0 76 h\u1ecdc sinh. H\u01b0\u1edfng \u1ee9ng phong tr\u00e0o \u1ee7ng h\u1ed9 trang thi\u1ebft b\u1ecb y t\u1ebf trong \u0111\u1ee3t ph\u00f2ng d\u1ecbch Covid-19, c\u1ea3 hai l\u1edbp \u0111\u00e3 quy\u00ean g\u00f3p \u1ee7ng h\u1ed9 189 chi\u1ebfc kh\u1ea9u trang. Bi\u1ebft r\u1eb1ng m\u1ed7i h\u1ecdc sinh l\u1edbp 9A \u1ee7ng h\u1ed9 3 chi\u1ebfc kh\u1ea9u trang, m\u1ed7i h\u1ecdc sinh l\u1edbp 9B \u1ee7ng h\u1ed9 2 chi\u1ebfc kh\u1ea9u trang. T\u00ednh s\u1ed1 h\u1ecdc sinh c\u1ee7a m\u1ed7i l\u1edbp. C\u00e2u 4. (1 \u0111i\u1ec3m). H\u00e3ng taxi A quy \u0111\u1ecbnh gi\u00e1 thu\u00ea xe cho nh\u1eefng chuy\u1ebfn \u0111i \u0111\u01b0\u1eddng d\u00e0i (tr\u00ean 50 km ). M\u1ed7i km l\u00e0 16 ngh\u00ecn \u0111\u1ed3ng \u0111\u1ed1i v\u1edbi 50 km \u0111\u1ea7u ti\u00ean v\u00e0 9 ngh\u00ecn 5 tr\u0103m \u0111\u1ed3ng \u0111\u1ed1i v\u1edbi c\u00e1c km ti\u1ebfp theo. a) M\u1ed9t kh\u00e1ch thu\u00ea xe taxi \u0111i qu\u00e3ng \u0111\u01b0\u1eddng 75 km th\u00ec ph\u1ea3i tr\u1ea3 s\u1ed1 ti\u1ec1n thu\u00ea xe l\u00e0 bao nhi\u00eau ngh\u00ecn \u0111\u1ed3ng? b) G\u1ecdi y (ngh\u00ecn \u0111\u1ed3ng) l\u00e0 s\u1ed1 ti\u1ec1n kh\u00e1ch thu\u00ea xe taxi ph\u1ea3i tr\u1ea3 sau khi \u0111i x km . Khi \u1ea5y m\u1ed1i li\u00ean h\u1ec7 gi\u1eefa hai \u0111\u1ea1i l\u01b0\u1ee3ng n\u00e0y l\u00e0 m\u1ed9t h\u00e0m s\u1ed1 b\u1eadc nh\u1ea5t y = ax + b . H\u00e3y x\u00e1c \u0111\u1ecbnh h\u00e0m s\u1ed1 n\u00e0y khi x \uf03e 50 ? C\u00e2u 5. (1 \u0111i\u1ec3m). M\u1ed9t tr\u01b0\u1eddng h\u1ecdc c\u1ea7n \u0111\u01b0a 510 h\u1ecdc sinh \u0111i tham quan V\u0169ng T\u00e0u. C\u00f3 hai c\u00e1ch \u0111\u1ec3 thu\u00ea xe: C\u00e1ch 1 l\u00e0 thu\u00ea xe 45 ch\u1ed7, gi\u00e1 thu\u00ea \u0111i v\u00e0 v\u1ec1 cho m\u1ed7i xe l\u00e0 1800000 \u0111\u1ed3ng; c\u00e1ch 2 l\u00e0 thu\u00ea xe 29 ch\u1ed7, gi\u00e1 thu\u00ea \u0111i v\u1ec1 cho m\u1ed7i xe l\u00e0 950000 \u0111\u1ed3ng. N\u1ebfu ch\u1ec9 thu\u00ea m\u1ed9t lo\u1ea1i xe cho c\u1ea3 \u0111o\u00e0n th\u00ec nh\u00e0 tr\u01b0\u1eddng thu\u00ea lo\u1ea1i xe n\u00e0o s\u1ebd ti\u1ebft ki\u1ec7m h\u01a1n? C\u00e2u 6. (0,75 \u0111i\u1ec3m). M\u1ed9t v\u00e9 xem phim c\u00f3 gi\u00e1 60000 \u0111\u1ed3ng. Khi c\u00f3 \u0111\u1ee3t gi\u1ea3m gi\u00e1, m\u1ed7i ng\u00e0y s\u1ed1 l\u01b0\u1ee3ng ng\u01b0\u1eddi xem t\u0103ng l\u00ean 50% , do \u0111\u00f3 doanh thu c\u0169ng t\u0103ng 25% . H\u1ecfi gi\u00e1 v\u00e9 khi \u0111\u01b0\u1ee3c gi\u1ea3m l\u00e0 bao nhi\u00eau? 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 7. (1 \u0111i\u1ec3m). Nh\u00e0 h\u00e1t Cao V\u0103n L\u1ea7u, Trung t\u00e2m tri\u1ec3n l\u00e3m v\u0103n h\u00f3a ngh\u1ec7 thu\u1eadt t\u1ec9nh B\u1ea1c Li\u00eau c\u00f3 h\u00ecnh d\u00e1ng ba chi\u1ebfc n\u00f3n l\u00e1 l\u1edbn nh\u1ea5t Vi\u1ec7t Nam, m\u00e1i nh\u00e0 h\u00ecnh n\u00f3n l\u00e0m b\u1eb1ng v\u1eadt li\u1ec7u composite v\u00e0 \u0111\u01b0\u1ee3c \u0111\u1eb7t h\u01b0\u1edbng v\u00e0o nhau. Em h\u00e3y t\u00ednh th\u1ec3 t\u00edch c\u1ee7a m\u1ed9t m\u00e1i nh\u00e0 h\u00ecnh n\u00f3n bi\u1ebft \u0111\u01b0\u1eddng k\u00ednh l\u00e0 45m v\u00e0 chi\u1ec1u cao l\u00e0 45m (l\u1ea5y \uf070 \uf0bb 3,14 , k\u1ebft qu\u1ea3 l\u00e0m tr\u00f2n \u0111\u1ebfn h\u00e0ng \u0111\u01a1n v\u1ecb, ba h\u00ecnh n\u00f3n c\u00f3 b\u00e1n k\u00ednh b\u1eb1ng nhau). C\u00e2u 8. (3 \u0111i\u1ec3m) Cho tam gi\u00e1c nh\u1ecdn ABC ( AB \uf03c AC) n\u1ed9i ti\u1ebfp \u0111\u01b0\u1eddng tr\u00f2n t\u00e2m (O) . Hai ti\u1ebfp tuy\u1ebfn t\u1ea1i B v\u00e0 C c\u1ee7a \u0111\u01b0\u1eddng tr\u00f2n (O) c\u1eaft nhau t\u1ea1i M , tia AM c\u1eaft \u0111\u01b0\u1eddng tr\u00f2n (O) t\u1ea1i \u0111i\u1ec3m D . a) Ch\u1ee9ng minh r\u1eb1ng t\u1ee9 gi\u00e1c OBMC n\u1ed9i ti\u1ebfp \u0111\u01b0\u1ee3c \u0111\u01b0\u1eddng tr\u00f2n. b) Ch\u1ee9ng minh MB2 = MD.MA . c) G\u1ecdi E l\u00e0 trung \u0111i\u1ec3m c\u1ee7a \u0111o\u1ea1n th\u1eb3ng AD ; tia CE c\u1eaft \u0111\u01b0\u1eddng tr\u00f2n (O) t\u1ea1i \u0111i\u1ec3m F Ch\u1ee9ng minh r\u1eb1ng: BF \/ \/AM . ----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) Trong m\u1eb7t ph\u1eb3ng t\u1ecda \u0111\u1ed9 Oxy, cho parabol (P) : y = x2 v\u00e0 \u0111\u01b0\u1eddng th\u1eb3ng (d) : y = mx + 3 (m l\u00e0 tham s\u1ed1). a. V\u1ebd parabol (P) . b. Khi m = 2 , 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 (P) a) V\u1ebd \u0111\u1ed3 th\u1ecb (P) v\u00e0 (d) tr\u00ean c\u00f9ng h\u1ec7 tr\u1ee5c t\u1ecda \u0111\u1ed9. 4 BGT: x \u22122 \u22121 0 1 2 y = x2 4 1 0 1 4 1 b) Khi m = 2 , t\u00ecm t\u1ecda \u0111\u1ed9 giao \u0111i\u1ec3m c\u1ee7a (P) v\u00e0 (d) x -2 -1 O 1 2 b\u1eb1ng ph\u00e9p to\u00e1n. Ph\u01b0\u01a1ng tr\u00ecnh ho\u00e0nh \u0111\u1ed9 giao \u0111i\u1ec3m c\u1ee7a (P) v\u00e0 (d) : x2 = 2x + 3 \uf0db x2 \u2212 2x \u2212 3 = 0 \uf0db \uf0e9x = \u22121 \uf0ea\uf0ebx = 3 Thay x = \u22121 v\u00e0o y = x2 , ta \u0111\u01b0\u1ee3c: y = (\u22121)2 = 1 . Thay x = 3 v\u00e0o y = x2 , ta \u0111\u01b0\u1ee3c: y = (3)2 = 9 . V\u1eady (\u22121; 1) , (3; 9) l\u00e0 hai giao \u0111i\u1ec3m c\u1ea7n t\u00ecm. C\u00e2u 2. (1 \u0111i\u1ec3m) Cho ph\u01b0\u01a1ng tr\u00ecnh x2 \u2212 x \u2212 13 = 0 c\u00f3 2 nghi\u1ec7m l\u00e0 x1 ,x2 . Kh\u00f4ng gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh, h\u00e3y t\u00ednh gi\u00e1 tr\u1ecb c\u1ee7a bi\u1ec3u th\u1ee9c A = (x1 \u2212 2x2 )(2x1 \u2212 x2 ) . L\u1eddi gi\u1ea3i V\u00ec \uf044 = b2 \u2212 4ac = (\u22121)2 \u2212 4.1.(\u221213) = 53 \uf03e 0 N\u00ean ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m ph\u00e2n bi\u1ec7t x1 ,x2 . 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 \uf0ef\uf0ec\uf0efS = x1 + x2 = \u2212b = 1 \uf0ed = x1 a Theo \u0111\u1ecbnh l\u00ed Vi-et, ta c\u00f3: \uf0ef\uf0ee\uf0efP c = \u221213 a .x2 = Ta c\u00f3: A = (x1 \u2212 2x2 )(2x1 \u2212 x2 ) A = 2x12 \u2212 x1x2 \u2212 4x1x2 + 2x22 ( )A = 2 x12 + x22 \u2212 5x1x2 ( )A \uf0e9 2 \uf0f9 = 2 \uf0ea\uf0eb x1 + x2 \u2212 2x1x2 \uf0fa\uf0fb \u2212 5x1x2 ( )A = 2 x1 + x2 2 \u2212 9x1x2 A = 2(1)2 \u2212 9.(\u221213) = 119 C\u00e2u 3. (1 \u0111i\u1ec3m) T\u1ed5ng s\u1ed1 h\u1ecdc sinh c\u1ee7a hai l\u1edbp 9A v\u00e0 9B \u1edf m\u1ed9t tr\u01b0\u1eddng THCS l\u00e0 76 h\u1ecdc sinh. H\u01b0\u1edfng \u1ee9ng phong tr\u00e0o \u1ee7ng h\u1ed9 trang thi\u1ebft b\u1ecb y t\u1ebf trong \u0111\u1ee3t ph\u00f2ng d\u1ecbch Covid-19, c\u1ea3 hai l\u1edbp \u0111\u00e3 quy\u00ean g\u00f3p \u1ee7ng h\u1ed9 189 chi\u1ebfc kh\u1ea9u trang. Bi\u1ebft r\u1eb1ng m\u1ed7i h\u1ecdc sinh l\u1edbp 9A \u1ee7ng h\u1ed9 3 chi\u1ebfc kh\u1ea9u trang, m\u1ed7i h\u1ecdc sinh l\u1edbp 9B \u1ee7ng h\u1ed9 2 chi\u1ebfc kh\u1ea9u trang. T\u00ednh s\u1ed1 h\u1ecdc sinh c\u1ee7a m\u1ed7i l\u1edbp. L\u1eddi gi\u1ea3i G\u1ecdi s\u1ed1 h\u1ecdc sinh c\u1ee7a l\u1edbp 9A , 9B l\u1ea7n l\u01b0\u1ee3t l\u00e0 x, y ( x, y \uf0ce * ) S\u1ed1 chi\u1ebfc kh\u1ea9u trang l\u1edbp 9A \u0111\u00e3 \u1ee7ng h\u1ed9 l\u00e0: 3x (chi\u1ebfc) S\u1ed1 chi\u1ebfc kh\u1ea9u trang l\u1edbp 9B \u0111\u00e3 \u1ee7ng h\u1ed9 l\u00e0: 2y (chi\u1ebfc) V\u00ec t\u1ed5ng s\u1ed1 h\u1ecdc sinh c\u1ee7a hai l\u1edbp l\u00e0 76 h\u1ecdc sinh n\u00ean ta c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh x + y = 76 (1) V\u00ec c\u1ea3 hai l\u1edbp \u1ee7ng h\u1ed9 189 chi\u1ebfc kh\u1ea9u trang n\u00ean ta c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh: 3x + 2y = 189 (2) T\u1eeb (1) v\u00e0 (2) , ta c\u00f3 h\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh: \uf0ecx + y = 76 \uf0db \uf0ecx = 37 (th\u1ecfa m\u00e3n) \uf0ed\uf0ee3x + 2y = 189 \uf0ed\uf0eey = 39 V\u1eady s\u1ed1 h\u1ecdc sinh c\u1ee7a l\u1edbp 9A v\u00e0 9B l\u1ea7n l\u01b0\u1ee3t l\u00e0 37 h\u1ecdc sinh v\u00e0 39 h\u1ecdc sinh C\u00e2u 4. (0,75 \u0111i\u1ec3m). H\u00e3ng taxi A quy \u0111\u1ecbnh gi\u00e1 thu\u00ea xe cho nh\u1eefng chuy\u1ebfn \u0111i \u0111\u01b0\u1eddng d\u00e0i (tr\u00ean 50 km ). M\u1ed7i km l\u00e0 16 ngh\u00ecn \u0111\u1ed3ng \u0111\u1ed1i v\u1edbi 50 km \u0111\u1ea7u ti\u00ean v\u00e0 9 ngh\u00ecn 5 tr\u0103m \u0111\u1ed3ng \u0111\u1ed1i v\u1edbi c\u00e1c km ti\u1ebfp theo. a) M\u1ed9t kh\u00e1ch thu\u00ea xe taxi \u0111i qu\u00e3ng \u0111\u01b0\u1eddng 75 km th\u00ec ph\u1ea3i tr\u1ea3 s\u1ed1 ti\u1ec1n thu\u00ea xe l\u00e0 bao nhi\u00eau ngh\u00ecn \u0111\u1ed3ng? b) G\u1ecdi y (ngh\u00ecn \u0111\u1ed3ng) l\u00e0 s\u1ed1 ti\u1ec1n kh\u00e1ch thu\u00ea xe taxi ph\u1ea3i tr\u1ea3 sau khi \u0111i x km . Khi \u1ea5y m\u1ed1i li\u00ean h\u1ec7 gi\u1eefa hai \u0111\u1ea1i l\u01b0\u1ee3ng n\u00e0y l\u00e0 m\u1ed9t h\u00e0m s\u1ed1 b\u1eadc nh\u1ea5t y = ax + b . H\u00e3y x\u00e1c \u0111\u1ecbnh h\u00e0m s\u1ed1 n\u00e0y khi x \uf03e 50 ? T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N 4","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH L\u1eddi gi\u1ea3i a) S\u1ed1 ti\u1ec1n thu\u00ea xe \u0111i qu\u00e3ng \u0111\u01b0\u1eddng 75 km : 16000.50 + 9500.25 = 1037500 (\u0111\u1ed3ng) b) H\u00e0m s\u1ed1 bi\u1ec3u th\u1ecb khi x \uf03e 50 : y = 50.16000 + (x \u2013 50).9500 = 9500x + 325000 C\u00e2u 5. (1 \u0111i\u1ec3m) M\u1ed9t tr\u01b0\u1eddng h\u1ecdc c\u1ea7n \u0111\u01b0a 510 h\u1ecdc sinh \u0111i tham quan V\u0169ng T\u00e0u. C\u00f3 hai c\u00e1ch \u0111\u1ec3 thu\u00ea xe: C\u00e1ch 1 l\u00e0 thu\u00ea xe 45 ch\u1ed7, gi\u00e1 thu\u00ea \u0111i v\u00e0 v\u1ec1 cho m\u1ed7i xe l\u00e0 1800000 \u0111\u1ed3ng; c\u00e1ch 2 l\u00e0 thu\u00ea xe 29 ch\u1ed7, gi\u00e1 thu\u00ea \u0111i v\u1ec1 cho m\u1ed7i xe l\u00e0 950000 \u0111\u1ed3ng. N\u1ebfu ch\u1ec9 thu\u00ea m\u1ed9t lo\u1ea1i xe cho c\u1ea3 \u0111o\u00e0n th\u00ec nh\u00e0 tr\u01b0\u1eddng thu\u00ea lo\u1ea1i xe n\u00e0o s\u1ebd ti\u1ebft ki\u1ec7m h\u01a1n? L\u1eddi gi\u1ea3i S\u1ed1 xe 45 ch\u1ed7 c\u1ea7n thu\u00ea \u0111\u1ec3 \u0111\u01b0a 510 h\u1ecdc sinh \u0111i tham quan: 510 \uf0bb 11,33 (xe) 45 V\u1eady nh\u00e0 tr\u01b0\u1eddng ph\u1ea3i thu\u00ea 12 (xe) S\u1ed1 ti\u1ec1n thu\u00ea xe 45 ch\u1ed7 l\u00e0: 12.1800000 = 21600000 (\u0111\u1ed3ng) S\u1ed1 xe 29 ch\u1ed7 c\u1ea7n thu\u00ea \u0111\u1ec3 \u0111\u01b0a 510 h\u1ecdc sinh \u0111i tham quan: 510 \uf0bb 17,58 (xe) 29 V\u1eady nh\u00e0 tr\u01b0\u1eddng ph\u1ea3i thu\u00ea 18 (xe) S\u1ed1 ti\u1ec1n thu\u00ea xe 29 ch\u1ed7 l\u00e0: 18.950000 =17100000 (\u0111\u1ed3ng) V\u00ec 17100000 \uf03c 21600000 n\u00ean thu\u00ea xe lo\u1ea1i 29 ch\u1ed7 s\u1ebd ti\u1ebft ki\u1ec7m h\u01a1n cho nh\u00e0 tr\u01b0\u1eddng. C\u00e2u 6. (1 \u0111i\u1ec3m) M\u1ed9t v\u00e9 xem phim c\u00f3 gi\u00e1 60000 \u0111\u1ed3ng. Khi c\u00f3 \u0111\u1ee3t gi\u1ea3m gi\u00e1, m\u1ed7i ng\u00e0y s\u1ed1 l\u01b0\u1ee3ng ng\u01b0\u1eddi xem t\u0103ng l\u00ean 50% , do \u0111\u00f3 doanh thu c\u0169ng t\u0103ng 25% . H\u1ecfi gi\u00e1 v\u00e9 khi \u0111\u01b0\u1ee3c gi\u1ea3m l\u00e0 bao nhi\u00eau? L\u1eddi gi\u1ea3i G\u1ecdi x l\u00e0 s\u1ed1 l\u01b0\u1ee3ng kh\u00e1n gi\u1ea3 \u0111i xem phim l\u00fac ch\u01b0a gi\u1ea3m gi\u00e1 ( x \uf0ce * ) S\u1ed1 ti\u1ec1n thu \u0111\u01b0\u1ee3c l\u00fac ch\u01b0a gi\u1ea3m gi\u00e1 l\u00e0 60000x (\u0111\u1ed3ng) S\u1ed1 l\u01b0\u1ee3ng kh\u00e1n gi\u1ea3 sau khi gi\u1ea3m gi\u00e1 l\u00e0: x.150% S\u1ed1 ti\u1ec1n thu \u0111\u01b0\u1ee3c sau khi gi\u1ea3m gi\u00e1 l\u00e0: 60000x.125% V\u1eady gi\u00e1 ti\u1ec1n s\u1ed1 v\u00e9 l\u00fac gi\u1ea3m: 60000x.125% = 50000 (\u0111\u1ed3ng) x.150% 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 7. (1 \u0111i\u1ec3m) Nh\u00e0 h\u00e1t Cao V\u0103n L\u1ea7u, Trung t\u00e2m tri\u1ec3n l\u00e3m v\u0103n h\u00f3a ngh\u1ec7 thu\u1eadt t\u1ec9nh B\u1ea1c Li\u00eau c\u00f3 h\u00ecnh d\u00e1ng ba chi\u1ebfc n\u00f3n l\u00e1 l\u1edbn nh\u1ea5t Vi\u1ec7t Nam, m\u00e1i nh\u00e0 h\u00ecnh n\u00f3n l\u00e0m b\u1eb1ng v\u1eadt li\u1ec7u composite v\u00e0 \u0111\u01b0\u1ee3c \u0111\u1eb7t h\u01b0\u1edbng v\u00e0o nhau. Em h\u00e3y t\u00ednh th\u1ec3 t\u00edch c\u1ee7a m\u1ed9t m\u00e1i nh\u00e0 h\u00ecnh n\u00f3n bi\u1ebft \u0111\u01b0\u1eddng k\u00ednh l\u00e0 45m v\u00e0 chi\u1ec1u cao l\u00e0 45m (l\u1ea5y \uf070 \uf0bb 3,14 , k\u1ebft qu\u1ea3 l\u00e0m tr\u00f2n \u0111\u1ebfn h\u00e0ng \u0111\u01a1n v\u1ecb, ba h\u00ecnh n\u00f3n c\u00f3 b\u00e1n k\u00ednh b\u1eb1ng nhau). L\u1eddi gi\u1ea3i M\u00e1i nh\u00e0 h\u00ecnh n\u00f3n \u0111\u01b0\u1eddng k\u00ednh l\u00e0 45m suy ra b\u00e1n k\u00ednh R = 45 m . 2 Th\u1ec3 t\u00edch c\u1ee7a m\u1ed9t m\u00e1i nh\u00e0 h\u00ecnh n\u00f3n l\u00e0 1 1 3,14.\uf0e6\uf0e7\uf0e8 45 \uf0f62 ( )V3 3 2 \uf0f7\uf0f8 = \uf070 R2h \uf0bb \uf0d7 .24 = 12717 m3 C\u00e2u 8. (3 \u0111i\u1ec3m) Cho tam gi\u00e1c nh\u1ecdn ABC ( AB \uf03c AC) n\u1ed9i ti\u1ebfp \u0111\u01b0\u1eddng tr\u00f2n t\u00e2m (O) . Hai ti\u1ebfp tuy\u1ebfn t\u1ea1i B v\u00e0 C c\u1ee7a \u0111\u01b0\u1eddng tr\u00f2n (O) c\u1eaft nhau t\u1ea1i M , tia AM c\u1eaft \u0111\u01b0\u1eddng tr\u00f2n (O) t\u1ea1i \u0111i\u1ec3m D . a) Ch\u1ee9ng minh r\u1eb1ng t\u1ee9 gi\u00e1c OBMC n\u1ed9i ti\u1ebfp \u0111\u01b0\u1ee3c \u0111\u01b0\u1eddng tr\u00f2n. b) Ch\u1ee9ng minh MB2 = MD.MA . c) G\u1ecdi E l\u00e0 trung \u0111i\u1ec3m c\u1ee7a \u0111o\u1ea1n th\u1eb3ng AD ; tia CE c\u1eaft \u0111\u01b0\u1eddng tr\u00f2n (O) t\u1ea1i \u0111i\u1ec3m F Ch\u1ee9ng minh r\u1eb1ng: BF \/\/ AM . L\u1eddi gi\u1ea3i a) Ch\u1ee9ng minh r\u1eb1ng t\u1ee9 gi\u00e1c OBMC n\u1ed9i ti\u1ebfp \u0111\u01b0\u1ee3c \u0111\u01b0\u1eddng tr\u00f2n. X\u00e9t (O) c\u00f3: MB,MC l\u00e0 c\u00e1c ti\u1ebfp tuy\u1ebfn c\u1ee7a \u0111\u01b0\u1eddng tr\u00f2n (O) t\u1ea1i B v\u00e0 C \uf0de MBO = MCO = 90\uf0b0 X\u00e9t t\u1ee9 gi\u00e1c OBMC , c\u00f3: MBO + MCO = 180\uf0b0 \uf0de T\u1ee9 gi\u00e1c OBMC n\u1ed9i ti\u1ebfp v\u00ec c\u00f3 hai g\u00f3c \u0111\u1ed1i b\u00f9 nhau. 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 A F C O E B D M b) Ch\u1ee9ng minh MB2 = MD.MA X\u00e9t \uf044MBD v\u00e0 \uf044MAB MBD = MAB = 1 sdBD 2 BMA chung \uf0de \uf044MBD \uf044MAB (g \u2013 g \uf0de MB = MD \uf0de MB2 = MA.MD . MA MB c) G\u1ecdi E l\u00e0 trung \u0111i\u1ec3m c\u1ee7a \u0111o\u1ea1n th\u1eb3ng AD ; tia CE c\u1eaft \u0111\u01b0\u1eddng tr\u00f2n (O) t\u1ea1i \u0111i\u1ec3m F Ch\u1ee9ng minh r\u1eb1ng: BF \/ \/AM . (O) c\u00f3 E l\u00e0 trung \u0111i\u1ec3m c\u1ee7a d\u00e2y AD kh\u00f4ng qua t\u00e2m (gt) \uf0de OE \u22a5 AD t\u1ea1i E \uf0de OEM = 90\uf0b0 X\u00e9t t\u1ee9 gi\u00e1c OEMC , c\u00f3: MEO = MCO = 90\uf0b0 \uf0de MEO + MCO = 180\uf0b0 \uf0de T\u1ee9 gi\u00e1c OEMC n\u1ed9i ti\u1ebfp v\u00ec c\u00f3 hai g\u00f3c \u0111\u1ed1i b\u00f9 nhau. \uf0de CEM = COM (c\u00f9ng ch\u1eafn MC ) 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 M\u00e0 BOM = COM = 1 sd BC (t\u00ednh ch\u1ea5t hai ti\u1ebfp tuy\u1ebfn c\u1eaft nhau) 2 V\u00e0 BFC = 1 sd BC (t\u00ednh ch\u1ea5t g\u00f3c n\u1ed9i ti\u1ebfp) 2 Do \u0111\u00f3: CEM = BFC m\u00e0 hai g\u00f3c n\u00e0y \u1edf v\u1ecb tr\u00ed \u0111\u1ed3ng v\u1ecb \uf0de BF \/\/ AM (\u0111pcm) ----H\u1ebeT---- 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 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 7 NA\u00caM HO\u00cfC: 2022 - 2023 M\u00d4N: TO\u00c1N 9 \u0110\u1ec0 THAM KH\u1ea2O M\u00c3 \u0110\u1ec0: Qu\u1eadn 7 - 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. Cho y = \u2212x2 (P) v\u00e0 y = x \u2212 2 (D) . 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 to\u00e1n. C\u00e2u 2. Cho ph\u01b0\u01a1ng tr\u00ecnh x2 \u2212 2x \u2212 3 + 1 = 0 . Kh\u00f4ng gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh, h\u00e3y t\u00ednh gi\u00e1 tr\u1ecb c\u1ee7a bi\u1ec3u th\u1ee9c M = x12x22 \u2212 2x1x2 \u2212 x1 \u2212 x2 C\u00e2u 3. B\u1ea1n B\u00ecnh v\u00e0 m\u1eb9 d\u1ef1 \u0111\u1ecbnh di du l\u1ecbch H\u1ed9i An v\u00e0 B\u00e0 N\u00e0 (\u0110\u00e0 N\u1eb5ng) trong 6 ng\u00e0y. Bi\u1ebft r\u1eb1ng chi ph\u00ed trung b\u00ecnh m\u1ed7i ng\u00e0y t\u1ea1i H\u1ed9i An l\u00e0 1500000 \u0111\u1ed3ng, c\u00f2n t\u1ea1i B\u00e0 N\u00e0 l\u00e0 2000000 \u0111\u1ed3ng. T\u00ecm s\u1ed1 ng\u00e0y ngh\u1ec9 t\u1ea1i m\u1ed7i \u0111\u1ecba \u0111i\u1ec3m, bi\u1ebft s\u1ed1 ti\u1ec1n m\u00e0 h\u1ecd ph\u1ea3i chi cho to\u00e0n b\u1ed9 chuy\u1ebfn \u0111i l\u00e0 10000000 \u0111\u1ed3ng. C\u00e2u 4. Trong 130 con b\u00f2 c\u1ee7a gia \u0111\u00ecnh \u00f4ng Hi\u1ec7p \u1edf huy\u1ec7n M\u1ed9c Ch\u00e2u, t\u1ec9nh S\u01a1n La c\u00f3 75 con sinh s\u1ea3n (hi\u1ec7n c\u00f3 42 con \u0111ang v\u1eaft s\u1eefa), c\u00f2n l\u1ea1i b\u00ea v\u00e0 b\u00f2 t\u01a1, t\u1ed5ng gi\u00e1 tr\u1ecb \u0111\u00e0n b\u00f2 kh\u00f4ng d\u01b0\u1edbi 6 t\u1ef7 \u0111\u1ed3ng. S\u1ea3n l\u01b0\u1ee3ng s\u1eefa h\u1eb1ng ng\u00e0y \u00f4ng thu kho\u1ea3ng 1,2 t\u1ea5n, b\u00e1n cho nh\u00e0 m\u00e1y \u0111\u01b0\u1ee3c 15 tri\u1ec7u \u0111\u1ed3ng, tr\u1eeb chi ph\u00ed \u00f4ng Hi\u1ec7p c\u00f2n l\u00e3i 40% so v\u1edbi doanh thu. a) H\u1ecfi m\u1ed7i ng\u00e0y 1 con b\u00f2 v\u1eaft \u0111\u01b0\u1ee3c bao nhi\u00eau kg s\u1eefa? b) M\u1ed7i th\u00e1ng (kho\u1ea3ng 30 ng\u00e0y) gia \u0111\u00ecnh \u00f4ng Hi\u1ec7p thu \u0111\u01b0\u1ee3c ti\u1ec1n l\u00e3i l\u00e0 bao nhi\u00eau t\u1eeb s\u1ea3n l\u01b0\u1ee3ng s\u1eefa b\u00f2? C\u00e2u 5. M\u1ed9t xe t\u1ea3i \u0111\u00f4ng l\u1ea1nh ch\u1edf h\u00e0ng c\u00f3 th\u00f9ng xe d\u1ea1ng h\u00ecnh h\u1ed9p ch\u1eef nh\u1eadt v\u1edbi k\u00edch th\u01b0\u1edbc nh\u01b0 h\u00ecnh b\u00ean. B\u1ea1n h\u00e3y t\u00ednh gi\u00fap th\u1ec3 t\u00edch c\u1ee7a th\u00f9ng xe v\u00e0 di\u1ec7n t\u00edch ph\u1ea7n Inox \u0111\u00f3ng th\u00f9ng xe (t\u00ednh lu\u00f4n s\u00e0n). 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. D\u01b0\u1edbi \u0111\u00e2y l\u00e0 \u0111\u1ed3 th\u1ecb bi\u1ec3u di\u1ec5n qu\u00e3ng \u0111\u01b0\u1eddng \u0111i \u0111\u01b0\u1ee3c gi\u00e1 ti\u1ec1n t\u01b0\u01a1ng \u1ee9ng m\u00e0 kh\u00e1ch h\u00e0ng ph\u1ea3i tr\u1ea3 cho h\u00e3ng taxi Blue Cap v\u00e0 Yellow Cab. Tr\u1ee5c ho\u00e0ng bi\u1ec3u di\u1ec5n s\u1ed1 km m\u1ed7i xe \u0111i \u0111\u01b0\u1ee3c (m\u1ed7i \u0111\u01a1n v\u1ecb: 1km ), tr\u1ee5c tung bi\u1ec3u di\u1ec5n s\u1ed1 ti\u1ec1n ph\u1ea3i tr\u1ea3 t\u01b0\u01a1ng \u1ee9ng (m\u1ed7i \u0111\u01a1n v\u1ecb: 7 ng\u00e0n \u0111\u1ed3ng). Quan s\u00e1t \u0111\u1ed3 th\u1ecb v\u00e0 cho bi\u1ebft: a) Anh Du di chuy\u1ec3n qu\u00e3ng \u0111\u01b0\u1eddng 3 km v\u1edbi xe c\u1ee7a h\u00e3ng Yellow Cab, anh ph\u1ea3i tr\u1ea3 bao nhi\u00eau ti\u1ec1n? b) C\u00f4 H\u1ea1 c\u1ea7n di chuy\u1ec3n qu\u00e3ng \u0111\u01b0\u1eddng 8 km , c\u00f4 n\u00ean ch\u1ecdn h\u00e3ng n\u00e0o \u0111\u1ec3 ti\u1ebft ki\u1ec7m chi ph\u00ed? C\u00e2u 7. Gen B c\u00f3 3600 li\u00ean k\u1ebft Hidro v\u00e0 c\u00f3 hi\u1ec7u gi\u1eefa Nucleotit lo\u1ea1i T v\u1edbi lo\u1ea1i Nucleotit kh\u00f4ng b\u1ed5 sung v\u1edbi n\u00f3 l\u00e0 300 Nucleotit. T\u00ednh s\u1ed1 Nucleotit t\u1eebng lo\u1ea1i c\u1ee7a gen B . Bi\u1ebft r\u1eb1ng, \u0111\u1ec3 t\u00ednh s\u1ed1 l\u01b0\u1ee3ng Nucleotit ( A,T,G,X ) trong ph\u00e2n t\u1eed ADN, ta \u00e1p d\u1ee5ng nguy\u00ean t\u1eafc b\u1ed5 sung: \u201c A li\u00ean k\u1ebft v\u1edbi T b\u1eb1ng 2 li\u00ean k\u1ebft Hidro v\u00e0 G li\u00ean k\u1ebft v\u1edbi X b\u1eb1ng 3 li\u00ean k\u1ebft Hidro\u201d v\u00e0 %A = %T , %G = %X . T\u1ed5ng s\u1ed1 Nucleotit trong gen B : N = A + T + G + X = 2A + 2G = 2T + 2X . C\u00e2u 8. Cho \u0111\u01b0\u1eddng tr\u00f2n (O; R) , M l\u00e0 m\u1ed9t \u0111i\u1ec3m n\u1eb1m ngo\u00e0i \u0111\u01b0\u1eddng tr\u00f2n sao cho OM = 2R . T\u1eeb M k\u1ebb 2 ti\u1ebfp tuy\u1ebfn MC, MD \u0111\u1ebfn \u0111\u01b0\u1eddng tr\u00f2n ( C, D l\u00e0 ti\u1ebfp \u0111i\u1ec3m) v\u00e0 c\u00e1t tuy\u1ebfn MAB . a) Ch\u1ee9ng minh: MC2 = MA.MB b) G\u1ecdi K l\u00e0 trung \u0111i\u1ec3m AB , ch\u1ee9ng minh 5 \u0111i\u1ec3m M,C,K,O, D c\u00f9ng thu\u1ed9c m\u1ed9t \u0111\u01b0\u1eddng tr\u00f2n. c) Cho AB = R 3 . T\u00ednh MA theo R . ----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. Cho y = \u2212x2 (P) v\u00e0 y = x \u2212 2 (D) . 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 to\u00e1n. L\u1eddi gi\u1ea3i a) V\u1ebd \u0111\u1ed3 th\u1ecb (P) v\u00e0 (D) tr\u00ean c\u00f9ng h\u1ec7 tr\u1ee5c t\u1ecda \u0111\u1ed9. BGT: x \u22124 \u22122 0 2 4 y = \u2212x2 \u22124 \u22121 0 \u22121 \u22124 4 x 02 y= x\u22122 \u22122 \u22121 2 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) : \u2212x2 = x \u2212 2 42 \uf0db x2 + 2x \u2212 8 = 0 \uf0db \uf0e9x = 2 \uf0ea\uf0ebx = \u22124 Thay x = 2 v\u00e0o y = \u2212x2 , ta \u0111\u01b0\u1ee3c: y = \u221222 = \u22121 . 44 Thay x = \u22124 v\u00e0o y = \u2212x2 , ta \u0111\u01b0\u1ee3c: y = \u2212(\u22124)2 = \u22124 . 44 V\u1eady (2; \u2212 1) , (\u22124; \u2212 4) l\u00e0 hai giao \u0111i\u1ec3m c\u1ea7n t\u00ecm. C\u00e2u 2. Cho ph\u01b0\u01a1ng tr\u00ecnh x2 \u2212 2x \u2212 3 + 1 = 0 . Kh\u00f4ng gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh, h\u00e3y t\u00ednh gi\u00e1 tr\u1ecb c\u1ee7a bi\u1ec3u th\u1ee9c M = x12x22 \u2212 2x1x2 \u2212 x1 \u2212 x2 L\u1eddi gi\u1ea3i ( )V\u00ec \uf044 = b2 \u2212 4ac = (\u22122)2 \u2212 4.1. \u2212 3 + 1 = 4 3 \uf03e 0 N\u00ean ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m ph\u00e2n bi\u1ec7t x1 ,x2 . 3 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 ( )Theo \u0111\u1ecbnh l\u00ed Vi-et, ta c\u00f3: \uf0ec \u2212b \u2212 \u22122 \uf0ef\uf0efS = x1 + x2 = a = 1 =2 \uf0ed \uf0ee\uf0ef\uf0efP c \u2212 3 +1 =1\u2212 = x1 .x2 = a = 1 3 Ta c\u00f3: M = x12x22 \u2212 2x1x2 \u2212 x1 \u2212 x2 ( ) ( )M = x1x2 2 \u2212 2x1x2 \u2212 x1 + x2 ( ) ( )2 M = 1 \u2212 3 \u2212 2. 1 \u2212 3 \u2212 2 M=0 C\u00e2u 3. B\u1ea1n B\u00ecnh v\u00e0 m\u1eb9 d\u1ef1 \u0111\u1ecbnh di du l\u1ecbch H\u1ed9i An v\u00e0 B\u00e0 N\u00e0 (\u0110\u00e0 N\u1eb5ng) trong 6 ng\u00e0y. Bi\u1ebft r\u1eb1ng chi ph\u00ed trung b\u00ecnh m\u1ed7i ng\u00e0y t\u1ea1i H\u1ed9i An l\u00e0 1500000 \u0111\u1ed3ng, c\u00f2n t\u1ea1i B\u00e0 N\u00e0 l\u00e0 2000000 \u0111\u1ed3ng. T\u00ecm s\u1ed1 ng\u00e0y ngh\u1ec9 t\u1ea1i m\u1ed7i \u0111\u1ecba \u0111i\u1ec3m, bi\u1ebft s\u1ed1 ti\u1ec1n m\u00e0 h\u1ecd ph\u1ea3i chi cho to\u00e0n b\u1ed9 chuy\u1ebfn \u0111i l\u00e0 10000000 \u0111\u1ed3ng. L\u1eddi gi\u1ea3i G\u1ecdi th\u1eddi gian ngh\u1ec9 t\u1ea1i H\u1ed9i An l\u00e0 x (ng\u00e0y), th\u1eddi gian ngh\u1ec9 t\u1ea1i B\u00e0 N\u00e0 l\u00e0 y (ng\u00e0y) (0 \uf03c x \uf03c 6;0 \uf03c y \uf03c 6) . V\u00ec b\u1ea1n B\u00ecnh v\u00e0 m\u1eb9 d\u1ef1 \u0111\u1ecbnh \u0111i du l\u1ecbch t\u1ea1i H\u1ed9i An v\u00e0 B\u00e0 N\u00e0 (\u0110\u00e0 N\u1eb5ng) trong 6 ng\u00e0y n\u00ean ta c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh x + y = 6 (1) Chi ph\u00ed \u1edf H\u1ed9i An l\u00e0 1500000x (\u0111\u1ed3ng), chi ph\u00ed \u1edf B\u00e0 N\u00e0 l\u00e0 2000000y (\u0111\u1ed3ng) V\u00ec t\u1ed5ng chi ph\u00ed cho to\u00e0n b\u1ed9 chuy\u1ebfn \u0111i l\u00e0 10000000 \u0111\u1ed3ng n\u00ean ta c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh ( )1500000x + 2000000y = 10000000 \uf0db 3x + 4y = 20 2 T\u1eeb (1) v\u00e0 (2) ta c\u00f3 h\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh \uf0ec x+y = 6 20 \uf0db \uf0ecx = 4 (th\u1ecfa m\u00e3n \u0111i\u1ec1u ki\u1ec7n) \uf0ed\uf0ee3x + 4y = \uf0ed\uf0eey = 2 V\u1eady B\u00ecnh v\u00e0 m\u1eb9 ngh\u1ec9 t\u1ea1i H\u1ed9i An 4 ng\u00e0y v\u00e0 ngh\u1ec9 t\u1ea1i B\u00e0 N\u00e0 2 ng\u00e0y. C\u00e2u 4. Trong 130 con b\u00f2 c\u1ee7a gia \u0111\u00ecnh \u00f4ng Hi\u1ec7p \u1edf huy\u1ec7n M\u1ed9c Ch\u00e2u, t\u1ec9nh S\u01a1n La c\u00f3 75 con sinh s\u1ea3n (hi\u1ec7n c\u00f3 42 con \u0111ang v\u1eaft s\u1eefa), c\u00f2n l\u1ea1i b\u00ea v\u00e0 b\u00f2 t\u01a1, t\u1ed5ng gi\u00e1 tr\u1ecb \u0111\u00e0n b\u00f2 kh\u00f4ng d\u01b0\u1edbi 6 t\u1ef7 \u0111\u1ed3ng. S\u1ea3n l\u01b0\u1ee3ng s\u1eefa h\u1eb1ng ng\u00e0y \u00f4ng thu kho\u1ea3ng 1,2 t\u1ea5n, b\u00e1n cho nh\u00e0 m\u00e1y \u0111\u01b0\u1ee3c 15 tri\u1ec7u \u0111\u1ed3ng, tr\u1eeb chi ph\u00ed \u00f4ng Hi\u1ec7p c\u00f2n l\u00e3i 40% so v\u1edbi doanh thu. a) H\u1ecfi m\u1ed7i ng\u00e0y 1 con b\u00f2 v\u1eaft \u0111\u01b0\u1ee3c bao nhi\u00eau kg s\u1eefa? b) M\u1ed7i th\u00e1ng (kho\u1ea3ng 30 ng\u00e0y) gia \u0111\u00ecnh \u00f4ng Hi\u1ec7p thu \u0111\u01b0\u1ee3c ti\u1ec1n l\u00e3i l\u00e0 bao nhi\u00eau t\u1eeb s\u1ea3n l\u01b0\u1ee3ng s\u1eefa b\u00f2? L\u1eddi gi\u1ea3i a) 1,2 t\u1ea5n = 1200 kg . T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N 4","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH Kh\u1ed1i l\u01b0\u1ee3ng s\u1eefa m\u1ed7i ng\u00e0y 1 con b\u00f2 v\u1eaft \u0111\u01b0\u1ee3c: 1200: 42 = 200 kg \uf0bb 28,6kg 7 b) S\u1ed1 ti\u1ec1n l\u00e3i gia \u0111\u00ecnh \u00f4ng hi\u1ec7p thu \u0111\u01b0\u1ee3c m\u1ed7i th\u00e1ng t\u1eeb s\u1ea3n l\u01b0\u1ee3ng s\u1eefa b\u00f2: 15.40%.30 = 180 tri\u1ec7u \u0111\u1ed3ng. C\u00e2u 5. M\u1ed9t xe t\u1ea3i \u0111\u00f4ng l\u1ea1nh ch\u1edf h\u00e0ng c\u00f3 th\u00f9ng xe d\u1ea1ng h\u00ecnh h\u1ed9p ch\u1eef nh\u1eadt v\u1edbi k\u00edch th\u01b0\u1edbc nh\u01b0 h\u00ecnh b\u00ean. B\u1ea1n h\u00e3y t\u00ednh gi\u00fap th\u1ec3 t\u00edch c\u1ee7a th\u00f9ng xe v\u00e0 di\u1ec7n t\u00edch ph\u1ea7n Inox \u0111\u00f3ng th\u00f9ng xe (t\u00ednh lu\u00f4n s\u00e0n). L\u1eddi gi\u1ea3i Th\u1ec3 t\u00edch th\u00f9ng xe t\u1ea3i: V = 3.2.1,5 = 9m3 ( )Di\u1ec7n t\u00edch ph\u1ea7n Inox \u0111\u00f3ng th\u00f9ng xe: Stp = Sxq + 2.Sday = 3 + 2 .2.1,5 + 2.3.2 = 27m2 C\u00e2u 6. D\u01b0\u1edbi \u0111\u00e2y l\u00e0 \u0111\u1ed3 th\u1ecb bi\u1ec3u di\u1ec5n qu\u00e3ng \u0111\u01b0\u1eddng \u0111i \u0111\u01b0\u1ee3c gi\u00e1 ti\u1ec1n t\u01b0\u01a1ng \u1ee9ng m\u00e0 kh\u00e1ch h\u00e0ng ph\u1ea3i tr\u1ea3 cho h\u00e3ng taxi Blue Cap v\u00e0 Yellow Cab. Tr\u1ee5c ho\u00e0ng bi\u1ec3u di\u1ec5n s\u1ed1 km m\u1ed7i xe \u0111i \u0111\u01b0\u1ee3c (m\u1ed7i \u0111\u01a1n v\u1ecb: 1 km), tr\u1ee5c tung bi\u1ec3u di\u1ec5n s\u1ed1 ti\u1ec1n ph\u1ea3i tr\u1ea3 t\u01b0\u01a1ng \u1ee9ng (m\u1ed7i \u0111\u01a1n v\u1ecb: 7 ng\u00e0n \u0111\u1ed3ng). Quan s\u00e1t \u0111\u1ed3 th\u1ecb v\u00e0 cho bi\u1ebft: a) Anh Du di chuy\u1ec3n qu\u00e3ng \u0111\u01b0\u1eddng 3 km v\u1edbi xe c\u1ee7a h\u00e3ng Yellow Cab, anh ph\u1ea3i tr\u1ea3 bao nhi\u00eau ti\u1ec1n? b) C\u00f4 H\u1ea1 c\u1ea7n di chuy\u1ec3n qu\u00e3ng \u0111\u01b0\u1eddng 8 km, c\u00f4 n\u00ean ch\u1ecdn h\u00e3ng n\u00e0o \u0111\u1ec3 ti\u1ebft ki\u1ec7m chi ph\u00ed? L\u1eddi gi\u1ea3i a) S\u1ed1 ti\u1ec1n anh Du ph\u1ea3i tr\u1ea3 khi di chuy\u1ec3n qu\u00e3ng \u0111\u01b0\u1eddng 3 km v\u1edbi xe c\u1ee7a h\u00e3ng Yellow Cab: 8.7 = 56 ngh\u00ecn \u0111\u1ed3ng. 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 b) Quan s\u00e1t \u0111\u1ed3 th\u1ecb, ta th\u1ea5y khi di chuy\u1ec3n qu\u00e3ng \u0111\u01b0\u1eddng l\u1edbn h\u01a1n 3 km th\u00ec s\u1ed1 ti\u1ec1n ph\u1ea3i tr\u1ea3 khi \u0111i xe Blue Cab s\u1ebd nh\u1ecf h\u01a1n xe Yellow Cab. C\u00e2u 7. Gen B c\u00f3 3600 li\u00ean k\u1ebft Hidro v\u00e0 c\u00f3 hi\u1ec7u gi\u1eefa Nucleotit lo\u1ea1i T v\u1edbi lo\u1ea1i Nucleotit kh\u00f4ng b\u1ed5 sung v\u1edbi n\u00f3 l\u00e0 300 Nucleotit. T\u00ednh s\u1ed1 Nucleotit t\u1eebng lo\u1ea1i c\u1ee7a gen B. Bi\u1ebft r\u1eb1ng, \u0111\u1ec3 t\u00ednh s\u1ed1 l\u01b0\u1ee3ng Nucleotit (A, T, G, X) trong ph\u00e2n t\u1eed ADN, ta \u00e1p d\u1ee5ng nguy\u00ean t\u1eafc b\u1ed5 sung: \u201cA li\u00ean k\u1ebft v\u1edbi T b\u1eb1ng 2 li\u00ean k\u1ebft Hidro v\u00e0 G li\u00ean k\u1ebft v\u1edbi X b\u1eb1ng 3 li\u00ean k\u1ebft Hidro\u201d v\u00e0 %A = %T , %G = %X . T\u1ed5ng s\u1ed1 Nucleotit trong gen B: N = A +T + G + X = 2A + 2G = 2T + 2X . L\u1eddi gi\u1ea3i G\u1ecdi A,G l\u1ea7n l\u01b0\u1ee3t l\u00e0 s\u1ed1 Nucleotit lo\u1ea1i A , lo\u1ea1i G ( A,G \uf0ce *) . S\u1ed1 li\u00ean k\u1ebft hidro l\u00e0 3600 n\u00ean ta c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh 2A + 3G = 3600 (1) S\u1ed1 Nucleotit lo\u1ea1i T b\u1eb1ng s\u1ed1 Nucleotit lo\u1ea1i A . Nucleotit kh\u00f4ng b\u1ed5 sung v\u1edbi T l\u00e0 G ho\u1eb7c X S\u1ed1 Nucleotit lo\u1ea1i T tr\u1eeb s\u1ed1 Nucleotit lo\u1ea1i kh\u00f4ng b\u1ed5 sung b\u1eb1ng 300 n\u00ean ta c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh: A \u2212 G = 300 (2) ( ) ( )T\u1eeb \uf0ec2A + 3G = 3600 \uf0ecA = 900 (th\u1ecfa m\u00e3n \u0111i\u1ec1u ki\u1ec7n) 1, 2 ta c\u00f3 h\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh: \uf0ed \uf0db \uf0ed\uf0eeG \uf0ee A \u2212 G = 300 = 600 V\u1eady s\u1ed1 nu lo\u1ea1i A,T l\u00e0 900, lo\u1ea1i G,X l\u00e0 600. C\u00e2u 8. Cho \u0111\u01b0\u1eddng tr\u00f2n (O; R) , M l\u00e0 m\u1ed9t \u0111i\u1ec3m n\u1eb1m ngo\u00e0i \u0111\u01b0\u1eddng tr\u00f2n sao cho OM = 2R . T\u1eeb M k\u1ebb 2 ti\u1ebfp tuy\u1ebfn MC, MD \u0111\u1ebfn \u0111\u01b0\u1eddng tr\u00f2n ( C, D l\u00e0 ti\u1ebfp \u0111i\u1ec3m) v\u00e0 c\u00e1t tuy\u1ebfn MAB . a) Ch\u1ee9ng minh: MC2 = MA.MB b) G\u1ecdi K l\u00e0 trung \u0111i\u1ec3m AB , ch\u1ee9ng minh 5 \u0111i\u1ec3m M,C,K,O, D c\u00f9ng thu\u1ed9c m\u1ed9t \u0111\u01b0\u1eddng tr\u00f2n. T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N 6","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH c) Cho AB = R 3 . T\u00ednh MA theo R . L\u1eddi gi\u1ea3i a) Ch\u1ee9ng minh : MC2 = MA.MB X\u00e9t tam gi\u00e1c MAC v\u00e0 tam gi\u00e1c MCB c\u00f3: MCA = MBC (c\u00f9ng b\u1eb1ng 1 s\u0111 AC ). 2 M chung MCB (g-g) \uf0de MAC \uf0de MC = MB \uf0db MC2 = MA.MB . MA MC b) G\u1ecdi K l\u00e0 trung \u0111i\u1ec3m AB , ch\u1ee9ng minh 5 \u0111i\u1ec3m M,C,K,O, D c\u00f9ng thu\u1ed9c m\u1ed9t \u0111\u01b0\u1eddng tr\u00f2n. G\u1ecdi K l\u00e0 trung \u0111i\u1ec3m AB \uf0de OK \u22a5 AB \uf0de MKO = 90\uf0b0 \uf0de K thu\u1ed9c \u0111\u01b0\u1eddng tr\u00f2n \u0111\u01b0\u1eddng k\u00ednh MO (1) . M\u00e0 MCO = MDO = 90\uf0b0 ( MC, MD l\u00e0 ti\u1ebfp tuy\u1ebfn c\u1ee7a (O; R) ) ( )\uf0de C,D thu\u1ed9c \u0111\u01b0\u1eddng tr\u00f2n \u0111\u01b0\u1eddng k\u00ednh MO 2 . ( ) ( )T\u1eeb 1 , 2 suy ra: 5 \u0111i\u1ec3m M,C,K,O, D c\u00f9ng thu\u1ed9c m\u1ed9t \u0111\u01b0\u1eddng tr\u00f2n. c) Cho AB = R 3 . T\u00ednh MA theo R . Ta c\u00f3 AB = R 3 \uf0de AK = R 3 v\u00e0 OM = 2R. 2 \uf044OAK vu\u00f4ng t\u1ea1i K c\u00f3: OK 2= AO2 \u2212 AK 2 = R2 \uf0e6 R3 \uf0f62 = R2 \uf0de OK = R . \u2212 \uf0e7\uf0e7\uf0e8 2 \uf0f7\uf0f7\uf0f8 4 2 ( )\uf044MOK vu\u00f4ng t\u1ea1i K c\u00f3: 2 \uf0e6 R \uf0f62 15R2 \uf0de MK = R 15 . MK 2= MO2 \u2212 OK2 = 2R \u2212 \uf0e7 = 4 2 \uf0f7 2 \uf0e8 \uf0f8 ( )Suy ra: MA = MK \u2212 AK = R 15 \u2212 R 3 = R 3 . 5 \u2212 1 . 2 22 ----H\u1ebeT--- 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 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 8 NA\u00caM HO\u00cfC: 2023 - 2024 M\u00d4N: TO\u00c1N 9 \u0110\u1ec0 THAM KH\u1ea2O M\u00c3 \u0110\u1ec0: Qu\u1eadn 8 - 1 \u0110\u00ea thi g\u1ed3m 8 c\u00e2u ho\u0309i t\u01b0\u0323 lu\u00e2\u0323n. Th\u01a1\u0300i gian: 120 phu\u0301t (kh\u00f4ng k\u00ea\u0309 th\u01a1\u0300i gian pha\u0301t \u0111\u00ea\u0300) C\u00e2u 1. (1,5 \u0111i\u1ec3m). Cho h\u00e0m s\u1ed1 y = \u22122x2 c\u00f3 \u0111\u1ed3 th\u1ecb (P) v\u00e0 h\u00e0m s\u1ed1 y = \u2212x \u2212 3 c\u00f3 \u0111\u1ed3 th\u1ecb (d) . a) V\u1ebd (P) v\u00e0 (d) tr\u00ean c\u00f9ng m\u1ed9t h\u1ec7 tr\u1ee5c t\u1ecda \u0111\u1ed9. b) T\u00ecm to\u1ea1 \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: 20x2 \u2013 23x \u2013 24 = 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: A = x12x2 + x22x1 . C\u00e2u 3. (0,75 \u0111i\u1ec3m). M\u1ed9t v\u1eadt chuy\u1ec3n \u0111\u1ed9ng \u0111\u1ec1u, qu\u00e3ng \u0111\u01b0\u1eddng chuy\u1ec3n \u0111\u1ed9ng s(m\u00e9t) c\u1ee7a v\u1eadt trong th\u1eddi gian t(gi\u00e2y) \u0111\u01b0\u1ee3c bi\u1ec3u di\u1ec5n theo h\u00e0m s\u1ed1 s = a.t . Bi\u1ebft \u0111\u1ed3 th\u1ecb chuy\u1ec3n \u0111\u1ed9ng c\u1ee7a v\u1eadt \u0111\u01b0\u1ee3c cho nh\u01b0 h\u00ecnh b\u00ean. a) H\u00e3y t\u00ecm h\u1ec7 s\u1ed1 a . b) Trong bao l\u00e2u th\u00ec v\u1eadt chuy\u1ec3n \u0111\u1ed9ng \u0111\u01b0\u1ee3c 5 m\u00e9t C\u00e2u 4. (1 \u0111i\u1ec3m). Trong m\u1ed9t bu\u1ed5i ho\u1ea1t \u0111\u1ed9ng tr\u1ea3i nghi\u1ec7m \u1ee9ng d\u1ee5ng th\u1ef1c t\u1ebf t\u1ec9 s\u1ed1 l\u01b0\u1ee3ng gi\u00e1c c\u1ee7a g\u00f3c nh\u1ecdn, m\u1ed9t nh\u00f3m h\u1ecdc sinh l\u1edbp 9A c\u00f3 th\u1ec3 t\u00ednh \u0111\u01b0\u1ee3c kho\u1ea3ng c\u00e1ch gi\u1eefa hai thuy\u1ec1n tr\u00ean bi\u1ec3n b\u1eb1ng c\u00e1ch d\u00f9ng th\u01b0\u1edbc cu\u1ed9n, eke, c\u1ecdc v\u00e0 gi\u00e1c k\u1ebf \u0111\u1ec3 x\u00e1c \u0111\u1ecbnh \u0111\u01b0\u1ee3c c\u00e1c v\u1ecb tr\u00ed G,F,H,E nh\u01b0 h\u00ecnh v\u1ebd b\u00ean d\u01b0\u1edbi. H\u1ecdc sinh \u0111\u00e3 \u0111o \u0111o\u1ea1n FG = 20 m\u00e9t, g\u00f3c FGH b\u1eb1ng 70\uf0b0 , g\u00f3c FGE b\u1eb1ng 77\uf0b0 . Em h\u00e3y cho bi\u1ebft h\u1ecdc sinh l\u1edbp 9A \u0111\u00e3 t\u00ednh \u0111\u01b0\u1ee3c kho\u1ea3ng c\u00e1ch gi\u1eefa hai thuy\u1ec1n l\u00e0 bao nhi\u00eau ? (L\u00e0m tr\u00f2n k\u1ebft qu\u1ea3 \u0111\u1ebfn ch\u1eef s\u1ed1 h\u00e0ng \u0111\u01a1n v\u1ecb) 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 c\u1ea7n b\u00e1n m\u1ed9t l\u00f4 h\u00e0ng g\u1ed3m 32 s\u1ea3n ph\u1ea9m c\u00f9ng lo\u1ea1i v\u1edbi gi\u00e1 b\u00e1n ban \u0111\u1ea7u l\u00e0 2 400 000 \u0111\u1ed3ng. Nh\u00e2n d\u1ecbp l\u1ec5 Noel, c\u1eeda h\u00e0ng gi\u1ea3m 10% so v\u1edbi gi\u00e1 b\u00e1n ban \u0111\u1ea7u th\u00ec b\u00e1n \u0111\u01b0\u1ee3c 12 s\u1ea3n ph\u1ea9m. V\u00e0o d\u1ecbp t\u1ebft T\u00e2y, m\u1ed7i s\u1ea3n ph\u1ea9m \u0111\u01b0\u1ee3c gi\u1ea3m 200000 \u0111\u1ed3ng (so v\u1edbi gi\u00e1 \u0111\u00e3 gi\u1ea3m \u1edf d\u1ecbp l\u1ec5 Noel) th\u00ec c\u1eeda h\u00e0ng b\u00e1n \u0111\u01b0\u1ee3c h\u1ebft s\u1ed1 s\u1ea3n ph\u1ea9m c\u00f2n l\u1ea1i. Sau khi b\u00e1n h\u1ebft th\u00ec c\u1eeda h\u00e0ng l\u00e3i \u0111\u01b0\u1ee3c 60% so v\u1edbi t\u1ed5ng s\u1ed1 ti\u1ec1n b\u1ecf ra g\u1ed3m gi\u00e1 v\u1ed1n c\u1ee7a c\u00e1c s\u1ea3n ph\u1ea9m v\u00e0 gi\u00e1 v\u1eadn chuy\u1ec3n 2000000 \u0111\u1ed3ng. H\u1ecfi gi\u00e1 v\u1ed1n c\u1ee7a m\u1ed7i s\u1ea3n ph\u1ea9m trong l\u00f4 h\u00e0ng c\u1ea7n b\u00e1n l\u00e0 bao nhi\u00eau ti\u1ec1n? C\u00e2u 6. (1 \u0111i\u1ec3m). Trong m\u1ed9t tr\u00f2 ch\u01a1i To\u00e1n h\u1ecdc c\u00f3 50 c\u00e2u tr\u1eafc nghi\u1ec7m, m\u1ed7i c\u00e2u c\u00f3 4 \u0111\u00e1p \u00e1n A,B,C,D v\u00e0 m\u1ed7i c\u00e2u ch\u1ec9 c\u00f3 m\u1ed9t \u0111\u00e1p \u00e1n \u0111\u00fang. Khi ng\u01b0\u1eddi ch\u01a1i ch\u1ecdn \u0111\u01b0\u1ee3c \u0111\u00e1p \u00e1n \u0111\u00fang th\u00ec c\u00e2u \u0111\u00f3 s\u1ebd \u0111\u01b0\u1ee3c 20 \u0111i\u1ec3m, khi \u0111\u00e1p \u00e1n sai th\u00ec c\u00e2u \u0111\u00f3 s\u1ebd b\u1ecb tr\u1eeb \u0111i 5 \u0111i\u1ec3m. B\u1ea1n An \u0111\u00e3 tham gia tr\u00f2 ch\u01a1i To\u00e1n h\u1ecdc \u0111\u00f3. Sau khi k\u1ebft th\u00fac tr\u00f2 ch\u01a1i, b\u1ea1n An \u0111\u01b0\u1ee3c 550 \u0111i\u1ec3m. H\u1ecfi b\u1ea1n An \u0111\u00e3 tr\u1ea3 l\u1eddi bao nhi\u00eau c\u00e2u \u0111\u00fang, bao nhi\u00eau c\u00e2u sai ? C\u00e2u 7. (0,75 \u0111i\u1ec3m). Ng\u01b0\u1eddi ta \u0111\u00e0o m\u1ed9t \u0111o\u1ea1n m\u01b0\u01a1ng d\u00e0i 20m , s\u00e2u 1,5m . Tr\u00ean b\u1ec1 m\u1eb7t c\u00f3 chi\u1ec1u r\u1ed9ng 1,8m v\u00e0 \u0111\u00e1y m\u01b0\u01a1ng l\u00e0 1,2m (xem h\u00ecnh minh h\u1ecda b\u00ean) . Ng\u01b0\u1eddi ta chuy\u1ec3n to\u00e0n b\u1ed9 kh\u1ed1i \u0111\u1ea5t \u0111i \u0111\u1ec3 r\u1ea3i l\u00ean m\u1ed9t mi\u1ebfng \u0111\u1ea5t h\u00ecnh ch\u1eef nh\u1eadt c\u00f3 k\u00edch th\u01b0\u1edbc 12m v\u00e0 15m . S\u1ed1 \u0111\u1ea5t \u0111\u01b0\u1ee3c chuy\u1ec3n b\u1eb1ng m\u1ed9t chi\u1ebfc \u00f4 t\u00f4 ch\u1edf m\u1ed7i chuy\u1ebfn 6m3 . a) T\u00ednh b\u1ec1 d\u00e0y c\u1ee7a l\u1edbp \u0111\u1ea5t r\u1ea3i l\u00ean mi\u1ebfng \u0111\u1ea5t h\u00ecnh ch\u1eef nh\u1eadt (bi\u1ebft l\u1edbp \u0111\u1ea5t \u0111\u01b0\u1ee3c r\u1ea3i \u0111\u1ec1u, kh\u00f4ng c\u00f3 ch\u1ed5 cao ch\u1ed5 th\u1ea5p). b) C\u1ea7n bao nhi\u00eau chuy\u1ebfn \u00f4 t\u00f4 \u0111\u1ec3 t\u1ea3i h\u1ebft kh\u1ed1i \u0111\u1ea5t. C\u00e2u 8. (3 \u0111i\u1ec3m) Cho tam gi\u00e1c ABC c\u00f3 ba g\u00f3c nh\u1ecdn n\u1ed9i ti\u1ebfp (O,R) . Ti\u1ebfp tuy\u1ebfn t\u1ea1i B v\u00e0 C c\u1ee7a (O) c\u1eaft nhau t\u1ea1i I . \u0110\u01b0\u1eddng th\u1eb3ng AI c\u1eaft (O) t\u1ea1i \u0111i\u1ec3m th\u1ee9 hai l\u00e0 D (kh\u00e1c A ). \u0110o\u1ea1n th\u1eb3ng OI c\u1eaft BC t\u1ea1i H . a) Ch\u1ee9ng minh : OI vu\u00f4ng g\u00f3c v\u1edbi BC v\u00e0 HB.HC = HO.HI b) V\u1ebd OK vu\u00f4ng g\u00f3c v\u1edbi AD . Ch\u1ee9ng minh 5 \u0111i\u1ec3m I,B,K,O,C c\u00f9ng thu\u1ed9c m\u1ed9t \u0111\u01b0\u1eddng tr\u00f2n c) T\u1eeb D k\u1ebb \u0111\u01b0\u1eddng th\u1eb3ng vu\u00f4ng g\u00f3c v\u1edbi OB , \u0111\u01b0\u1eddng th\u1eb3ng n\u00e0y c\u1eaft BC t\u1ea1i M v\u00e0 c\u1eaft AB t\u1ea1i N . Ch\u1ee9ng minh : M l\u00e0 trung \u0111i\u1ec3m c\u1ee7a DN . ----H\u1ebeT--- T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N 2","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH H\u01af\u1edaNG D\u1eaaN GI\u1ea2I C\u00e2u 1. (1,5 \u0111i\u1ec3m). Cho h\u00e0m s\u1ed1 y = \u22122x2 c\u00f3 \u0111\u1ed3 th\u1ecb (P) v\u00e0 h\u00e0m s\u1ed1 y = \u2212x \u2212 3 c\u00f3 \u0111\u1ed3 th\u1ecb (d) . a) V\u1ebd (P) v\u00e0 (d) tr\u00ean c\u00f9ng m\u1ed9t h\u1ec7 tr\u1ee5c t\u1ecda \u0111\u1ed9. b) T\u00ecm to\u1ea1 \u0111\u1ed9 giao \u0111i\u1ec3m c\u1ee7a (P) v\u00e0 (d) b\u1eb1ng ph\u00e9p t\u00ednh. L\u1eddi gi\u1ea3i a) V\u1ebd \u0111\u1ed3 th\u1ecb (P) v\u00e0 (d) tr\u00ean c\u00f9ng h\u1ec7 tr\u1ee5c t\u1ecda \u0111\u1ed9. BGT: x \u22122 \u22121 0 1 2 y = \u22122x2 \u22128 \u22122 0 \u22122 \u22128 x 01 y = \u2212x \u2212 3 \u22123 \u22124 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) : \u22122x2 = \u2212x \u2212 3 \uf0db \u22122x2 + x + 3 = 0 \uf0db \uf0e9\uf0eax = 3 \uf0ea 2 \uf0ea\uf0ebx = \u22121 Thay x= 3 v\u00e0o y = \u22122x2 , ta \u0111\u01b0\u1ee3c: y = \u22122 \uf0e6 3 \uf0f62 = \u22129 . 2 \uf0e7 2 \uf0f7 2 \uf0e8 \uf0f8 Thay x = \u22121 v\u00e0o y = \u22122x2 , ta \u0111\u01b0\u1ee3c: y = \u22122(\u22121)2 = \u22122 . V\u1eady \uf0e6 3 ; \u2212 9 \uf0f6 , (\u22121; \u2212 2) l\u00e0 hai giao \u0111i\u1ec3m c\u1ea7n t\u00ecm. \uf0e7 2 2 \uf0f7 \uf0e8 \uf0f8 C\u00e2u 2. (1 \u0111i\u1ec3m). Cho ph\u01b0\u01a1ng tr\u00ecnh: 20x2 \u2013 23x \u2013 24 = 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: A = x12x2 + x22x1 L\u1eddi gi\u1ea3i 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 V\u00ec \uf044 = b2 \u2212 4ac = (23)2 \u2212 4.20.(\u221224) = 2449 \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 = \u2212 23 \uf0ed = x1 a 20 Theo \u0111\u1ecbnh l\u00ed Vi-et, ta c\u00f3: \uf0ef\uf0ef\uf0eeP c = \u2212 24 6 a =\u2212 5 .x2 = 20 Ta c\u00f3: A = x12 x2 + x22x1 = x1x2 (x1 + x2 ) = P.S = 23 \uf0d7 \u22126 = \u221269 20 5 50 C\u00e2u 3. (0,75 \u0111i\u1ec3m). M\u1ed9t v\u1eadt chuy\u1ec3n \u0111\u1ed9ng \u0111\u1ec1u, qu\u00e3ng \u0111\u01b0\u1eddng chuy\u1ec3n \u0111\u1ed9ng s(m\u00e9t) c\u1ee7a v\u1eadt trong th\u1eddi gian t(gi\u00e2y) \u0111\u01b0\u1ee3c bi\u1ec3u di\u1ec5n theo h\u00e0m s\u1ed1 s = a.t . Bi\u1ebft \u0111\u1ed3 th\u1ecb chuy\u1ec3n \u0111\u1ed9ng c\u1ee7a v\u1eadt \u0111\u01b0\u1ee3c cho nh\u01b0 h\u00ecnh b\u00ean. a) H\u00e3y t\u00ecm h\u1ec7 s\u1ed1 a . b) Trong bao l\u00e2u th\u00ec v\u1eadt chuy\u1ec3n \u0111\u1ed9ng \u0111\u01b0\u1ee3c 5 m\u00e9t L\u1eddi gi\u1ea3i a) Thay t = 4 ; s = 2 ; v\u00e0o ph\u01b0\u01a1ng tr\u00ecnh s = a.t 2 = a.4 \uf0de a = 1 2 b) s = 1 .t 2 \uf0de5= 1t 2 \uf0db t = 10 V\u1eady trong 10 gi\u00e2y th\u00ec v\u1eadt chuy\u1ec3n \u0111\u1ed9ng \u0111\u01b0\u1ee3c 5 m\u00e9t C\u00e2u 4. (1 \u0111i\u1ec3m). Trong m\u1ed9t bu\u1ed5i ho\u1ea1t \u0111\u1ed9ng tr\u1ea3i nghi\u1ec7m \u1ee9ng d\u1ee5ng th\u1ef1c t\u1ebf t\u1ec9 s\u1ed1 l\u01b0\u1ee3ng gi\u00e1c c\u1ee7a g\u00f3c nh\u1ecdn, m\u1ed9t nh\u00f3m h\u1ecdc sinh l\u1edbp 9A c\u00f3 th\u1ec3 t\u00ednh \u0111\u01b0\u1ee3c kho\u1ea3ng c\u00e1ch gi\u1eefa hai thuy\u1ec1n tr\u00ean bi\u1ec3n b\u1eb1ng c\u00e1ch d\u00f9ng th\u01b0\u1edbc cu\u1ed9n, eke, c\u1ecdc v\u00e0 gi\u00e1c k\u1ebf \u0111\u1ec3 x\u00e1c \u0111\u1ecbnh \u0111\u01b0\u1ee3c c\u00e1c v\u1ecb tr\u00ed G,F,H,E nh\u01b0 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 h\u00ecnh v\u1ebd b\u00ean d\u01b0\u1edbi. H\u1ecdc sinh \u0111\u00e3 \u0111o \u0111o\u1ea1n FG = 20 m\u00e9t, g\u00f3c FGH b\u1eb1ng 700 , g\u00f3c FGE b\u1eb1ng 770 . Em h\u00e3y cho bi\u1ebft h\u1ecdc sinh l\u1edbp 9A \u0111\u00e3 t\u00ednh \u0111\u01b0\u1ee3c kho\u1ea3ng c\u00e1ch gi\u1eefa hai thuy\u1ec1n l\u00e0 bao nhi\u00eau ? (L\u00e0m tr\u00f2n k\u1ebft qu\u1ea3 \u0111\u1ebfn ch\u1eef s\u1ed1 h\u00e0ng \u0111\u01a1n v\u1ecb) L\u1eddi gi\u1ea3i X\u00e9t \uf044FGH vu\u00f4ng t\u1ea1i F tan(FGH) = FH FG tan(70\uf0b0) = FH 20 FH = 20.tan(70\uf0b0) \uf0bb 54,95m X\u00e9t \uf044FGE vu\u00f4ng t\u1ea1i F tan(FGE) = FE FG tan(77\uf0b0) = FE 20 FE = 20.tan(77\uf0b0) \uf0bb 86,63m HE = EF \u2212 HF = 86,63 \u2212 54,95 = 32m Kho\u1ea3ng c\u00e1ch gi\u1eefa hai thuy\u1ec1n l\u00e0 32 m\u00e9t C\u00e2u 5. (1 \u0111i\u1ec3m). M\u1ed9t c\u1eeda h\u00e0ng c\u1ea7n b\u00e1n m\u1ed9t l\u00f4 h\u00e0ng g\u1ed3m 32 s\u1ea3n ph\u1ea9m c\u00f9ng lo\u1ea1i v\u1edbi gi\u00e1 b\u00e1n ban \u0111\u1ea7u l\u00e0 2 400 000 \u0111\u1ed3ng. Nh\u00e2n d\u1ecbp l\u1ec5 Noel, c\u1eeda h\u00e0ng gi\u1ea3m 10% so v\u1edbi gi\u00e1 b\u00e1n ban \u0111\u1ea7u th\u00ec b\u00e1n \u0111\u01b0\u1ee3c 12 s\u1ea3n ph\u1ea9m. V\u00e0o d\u1ecbp t\u1ebft T\u00e2y, m\u1ed7i s\u1ea3n ph\u1ea9m \u0111\u01b0\u1ee3c gi\u1ea3m 200000 \u0111\u1ed3ng (so v\u1edbi gi\u00e1 \u0111\u00e3 gi\u1ea3m \u1edf d\u1ecbp l\u1ec5 Noel) th\u00ec c\u1eeda h\u00e0ng b\u00e1n \u0111\u01b0\u1ee3c h\u1ebft s\u1ed1 s\u1ea3n ph\u1ea9m c\u00f2n l\u1ea1i. Sau khi b\u00e1n h\u1ebft th\u00ec c\u1eeda h\u00e0ng l\u00e3i \u0111\u01b0\u1ee3c 60% so v\u1edbi t\u1ed5ng s\u1ed1 ti\u1ec1n b\u1ecf ra g\u1ed3m gi\u00e1 v\u1ed1n c\u1ee7a c\u00e1c s\u1ea3n ph\u1ea9m v\u00e0 gi\u00e1 v\u1eadn chuy\u1ec3n 2000000 \u0111\u1ed3ng. H\u1ecfi gi\u00e1 v\u1ed1n c\u1ee7a m\u1ed7i s\u1ea3n ph\u1ea9m trong l\u00f4 h\u00e0ng c\u1ea7n b\u00e1n l\u00e0 bao nhi\u00eau ti\u1ec1n? L\u1eddi gi\u1ea3i Gi\u00e1 ti\u1ec1n 1 s\u1ea3n ph\u1ea9m khi \u0111\u01b0\u1ee3c gi\u1ea3m 10% 2400000(1 \u2212 10%) = 2160000 (\u0111\u1ed3ng) S\u1ed1 ti\u1ec1n b\u00e1n \u0111\u01b0\u1ee3c sau khi b\u00e1n h\u1ebft 32 s\u1ea3n ph\u1ea9m: 12 + 2160000 + (32 \u2212 12)(2160000 \u2212 200000) = 65120000 (\u0111\u1ed3ng) G\u1ecdi x (\u0111\u1ed3ng) l\u00e0 s\u1ed1 ti\u1ec1n v\u1ed1n c\u1ee7a m\u1ed7i s\u1ea3n ph\u1ea9m, x \uf03e 0 S\u1ed1 ti\u1ec1n v\u1ed1n v\u00e0 l\u00e3i sau khi b\u00e1n 32 s\u1ea3n ph\u1ea9m l\u00e0: (32x + 2000000)(1+ 60%) = 1,6(32x + 2000000) 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 Ta c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh: 1,6(32x + 2000000) = 65120000 x = 1209 375 (\u0111\u1ed3ng) V\u1eady gi\u00e1 v\u1ed1n c\u1ee7a m\u1ed7i s\u1ea3n ph\u1ea9m l\u00e0 1209 375 \u0111\u1ed3ng C\u00e2u 6. (1 \u0111i\u1ec3m). Trong m\u1ed9t tr\u00f2 ch\u01a1i To\u00e1n h\u1ecdc c\u00f3 50 c\u00e2u tr\u1eafc nghi\u1ec7m, m\u1ed7i c\u00e2u c\u00f3 4 \u0111\u00e1p \u00e1n A,B,C,D v\u00e0 m\u1ed7i c\u00e2u ch\u1ec9 c\u00f3 m\u1ed9t \u0111\u00e1p \u00e1n \u0111\u00fang. Khi ng\u01b0\u1eddi ch\u01a1i ch\u1ecdn \u0111\u01b0\u1ee3c \u0111\u00e1p \u00e1n \u0111\u00fang th\u00ec c\u00e2u \u0111\u00f3 s\u1ebd \u0111\u01b0\u1ee3c 20 \u0111i\u1ec3m, khi \u0111\u00e1p \u00e1n sai th\u00ec c\u00e2u \u0111\u00f3 s\u1ebd b\u1ecb tr\u1eeb \u0111i 5 \u0111i\u1ec3m. B\u1ea1n An \u0111\u00e3 tham gia tr\u00f2 ch\u01a1i To\u00e1n h\u1ecdc \u0111\u00f3. Sau khi k\u1ebft th\u00fac tr\u00f2 ch\u01a1i, b\u1ea1n An \u0111\u01b0\u1ee3c 550 \u0111i\u1ec3m. H\u1ecfi b\u1ea1n An \u0111\u00e3 tr\u1ea3 l\u1eddi bao nhi\u00eau c\u00e2u \u0111\u00fang, bao nhi\u00eau c\u00e2u sai ? L\u1eddi gi\u1ea3i G\u1ecdi x l\u00e0 s\u1ed1 c\u00e2u tr\u1ea3 l\u1eddi \u0111\u00fang, y l\u00e0 s\u1ed1 c\u00e2u tr\u1ea3 l\u1eddi sai x, y \uf0ce * C\u00f3 50 c\u00e2u tr\u1eafc nghi\u1ec7m: x + y = 50 B\u1ea1n An \u0111\u01b0\u1ee3c 550 \u0111i\u1ec3m: 20x \u2212 5y = 550 Ta c\u00f3 h\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh: \uf0ecx + y = 50 \uf0ee\uf0ed20x \u2212 5y = 550 \uf0db \uf0ec\uf0efx = 32(n) \uf0ed = 18(n) \uf0ef\uf0eey V\u1eady b\u1ea1n An tr\u1ea3 l\u1eddi \u0111\u01b0\u1ee3c 32 c\u00e2u \u0111\u00fang, 18 c\u00e2u sai C\u00e2u 7. (0,75 \u0111i\u1ec3m). Ng\u01b0\u1eddi ta \u0111\u00e0o m\u1ed9t \u0111o\u1ea1n m\u01b0\u01a1ng d\u00e0i 20m , s\u00e2u 1,5m . Tr\u00ean b\u1ec1 m\u1eb7t c\u00f3 chi\u1ec1u r\u1ed9ng 1,8m v\u00e0 \u0111\u00e1y m\u01b0\u01a1ng l\u00e0 1,2m (xem h\u00ecnh minh h\u1ecda b\u00ean) . Ng\u01b0\u1eddi ta chuy\u1ec3n to\u00e0n b\u1ed9 kh\u1ed1i \u0111\u1ea5t \u0111i \u0111\u1ec3 r\u1ea3i l\u00ean m\u1ed9t mi\u1ebfng \u0111\u1ea5t h\u00ecnh ch\u1eef nh\u1eadt c\u00f3 k\u00edch th\u01b0\u1edbc 12m v\u00e0 15m . S\u1ed1 \u0111\u1ea5t \u0111\u01b0\u1ee3c chuy\u1ec3n b\u1eb1ng m\u1ed9t chi\u1ebfc \u00f4 t\u00f4 ch\u1edf m\u1ed7i chuy\u1ebfn 6m3 . a) T\u00ednh b\u1ec1 d\u00e0y c\u1ee7a l\u1edbp \u0111\u1ea5t r\u1ea3i l\u00ean mi\u1ebfng \u0111\u1ea5t h\u00ecnh ch\u1eef nh\u1eadt (bi\u1ebft l\u1edbp \u0111\u1ea5t \u0111\u01b0\u1ee3c r\u1ea3i \u0111\u1ec1u, kh\u00f4ng c\u00f3 ch\u1ed5 cao ch\u1ed5 th\u1ea5p). b) C\u1ea7n bao nhi\u00eau chuy\u1ebfn \u00f4 t\u00f4 \u0111\u1ec3 t\u1ea3i h\u1ebft kh\u1ed1i \u0111\u1ea5t. L\u1eddi gi\u1ea3i a) Th\u1ec3 t\u00edch c\u1ee7a kh\u1ed1i \u0111\u1ea5t \u0111\u01b0\u1ee3c \u0111\u00e0o l\u00ean: T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N 6","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH ( )(1,2 + 1,8).1,5 m3 20. 2 = 45 B\u1ec1 d\u00e0y c\u1ee7a l\u1edbp \u0111\u1ea5t: 45 : (12.15) = 0,25(m) b) S\u1ed1 chuy\u1ebfn \u00f4 t\u00f4 ch\u1edf \u0111\u1ea5t l\u00e0: 45 : 6 = 7,5 V\u1eady c\u1ea7n ch\u1edf 8 chuy\u1ebfn \u00f4 t\u00f4 \u0111\u1ec3 t\u1ea3i h\u1ebft ch\u1ed5 \u0111\u1ea5t. C\u00e2u 8. (3 \u0111i\u1ec3m) Cho tam gi\u00e1c ABC c\u00f3 ba g\u00f3c nh\u1ecdn n\u1ed9i ti\u1ebfp (O,R) . Ti\u1ebfp tuy\u1ebfn t\u1ea1i B v\u00e0 C c\u1ee7a (O) c\u1eaft nhau t\u1ea1i I . \u0110\u01b0\u1eddng th\u1eb3ng AI c\u1eaft (O) t\u1ea1i \u0111i\u1ec3m th\u1ee9 hai l\u00e0 D (kh\u00e1c A ). \u0110o\u1ea1n th\u1eb3ng OI c\u1eaft BC t\u1ea1i H . a) Ch\u1ee9ng minh : OI vu\u00f4ng g\u00f3c v\u1edbi BC v\u00e0 HB.HC = HO.HI b) V\u1ebd OK vu\u00f4ng g\u00f3c v\u1edbi AD . Ch\u1ee9ng minh 5 \u0111i\u1ec3m I,B,K,O,C c\u00f9ng thu\u1ed9c m\u1ed9t \u0111\u01b0\u1eddng tr\u00f2n c) T\u1eeb D k\u1ebb \u0111\u01b0\u1eddng th\u1eb3ng vu\u00f4ng g\u00f3c v\u1edbi OB , \u0111\u01b0\u1eddng th\u1eb3ng n\u00e0y c\u1eaft BC t\u1ea1i M v\u00e0 c\u1eaft AB t\u1ea1i N . Ch\u1ee9ng minh : M l\u00e0 trung \u0111i\u1ec3m c\u1ee7a DN . L\u1eddi gi\u1ea3i a) Ch\u1ee9ng minh OI vu\u00f4ng g\u00f3c v\u1edbi BC OB OC R IB IC (t\u00ednh ch\u1ea5t hai ti\u1ebfp tuy\u1ebfn c\u1eaft nhau t\u1ea1i I ) OI thu\u1ed9c \u0111\u01b0\u1eddng trung tr\u1ef1c c\u1ee7a BC OI BC t\u1ea1i H . X\u00e9t OBI vu\u00f4ng t\u1ea1i B \u0111\u01b0\u1eddng cao BH BH 2 HO.HI M\u00e0 BH CH ( H OI ) BH.CH HO.HI b) Ch\u1ee9ng minh \u0111\u01b0\u1ee3c ch\u1ee9ng minh 5 \u0111i\u1ec3m I,B,K,O,C c\u00f9ng thu\u1ed9c m\u1ed9t \u0111\u01b0\u1eddng tr\u00f2n. OK AD OIK 900 OBI OKI OCI 900 V\u1eady 5 \u0111i\u1ec3m I,B,K,O,C c\u00f9ng thu\u1ed9c m\u1ed9t \u0111\u01b0\u1eddng tr\u00f2n. c) Ch\u1ee9ng minh : M l\u00e0 trung \u0111i\u1ec3m c\u1ee7a DN . Ta c\u00f3 DN \/ \/BI (c\u00f9ng vu\u00f4ng g\u00f3c OB ) ADN KIB (\u0111\u1ed3ng v\u1ecb) 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 M\u00e0 KCB KIB (c\u00f9ng ch\u1eafn BK ) KCM KDM (c\u00f9ng nh\u00ecn MK ) T\u1ee9 gi\u00e1c KCDM n\u1ed9i ti\u1ebfp KMC KDC ABC (\u0111\u1ed3ng v\u1ecb) KM \/ \/AB KM \/ \/AN M\u00e0 K l\u00e0 trung \u0111i\u1ec3m AD M l\u00e0 trung \u0111i\u1ec3m DN ----H\u1ebeT--- 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 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 8 NA\u00caM HO\u00cfC: 2023 - 2024 M\u00d4N: TO\u00c1N 9 \u0110\u1ec0 THAM KH\u1ea2O M\u00c3 \u0110\u1ec0: Qu\u1eadn 8 - 2 \u0110\u1ec1 thi g\u1ed3m 8 c\u00e2u ho\u0309i t\u01b0\u0323 lu\u00e2\u0323n. Th\u01a1\u0300i gian: 120 phu\u0301t (kh\u00f4ng k\u00ea\u0309 th\u01a1\u0300i gian pha\u0301t \u0111\u1ec1) C\u00e2u 1. (1,5 \u0111i\u1ec3m). Cho (P) : y = 1 x2 v\u00e0 \u0111\u01b0\u1eddng th\u1eb3ng (d) : y = 1 x + 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). Cho ph\u01b0\u01a1ng tr\u00ecnh x2 \u2212 x \u2212 12 = 0 c\u00f3 2 nghi\u1ec7m l\u00e0 x1 , x2 . Kh\u00f4ng gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh, h\u00e3y t\u00ednh gi\u00e1 tr\u1ecb c\u1ee7a bi\u1ec3u th\u1ee9c A = x1 + 1 + x2 + 1 . x2 x1 L\u01b0u \u00fd: T\u1eeb b\u00e0i n\u00e0y, c\u00e1c s\u1ed1 li\u1ec7u t\u00ednh to\u00e1n v\u1ec1 \u0111\u1ed9 d\u00e0i khi l\u00e0m tr\u00f2n (n\u1ebfu c\u00f3) l\u1ea5y \u0111\u1ebfn m\u1ed9t ch\u1eef s\u1ed1 th\u1eadp ph\u00e2n, s\u1ed1 \u0111o g\u00f3c l\u00e0m tr\u00f2n \u0111\u1ebfn ph\u00fat. C\u00e2u 3. (0,75 \u0111i\u1ec3m). M\u1ed1i quan h\u1ec7 gi\u1eefa l\u1ee3i nhu\u1eadn v\u00e0 s\u1ed1 s\u1ea3n ph\u1ea9m b\u00e1n \u0111\u01b0\u1ee3c trong m\u1ed9t th\u00e1ng t\u1ea1i m\u1ed9t c\u1eeda h\u00e0ng t\u00ednh theo c\u00f4ng th\u1ee9c T = 20n \u2212 500 . Trong \u0111\u00f3 T l\u00e0 s\u1ed1 ti\u1ec1n l\u1ee3i nhu\u1eadn t\u00ednh theo ng\u00e0n \u0111\u1ed3ng, n l\u00e0 s\u1ed1 s\u1ea3n ph\u1ea9m b\u00e1n \u0111\u01b0\u1ee3c trong th\u00e1ng. a) N\u1ebfu trong th\u00e1ng 9 c\u1eeda h\u00e0ng b\u00e1n \u0111\u01b0\u1ee3c 5000 s\u1ea3n ph\u1ea9m th\u00ec l\u1ee3i nhu\u1eadn thu v\u1ec1 l\u00e0 bao nhi\u00eau? b) M\u1ed1i quan h\u1ec7 gi\u1eefa s\u1ed1 ti\u1ec1n l\u1ee3i nhu\u1eadn v\u00e0 s\u1ed1 nh\u00e2n vi\u00ean l\u00e0m vi\u1ec7c l\u00e0 T = 9000.k v\u1edbi k l\u00e0 s\u1ed1 nh\u00e2n ( )vi\u00ean k \uf0ce * , T l\u00e0 l\u1ee3i nhu\u1eadn t\u00ednh theo \u0111\u01a1n v\u1ecb ng\u00e0n \u0111\u1ed3ng. V\u1eady n\u1ebfu c\u1eeda h\u00e0ng c\u00f3 8 nh\u00e2n vi\u00ean th\u00ec m\u1ed9t th\u00e1ng b\u00e1n \u0111\u01b0\u1ee3c bao nhi\u00eau s\u1ea3n ph\u1ea9m? C\u00e2u 4. (1 \u0111i\u1ec3m). Nh\u1eb1m \u0111\u1ed9ng vi\u00ean, khen th\u01b0\u1edfng c\u00e1c em \u0111\u1ea1t danh hi\u1ec7u \u201ch\u1ecdc sinh gi\u1ecfi c\u1ea5p th\u00e0nh ph\u1ed1\u201d n\u0103m h\u1ecdc 2021 \u2212 2022 , tr\u01b0\u1eddng THCS A t\u1ed5 ch\u1ee9c chuy\u1ebfn tham quan ngo\u1ea1i kh\u00f3a t\u1ea1i m\u1ed9t \u0111i\u1ec3m du l\u1ecbch v\u1edbi m\u1ee9c gi\u00e1 ban \u0111\u1ea7u l\u00e0 375 000 \u0111\u1ed3ng\/ng\u01b0\u1eddi. Bi\u1ebft c\u00f4ng ty du l\u1ecbch gi\u1ea3m 10% chi ph\u00ed cho m\u1ed7i gi\u00e1o vi\u00ean v\u00e0 gi\u1ea3m 30% chi ph\u00ed cho m\u1ed7i h\u1ecdc sinh. S\u1ed1 h\u1ecdc sinh tham gia g\u1ea5p 4 l\u1ea7n s\u1ed1 gi\u00e1o vi\u00ean v\u00e0 t\u1ed5ng chi ph\u00ed tham quan (sau khi gi\u1ea3m gi\u00e1) l\u00e0 12 487 500 \u0111\u1ed3ng. T\u00ednh s\u1ed1 gi\u00e1o vi\u00ean v\u00e0 s\u1ed1 h\u1ecdc sinh \u0111\u00e3 tham gia chuy\u1ebfn \u0111i? C\u00e2u 5. (1 \u0111i\u1ec3m). B\u00e0 N\u0103m nu\u00f4i t\u1ea5t c\u1ea3 10 con heo, m\u1ed7i con 70 kg. \u0110\u1ebfn ng\u00e0y xu\u1ea5t chu\u1ed3ng b\u00e0 b\u00e1n \u0111\u01b0\u1ee3c 4 con v\u1edbi gi\u00e1 l\u00fac \u1ed5n \u0111\u1ecbnh l\u00e0 58 000 \u0111\u1ed3ng\/1kg heo h\u01a1i (c\u00e2n heo h\u01a1i l\u00e0 heo v\u1eabn c\u00f2n s\u1ed1ng, c\u00e2n t\u1ea1i chu\u1ed3ng). Th\u00e1ng sau, g\u1eb7p tr\u1eadn d\u1ecbch b\u1ec7nh b\u00e0 N\u0103m ph\u1ea3i b\u00e1n v\u1edbi gi\u00e1 gi\u1ea3m 40% so v\u1edbi gi\u00e1 khi \u1ed5n \u0111\u1ecbnh. Bi\u1ebft gi\u00e1 v\u1ed1n b\u00e0 \u0111\u1ea7u t\u01b0 l\u00fac ban \u0111\u1ea7u nu\u00f4i heo l\u00e0 32 000 \u0111\u1ed3ng\/ 1kg heo h\u01a1i. a) Sau khi b\u00e1n h\u1ebft 10 con heo th\u00ec s\u1ebd b\u00e0 N\u0103m l\u1eddi bao nhi\u00eau ti\u1ec1n? b) M\u1ed9t ng\u01b0\u1eddi bu\u00f4n b\u00e1n heo \u1edf ch\u1ee3 \u0111\u00e3 mua 2 con heo c\u1ee7a b\u00e0 N\u0103m v\u1edbi gi\u00e1 l\u00fac \u1ed5n \u0111\u1ecbnh. Ng\u01b0\u1eddi \u0111\u00f3 mu\u1ed1n l\u1eddi 60% so t\u1ed5ng s\u1ed1 ti\u1ec1n v\u1ed1n b\u1ecf ra g\u1ed3m s\u1ed1 ti\u1ec1n mua heo v\u00e0 chi ph\u00ed v\u1eadn chuy\u1ec3n heo l\u00e0 2 tri\u1ec7u \u0111\u1ed3ng, th\u00ec ng\u01b0\u1eddi \u0111\u00f3 c\u1ea7n b\u00e1n l\u1ebb m\u1ed7i kg heo bao nhi\u00eau ti\u1ec1n ? (L\u00e0m tr\u00f2n k\u1ebft qu\u1ea3 \u0111\u1ebfn ch\u1eef s\u1ed1 h\u00e0ng ngh\u00ecn) C\u00e2u 6. (1 \u0111i\u1ec3m). C\u00e1i m\u0169 c\u00f3 v\u00e0nh c\u1ee7a ch\u00fa h\u1ec1 v\u1edbi c\u00e1c k\u00edch th\u01b0\u1edbc cho theo h\u00ecnh v\u1ebd 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) H\u00e3y t\u00ednh t\u1ed5ng di\u1ec7n t\u00edch v\u1ea3i c\u1ea7n c\u00f3 \u0111\u1ec3 l\u00e0m n\u00ean c\u00e1i m\u0169 c\u1ee7a ch\u00fa h\u1ec1 (kh\u00f4ng k\u1ec3 ri\u1ec1m, m\u00e9p, ph\u1ea7n th\u1eeba). b) Ch\u00fa h\u1ec1 d\u1ef1 \u0111\u1ecbnh mua b\u1ed9t \u0111\u1ed5 \u0111\u1ea7y n\u00f3n \u0111\u1ec3 l\u00e0m \u1ea3o thu\u1eadt. Ch\u00fa h\u1ec1 c\u1ea7n mua kh\u1ed1i l\u01b0\u1ee3ng b\u1ed9t l\u00e0 bao nhi\u00eau (l\u00e0m tr\u00f2n \u0111\u1ebfn h\u00e0ng \u0111\u01a1n v\u1ecb)? (bi\u1ebft r\u1eb1ng kh\u1ed1i l\u01b0\u1ee3ng ri\u00eang c\u1ee7a lo\u1ea1i b\u1ed9t \u0111\u00f3 l\u00e0 1gam \/ cm3 ngh\u0129a l\u00e0 1cm3 t\u01b0\u01a1ng \u1ee9ng v\u1edbi 1gam ). Cho c\u00f4ng th\u1ee9c t\u00ednh th\u1ec3 t\u00edch h\u00ecnh n\u00f3n: V = 1 \uf070 R2.h 3 C\u00f4ng th\u1ee9c t\u00ednh d\u1ec7n t\u00edch xung quanh h\u00ecnh n\u00f3n l\u00e0 S = \uf070 .R.l Trong \u0111\u00f3 h l\u00e0 chi\u1ec1u cao h\u00ecnh n\u00f3n, R l\u00e0 b\u00e1n k\u00ednh \u0111\u00e1y, l l\u00e0 \u0111\u01b0\u1eddng sinh. L\u1ea5y \uf070 \uf0bb 3,14 C\u00e2u 7. (0,75 \u0111i\u1ec3m). \u0110\u1ec3 gi\u00fap xe l\u1eeda chuy\u1ec3n t\u1eeb m\u1ed9t \u0111\u01b0\u1eddng ray t\u1eeb h\u01b0\u1edbng n\u00e0y sang m\u1ed9t \u0111\u01b0\u1eddng ray theo h\u01b0\u1edbng kh\u00e1c, ng\u01b0\u1eddi ta l\u00e0m xen gi\u1eefa m\u1ed9t \u0111o\u1ea1n \u0111\u01b0\u1eddng ray h\u00ecnh v\u00f2ng cung (h\u00ecnh v\u1ebd minh h\u1ecda b\u00ean). Bi\u1ebft chi\u1ec1u r\u1ed9ng c\u1ee7a \u0111\u01b0\u1eddng ray l\u00e0 AB \uf0bb 1,1m , \u0111o\u1ea1n BC \uf0bb 28,4 m , BC vu\u00f4ng g\u00f3c v\u1edbi AO . H\u00e3y cho bi\u1ebft s\u1ed1 \u0111o cung AC (l\u00e0m tr\u00f2n k\u1ebft qu\u1ea3 \u0111\u1ebfn \u0111\u1ed9). C\u00e2u 8. (3 \u0111i\u1ec3m) Cho \u0111\u01b0\u1eddng tr\u00f2n (O,R) . Qua \u0111i\u1ec3m A \u1edf ngo\u00e0i \u0111\u01b0\u1eddng tr\u00f2n, ta v\u1ebd c\u00e1c ti\u1ebfp tuy\u1ebfn AB v\u00e0 AC t\u1edbi \u0111\u01b0\u1eddng tr\u00f2n ( B v\u00e0 C l\u00e0 c\u00e1c ti\u1ebfp \u0111i\u1ec3m), AO c\u1eaft BC t\u1ea1i H . V\u1ebd c\u00e1t tuy\u1ebfn AEF ( E,B c\u00f9ng thu\u1ed9c m\u1ed9t n\u1eeda m\u1eb7t ph\u1eb3ng b\u1edd OA ). G\u1ecdi D l\u00e0 trung \u0111i\u1ec3m c\u1ee7a EF . a) Ch\u1ee9ng minh: t\u1ee9 gi\u00e1c ODBC n\u1ed9i ti\u1ebfp. b) V\u1ebd \u0111\u01b0\u1eddng k\u00ednh BK c\u1ee7a (O) . G\u1ecdi M l\u00e0 h\u00ecnh chi\u1ebfu c\u1ee7a C tr\u00ean BK , AK c\u1eaft CM t\u1ea1i I . Ch\u1ee9ng minh I l\u00e0 trung \u0111i\u1ec3m c\u1ee7a CM . c) Tia CM c\u1eaft (O) t\u1ea1i \u0111i\u1ec3m th\u1ee9 hai N , AN c\u1eaft (O) t\u1ea1i \u0111i\u1ec3m th\u1ee9 hai J , CJ c\u1eaft AB t\u1ea1i Z . Ch\u1ee9ng minh ZH vu\u00f4ng g\u00f3c v\u1edbi OC . ----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 = 1 x2 v\u00e0 \u0111\u01b0\u1eddng th\u1eb3ng (d ) : y = 1 x + 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. BGT: x \u22124 \u22122 0 2 4 y = 1 x2 4 1 0 1 4 4 x0 2 3 y= 1x+2 2 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 = 1 x + 2 42 \uf0db 1 x2 \u2212 1 x \u2212 2 = 0 42 \uf0db \uf0e9 x = 4 \uf0ea\uf0eb x = \u22122 Thay x = 4 v\u00e0o y = 1 x + 2 , ta \u0111\u01b0\u1ee3c: y = 1 .4 + 2 = 4 . 22 Thay x = \u22122 v\u00e0o y = 1 x + 2 , ta \u0111\u01b0\u1ee3c: y = 1 .(\u22122) + 2 = 1. 22 V\u1eady (4;4) , (\u22122;1) l\u00e0 hai giao \u0111i\u1ec3m c\u1ea7n t\u00ecm. C\u00e2u 2. (1 \u0111i\u1ec3m). Cho ph\u01b0\u01a1ng tr\u00ecnh x2 \u2212 x \u221212 = 0 c\u00f3 2 nghi\u1ec7m l\u00e0 x1, x2 . Kh\u00f4ng gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh, h\u00e3y t\u00ednh gi\u00e1 tr\u1ecb c\u1ee7a bi\u1ec3u th\u1ee9c A = x1 +1 + x2 +1 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 = (\u22121)2 \u2212 4.1.(\u221212) = 49 \uf03e 0 N\u00ean ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m ph\u00e2n bi\u1ec7t x1, x2 . \uf0ec\uf0ef\uf0efS = x1 + x2 = \u2212b = 1 \uf0ed = a Theo \u0111\u1ecbnh l\u00ed Vi-et, ta c\u00f3: \uf0ef\uf0ef\uf0eeP c = \u221212 a x1.x2 = Ta c\u00f3: A = x1 +1 + x2 +1 x2 x1 A = ( x1 +1). x1 + ( x2 +1). x2 x1. x2 A = x12 + x1 + x22 + x2 x1. x2 A = x12 + x22 + x1 + x2 x1. x2 A = S2 \u2212 2P + S P A = 12 \u2212 2.(\u221212) +1 = \u221213 \u221212 6 L\u01b0u \u00fd: T\u1eeb b\u00e0i n\u00e0y, c\u00e1c s\u1ed1 li\u1ec7u t\u00ednh to\u00e1n v\u1ec1 \u0111\u1ed9 d\u00e0i khi l\u00e0m tr\u00f2n (n\u1ebfu c\u00f3) l\u1ea5y \u0111\u1ebfn m\u1ed9t ch\u1eef s\u1ed1 th\u1eadp ph\u00e2n, s\u1ed1 \u0111o g\u00f3c l\u00e0m tr\u00f2n \u0111\u1ebfn ph\u00fat. C\u00e2u 3. (0,75 \u0111i\u1ec3m) M\u1ed1i quan h\u1ec7 gi\u1eefa l\u1ee3i nhu\u1eadn v\u00e0 s\u1ed1 s\u1ea3n ph\u1ea9m b\u00e1n \u0111\u01b0\u1ee3c trong m\u1ed9t th\u00e1ng t\u1ea1i m\u1ed9t c\u1eeda h\u00e0ng t\u00ednh theo c\u00f4ng th\u1ee9c T = 20n \u2212 500 . Trong \u0111\u00f3 T l\u00e0 s\u1ed1 ti\u1ec1n l\u1ee3i nhu\u1eadn t\u00ednh theo ng\u00e0n \u0111\u1ed3ng, n l\u00e0 s\u1ed1 s\u1ea3n ph\u1ea9m b\u00e1n \u0111\u01b0\u1ee3c trong th\u00e1ng. a) N\u1ebfu trong th\u00e1ng 9 c\u1eeda h\u00e0ng b\u00e1n \u0111\u01b0\u1ee3c 5000 s\u1ea3n ph\u1ea9m th\u00ec l\u1ee3i nhu\u1eadn thu v\u1ec1 l\u00e0 bao nhi\u00eau? b) M\u1ed1i quan h\u1ec7 gi\u1eefa s\u1ed1 ti\u1ec1n l\u1ee3i nhu\u1eadn v\u00e0 s\u1ed1 nh\u00e2n vi\u00ean l\u00e0m vi\u1ec7c l\u00e0 T = 9000.k v\u1edbi k l\u00e0 s\u1ed1 ( )nh\u00e2n vi\u00ean k \uf0ce * , T l\u00e0 l\u1ee3i nhu\u1eadn t\u00ednh theo \u0111\u01a1n v\u1ecb ng\u00e0n \u0111\u1ed3ng. V\u1eady n\u1ebfu c\u1eeda h\u00e0ng c\u00f3 8 nh\u00e2n vi\u00ean th\u00ec m\u1ed9t th\u00e1ng b\u00e1n \u0111\u01b0\u1ee3c bao nhi\u00eau s\u1ea3n ph\u1ea9m? L\u1eddi gi\u1ea3i a) Ta c\u00f3 n = 5000 \uf0de T = 20.5000 \u2212 500 = 99 500 (ng\u00e0n \u0111\u1ed3ng) 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\u1eady trong th\u00e1ng 9 , l\u1ee3i nhu\u1eadn c\u1ee7a c\u1eeda h\u00e0ng l\u00e0 99 500 (ng\u00e0n \u0111\u1ed3ng). b) Ta c\u00f3 k = 8 \uf0de T = 9000.8 = 72 000 M\u00e0 T = 20n \u2212 500 \uf0de 72 000 = 20n \u2212 500 \uf0db n = 3625 V\u1eady n\u1ebfu c\u1eeda h\u00e0ng c\u00f3 8 nh\u00e2n vi\u00ean, m\u1ed7i th\u00e1ng s\u1ebd b\u00e1n \u0111\u01b0\u1ee3c 3625 s\u1ea3n ph\u1ea9m. C\u00e2u 4. (1 \u0111i\u1ec3m). Nh\u1eb1m \u0111\u1ed9ng vi\u00ean, khen th\u01b0\u1edfng c\u00e1c em \u0111\u1ea1t danh hi\u1ec7u \u201ch\u1ecdc sinh gi\u1ecfi c\u1ea5p th\u00e0nh ph\u1ed1\u201d n\u0103m h\u1ecdc 2021\u2212 2022, tr\u01b0\u1eddng THCS A t\u1ed5 ch\u1ee9c chuy\u1ebfn tham quan ngo\u1ea1i kh\u00f3a t\u1ea1i m\u1ed9t \u0111i\u1ec3m du l\u1ecbch v\u1edbi m\u1ee9c gi\u00e1 ban \u0111\u1ea7u l\u00e0 375 000 \u0111\u1ed3ng\/ng\u01b0\u1eddi. Bi\u1ebft c\u00f4ng ty du l\u1ecbch gi\u1ea3m 10% chi ph\u00ed cho m\u1ed7i gi\u00e1o vi\u00ean v\u00e0 gi\u1ea3m 30% chi ph\u00ed cho m\u1ed7i h\u1ecdc sinh. S\u1ed1 h\u1ecdc sinh tham gia g\u1ea5p 4 l\u1ea7n s\u1ed1 gi\u00e1o vi\u00ean v\u00e0 t\u1ed5ng chi ph\u00ed tham quan (sau khi gi\u1ea3m gi\u00e1) l\u00e0 12 487 500 \u0111\u1ed3ng. T\u00ednh s\u1ed1 gi\u00e1o vi\u00ean v\u00e0 s\u1ed1 h\u1ecdc sinh \u0111\u00e3 tham gia chuy\u1ebfn \u0111i? L\u1eddi gi\u1ea3i ( )G\u1ecdi x; y (ng\u01b0\u1eddi) l\u1ea7n l\u01b0\u1ee3t l\u00e0 s\u1ed1 h\u1ecdc sinh v\u00e0 s\u1ed1 gi\u00e1o vi\u00ean tham gia chuy\u1ebfn \u0111\u00ed x; y \uf0ce * Theo \u0111\u1ec1, ta c\u00f3: \uf0ec\uf0efx = 4 y + (1\u221210%).375000.y = 12 487500 \uf0ed\uf0ef\uf0ee(1\u2212 30%).375000.x \uf0db \uf0ecx = 36 (nh\u1eadn) \uf0ed \uf0ee y = 9 V\u1eady: S\u1ed1 h\u1ecdc sinh tham gia chuy\u1ebfn \u0111i l\u00e0 36 (h\u1ecdc sinh) S\u1ed1 gi\u00e1o vi\u00ean tham gia chuy\u1ebfn \u0111i l\u00e0 9 (gi\u00e1o vi\u00ean). C\u00e2u 5. (1 \u0111i\u1ec3m) B\u00e0 N\u0103m nu\u00f4i t\u1ea5t c\u1ea3 10 con heo, m\u1ed7i con 70 kg. \u0110\u1ebfn ng\u00e0y xu\u1ea5t chu\u1ed3ng b\u00e0 b\u00e1n \u0111\u01b0\u1ee3c 4 con v\u1edbi gi\u00e1 l\u00fac \u1ed5n \u0111\u1ecbnh l\u00e0 58 000 \u0111\u1ed3ng\/1kg heo h\u01a1i (c\u00e2n heo h\u01a1i l\u00e0 heo v\u1eabn c\u00f2n s\u1ed1ng, c\u00e2n t\u1ea1i chu\u1ed3ng). Th\u00e1ng sau, g\u1eb7p tr\u1eadn d\u1ecbch b\u1ec7nh b\u00e0 N\u0103m ph\u1ea3i b\u00e1n v\u1edbi gi\u00e1 gi\u1ea3m 40% so v\u1edbi gi\u00e1 khi \u1ed5n \u0111\u1ecbnh. Bi\u1ebft gi\u00e1 v\u1ed1n b\u00e0 \u0111\u1ea7u t\u01b0 l\u00fac ban \u0111\u1ea7u nu\u00f4i heo l\u00e0 32 000 \u0111\u1ed3ng\/1kg heo h\u01a1i. a) Sau khi b\u00e1n h\u1ebft 10 con heo th\u00ec s\u1ebd b\u00e0 N\u0103m l\u1eddi bao nhi\u00eau ti\u1ec1n? b) M\u1ed9t ng\u01b0\u1eddi bu\u00f4n b\u00e1n heo \u1edf ch\u1ee3 \u0111\u00e3 mua 2 con heo c\u1ee7a b\u00e0 N\u0103m v\u1edbi gi\u00e1 l\u00fac \u1ed5n \u0111\u1ecbnh. Ng\u01b0\u1eddi \u0111\u00f3 mu\u1ed1n l\u1eddi 60% so t\u1ed5ng s\u1ed1 ti\u1ec1n v\u1ed1n b\u1ecf ra g\u1ed3m s\u1ed1 ti\u1ec1n mua heo v\u00e0 chi ph\u00ed v\u1eadn chuy\u1ec3n heo l\u00e0 2 tri\u1ec7u \u0111\u1ed3ng, th\u00ec ng\u01b0\u1eddi \u0111\u00f3 c\u1ea7n b\u00e1n l\u1ebb m\u1ed7i kg heo bao nhi\u00eau ti\u1ec1n ? (L\u00e0m tr\u00f2n k\u1ebft qu\u1ea3 \u0111\u1ebfn ch\u1eef s\u1ed1 h\u00e0ng ngh\u00ecn) 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 L\u1eddi gi\u1ea3i a) S\u1ed1 ti\u1ec1n b\u00e0 N\u0103m thu \u0111\u01b0\u1ee3c sau khi b\u00e1n 10 con heo l\u00e0: 4.70.58000 + 6.70.(1\u2212 40%).58000 = 30856000 (\u0111\u1ed3ng) S\u1ed1 ti\u1ec1n \u0111\u1ea7u t\u01b0 l\u00fac \u0111\u1ea7u l\u00e0: 10.70.32000 = 22 400000 (\u0111\u1ed3ng) S\u1ed1 ti\u1ec1n l\u1eddi l\u00e0: 30856000 \u2212 22 400000 = 8456000 (\u0111\u1ed3ng) b) S\u1ed1 ti\u1ec1n mua hai con heo v\u00e0 chi ph\u00ed v\u1eadn chuy\u1ec3n l\u00e0: 2.70.58000 + 2000000 = 10120000 (\u0111\u1ed3ng) S\u1ed1 ti\u1ec1n b\u00e1n m\u1ed7i kg heo l\u00e0: 10120000.(1+ 60%) : (2.70) \uf0bb 116000 (\u0111\u1ed3ng) C\u00e2u 6. (1 \u0111i\u1ec3m) C\u00e1i m\u0169 c\u00f3 v\u00e0nh c\u1ee7a ch\u00fa h\u1ec1 v\u1edbi c\u00e1c k\u00edch th\u01b0\u1edbc cho theo h\u00ecnh v\u1ebd a) H\u00e3y t\u00ednh t\u1ed5ng di\u1ec7n t\u00edch v\u1ea3i c\u1ea7n c\u00f3 \u0111\u1ec3 l\u00e0m n\u00ean c\u00e1i m\u0169 c\u1ee7a ch\u00fa h\u1ec1 (kh\u00f4ng k\u1ec3 ri\u1ec1m, m\u00e9p, ph\u1ea7n th\u1eeba). b) Ch\u00fa h\u1ec1 d\u1ef1 \u0111\u1ecbnh mua b\u1ed9t \u0111\u1ed5 \u0111\u1ea7y n\u00f3n \u0111\u1ec3 l\u00e0m \u1ea3o thu\u1eadt. Ch\u00fa h\u1ec1 c\u1ea7n mua kh\u1ed1i l\u01b0\u1ee3ng b\u1ed9t l\u00e0 bao nhi\u00eau (l\u00e0m tr\u00f2n \u0111\u1ebfn h\u00e0ng \u0111\u01a1n v\u1ecb)? (bi\u1ebft r\u1eb1ng kh\u1ed1i l\u01b0\u1ee3ng ri\u00eang c\u1ee7a lo\u1ea1i b\u1ed9t \u0111\u00f3 l\u00e0 1gam \/ cm3 ngh\u0129a l\u00e0 1cm3 t\u01b0\u01a1ng \u1ee9ng v\u1edbi 1gam ). Cho c\u00f4ng th\u1ee9c t\u00ednh th\u1ec3 t\u00edch h\u00ecnh n\u00f3n: V = 1 \uf070 R2.h 3 C\u00f4ng th\u1ee9c t\u00ednh d\u1ec7n t\u00edch xung quanh h\u00ecnh n\u00f3n l\u00e0 S = \uf070 .R.l Trong \u0111\u00f3 h l\u00e0 chi\u1ec1u cao h\u00ecnh n\u00f3n, R l\u00e0 b\u00e1n k\u00ednh \u0111\u00e1y, l l\u00e0 \u0111\u01b0\u1eddng sinh. L\u1ea5y \uf070 \uf0bb 3,14 L\u1eddi gi\u1ea3i a) B\u00e1n k\u00ednh h\u00ecnh n\u00f3n: r = 86 \u2212 21.2 = 22(cm) 2 ( )Di\u1ec7n t\u00edch xung quanh: Sxq = \uf070 rl = 3,14.22.72 = 4973,76 cm2 ( )Di\u1ec7n .\uf0e6\uf0e7\uf0e8 86 \uf0f62 t\u00edch v\u00e0nh n\u00f3n: Svn = \uf070 2 \uf0f7\uf0f8 \u2212 \uf070 .222 = 4286,1 cm2 ( )Di\u1ec7n t\u00edch c\u1ea7n t\u00ecm: S = Sxq + Svn = 4973,76 + 4286,1 = 9259,86 cm2 b) Chi\u1ec1u cao h\u00ecnh n\u00f3n: h = 722 \u2212 222 = 10 47 (cm) 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\u1ec3 t\u00edch h\u00ecnh n\u00f3n: V = 1 .\uf070 R2h = 1 .3,14.222.10 47 \uf0bb 34729,8 cm3 33 V\u1eady s\u1ed1 b\u1ed9t c\u1ea7n \u0111\u1ec3 \u0111\u1ed5 \u0111\u1ea7y n\u00f3n l\u00e0: 34730( g ) C\u00e2u 7. (0,75 \u0111i\u1ec3m) \u0110\u1ec3 gi\u00fap xe l\u1eeda chuy\u1ec3n t\u1eeb m\u1ed9t \u0111\u01b0\u1eddng ray t\u1eeb h\u01b0\u1edbng n\u00e0y sang m\u1ed9t \u0111\u01b0\u1eddng ray theo h\u01b0\u1edbng kh\u00e1c, ng\u01b0\u1eddi ta l\u00e0m xen gi\u1eefa m\u1ed9t \u0111o\u1ea1n \u0111\u01b0\u1eddng ray h\u00ecnh v\u00f2ng cung (h\u00ecnh v\u1ebd minh h\u1ecda b\u00ean). Bi\u1ebft chi\u1ec1u r\u1ed9ng c\u1ee7a \u0111\u01b0\u1eddng ray l\u00e0 AB \uf0bb 1,1m , \u0111o\u1ea1n BC \uf0bb 28, 4 m , BC vu\u00f4ng g\u00f3c v\u1edbi AO . H\u00e3y cho bi\u1ebft s\u1ed1 \u0111o cung AC (l\u00e0m tr\u00f2n k\u1ebft qu\u1ea3 \u0111\u1ebfn \u0111\u1ed9). L\u1eddi gi\u1ea3i - Thanh ray tr\u00f9ng v\u1edbi BC ti\u1ebfp x\u00fac v\u1edbi \u0111\u01b0\u1eddng tr\u00f2n (O,OB) t\u1ea1i B n\u00ean l\u00e0 ti\u1ebfp tuy\u1ebfn c\u1ee7a \u0111\u01b0\u1eddng tr\u00f2n (O,OB) \uf0de BC \u22a5 OB - OA c\u1eaft \u0111\u01b0\u1eddng tr\u00f2n (O,OA) t\u1ea1i \u0111i\u1ec3m D( D \uf0b9 A) \uf0de AD = 2R - Tam gi\u00e1c ACD n\u1ed9i ti\u1ebfp \u0111\u01b0\u1eddng tr\u00f2n (O,OA) c\u00f3 \u0111\u01b0\u1eddng k\u00ednh AD n\u00ean l\u00e0 tam gi\u00e1c vu\u00f4ng t\u1ea1i C . - X\u00e9t tam gi\u00e1c ACD vu\u00f4ng t\u1ea1i C , \u0111\u01b0\u1eddng cao BC , ta c\u00f3: CB2 = AB.BD (h\u1ec7 th\u1ee9c l\u01b0\u1ee3ng trong tam gi\u00e1c vu\u00f4ng) \uf0db CB2 = AB.( AD \u2212 AB) \uf0db CB2 = AB.(2R \u2212 AB) - Thay s\u1ed1, ta c\u00f3: (28, 4)2 \uf0bb 1,1.(2R \u22121,1) \uf0db 2R \uf0bb 807,77 \uf0db R \uf0bb 367, 2(m) C\u00e2u 8. (3 \u0111i\u1ec3m) Cho \u0111\u01b0\u1eddng tr\u00f2n (O, R) . Qua \u0111i\u1ec3m A \u1edf ngo\u00e0i \u0111\u01b0\u1eddng tr\u00f2n, ta v\u1ebd c\u00e1c ti\u1ebfp tuy\u1ebfn AB v\u00e0 AC t\u1edbi \u0111\u01b0\u1eddng tr\u00f2n ( B v\u00e0 C l\u00e0 c\u00e1c ti\u1ebfp \u0111i\u1ec3m), AO c\u1eaft BC t\u1ea1i H . V\u1ebd c\u00e1t tuy\u1ebfn AEF ( E, B c\u00f9ng thu\u1ed9c m\u1ed9t n\u1eeda m\u1eb7t ph\u1eb3ng b\u1edd OA). G\u1ecdi D l\u00e0 trung \u0111i\u1ec3m c\u1ee7a EF . a) Ch\u1ee9ng minh: t\u1ee9 gi\u00e1c ODBC n\u1ed9i ti\u1ebfp. b) V\u1ebd \u0111\u01b0\u1eddng k\u00ednh BK c\u1ee7a (O) . G\u1ecdi M l\u00e0 h\u00ecnh chi\u1ebfu c\u1ee7a C tr\u00ean BK , AK c\u1eaft CM t\u1ea1i I . Ch\u1ee9ng minh I l\u00e0 trung \u0111i\u1ec3m c\u1ee7a CM . c) Tia CM c\u1eaft (O) t\u1ea1i \u0111i\u1ec3m th\u1ee9 hai N , AN c\u1eaft (O) t\u1ea1i \u0111i\u1ec3m th\u1ee9 hai J , CJ c\u1eaft AB t\u1ea1i Z . Ch\u1ee9ng minh ZH vu\u00f4ng g\u00f3c v\u1edbi OC . 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) Ch\u1ee9ng minh t\u1ee9 gi\u00e1c ODBC n\u1ed9i ti\u1ebfp X\u00e9t (O) c\u00f3: FE l\u00e0 d\u00e2y cung D l\u00e0 trung \u0111i\u1ec3m c\u1ee7a FE \uf0de OD \u22a5 FE X\u00e9t t\u1ee9 gi\u00e1c ODBC c\u00f3: ODA = OBA = 90\uf0b0 M\u00e0 hai g\u00f3c c\u00f9ng nh\u00ecn c\u1ea1nh OA \uf0de T\u1ee9 gi\u00e1c ODBC n\u1ed9i ti\u1ebfp (Hai \u0111\u1ec9nh li\u00ean ti\u1ebfp c\u00f9ng nh\u00ecn m\u1ed9t c\u1ea1nh d\u01b0\u1edbi hai g\u00f3c b\u1eb1ng nhau) b) Ch\u1ee9ng minh: I l\u00e0 trung \u0111i\u1ec3m MC Ta c\u00f3: \uf0ecOB = OC = R \uf0ed = AC \uf0ee AB \uf0de O, A c\u00e1ch \u0111\u1ec1u B,C \uf0de OA l\u00e0 \u0111\u01b0\u1eddng trung tr\u1ef1c c\u1ee7a BC \uf0de OA \u22a5 BC t\u1ea1i H Do M l\u00e0 h\u00ecnh chi\u1ebfu c\u1ee7a C tr\u00ean BK \uf0de CM \u22a5 BK t\u1ea1i M X\u00e9t t\u1ee9 gi\u00e1c MOHC c\u00f3: \uf0ec\uf0efOMC = 90\uf0b0(CM \u22a5 BK ) \uf0ed \uf0ef\uf0eeOHC = 90\uf0b0 (OA \u22a5 BC = \uf07bH\uf07d) \uf0de OMC + OHC = 90\uf0b0 + 90\uf0b0 = 180\uf0b0 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 9 \uf0de T\u1ee9 gi\u00e1c MOHC n\u1ed9i ti\u1ebfp (T\u1ed5ng hai g\u00f3c \u0111\u1ed1i b\u1eb1ng 180\uf0b0 ) \uf0de OMH = OCH (ch\u1eafn OH ) M\u00e0 OCH = OBH ( \uf044OCB c\u00e2n t\u1ea1i O , OB = OC = R ) \uf0de OMH = OBH \uf0de \uf044BHM c\u00e2n t\u1ea1i H \uf0de BH = HM M\u00e0 BH = HC ( OA l\u00e0 \u0111\u01b0\u1eddng trung tr\u1ef1c BC ) \uf0de HM = HC \uf0de \uf044MHC c\u00e2n t\u1ea1i H \uf0de HMC = HCM X\u00e9t \uf044HIM v\u00e0 \uf044HIC c\u00f3: HI l\u00e0 c\u1ea1nh chung HM = HC (cmt) HMC = HCM (cmt) \uf0de \uf044HIM = \uf044HIC (c \u2212 g \u2212 c) \uf0de IM = IC \uf0de I l\u00e0 trung \u0111i\u1ec3m MC . c) Ch\u1ee9ng minh: ZH vu\u00f4ng g\u00f3c v\u1edbi OC Ta c\u00f3: AB \u22a5 BK ( AB l\u00e0 ti\u1ebfp tuy\u1ebfn c\u1ee7a (O) ) CN \u22a5 BK t\u1ea1i M (cmt) \uf0de AB \/ \/ CN \uf0de CNA = NAB (sole trong) M\u00e0 CNA = ZCA (ch\u1eafn JC ) \uf0de NAB = ZCA X\u00e9t \uf044ZJA v\u00e0 \uf044ZAC c\u00f3: CZA chung NAB = ZCA (cmt) \uf0de \uf044ZJA \uf044ZAC ( g \u2212 g ) 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 \uf0de ZJ = ZA \uf0db ZA2 = ZJ.ZC (1) ZA ZC X\u00e9t \uf044ZBJ v\u00e0 \uf044ZJC c\u00f3 BZJ chung JBZ = BCZ (ch\u1eafn BJ ) \uf0de \uf044ZBJ \uf044ZCB( g \u2212 g ) \uf0de ZB = ZJ \uf0db ZB2 = ZJ.ZC (2) ZC ZB T\u1eeb (1) v\u00e0 (2) \uf0de ZA2 = ZB2 (ZJ.ZC ) \uf0de ZA = ZB \uf0de Z l\u00e0 trung \u0111i\u1ec3m AB X\u00e9t \uf044ABC c\u00f3: Z l\u00e0 trung \u0111i\u1ec3m AB (cmt) H l\u00e0 trung \u0111i\u1ec3m BC (cmt) \uf0de ZH l\u00e0 \u0111\u01b0\u1eddng trung b\u00ecnh \uf0de ZH \/ \/ AC M\u00e0 AC \u22a5 OC ( AC l\u00e0 ti\u1ebfp tuy\u1ebfn c\u1ee7a (O) ) \uf0de ZH \u22a5 OC . ----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\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 8 NA\u00caM HO\u00cfC: 2023 - 2024 M\u00d4N: TO\u00c1N 9 \u0110\u1ec0 THAM KH\u1ea2O M\u00c3 \u0110\u1ec0: Qu\u1eadn 8 - 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 = 1 x2 v\u00e0 \u0111\u01b0\u1eddng th\u1eb3ng (D) : y = \u2212 1 x + 1 . 22 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 20x2 + 5x \u2212 2020 = 0 c\u00f3 2 nghi\u1ec7m l\u00e0 x1 , x2 . Kh\u00f4ng gi\u1ea3i ph\u01b0\u01a1ng ( ) ( )tr\u00ecnh, h\u00e3y t\u00ednh gi\u00e1 tr\u1ecb c\u1ee7a bi\u1ec3u th\u1ee9c A = x1+ x2 x1 x2 1 \u2212 x2 1 \u2212 x1 . C\u00e2u 3. (1 \u0111i\u1ec3m). M\u1ed9t \u00f4 t\u00f4 c\u00f3 b\u00ecnh ch\u1ee9a x\u0103ng ch\u1ee9a \u0111\u01b0\u1ee3c nhi\u1ec1u nh\u1ea5t l\u00e0 40 l\u00edt x\u0103ng. C\u1ee9 ch\u1ea1y 100 km th\u00ec \u00f4 t\u00f4 ti\u00eau th\u1ee5 h\u1ebft 8 l\u00edt x\u0103ng. G\u1ecdi x(km) l\u00e0 qu\u00e3ng \u0111\u01b0\u1eddng \u00f4 t\u00f4 \u0111i \u0111\u01b0\u1ee3c v\u00e0 y (l\u00edt) l\u00e0 s\u1ed1 l\u00edt x\u0103ng ti\u00eau th\u1ee5. a) H\u00e3y l\u1eadp c\u00f4ng th\u1ee9c t\u00ednh y theo x. b) Khi \u00f4 t\u00f4 ch\u1ea1y t\u1eeb TPHCM \u0111\u1ebfn \u0110\u00e0 L\u1ea1t qu\u00e3ng \u0111\u01b0\u1eddng d\u00e0i 290km th\u00ec s\u1ed1 l\u00edt x\u0103ng trong b\u00ecnh c\u00f2n l\u1ea1i bao nhi\u00eau l\u00edt n\u1ebfu l\u00fac \u0111\u1ea7u b\u00ecnh \u0111\u1ea7y. C\u00e2u 4. (1 \u0111i\u1ec3m). M\u1ed9t n\u1ec1n nh\u00e0 h\u00ecnh nh\u1eadt c\u00f3 l\u00edch th\u01b0\u1edbc 4m v\u00e0 12m .Ng\u01b0\u1eddi ta nh\u1edd th\u1ee3 x\u00e2y d\u1ef1ng l\u00e1t h\u1ebft n\u1ec1n nh\u00e0 b\u1eb1ng lo\u1ea1i g\u1ea1ch h\u00ecnh vu\u00f4ng c\u1ea1nh 60cm . Khi l\u00e1t g\u1ea1ch n\u1ec1n, do t\u00ednh th\u1ea9m m\u0129 c\u1ee7a th\u1ee3 x\u00e2y ph\u1ea3i d\u00f9ng m\u00e1y c\u1eaft b\u1ecf m\u1ed9t ph\u1ea7n c\u1ee7a nh\u1eefng vi\u00ean g\u1ea1ch l\u00e1t cu\u1ed1i trong tr\u01b0\u1eddng h\u1ee3p vi\u00ean g\u1ea1ch \u0111\u00f3 b\u1ecb h\u01b0 v\u00e0 kh\u00f4ng s\u1eed d\u1ee5ng ph\u1ea7n c\u1eaft b\u1ecf c\u1ee7a vi\u00ean g\u1ea1ch \u0111\u00f3. Cho r\u1eb1ng hao ph\u00ed khi l\u00e1t g\u1ea1ch l\u00e0 3% tr\u00ean s\u1ed1 t\u1ed5ng g\u1ea1ch l\u00e1t n\u1ec1n v\u00e0 ph\u1ea3i \u0111\u1ec3 d\u00e0nh 5 vi\u00ean g\u1ea1ch \u0111\u1ec3 d\u1ef1 tr\u1eef sau n\u00e0y \u0111\u1ec3 thay th\u1ebf c\u00e1c vi\u00ean g\u1ea1ch b\u1ecb h\u1ecfng (n\u1ebfu c\u00f3). H\u1ecfi ng\u01b0\u1eddi ta c\u1ea7n ph\u1ea3i mua bao nhi\u00eau vi\u00ean g\u1ea1ch lo\u1ea1i n\u00f3i tr\u00ean? C\u00e2u 5. (1 \u0111i\u1ec3m). Trong \u0111\u1ee3t d\u1ecbch Covid-19, h\u1ecdc sinh hai l\u1edbp 9A v\u00e0 9B tr\u01b0\u1eddng THCS X \u1ee7ng h\u1ed9 217 chi\u1ebfc kh\u1ea9u trang cho nh\u1eefng n\u01a1i c\u00e1ch ly t\u1eadp trung. Bi\u1ebft r\u1eb1ng s\u1ed1 h\u1ecdc sinh l\u1edbp 9A nhi\u1ec1u h\u01a1n s\u1ed1 h\u1ecdc sinh l\u1edbp 9B l\u00e0 4 h\u1ecdc sinh v\u00e0 m\u1ed7i h\u1ecdc sinh l\u1edbp 9A \u1ee7ng h\u1ed9 3 chi\u1ebfc kh\u1ea9u trang, m\u1ed7i h\u1ecdc sinh l\u1edbp 9B \u1ee7ng h\u1ed9 2 chi\u1ebfc kh\u1ea9u trang. T\u00ednh s\u1ed1 h\u1ecdc sinh m\u1ed7i l\u1edbp. C\u00e2u 6. (1 \u0111i\u1ec3m). M\u1ed9t kh\u1ed1i g\u1ed7 d\u1ea1ng h\u00ecnh tr\u1ee5, b\u00e1n k\u00ednh \u0111\u01b0\u1eddng tr\u00f2n \u0111\u00e1y l\u00e0 r = 10cm , chi\u1ec1u cao g\u1ea5p hai l\u1ea7n b\u00e1n k\u00ednh (\u0111\u01a1n v\u1ecb: cm ). Ng\u01b0\u1eddi ta kho\u00e9t r\u1ed7ng hai n\u1eefa h\u00ecnh c\u1ea7u c\u00f3 k\u00edch th\u01b0\u1edbc b\u1eb1ng nhau nh\u01b0 h\u00ecnh d\u01b0\u1edbi (ph\u1ea7n ti\u1ebfp x\u00fac c\u00f3 b\u1ec1 d\u00e0y kh\u00f4ng \u0111\u00e1ng k\u1ec3). H\u00e3y t\u00ednh di\u1ec7n t\u00edch b\u1ec1 m\u1eb7t c\u1ee7a kh\u1ed1i g\u1ed7 c\u00f2n l\u1ea1i (di\u1ec7n t\u00edch c\u1ea3 trong l\u1eabn ngo\u00e0i). (L\u00e0m tr\u00f2n k\u1ebft qu\u1ea3 \u0111\u1ebfn ch\u1eef s\u1ed1 h\u00e0ng \u0111\u01a1n v\u1ecb)? Bi\u1ebft di\u1ec7n t\u00edch xung quanh h\u00ecnh tr\u1ee5 l\u00e0 Sxq = 2\uf070 rh v\u1edbi r l\u00e0 b\u00e1n k\u00ednh \u0111\u01b0\u1eddng tr\u00f2n \u0111\u00e1y, h l\u00e0 chi\u1ec1u cao. Di\u1ec7n t\u00edch h\u00ecnh c\u1ea7u c\u00f3 b\u00e1n k\u00ednh l\u00e0 r l\u00e0 S = 4\uf070 r2 . 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 7. (1 \u0111i\u1ec3m). L\u00fac 8 gi\u1edd s\u00e1ng, m\u1ed9t xe m\u00e1y \u0111i t\u1eeb t\u1ec9nh A \u0111\u1ebfn t\u1ec9nh B c\u00e1ch nhau 270km v\u1edbi v\u1eadn t\u1ed1c trung b\u00ecnh l\u00e0 40km \/ h . Sau khi xe m\u00e1y \u0111i \u0111\u01b0\u1ee3c 90 ph\u00fat th\u00ec m\u1ed9t \u00f4 t\u00f4 xu\u1ea5t ph\u00e1t \u0111i t\u1eeb B v\u1ec1 A v\u1edbi v\u1eadn t\u1ed1c trung b\u00ecnh l\u00e0 50km \/ h . H\u1ecfi 2 xe g\u1eb7p nhau tr\u00ean qu\u00e3ng \u0111\u01b0\u1eddng AB l\u00fac m\u1ea5y gi\u1edd? C\u00e2u 8. (3 \u0111i\u1ec3m). T\u1eeb \u0111i\u1ec3m A n\u1eb1m ngo\u00e0i \u0111\u01b0\u1eddng tr\u00f2n (O) (OA \uf03e 2R) . V\u1ebd hai ti\u1ebfp tuy\u1ebfn AB v\u00e0 AC v\u00e0 c\u00e1t tuy\u1ebfn ADE c\u1ee7a (O) . G\u1ecdi K trung \u0111i\u1ec3m AC , OA c\u1eaft BC t\u1ea1i H . S\u1eeda \u0111\u1ec1: T\u1eeb \u0111i\u1ec3m A n\u1eb1m ngo\u00e0i \u0111\u01b0\u1eddng tr\u00f2n (O) (OA \uf03e 2R) . V\u1ebd hai ti\u1ebfp tuy\u1ebfn AB v\u00e0 AC c\u1ee7a (O) . G\u1ecdi K trung \u0111i\u1ec3m AC , BK c\u1eaft (O) t\u1ea1i D , OA c\u1eaft BC t\u1ea1i H . a) Ch\u1ee9ng minh HK\u2225 AB v\u00e0 t\u1ee9 gi\u00e1c CHDK n\u1ed9i ti\u1ebfp. b) Tia AD c\u1eaft (O) t\u1ea1i E . Ch\u1ee9ng minh KC2 = KD.KB v\u00e0 BE\u2225 AC . c) G\u1ecdi I l\u00e0 giao \u0111i\u1ec3m c\u1ee7a BC v\u00e0 AE , tia KI c\u1eaft BE t\u1ea1i S . Ch\u1ee9ng minh BD.BK = 2HS2 . ----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 = 1 x2 v\u00e0 \u0111\u01b0\u1eddng th\u1eb3ng (D) : y = \u2212 1 x + 1 . 22 a) V\u1ebd \u0111\u1ed3 th\u1ecb (P) v\u00e0 (D) tr\u00ean c\u00f9ng h\u1ec7 tr\u1ee5c t\u1ecda \u0111\u1ed9. b) T\u00ecm t\u1ecda \u0111\u1ed9 giao \u0111i\u1ec3m c\u1ee7a (P) v\u00e0 (D) b\u1eb1ng ph\u00e9p t\u00ednh. L\u1eddi gi\u1ea3i a) V\u1ebd \u0111\u1ed3 th\u1ecb (P) v\u00e0 (D) tr\u00ean c\u00f9ng h\u1ec7 tr\u1ee5c t\u1ecda \u0111\u1ed9. BGT: x \u22124 \u22122 0 2 4 (P) : y = 1 x2 8 2 0 2 8 2 x 0 2 1 0 (D) : y = \u2212 1 x +1 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 = \u2212 1 x + 1 22 \uf0db 1 x2 + 1 x \u22121 = 0 22 \uf0db \uf0e9x = 1 \uf0ea\uf0ebx = \u22122 Thay x = 1 v\u00e0o y = 1 x2 , ta \u0111\u01b0\u1ee3c: y = 1 .12 = 1 . 2 22 Thay x = \u22122 v\u00e0o y = 1 x2 , ta \u0111\u01b0\u1ee3c: y = 1 .(\u22122)2 = 2 . 22 V\u1eady \uf0e6\uf0e7\uf0e8 1; 1 \uf0f6 , (\u22122; 2) l\u00e0 hai giao \u0111i\u1ec3m c\u1ea7n t\u00ecm. 2 \uf0f7\uf0f8 C\u00e2u 2. (1 \u0111i\u1ec3m) Cho ph\u01b0\u01a1ng tr\u00ecnh 20x2 + 5x \u2212 2020 = 0 c\u00f3 2 nghi\u1ec7m l\u00e0 x1 , x2 . Kh\u00f4ng gi\u1ea3i ph\u01b0\u01a1ng A = x1 + x2 x2 x1 ( ) ( )tr\u00ecnh, h\u00e3y t\u00ednh gi\u00e1 tr\u1ecb c\u1ee7a bi\u1ec3u th\u1ee9c 1 \u2212 x2 1 \u2212 x1 . L\u1eddi gi\u1ea3i 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 V\u00ec \uf044 = b2 \u2212 4ac = 52 \u2212 4.20.(\u22122020) = 161625 \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 = \u2212 1 \uf0ed = x1 a4 Theo \u0111\u1ecbnh l\u00ed Vi-et, ta c\u00f3: \uf0ef\uf0ef\uf0eeP c = \u2212101 a .x2 = ( ) ( )Ta c\u00f3: A = x1 + x2 x1 x2 1 \u2212 x2 1 \u2212 x1 ( )A = x1 + x2 \u2212 x2 x1 x1 + x2 ( )A = x12 + x22 \u2212 x1x2 x1 + x2 x1 + x2 2 \u2212 2x1x2 \u2212 x1x2 ( ) ( )A = x1 + x2 \uf0e6 \u2212 1 \uf0f62 \u2212 2.(\u2212101) \uf0e6 1 \uf0f6 2829 \uf0e7\uf0e8 4 \uf0f7 \uf0e7\uf0e8 4 \uf0f7\uf0f8 1616 \uf0f8 A = \u2212 \u2212 = \u2212 . \u2212101 C\u00e2u 3. (1 \u0111i\u1ec3m) M\u1ed9t \u00f4 t\u00f4 c\u00f3 b\u00ecnh ch\u1ee9a x\u0103ng ch\u1ee9a \u0111\u01b0\u1ee3c nhi\u1ec1u nh\u1ea5t l\u00e0 40 l\u00edt x\u0103ng. C\u1ee9 ch\u1ea1y 100km th\u00ec \u00f4 t\u00f4 ti\u00eau th\u1ee5 h\u1ebft 8 l\u00edt x\u0103ng. G\u1ecdi x(km) l\u00e0 qu\u00e3ng \u0111\u01b0\u1eddng \u00f4 t\u00f4 \u0111i \u0111\u01b0\u1ee3c v\u00e0 y (l\u00edt) l\u00e0 s\u1ed1 l\u00edt x\u0103ng ti\u00eau thu. a) H\u00e3y l\u1eadp c\u00f4ng th\u1ee9c t\u00ednh y theo x. b) Khi \u00f4 t\u00f4 ch\u1ea1y t\u1eeb TPHCM \u0111\u1ebfn \u0110\u00e0 L\u1ea1t qu\u00e3ng \u0111\u01b0\u1eddng d\u00e0i 290km th\u00ec s\u1ed1 l\u00edt x\u0103ng trong b\u00ecnh c\u00f2n l\u1ea1i bao nhi\u00eau l\u00edt n\u1ebfu l\u00fac \u0111\u1ea7u b\u00ecnh \u0111\u1ea7y. L\u1eddi gi\u1ea3i a) H\u00e3y l\u1eadp c\u00f4ng th\u1ee9c t\u00ednh y theo x. Theo \u0111\u1ec1 b\u00e0i, ta c\u00f3: V\u1edbi \uf0ecx = 0 \uf0de 0 = 0.a + b . (1) \uf0ed\uf0eey = 0 V\u1edbi \uf0ecx = 100 \uf0de 8 = 100.a + b .(2) \uf0ed\uf0eey =8 T\u1eeb (1) v\u00e0 (2) ta c\u00f3 h\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh: \uf0ec0a + b = 0 \uf0db \uf0ec\uf0efa = 2 \uf0ed\uf0ee100a + b = 8 \uf0ed = 25 . \uf0ef\uf0eeb 0 V\u1eady: a = 2 , b = 0 v\u00e0 y = 2 x . 25 25 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) Khi \u00f4 t\u00f4 ch\u1ea1y t\u1eeb TPHCM \u0111\u1ebfn \u0110\u00e0 L\u1ea1t qu\u00e3ng \u0111\u01b0\u1eddng d\u00e0i 290km th\u00ec s\u1ed1 l\u00edt x\u0103ng trong b\u00ecnh c\u00f2n l\u1ea1i bao nhi\u00eau l\u00edt n\u1ebfu l\u00fac \u0111\u1ea7u b\u00ecnh \u0111\u1ea7y. S\u1ed1 l\u00edt x\u0103ng trong b\u00ecnh c\u00f2n l\u1ea1i l\u00e0: 40 \u2212 2 .290 = 16,8 (l\u00edt). 25 C\u00e2u 4. (1 \u0111i\u1ec3m) M\u1ed9t n\u1ec1n nh\u00e0 h\u00ecnh nh\u1eadt c\u00f3 l\u00edch th\u01b0\u1edbc 4m v\u00e0 12m . Ng\u01b0\u1eddi ta nh\u1edd th\u1ee3 x\u00e2y d\u1ef1ng l\u00e1t h\u1ebft n\u1ec1n nh\u00e0 b\u1eb1ng lo\u1ea1i g\u1ea1ch h\u00ecnh vu\u00f4ng c\u1ea1nh 60cm . Khi l\u00e1t g\u1ea1ch n\u1ec1n, do t\u00ednh th\u1ea9m m\u0129 c\u1ee7a th\u1ee3 x\u00e2y ph\u1ea3i d\u00f9ng m\u00e1y c\u1eaft b\u1ecf m\u1ed9t ph\u1ea7n c\u1ee7a nh\u1eefng vi\u00ean g\u1ea1ch l\u00e1t cu\u1ed1i trong tr\u01b0\u1eddng h\u1ee3p vi\u00ean g\u1ea1ch \u0111\u00f3 b\u1ecb h\u01b0 v\u00e0 kh\u00f4ng s\u1eed d\u1ee5ng ph\u1ea7n c\u1eaft b\u1ecf c\u1ee7a vi\u00ean g\u1ea1ch \u0111\u00f3. Cho r\u1eb1ng hao ph\u00ed khi l\u00e1t g\u1ea1ch l\u00e0 3% tr\u00ean s\u1ed1 t\u1ed5ng g\u1ea1ch l\u00e1t n\u1ec1n v\u00e0 ph\u1ea3i \u0111\u1ec3 d\u00e0nh 5 vi\u00ean g\u1ea1ch \u0111\u1ec3 d\u1ef1 tr\u1eef sau n\u00e0y \u0111\u1ec3 thay th\u1ebf c\u00e1c vi\u00ean g\u1ea1ch b\u1ecb h\u1ecfng (n\u1ebfu c\u00f3). H\u1ecfi ng\u01b0\u1eddi ta c\u1ea7n ph\u1ea3i mua bao nhi\u00eau vi\u00ean g\u1ea1ch lo\u1ea1i n\u00f3i tr\u00ean? L\u1eddi gi\u1ea3i Di\u1ec7n t\u00edch n\u1ec1n nh\u00e0 l\u00e0: 4.12 = 48m2. Di\u1ec7n t\u00edch vi\u00ean g\u1ea1ch l\u00e0: 602 = 3600cm2 = 0,36m2. S\u1ed1 vi\u00ean g\u1ea1ch l\u00e1t n\u1ec1n nh\u00e0 ch\u01b0a t\u00ednh hao ph\u00ed l\u00e0: 48 : 0,36 \uf0bb 134 (vi\u00ean). S\u1ed1 vi\u00ean g\u1ea1ch l\u00e1t n\u1ec1n nh\u00e0 \u0111\u00e3 t\u00ednh hao ph\u00ed l\u00e0: 134.(1 + 3%) \uf0bb 139 (vi\u00ean). S\u1ed1 vi\u00ean g\u1ea1ch c\u1ea7n mua l\u00e0: 139 + 5 = 144 (vi\u00ean). C\u00e2u 5. (1 \u0111i\u1ec3m) Trong \u0111\u1ee3t d\u1ecbch Covid-19, h\u1ecdc sinh hai l\u1edbp 9A v\u00e0 9B tr\u01b0\u1eddng THCS X \u1ee7ng h\u1ed9 217 chi\u1ebfc kh\u1ea9u trang cho nh\u1eefng n\u01a1i c\u00e1ch ly t\u1eadp trung. Bi\u1ebft r\u1eb1ng s\u1ed1 h\u1ecdc sinh l\u1edbp 9A nhi\u1ec1u h\u01a1n s\u1ed1 h\u1ecdc sinh l\u1edbp 9B l\u00e0 4 h\u1ecdc sinh v\u00e0 m\u1ed7i h\u1ecdc sinh l\u1edbp 9A \u1ee7ng h\u1ed9 3 chi\u1ebfc kh\u1ea9u trang, m\u1ed7i h\u1ecdc sinh l\u1edbp 9B \u1ee7ng h\u1ed9 2 chi\u1ebfc kh\u1ea9u trang. T\u00ednh s\u1ed1 h\u1ecdc sinh m\u1ed7i l\u1edbp. L\u1eddi gi\u1ea3i G\u1ecdi x v\u00e0 y l\u1ea7n l\u01b0\u1ee3t l\u00e0 s\u1ed1 h\u1ecdc sinh l\u1edbp 9A v\u00e0 9B ( x, y \uf0ce * ) S\u1ed1 h\u1ecdc sinh l\u1edbp 9A nhi\u1ec1u h\u01a1n s\u1ed1 h\u1ecdc sinh l\u1edbp 9B l\u00e0 4 h\u1ecdc sinh n\u00ean ta c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh: x \u2212 y = 4. (1) T\u1ed5ng s\u1ed1 kh\u1ea9u trang \u1ee7ng h\u1ed9 c\u1ee7a hai l\u1edbp 9A v\u00e0 9B l\u00e0 217 chi\u1ebfc kh\u1ea9u trang n\u00ean ta c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh: 3x + 2y = 217. (2) T\u1eeb (1) v\u00e0 (2) ta c\u00f3 h\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh: \uf0ecx \u2212 y = 4 217 \uf0db \uf0ecx = 45 (n) \uf0ed\uf0ee3x + 2y = \uf0ed\uf0eey = . 41 V\u1eady s\u1ed1 h\u1ecdc sinh l\u1edbp 9A v\u00e0 9B l\u1ea7n l\u01b0\u1ee3t l\u00e0 45 v\u00e0 41 h\u1ecdc sinh. C\u00e2u 6. (1 \u0111i\u1ec3m) M\u1ed9t kh\u1ed1i g\u1ed7 d\u1ea1ng h\u00ecnh tr\u1ee5, b\u00e1n k\u00ednh \u0111\u01b0\u1eddng tr\u00f2n \u0111\u00e1y l\u00e0 r = 10cm , chi\u1ec1u cao g\u1ea5p hai l\u1ea7n b\u00e1n k\u00ednh (\u0111\u01a1n v\u1ecb: cm ). Ng\u01b0\u1eddi ta kho\u00e9t r\u1ed7ng hai n\u1eefa h\u00ecnh c\u1ea7u c\u00f3 k\u00edch th\u01b0\u1edbc b\u1eb1ng nhau nh\u01b0 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 h\u00ecnh d\u01b0\u1edbi (ph\u1ea7n ti\u1ebfp x\u00fac c\u00f3 b\u1ec1 d\u00e0y kh\u00f4ng \u0111\u00e1ng k\u1ec3). H\u00e3y t\u00ednh di\u1ec7n t\u00edch b\u1ec1 m\u1eb7t c\u1ee7a kh\u1ed1i g\u1ed7 c\u00f2n l\u1ea1i (di\u1ec7n t\u00edch c\u1ea3 trong l\u1eabn ngo\u00e0i). (L\u00e0m tr\u00f2n k\u1ebft qu\u1ea3 \u0111\u1ebfn ch\u1eef s\u1ed1 h\u00e0ng \u0111\u01a1n v\u1ecb)? Bi\u1ebft di\u1ec7n t\u00edch xung quanh h\u00ecnh tr\u1ee5 l\u00e0 Sxq = 2\uf070 rh v\u1edbi r l\u00e0 b\u00e1n k\u00ednh \u0111\u01b0\u1eddng tr\u00f2n \u0111\u00e1y, h l\u00e0 chi\u1ec1u cao. Di\u1ec7n t\u00edch h\u00ecnh c\u1ea7u c\u00f3 b\u00e1n k\u00ednh l\u00e0 r l\u00e0 S = 4\uf070 r2 . L\u1eddi gi\u1ea3i ( )( )Di\u1ec7n t\u00edch xung quanh kh\u1ed1i g\u1ed7: Sxq = 2\uf070 rh = 2\uf070 .10. 2.10 = 400\uf070 cm2 . ( )Di\u1ec7n t\u00edch h\u00ecnh c\u1ea7u l\u00e0: S = 4\uf070 r2 = 4\uf070 .102 = 400\uf070 cm2 . ( )V\u1eady di\u1ec7n t\u00edch b\u1ec1 m\u1eb7t kh\u1ed1i g\u1ed7 c\u00f2n l\u1ea1i l\u00e0: S = 400\uf070 + 400\uf070 = 800\uf070 \uf0bb 2513 cm2 . C\u00e2u 7. (1 \u0111i\u1ec3m) L\u00fac 8 gi\u1edd s\u00e1ng, m\u1ed9t xe m\u00e1y \u0111i t\u1eeb t\u1ec9nh A \u0111\u1ebfn t\u1ec9nh B c\u00e1ch nhau 270km v\u1edbi v\u1eadn t\u1ed1c trung b\u00ecnh l\u00e0 40km \/ h . Sau khi xe m\u00e1y \u0111i \u0111\u01b0\u1ee3c 90 ph\u00fat th\u00ec m\u1ed9t \u00f4 t\u00f4 xu\u1ea5t ph\u00e1t \u0111i t\u1eeb B v\u1ec1 A v\u1edbi v\u1eadn t\u1ed1c trung b\u00ecnh l\u00e0 50km \/ h . H\u1ecfi 2 xe g\u1eb7p nhau tr\u00ean qu\u00e3ng \u0111\u01b0\u1eddng AB l\u00fac m\u1ea5y gi\u1edd? L\u1eddi gi\u1ea3i G\u1ecdi C l\u00e0 v\u1ecb tr\u00ed xe m\u00e1y sau khi \u0111i \u0111\u01b0\u1ee3c 90 ph\u00fat; D l\u00e0 v\u1ecb tr\u00ed hai xe g\u1eb7p nhau. G\u1ecdi x l\u00e0 th\u1eddi gian \u0111i \u0111\u01b0\u1ee3c c\u1ee7a hai xe (t\u00ednh t\u1eeb l\u00fac xe m\u00e1y \u0111\u00e3 \u0111i \u0111\u01b0\u1ee3c 90 ph\u00fat) \u0111\u1ebfn l\u00fac hai xe g\u1eb7p nhau. ( x \uf03e 0,h) . Qu\u00e3ng \u0111\u01b0\u1eddng xe m\u00e1y \u0111i t\u1eeb v\u1ecb tr\u00ed A \u0111\u1ebfn C sau 90 ph\u00fat l\u00e0: SA\u2192C = 40. 3 = 60km. 2 ( )Qu\u00e3ng \u0111\u01b0\u1eddng xe m\u00e1y \u0111i t\u1eeb v\u1ecb tr\u00ed C \u0111\u1ebfn D l\u00e0: SC\u2192D = 40.x km . ( )Qu\u00e3ng \u0111\u01b0\u1eddng \u00f4 t\u00f4 \u0111i t\u1eeb v\u1ecb tr\u00ed B \u0111\u1ebfn D l\u00e0: SB\u2192D = 50.x km . Khi hai xe g\u1eb7p nhau ta c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh: 40x + 50x = 270 \u2212 60 \uf0db x = 7 (n). 3 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 \u0110\u1ed5i 7 gi\u1edd = 2 gi\u1edd 20 ph\u00fat. 3 V\u1eady hai xe g\u1eb7p nhau l\u00fac 11 gi\u1edd 50 ph\u00fat. C\u00e2u 8. (3 \u0111i\u1ec3m) T\u1eeb \u0111i\u1ec3m A n\u1eb1m ngo\u00e0i \u0111\u01b0\u1eddng tr\u00f2n (O) (OA \uf03e 2R) . V\u1ebd hai ti\u1ebfp tuy\u1ebfn AB v\u00e0 AC v\u00e0 c\u00e1t tuy\u1ebfn ADE c\u1ee7a (O) . G\u1ecdi K trung \u0111i\u1ec3m AC , OA c\u1eaft BC t\u1ea1i H . S\u1eeda \u0111\u1ec1: T\u1eeb \u0111i\u1ec3m A n\u1eb1m ngo\u00e0i \u0111\u01b0\u1eddng tr\u00f2n (O) (OA \uf03e 2R) . V\u1ebd hai ti\u1ebfp tuy\u1ebfn AB v\u00e0 AC c\u1ee7a (O) . G\u1ecdi K trung \u0111i\u1ec3m AC , BK c\u1eaft (O) t\u1ea1i D , OA c\u1eaft BC t\u1ea1i H . a. Ch\u1ee9ng minh HK\u2225 AB v\u00e0 t\u1ee9 gi\u00e1c CHDK n\u1ed9i ti\u1ebfp. ( )b. Tia AD c\u1eaft O t\u1ea1i E . Ch\u1ee9ng minh KC2 = KD.KB v\u00e0 BE\u2225 AC . c. G\u1ecdi I l\u00e0 giao \u0111i\u1ec3m c\u1ee7a BC v\u00e0 AE , tia KI c\u1eaft BE t\u1ea1i S . Ch\u1ee9ng minh BD.BK = 2HS2 L\u1eddi gi\u1ea3i a) Ch\u1ee9ng minh HK\/\/AB v\u00e0 t\u1ee9 gi\u00e1c CHDK n\u1ed9i ti\u1ebfp. 7 Ta c\u00f3: OB = OC (b\u00e1n k\u00ednh); AB = AC (T\u00ednh ch\u1ea5t hai ti\u1ebfp tuy\u1ebfn c\u1eaft nhau) \uf0de OA l\u00e0 \u0111\u01b0\u1eddng trung tr\u1ef1c c\u1ee7a BC \uf0de H l\u00e0 trung \u0111i\u1ec3m c\u1ee7a BC . X\u00e9t tam gi\u00e1c ABC c\u00f3 HK l\u00e0 \u0111\u01b0\u1eddng trung b\u00ecnh \uf0de HK\/\/AB . Ta c\u00f3: HKD = ABD ( hai g\u00f3c so le trong) BCD = ABD (gnt v\u00e0 g\u00f3c t\u1ea1o b\u1edfi tia tt v\u00e0 d\u00e2y cung c\u00f9ng ch\u1eafn BD ) \uf0de HKD = BCD 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 \uf0de HKD = HCD X\u00e9t t\u1ee9 gi\u00e1c CHDK , c\u00f3: HKD = HCD (cmt) \uf0de T\u1ee9 gi\u00e1c CHDK n\u1ed9i ti\u1ebfp v\u00ec c\u00f3 hai \u0111\u1ec9nh k\u1ec1 c\u00f9ng nh\u00ecn m\u1ed9t c\u1ea1nh d\u01b0\u1edbi hai g\u00f3c b\u1eb1ng nhau. ( )b) Tia AD c\u1eaft O t\u1ea1i E . Ch\u1ee9ng minh KC2 = KD.KB v\u00e0 BE\/\/AC . Ta c\u00f3: KCD = DBC (gnt v\u00e0 g\u00f3c t\u1ea1o b\u1edfi tia tt v\u00e0 d\u00e2y cung c\u00f9ng ch\u1eafn DC ) \uf0de KCD = KBC X\u00e9t \uf044KCD v\u00e0 \uf044KBC c\u00f3: \uf0ec\uf0efK chung \uf0ed \uf0ef\uf0eeKCD = KBC (cmt) \uf0de \uf044KCD\u223d\uf044KBC (g \u2013 g) \uf0de KC = KD (t\u1ec9 l\u1ec7 c\u1ea1nh t\u01b0\u01a1ng \u1ee9ng) KB KC \uf0de KC2 = KD.KB Ta c\u00f3: K trung \u0111i\u1ec3m AC n\u00ean KB2 = KC2 = KD.KB D\u1ec5 d\u00e0ng ch\u1ee9ng minh \u0111\u01b0\u1ee3c \uf044KAD\u223d\uf044KBA (c \u2013 g- c) Suy ra: KAD = KBA (hai g\u00f3c t\u01b0\u01a1ng \u1ee9ng) Suy ra: KAD = DBA M\u00e0 BED = DBA (gnt v\u00e0 g\u00f3c t\u1ea1o b\u1edfi tia tt v\u00e0 d\u00e2y cung c\u00f9ng ch\u1eafn BD ) Suy ra: BED = KAD Suy ra: BE\u2225 AC (hai g\u00f3c so le trong). c) G\u1ecdi I l\u00e0 giao \u0111i\u1ec3m c\u1ee7a BC v\u00e0 AE , tia KI c\u1eaft BE t\u1ea1i S . Ch\u1ee9ng minh BD.BK = 2HS2 D\u1ec5 d\u00e0ng Cm: \uf044ISB\u223d\uf044IKC (g \u2013 g) \uf0de IS = SB (1) IK KC D\u1ec5 d\u00e0ng Cm: \uf044ISE\u223d\uf044IKA (g \u2013 g) \uf0de IS = SE (2) IK KA Suy ra: SB = SE KC KA Suy ra: S trung \u0111i\u1ec3m BE . Suy ra: OS \u22a5 EB (li\u00ean h\u1ec7 \u0111\u01b0\u1eddng k\u00ednh v\u00e0 d\u00e2y cung) Suy ra: C,S,O th\u1eb3ng h\u00e0ng (do OC \u22a5 AC; EB\u2225 AC ). Suy ra: \uf044CSB vu\u00f4ng t\u1ea1i S c\u00f3 SH l\u00e0 \u0111\u01b0\u1eddng trung tuy\u1ebfn T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N 8"]