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 Suy ra: HS = HB = HC (\u0111\u01b0\u1eddng trung tuy\u1ebfn \u1ee9ng v\u1edbi c\u1ea1nh huy\u1ec1n) D\u1ec5 d\u00e0ng cm: \uf044BHD\u223d\uf044BKC (g \u2013 g) \uf0de BH = BD \uf0de BH.BC = BK.BD BK BC Suy ra: HS.2HS = BK.BD\uf0de 2HS2 = BK.BD. ----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\u1ede GD&\u0110T TP H\u1ed2 CH\u00cd MINH \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 PH\u00d2NG GD&\u0110T QU\u1eacN 10 N\u0102M H\u1eccC: 2023 - 2024 M\u00c3 \u0110\u1ec0: Qu\u1eadn 10 - 1 M\u00d4N: TO\u00c1N 9 \u0110\u1ec0 THAM KH\u1ea2O \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) B\u00e0i 1. (1,5 \u0111i\u1ec3m). Cho y = 1 x2 c\u00f3 \u0111\u1ed3 th\u1ecb (P) v\u00e0 h\u00e0m s\u1ed1 y = \u22121 x + 2 c\u00f3 \u0111\u1ed3 th\u1ecb l\u00e0 \u0111\u01b0\u1eddng th\u1eb3ng (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 t\u00ednh. B\u00e0i 2. (1,0 \u0111i\u1ec3m). Cho ph\u01b0\u01a1ng tr\u00ecnh 3x2 + 5x \u2212 1 = 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 = x12 + x22 \u2212 x1x2 B\u00e0i 3. (1,0 \u0111i\u1ec3m). Nh\u00e2n d\u1ecbp T\u1ebft Nguy\u00ean \u0111\u00e1n, m\u1ed9t c\u1eeda h\u00e0ng b\u00e1n \u0111\u1ed3 d\u00f9ng th\u1ec3 thao c\u00f3 ch\u01b0\u01a1ng tr\u00ecnh gi\u1ea3m gi\u00e1 to\u00e0n b\u1ed9 s\u1ea3n ph\u1ea9m trong c\u1eeda h\u00e0ng nh\u01b0 sau: M\u1ed9t \u00e1o th\u1ec3 thao gi\u1ea3m 10% , m\u1ed9t qu\u1ea7n th\u1ec3 thao gi\u1ea3m 20% , m\u1ed9t \u0111\u00f4i gi\u00e0y th\u1ec3 thao gi\u1ea3m 30% . \u0110\u1eb7c bi\u1ec7t n\u1ebfu mua \u0111\u1ee7 3 s\u1ea3n ph\u1ea9m bao g\u1ed3m 1qu\u1ea7n, 1 \u00e1o, 1\u0111\u00f4i gi\u00e0y th\u00ec s\u1ebd \u0111\u01b0\u1ee3c gi\u1ea3m ti\u1ebfp 5% tr\u00ean gi\u00e1 \u0111\u00e3 gi\u1ea3m. B\u1ea1n An v\u00e0o c\u1eeda h\u00e0ng n\u00e0y v\u00e0 mua 3 \u00e1o v\u1edbi gi\u00e1 300 000 \u0111\u1ed3ng\/c\u00e1i, 2 qu\u1ea7n v\u1edbi gi\u00e1 250 000 \u0111\u1ed3ng\/c\u00e1i v\u00e0 1 \u0111\u00f4i gi\u00e0y v\u1edbi gi\u00e1 1 000 000 \u0111\u1ed3ng\/\u0111\u00f4i (gi\u00e1 tr\u00ean l\u00e0 gi\u00e1 ch\u01b0a gi\u1ea3m). V\u1eady s\u1ed1 ti\u1ec1n b\u1ea1n An ph\u1ea3i tr\u1ea3 l\u00e0 bao nhi\u00eau? B\u00e0i 4. (0,75 \u0111i\u1ec3m). M\u1ed9t v\u1eadt r\u01a1i t\u1ef1 do t\u1eeb \u0111\u1ed9 cao 150m so v\u1edbi m\u1eb7t \u0111\u1ea5t. B\u1ecf qua s\u1ee9c c\u1ea3n c\u1ee7a kh\u00f4ng kh\u00ed, qu\u00e3ng \u0111\u01b0\u1eddng chuy\u1ec3n \u0111\u1ed9ng s (m\u00e9t) c\u1ee7a v\u1eadt r\u01a1i sau th\u1eddi gian t (gi\u00e2y) \u0111\u01b0\u1ee3c bi\u1ec3u di\u1ec5n g\u1ea7n \u0111\u00fang b\u1edfi c\u00f4ng th\u1ee9c: s = 5t2 . a) Sau 3 gi\u00e2y (t\u00ednh t\u1eeb l\u00fac b\u1eaft \u0111\u1ea7u r\u01a1i) v\u1eadt n\u00e0y c\u00e1ch m\u1eb7t \u0111\u1ea5t bao nhi\u00eau m\u00e9t? b) T\u00ednh qu\u00e3ng \u0111\u01b0\u1eddng \u0111i \u0111\u01b0\u1ee3c c\u1ee7a v\u1eadt \u0111\u00f3 trong gi\u00e2y th\u1ee9 4 . B\u00e0i 5. (1,0 \u0111i\u1ec3m). C\u00f2n m\u1ed9t tu\u1ea7n n\u1eefa s\u1ebd \u0111\u1ebfn ng\u00e0y 20 \/ 11 , c\u00e1c b\u1ea1n h\u1ecdc sinh l\u1edbp 9A \u0111\u0103ng k\u00ed thi \u0111ua hoa \u0111i\u1ec3m 10 v\u1edbi mong mu\u1ed1n \u0111\u1ea1t th\u1eadt nhi\u1ec1u \u0111i\u1ec3m 10 \u0111\u1ec3 t\u1eb7ng th\u1ea7y c\u00f4 gi\u00e1o. \u0110\u1ebfn ng\u00e0y 19 \/ 11 , l\u1edbp tr\u01b0\u1edfng t\u1ed5ng k\u1ebft s\u1ed1 \u0111i\u1ec3m 10 c\u1ee7a c\u00e1c b\u1ea1n trong l\u1edbp v\u00e0 \u0111\u01b0\u1ee3c nh\u01b0 sau: \u2022 Kh\u00f4ng c\u00f3 b\u1ea1n n\u00e0o trong l\u1edbp kh\u00f4ng c\u00f3 \u0111i\u1ec3m 10 trong tu\u1ea7n v\u1eeba qua. \u2022 C\u00f3 20 b\u1ea1n c\u00f3 \u00edt nh\u1ea5t 2 \u0111i\u1ec3m 10 . \u2022 C\u00f3 10 b\u1ea1n c\u00f3 \u00edt nh\u1ea5t 3 \u0111i\u1ec3m 10 . \u2022 C\u00f3 5 b\u1ea1n c\u00f3 \u00edt nh\u1ea5t 4 \u0111i\u1ec3m 10 . \u2022 Kh\u00f4ng c\u00f3 ai c\u00f3 nhi\u1ec1u h\u01a1n 4 \u0111i\u1ec3m 10 . H\u1ecfi l\u1edbp 9A c\u00f3 bao nhi\u00eau \u0111i\u1ec3m 10 tu\u1ea7n v\u1eeba qua? Bi\u1ebft r\u1eb1ng l\u1edbp 9A c\u00f3 35 h\u1ecdc sinh. 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 B\u00e0i 6. (1,0 \u0111i\u1ec3m). M\u1ed9t x\u00f4 \u0111\u1ef1ng n\u01b0\u1edbc c\u00f3 d\u1ea1ng h\u00ecnh n\u00f3n c\u1ee5t (c\u00f3 c\u00e1c k\u00edch th\u01b0\u1edbc nh\u01b0 h\u00ecnh). \u0110\u00e1y x\u00f4 c\u00f3 \u0111\u01b0\u1eddng k\u00ednh l\u00e0 20cm , mi\u1ec7ng x\u00f4 l\u00e0 \u0111\u00e1y l\u1edbn c\u1ee7a h\u00ecnh n\u00f3n c\u1ee5t c\u00f3 \u0111\u01b0\u1eddng k\u00ednh 30cm v\u00e0 chi\u1ec1u cao c\u1ee7a x\u00f4 l\u00e0 22cm . a) X\u00f4 c\u00f3 th\u1ec3 ch\u1ee9a t\u1ed1i \u0111a bao nhi\u00eau l\u00edt n\u01b0\u1edbc? Bi\u1ebft r\u1eb1ng th\u1ec3 t\u00edch c\u1ee7a h\u00ecnh n\u00f3n c\u1ee5t c\u00f3 R , r , h ( )l\u1ea7n l\u01b0\u1ee3t l\u00e0 b\u00e1n k\u00ednh \u0111\u00e1y l\u1edbn, b\u00e1n k\u00ednh \u0111\u00e1y nh\u1ecf v\u00e0 chi\u1ec1u cao l\u00e0: V = 1 \uf070 h R2 + Rr + r2 ( 3 k\u1ebft qu\u1ea3 l\u00e0m tr\u00f2n \u0111\u1ebfn h\u00e0ng \u0111\u01a1n v\u1ecb) b) B\u00e1c N\u0103m d\u00f9ng hai x\u00f4 n\u01b0\u1edbc \u0111\u1ec3 l\u1ea5y n\u01b0\u1edbc t\u1eeb m\u1ed9t h\u1ed3 \u0111\u1ec3 s\u1eed d\u1ee5ng trong sinh ho\u1ea1t v\u00e0 tr\u1ed3ng tr\u1ecdt. Gia \u0111\u00ecnh b\u00e1c s\u1eed d\u1ee5ng trung b\u00ecnh m\u1ed7i ng\u00e0y 150 l\u00edt n\u01b0\u1edbc. H\u1ecfi b\u00e1c N\u0103m c\u1ea7n ph\u1ea3i l\u1ea5y \u00edt nh\u1ea5t bao nhi\u00eau l\u1ea7n m\u1ed7i ng\u00e0y (m\u1ed7i l\u1ea7n x\u00e1ch 2 x\u00f4) \u0111\u1ec3 ph\u1ee5c v\u1ee5 cho sinh ho\u1ea1t v\u00e0 tr\u1ed3ng tr\u1ecdt, bi\u1ebft r\u1eb1ng m\u1ed7i l\u1ea7n x\u00e1ch n\u01b0\u1edbc v\u1ec1 th\u00ec l\u01b0\u1ee3ng n\u01b0\u1edbc b\u1ecb hao h\u1ee5t kho\u1ea3ng 5% . B\u00e0i 7. (1,0 \u0111i\u1ec3m)T\u1ed5ng s\u1ed1 h\u1ecdc sinh c\u1ee7a l\u1edbp 9A v\u00e0 9B v\u00e0o \u0111\u1ea7u n\u0103m h\u1ecdc l\u00e0 90 h\u1ecdc sinh. \u0110\u1ebfn \u0111\u1ea7u h\u1ecdc k\u00ec II , l\u1edbp 9A c\u00f3 2 h\u1ecdc sinh \u0111i du h\u1ecdc v\u00e0 4 h\u1ecdc sinh chuy\u1ec3n qua l\u1edbp 9B n\u00ean l\u00fac n\u00e0y s\u1ed1 h\u1ecdc B\u00e0i 8. sinh l\u1edbp 9A ch\u1ec9 b\u1eb1ng 5 s\u1ed1 h\u1ecdc sinh l\u1edbp 9B . T\u00ednh s\u1ed1 h\u1ecdc sinh \u0111\u1ea7u n\u0103m c\u1ee7a l\u1edbp 9A v\u00e0 9B . 6 (3,0 \u0111i\u1ec3m) Cho \uf044ABC nh\u1ecdn n\u1ed9i ti\u1ebfp \u0111\u01b0\u1eddng tr\u00f2n (O) c\u00f3 c\u00e1c \u0111\u01b0\u1eddng cao AK, BD, CE c\u1eaft nhau t\u1ea1i H , tia AK c\u1eaft (O) t\u1ea1i Q . G\u1ecdi N l\u00e0 trung \u0111i\u1ec3m c\u1ee7a BC , F l\u00e0 trung \u0111i\u1ec3m AH . K\u1ebb \u0111\u01b0\u1eddng k\u00ednh AG c\u1ee7a (O) , \u0111\u01b0\u1eddng th\u1eb3ng qua Q song song v\u1edbi ED c\u1eaft (O) t\u1ea1i giao \u0111i\u1ec3m th\u1ee9 2 l\u00e0 T ( T kh\u00e1c Q ). G\u1ecdi J l\u00e0 giao \u0111i\u1ec3m c\u1ee7a NF v\u00e0 ED . a) Ch\u1ee9ng minh BEDC v\u00e0 AEHD l\u00e0 c\u00e1c t\u1ee9 gi\u00e1c n\u1ed9i ti\u1ebfp. b) Ch\u1ee9ng minh: FD \u22a5 ND . T\u1eeb \u0111\u00f3 suy ra ND2 = NJ.NF c) \u0110\u01b0\u1eddng tr\u00f2n \u0111\u01b0\u1eddng k\u00ednh AH c\u1eaft (O) t\u1ea1i giao \u0111i\u1ec3m th\u1ee9 hai l\u00e0 M ( M kh\u00e1c A ). Ch\u1ee9ng minh: ND2 = NH.NM v\u00e0 M, J, T th\u1eb3ng h\u00e0ng. ----H\u1ebeT--- T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N 2","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH H\u01af\u1edaNG D\u1eaaN GI\u1ea2I B\u00e0i 1. (1,5 \u0111i\u1ec3m) Cho y = 1 x2 c\u00f3 \u0111\u1ed3 th\u1ecb ( P) v\u00e0 h\u00e0m s\u1ed1 y = \u22121 x + 2 c\u00f3 \u0111\u1ed3 th\u1ecb l\u00e0 \u0111\u01b0\u1eddng th\u1eb3ng 42 (d) 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\u01a1\u0300i 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 1 014 4 x 02 y = \u22121 x + 2 2 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 = \u22121 x + 2 42 \uf0db 1 x2 + 1 x \u2212 2 = 0 42 \uf0db \uf0e9 x = 2 \uf0ea x = \u22124 \uf0eb Thay x = 2 v\u00e0o y = 1 x2 , ta \u0111\u01b0\u1ee3c: y = 1 .22 = 1 44 Thay x = \u2212 4 v\u00e0o y = 1 x2 , ta \u0111\u01b0\u1ee3c: y = 1 .(\u22124)2 = 4 44 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\u1eady (2; 1) , (\u2212 4; 4) l\u00e0 hai giao \u0111i\u1ec3m c\u1ea7n t\u00ecm. B\u00e0i 2. (1,0 \u0111i\u1ec3m). Cho ph\u01b0\u01a1ng tr\u00ecnh 3x2 + 5x \u22121 = 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 = x12 + x22 \u2212 x1x2 L\u01a1\u0300i gi\u1ea3i V\u00ec \uf044 = b2 \u2212 4ac = 52 \u2212 4.3.(\u22121) = 37 \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 5 \uf0ed a 3 Theo \u0111\u1ecbnh l\u00ed Vi-et, ta c\u00f3: \uf0ef\uf0ef\uf0eeP = c 1 x1.x2 = a = \u2212 3 Ta c\u00f3: A = x12 + x22 \u2212 x1x2 A = x12 + x22 \u2212 x1x2 A = ( x1 + x2 )2 \u2212 2x1x2 \u2212 x1x2 A = ( x1 + x2 )2 \u2212 3x1x2 A = \uf0e6 \u2212 5 \uf0f62 \u2212 3\uf0e6\uf0e7\uf0e8 \u2212 1 \uf0f6 \uf0e7\uf0e8 3 \uf0f7\uf0f8 3 \uf0f7\uf0f8 A = 34 9 B\u00e0i 3. (1,0 \u0111i\u1ec3m). Nh\u00e2n d\u1ecbp T\u1ebft Nguy\u00ean \u0111\u00e1n, m\u1ed9t c\u1eeda h\u00e0ng b\u00e1n \u0111\u1ed3 d\u00f9ng th\u1ec3 thao c\u00f3 ch\u01b0\u01a1ng tr\u00ecnh gi\u1ea3m gi\u00e1 to\u00e0n b\u1ed9 s\u1ea3n ph\u1ea9m trong c\u1eeda h\u00e0ng nh\u01b0 sau: M\u1ed9t \u00e1o th\u1ec3 thao gi\u1ea3m 10% , m\u1ed9t qu\u1ea7n th\u1ec3 thao gi\u1ea3m 20% , m\u1ed9t \u0111\u00f4i gi\u00e0y th\u1ec3 thao gi\u1ea3m 30% . \u0110\u1eb7c bi\u1ec7t n\u1ebfu mua \u0111\u1ee7 3 s\u1ea3n ph\u1ea9m bao g\u1ed3m 1qu\u1ea7n, 1 \u00e1o, 1\u0111\u00f4i gi\u00e0y th\u00ec s\u1ebd \u0111\u01b0\u1ee3c gi\u1ea3m ti\u1ebfp 5% tr\u00ean gi\u00e1 \u0111\u00e3 gi\u1ea3m. B\u1ea1n An v\u00e0o c\u1eeda h\u00e0ng n\u00e0y v\u00e0 mua 3 \u00e1o v\u1edbi gi\u00e1 300 000 \u0111\u1ed3ng\/c\u00e1i, 2 qu\u1ea7n v\u1edbi gi\u00e1 250 000\u0111\u1ed3ng\/c\u00e1i v\u00e0 1 \u0111\u00f4i gi\u00e0y v\u1edbi gi\u00e1 1 000 000 \u0111\u1ed3ng\/\u0111\u00f4i (gi\u00e1 tr\u00ean l\u00e0 gi\u00e1 ch\u01b0a gi\u1ea3m). V\u1eady s\u1ed1 ti\u1ec1n b\u1ea1n An ph\u1ea3i tr\u1ea3 l\u00e0 bao nhi\u00eau? L\u01a1\u0300i gi\u1ea3i Gi\u00e1 c\u1ee7a 1\u00e1o th\u1ec3 thao sau khi gi\u1ea3m 10% l\u00e0 : 300 000.(100% \u221210%) = 270 000 (\u0111\u1ed3ng) Gi\u00e1 c\u1ee7a 1qu\u1ea7n th\u1ec3 thao sau khi gi\u1ea3m 20% l\u00e0 : 250 000.(100% \u2212 20%) = 200 000 (\u0111\u1ed3ng) Gi\u00e1 c\u1ee7a 1 \u0111\u00f4i gi\u00e0y th\u1ec3 thao sau khi gi\u1ea3m 30% l\u00e0 : 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 1 000 000.(100% \u2212 30%) = 700 000 (\u0111\u1ed3ng) Gi\u00e1 c\u1ee7a 1 qu\u1ea7n, 1 \u00e1o, 1\u0111\u00f4i gi\u00e0y sau khi gi\u1ea3m ti\u1ebfp 5% l\u00e0: (270 000 + 200 000 + 700 000).(100% \u2212 5%) = 1 111 500 (\u0111\u1ed3ng) V\u1eady s\u1ed1 ti\u1ec1n b\u1ea1n An ph\u1ea3i tr\u1ea3 l\u00e0: 1 111 500 + 2 . 270 000 + 200 000 = 1 851 500(\u0111\u1ed3ng) B\u00e0i 4. (0,75 \u0111i\u1ec3m). M\u1ed9t v\u1eadt r\u01a1i t\u1ef1 do t\u1eeb \u0111\u1ed9 cao 150m so v\u1edbi m\u1eb7t \u0111\u1ea5t. B\u1ecf qua s\u1ee9c c\u1ea3n c\u1ee7a kh\u00f4ng kh\u00ed, qu\u00e3ng \u0111\u01b0\u1eddng chuy\u1ec3n \u0111\u1ed9ng s (m\u00e9t) c\u1ee7a v\u1eadt r\u01a1i sau th\u1eddi gian t (gi\u00e2y) \u0111\u01b0\u1ee3c bi\u1ec3u di\u1ec5n g\u1ea7n \u0111\u00fang b\u1edfi c\u00f4ng th\u1ee9c: s = 5t2 . a) Sau 3 gi\u00e2y (t\u00ednh t\u1eeb l\u00fac b\u1eaft \u0111\u1ea7u r\u01a1i) v\u1eadt n\u00e0y c\u00e1ch m\u1eb7t \u0111\u1ea5t bao nhi\u00eau m\u00e9t? b) T\u00ednh qu\u00e3ng \u0111\u01b0\u1eddng \u0111i \u0111\u01b0\u1ee3c c\u1ee7a v\u1eadt \u0111\u00f3 trong gi\u00e2y th\u1ee9 4 . L\u01a1\u0300i gi\u1ea3i a) Thay t = 3 v\u00e0o s = 5t2 ta c\u00f3: s = 5.32 \uf0db s = 45 V\u1eady sau 3 gi\u00e2y (t\u00ednh t\u1eeb l\u00fac b\u1eaft \u0111\u1ea7u r\u01a1i) v\u1eadt n\u00e0y c\u00e1ch m\u1eb7t \u0111\u1ea5t: 150 \u2212 45 = 105(m) b) Thay t = 4v\u00e0o s = 5t2 ta c\u00f3: s = 5.42 \uf0db s = 80 V\u1eady qu\u00e3ng \u0111\u01b0\u1eddng \u0111i \u0111\u01b0\u1ee3c c\u1ee7a v\u1eadt \u0111\u00f3 trong gi\u00e2y th\u1ee9 4 l\u00e0 80m B\u00e0i 5. (1,0 \u0111i\u1ec3m). C\u00f2n m\u1ed9t tu\u1ea7n n\u1eefa s\u1ebd \u0111\u1ebfn ng\u00e0y 20 \/ 11, c\u00e1c b\u1ea1n h\u1ecdc sinh l\u1edbp 9A \u0111\u0103ng k\u00ed thi \u0111ua hoa \u0111i\u1ec3m 10 v\u1edbi mong mu\u1ed1n \u0111\u1ea1t th\u1eadt nhi\u1ec1u \u0111i\u1ec3m 10 \u0111\u1ec3 t\u1eb7ng th\u1ea7y c\u00f4 gi\u00e1o. \u0110\u1ebfn ng\u00e0y 19 \/ 11 , l\u1edbp tr\u01b0\u1edfng t\u1ed5ng k\u1ebft s\u1ed1 \u0111i\u1ec3m 10 c\u1ee7a c\u00e1c b\u1ea1n trong l\u1edbp v\u00e0 \u0111\u01b0\u1ee3c nh\u01b0 sau: \u2022 Kh\u00f4ng c\u00f3 b\u1ea1n n\u00e0o trong l\u1edbp kh\u00f4ng c\u00f3 \u0111i\u1ec3m 10 trong tu\u1ea7n v\u1eeba qua. \u2022 C\u00f3 20 b\u1ea1n c\u00f3 \u00edt nh\u1ea5t 2 \u0111i\u1ec3m 10 . \u2022 C\u00f3 10 b\u1ea1n c\u00f3 \u00edt nh\u1ea5t 3 \u0111i\u1ec3m 10 . \u2022 C\u00f3 5 b\u1ea1n c\u00f3 \u00edt nh\u1ea5t 4 \u0111i\u1ec3m 10 . \u2022 Kh\u00f4ng c\u00f3 ai c\u00f3 nhi\u1ec1u h\u01a1n 4 \u0111i\u1ec3m 10 . H\u1ecfi l\u1edbp 9A c\u00f3 bao nhi\u00eau \u0111i\u1ec3m 10 tu\u1ea7n v\u1eeba qua? Bi\u1ebft r\u1eb1ng l\u1edbp 9A c\u00f3 35 h\u1ecdc sinh. L\u01a1\u0300i gi\u1ea3i V\u00ec kh\u00f4ng c\u00f3 b\u1ea1n n\u00e0o trong l\u1edbp kh\u00f4ng c\u00f3 \u0111i\u1ec3m 10 n\u00ean s\u1ed1 b\u1ea1n \u0111\u01b0\u1ee3c 1 \u0111i\u1ec3m 10 l\u00e0: 35 \u2212 20 = 15 (ba\u00efn) 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 S\u1ed1 b\u1ea1n \u0111\u01b0\u1ee3c 2 \u0111i\u1ec3m 10 l\u00e0: 20 \u221210 =10 (ba\u00efn) S\u1ed1 b\u1ea1n \u0111\u01b0\u1ee3c 3 \u0111i\u1ec3m 10 l\u00e0: 10 \u2212 5 = 5 (ba\u00efn) Do kh\u00f4ng c\u00f3 b\u1ea1n n\u00e0o \u0111\u01b0\u1ee3c nhi\u1ec1u h\u01a1n 4 \u0111i\u1ec3m 10 n\u00ean s\u1ed1 b\u1ea1n \u0111\u01b0\u1ee3c 4 \u0111i\u1ec3m 10 l\u00e0: 5 b\u1ea1n V\u1eady s\u1ed1 \u0111i\u1ec3m 10 trong tu\u1ea7n v\u1eeba qua c\u1ee7a l\u1edbp l\u00e0: 15 + 10.2 + 5.3 + 5.4 = 70(\u00f1ie\u00e5m) B\u00e0i 6. (1,0 \u0111i\u1ec3m). M\u1ed9t x\u00f4 \u0111\u1ef1ng n\u01b0\u1edbc c\u00f3 d\u1ea1ng h\u00ecnh n\u00f3n c\u1ee5t (c\u00f3 c\u00e1c k\u00edch th\u01b0\u1edbc nh\u01b0 h\u00ecnh). \u0110\u00e1y x\u00f4 c\u00f3 \u0111\u01b0\u1eddng k\u00ednh l\u00e0 20cm , mi\u1ec7ng x\u00f4 l\u00e0 \u0111\u00e1y l\u1edbn c\u1ee7a h\u00ecnh n\u00f3n c\u1ee5t c\u00f3 \u0111\u01b0\u1eddng k\u00ednh 30cm v\u00e0 chi\u1ec1u cao c\u1ee7a x\u00f4 l\u00e0 22cm . a) X\u00f4 c\u00f3 th\u1ec3 ch\u1ee9a t\u1ed1i \u0111a bao nhi\u00eau l\u00edt n\u01b0\u1edbc? Bi\u1ebft r\u1eb1ng th\u1ec3 t\u00edch c\u1ee7a h\u00ecnh n\u00f3n c\u1ee5t c\u00f3 R , r , h l\u1ea7n l\u01b0\u1ee3t l\u00e0 b\u00e1n k\u00ednh \u0111\u00e1y l\u1edbn, b\u00e1n k\u00ednh \u0111\u00e1y nh\u1ecf v\u00e0 chi\u1ec1u cao l\u00e0: ( )V = 1 \uf070 h R2 + Rr + r2 ( k\u1ebft qu\u1ea3 l\u00e0m tr\u00f2n \u0111\u1ebfn h\u00e0ng \u0111\u01a1n 3 v\u1ecb) b) B\u00e1c N\u0103m d\u00f9ng hai x\u00f4 n\u01b0\u1edbc \u0111\u1ec3 l\u1ea5y n\u01b0\u1edbc t\u1eeb m\u1ed9t h\u1ed3 \u0111\u1ec3 s\u1eed d\u1ee5ng trong sinh ho\u1ea1t v\u00e0 tr\u1ed3ng tr\u1ecdt. Gia \u0111\u00ecnh b\u00e1c s\u1eed d\u1ee5ng trung b\u00ecnh m\u1ed7i ng\u00e0y 150 l\u00edt n\u01b0\u1edbc. H\u1ecfi b\u00e1c N\u0103m c\u1ea7n ph\u1ea3i l\u1ea5y \u00edt nh\u1ea5t bao nhi\u00eau l\u1ea7n m\u1ed7i ng\u00e0y (m\u1ed7i l\u1ea7n x\u00e1ch 2 x\u00f4) \u0111\u1ec3 ph\u1ee5c v\u1ee5 cho sinh ho\u1ea1t v\u00e0 tr\u1ed3ng tr\u1ecdt, bi\u1ebft r\u1eb1ng m\u1ed7i l\u1ea7n x\u00e1ch n\u01b0\u1edbc v\u1ec1 th\u00ec l\u01b0\u1ee3ng n\u01b0\u1edbc b\u1ecb hao h\u1ee5t kho\u1ea3ng 5% . L\u01a1\u0300i gi\u1ea3i a) B\u00e1n k\u00ednh mi\u1ec7ng x\u00f4 l\u00e0: R = 30 = 15 (cm) 2 B\u00e1n k\u00ednh \u0111\u00e1y x\u00f4 l\u00e0: r = 20 = 10 (cm) 2 ( )Thay h = 22, R = 15, r = 10 v\u00e0o V = 1 \uf070 h R2 + Rr + r2 ta c\u00f3: 3 ( )V = 1 \uf070 .22. 152 +15.10 +102 3 V = 10450 \uf070 (cm3 ) 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 V = 209 \uf070 (l\u00edt ) 60 V \uf0bb 11 (l\u00edt ) V\u1eady x\u00f4 c\u00f3 th\u1ec3 ch\u1ee9a t\u1ed1i \u0111a 11 l\u00edt n\u01b0\u1edbc b) L\u01b0\u1ee3ng n\u01b0\u1edbc m\u1ed7i l\u1ea7n x\u00e1ch l\u00e0: 2. 209 \uf070 .(100% \u2212 5%) = 3971\uf070 (l\u00edt) 60 600 B\u00e1c N\u0103m c\u1ea7n ph\u1ea3i l\u1ea5y \u00edt nh\u1ea5t s\u1ed1 l\u1ea7n l\u00e0: 150 : 3971\uf070 \uf0bb 8 (la\u00e0n) 600 V\u1eady b\u00e1c N\u0103m c\u1ea7n ph\u1ea3i l\u1ea5y \u00edt nh\u1ea5t 8 l\u1ea7n B\u00e0i 7. (1,0 \u0111i\u1ec3m)T\u1ed5ng s\u1ed1 h\u1ecdc sinh c\u1ee7a l\u1edbp 9A v\u00e0 9B v\u00e0o \u0111\u1ea7u n\u0103m h\u1ecdc l\u00e0 90 h\u1ecdc sinh. \u0110\u1ebfn \u0111\u1ea7u h\u1ecdc k\u00ec II , l\u1edbp 9A c\u00f3 2 h\u1ecdc sinh \u0111i du h\u1ecdc v\u00e0 4 h\u1ecdc sinh chuy\u1ec3n qua l\u1edbp 9B n\u00ean l\u00fac n\u00e0y s\u1ed1 h\u1ecdc sinh l\u1edbp 9A ch\u1ec9 b\u1eb1ng 5 s\u1ed1 h\u1ecdc sinh l\u1edbp 9B . T\u00ednh s\u1ed1 h\u1ecdc sinh \u0111\u1ea7u n\u0103m c\u1ee7a l\u1edbp 9A v\u00e0 6 9B . *, x 90 L\u01a1\u0300i gi\u1ea3i G\u1ecdi s\u1ed1 h\u1ecdc sinh c\u1ee7a l\u1edbp 9A l\u00e0 x (ho\u00efc sinh) x G\u1ecdi s\u1ed1 h\u1ecdc sinh c\u1ee7a l\u1edbp 9B l\u00e0 y (ho\u00efc sinh) y *, y 90 V\u00ec t\u1ed5ng s\u1ed1 h\u1ecdc sinh c\u1ee7a l\u1edbp 9A v\u00e0 9B v\u00e0o \u0111\u1ea7u n\u0103m h\u1ecdc l\u00e0 90 h\u1ecdc sinh n\u00ean ta c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh: x + y = 90 S\u1ed1 h\u1ecdc sinh 9A \u0111\u1ea7u h\u1ecdc k\u00ec II l\u00e0: x - 2 - 4 = x - 6 (ho\u00efcsinh) S\u1ed1 h\u1ecdc sinh 9B \u0111\u1ea7u h\u1ecdc k\u00ec II l\u00e0: x + 4(ho\u00efcsinh) V\u00ec \u0111\u1ebfn \u0111\u1ea7u h\u1ecdc k\u00ec II s\u1ed1 h\u1ecdc sinh l\u1edbp 9A ch\u1ec9 b\u1eb1ng 5 s\u1ed1 h\u1ecdc sinh l\u1edbp 9B n\u00ean ta c\u00f3 ph\u01b0\u01a1ng 6 tr\u00ecnh: T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N 7","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH x - 6 = 5 .( y + 4) 6 \uf0db x - 6 = 5 y + 10 63 \uf0db x - 5 y = 10 + 6 63 \uf0db x - 5 y = 28 63 Ta c\u00f3 h\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh: \uf0ec\uf0efx + y = 90 \uf0ed\uf0ef\uf0eex - 5y = 28 6 3 Gi\u1ea3i h\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh ta \u0111\u01b0\u1ee3c: \uf0ec\uf0efx = 46 (nha\u00e4n) \uf0ed \uf0ef\uf0eey = 44 ( nha\u00e4n ) V\u1eady s\u1ed1 h\u1ecdc sinh \u0111\u1ea7u n\u0103m c\u1ee7a l\u1edbp 9A l\u00e0 46 h\u1ecdc sinh S\u1ed1 h\u1ecdc sinh \u0111\u1ea7u n\u0103m c\u1ee7a l\u1edbp 9B l\u00e0 44 h\u1ecdc sinh B\u00e0i 8. (3,0 \u0111i\u1ec3m) Cho \uf044ABC nh\u1ecdn n\u1ed9i ti\u1ebfp \u0111\u01b0\u1eddng tr\u00f2n (O) c\u00f3 c\u00e1c \u0111\u01b0\u1eddng cao AK, BD, CE c\u1eaft nhau t\u1ea1i H , tia AK c\u1eaft (O) t\u1ea1i Q . G\u1ecdi N l\u00e0 trung \u0111i\u1ec3m c\u1ee7a BC , F l\u00e0 trung \u0111i\u1ec3m AH . K\u1ebb \u0111\u01b0\u1eddng k\u00ednh AG c\u1ee7a (O) , \u0111\u01b0\u1eddng th\u1eb3ng qua Q song song v\u1edbi ED c\u1eaft (O) t\u1ea1i giao \u0111i\u1ec3m th\u1ee9 2 l\u00e0 T ( T kh\u00e1c Q ). G\u1ecdi J l\u00e0 giao \u0111i\u1ec3m c\u1ee7a NF v\u00e0 ED . a) Ch\u1ee9ng minh BEDC v\u00e0 AEHD l\u00e0 c\u00e1c t\u1ee9 gi\u00e1c n\u1ed9i ti\u1ebfp. b) Ch\u1ee9ng minh: FD \u22a5 ND . T\u1eeb \u0111\u00f3 suy ra ND2 = NJ.NF c) \u0110\u01b0\u1eddng tr\u00f2n \u0111\u01b0\u1eddng k\u00ednh AH c\u1eaft (O) t\u1ea1i giao \u0111i\u1ec3m th\u1ee9 hai l\u00e0 M ( M kh\u00e1c A ). Ch\u1ee9ng minh: ND2 = NH.NM v\u00e0 M , J , T th\u1eb3ng h\u00e0ng. L\u01a1\u0300i gi\u1ea3i T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N 8","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH a) Ch\u1ee9ng minh BEDC v\u00e0 AEHD l\u00e0 c\u00e1c t\u1ee9 gi\u00e1c n\u1ed9i ti\u1ebfp. X\u00e9t t\u1ee9 gi\u00e1c BEDC c\u00f3: \uf0ec\uf0efBEC = 90\uf0b0(CE \u22a5 AB) \uf0ed \uf0ef\uf0eeBDC = 90\uf0b0( BD \u22a5 AC ) \uf0de BEC = BDC \uf0de T\u1ee9 gi\u00e1c BEDC n\u1ed9i ti\u1ebfp ( 2 \u0111\u1ec9nh li\u1ec1n k\u1ec1 c\u00f9ng nh\u00ecn 1 c\u1ea1nh) X\u00e9t t\u1ee9 gi\u00e1c AEHD c\u00f3: \uf0ec\uf0efAEH = 90\uf0b0(CE \u22a5 AB) \uf0ed \uf0ef\uf0eeADH = 90\uf0b0( BD \u22a5 AC ) \uf0de AEH + ADH = 180\uf0b0 \uf0de T\u1ee9 gi\u00e1c AEHD n\u1ed9i ti\u1ebfp (t\u1ed5ng 2 g\u00f3c \u0111\u1ed1i = 180\uf0b0 ) b) Ch\u1ee9ng minh: FD \u22a5 ND . T\u1eeb \u0111\u00f3 suy ra ND2 = NJ.NF 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 V\u00ec t\u1ee9 gi\u00e1c AEHD n\u1ed9i ti\u1ebfp (cmt) c\u00f3 ADH l\u00e0 g\u00f3c n\u1ed9i ti\u1ebfp m\u00e0 ADH = 90\uf0b0 \uf0de AH l\u00e0 \u0111\u01b0\u1eddng k\u00ednh \uf0de F l\u00e0 t\u00e2m (v\u00ec F l\u00e0 trung \u0111i\u1ec3m c\u1ee7a AH ) \uf0de AF = FD (b\u00e1n k\u00ednh) \uf0de \uf044AFD c\u00e2n t\u1ea1i F \uf0de FAD = FDA V\u00ec t\u1ee9 gi\u00e1c BEDC n\u1ed9i ti\u1ebfp (cmt) c\u00f3 BDC l\u00e0 g\u00f3c n\u1ed9i ti\u1ebfp m\u00e0 BDC = 90\uf0b0 \uf0de BC l\u00e0 \u0111\u01b0\u1eddng k\u00ednh \uf0de N l\u00e0 t\u00e2m(v\u00ec N l\u00e0 trung \u0111i\u1ec3m c\u1ee7a BC ) \uf0de ND = NC (b\u00e1n k\u00ednh) \uf0de \uf044NDC c\u00e2n t\u1ea1i N \uf0de NDC = NCD X\u00e9t \uf044AKC vu\u00f4ng t\u1ea1i K (v\u00ec AK \u22a5 BC ) c\u00f3: KAC + KCA = 90\uf0b0 \uf0de FAD + NCD = 90\uf0b0 \uf0de FDA + NDC = 90\uf0b0 M\u00e0 FDA + NDC + FDN = 180\uf0b0 \uf0de FDN = 90\uf0b0 \uf0de FD \u22a5 ND \uf0de \uf044FDN vu\u00f4ng t\u1ea1i D V\u00ec FE = FD (b\u00e1n k\u00ednh, t\u1ee9 gi\u00e1c AEHD n\u1ed9i ti\u1ebfp (F ) ) NE = ND (b\u00e1n k\u00ednh, t\u1ee9 gi\u00e1c BEDC n\u1ed9i ti\u1ebfp ( N ) ) \uf0de NF l\u00e0 trung tr\u1ef1c c\u1ee7a ED \uf0de NF \u22a5 ED t\u1ea1i J X\u00e9t \uf044FDN vu\u00f4ng t\u1ea1i D c\u00f3 DJ l\u00e0 \u0111\u01b0\u1eddng cao \uf0de ND2 = NJ.NF (h\u1ec7 th\u1ee9c l\u01b0\u1ee3ng trong tam gi\u00e1c vu\u00f4ng) c) \u0110\u01b0\u1eddng tr\u00f2n \u0111\u01b0\u1eddng k\u00ednh AH c\u1eaft (O) t\u1ea1i giao \u0111i\u1ec3m th\u1ee9 hai l\u00e0 M ( M kh\u00e1c A ). Ch\u1ee9ng minh: ND2 = NH.NM v\u00e0 M , J , T th\u1eb3ng h\u00e0ng. T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N 10","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH Ta c\u00f3 HMA = 90\uf0b0 (g\u00f3c n\u1ed9i ti\u1ebfp ch\u1eafn n\u1eeda (F ) ) \uf0de HM \u22a5 AM ( )GMA = 90\uf0b0 (g\u00f3c n\u1ed9i ti\u1ebfp ch\u1eafn n\u1eeda O ) \uf0de GM \u22a5 AM \uf0de G, H , M th\u1eb3ng h\u00e0ng GCA = 90\uf0b0 (g\u00f3c n\u1ed9i ti\u1ebfp ch\u1eafn n\u1eeda (O) ) \uf0de GC \u22a5 AC M\u00e0 BD \u22a5 AC (gt) \uf0de GC \/ \/BD \uf0de GC \/ \/BH GBA = 90\uf0b0 (g\u00f3c n\u1ed9i ti\u1ebfp ch\u1eafn n\u1eeda (O) ) \uf0de GB \u22a5 AB M\u00e0 CE \u22a5 AB (gt) \uf0de GB \/ \/CE \uf0de GB \/ \/CH X\u00e9t t\u1ee9 gi\u00e1c BHCG c\u00f3: \uf0ec\uf0efGC \/ \/ BH (cmt) \uf0ed \uf0ef\uf0eeGB \/ \/CH (cmt ) \uf0de BHCG l\u00e0 h\u00ecnh b\u00ecnh h\u00e0nh 11 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 M\u00e0 N l\u00e0 trung \u0111i\u1ec3m BC \uf0de N l\u00e0 trung \u0111i\u1ec3m HG \uf0de H , N, G th\u1eb3ng h\u00e0ng M\u00e0 G, H , M th\u1eb3ng h\u00e0ng (cmt) \uf0de M , H , N, G th\u1eb3ng h\u00e0ng X\u00e9t \uf044NEH v\u00e0 \uf044NME c\u00f3: \uf0ec\uf0efMNE chung \uf0ed \uf0e6 1 \uf0f6 \uf0efNEH = NME \uf0e7 = 2 s\u00f1 ME \uf0f7 \uf0ee \uf0e8 \uf0f8 \uf0de \uf044NEH \uf044NME ( g \u2212 g ) \uf0de NE = NH NM NE \uf0de NE2 = NH.NM M\u00e0 NE = ND (cmt) \uf0de ND2 = NH.NM M\u00e0 ND2 = NJ.NF (ch\u1ee9ng minh c\u00e2u b) \uf0de NH.NM = NJ.NF \uf0de NH = NJ NF NM X\u00e9t \uf044NJH v\u00e0 \uf044NMF c\u00f3: \uf0ec\uf0efMNF chung \uf0ed NH NJ (cmt) \uf0ef\uf0ee NF = NM \uf0de \uf044NJH \uf044NMF (c \u2212 g \u2212 c) \uf0de NJH = NMF \uf0de t\u1ee9 gi\u00e1c MHJF n\u1ed9i ti\u1ebfp (g\u00f3c trong = g\u00f3c \u0111\u1ed1i ngo\u00e0i) \uf0de HMJ = HFJ ( 2 g\u00f3c n\u1ed9i ti\u1ebfp c\u00f9ng ch\u1eafn JH ) Ta c\u00f3 GMT = GAT ( 2 g\u00f3c n\u1ed9i ti\u1ebfp c\u00f9ng ch\u1eafn GT c\u1ee7a (O) ) V\u00ec t\u1ee9 gi\u00e1c BEDC n\u1ed9i ti\u1ebfp \uf0de ADE = ABC (g\u00f3c trong = g\u00f3c \u0111\u1ed1i ngo\u00e0i) ABC = 1 s\u00f1 AC = 1 AOC = 1 (180\uf0b0 \u2212 2.OAC) = 90\uf0b0 \u2212 OAC 2 22 T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N 12","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH \uf0de ABC + OAC = 90\uf0b0 \uf0de ADE + OAC = 90\uf0b0 \uf0de AG \u22a5 ED M\u00e0 NF \u22a5 ED (cmt ) \uf0de AG \/ \/ NF \uf0de QAG = HFJ (\u00f1o\u00e0ng v\u00f2) M\u1eb7t kh\u00e1c AG \u22a5 ED (cmt); ED \/ \/QT ( gt ) \uf0de AG \u22a5 QT M\u00e0 AG l\u00e0 \u0111\u01b0\u1eddng k\u00ednh \uf0de AG \u0111i qua trung \u0111i\u1ec3m c\u1ee7a QT (quan h\u1ec7 \u0111\u01b0\u1eddng k\u00ednh v\u00e0 d\u00e2y) \uf0de AG v\u1eeba l\u00e0 \u0111\u01b0\u1eddng cao, v\u1eeba l\u00e0 trung tuy\u1ebfn c\u1ee7a \uf044AQT \uf0de \uf044AQT c\u00e2n t\u1ea1i A \uf0de AG l\u00e0 ph\u00e2n gi\u00e1c QAT \uf0de QAG = GAT \uf0de GAT = HFJ (c\u00f9ng = QAG ) \uf0de GMT = HFJ (c\u00f9ng = GAT ) \uf0de GMT = HMJ (c\u00f9ng = HFJ ) M\u00e0 G, H , M th\u1eb3ng h\u00e0ng \uf0de M , J, T th\u1eb3ng h\u00e0ng (\u0111i\u1ec1u ph\u1ea3i ch\u1ee9ng minh) ----H\u1ebeT--- T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N 13","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH S\u00d4\u00db GD&\u00d1T TP HO\u00c0 CH\u00cd MINH \u00d1E\u00c0 THAM KHA\u00dbO TUYE\u00c5N SINH 10 PHO\u00d8NG G\u00d1&\u00d1T QUA\u00c4N 10 NA\u00caM HO\u00cfC: 2023 - 2024 \u0110\u1ec0 THAM KH\u1ea2O M\u00d4N: TO\u00c1N 9 M\u00c3 \u0110\u1ec0: Qu\u1eadn 10 2 \u0110\u00ea thi g\u1ed3m 8 c\u00e2u ho\u0309i t\u01b0\u0323 lu\u00e2\u0323n. Th\u01a1\u0300i gian: 120 phu\u0301t (kh\u00f4ng k\u00ea\u0309 th\u01a1\u0300i gian pha\u0301t \u0111\u00ea\u0300) C\u00e2u 1. Cho (P) : y = \u22122x2 v\u00e0 \u0111\u01b0\u1eddng th\u1eb3ng (d) : y = \u2212x \u2212 1 . a) V\u1ebd \u0111\u1ed3 th\u1ecb (P) v\u00e0 (d) tr\u00ean c\u00f9ng h\u1ec7 tr\u1ee5c t\u1ecda \u0111\u1ed9. b) T\u00ecm t\u1ecda \u0111\u1ed9 giao \u0111i\u1ec3m c\u1ee7a (P) v\u00e0 (d) b\u1eb1ng ph\u00e9p t\u00ednh. C\u00e2u 2. Cho ph\u01b0\u01a1ng tr\u00ecnh \u2212 2x2 + 2x + 3 = 0 c\u00f3 2 nghi\u1ec7m ph\u00e2n bi\u1ec7t 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 = x2 + 1 + x1 + 1 1 \u2212 x1 1 \u2212 x2 C\u00e2u 3. \u1ede B\u1ea1c Li\u00eau ng\u00e0nh n\u00f4ng nghi\u1ec7p khuy\u1ebfn kh\u00edch b\u00e0 con n\u00f4ng d\u00e2n ph\u01b0\u01a1ng th\u1ee9c nu\u00f4i tr\u1ed3ng : \u201ccon t\u00f4m, c\u00e2y l\u00faa\u201d, c\u1ea3i t\u1ea1o \u0111\u1ed3ng l\u00faa n\u0103ng su\u1ea5t th\u1ea5p th\u00e0nh c\u00e1c h\u1ed3 nu\u00f4i t\u00f4m n\u01b0\u1edbc m\u1eb7n. Dung d\u1ecbch n\u01b0\u1edbc mu\u1ed1i nu\u00f4i t\u00f4m c\u00f3 n\u1ed3ng \u0111\u1ed9 5% . Nh\u01b0ng n\u01a1i \u0111\u00e2y ch\u1ec9 c\u00f3 n\u01b0\u1edbc bi\u1ec3n n\u1ed3ng \u0111\u1ed9 10% v\u00e0 n\u01b0\u1edbc l\u1ee3 n\u1ed3ng \u0111\u1ed9 mu\u1ed1i 1% . \u0110\u1ec3 \u0111\u1ed5 \u0111\u1ea7y h\u1ed3 nu\u00f4i t\u00f4m c\u00f3 dung t\u00edch 1000 l\u00edt ph\u1ea3i c\u1ea7n b\u01a1m v\u00e0o h\u1ed3 m\u1ed7i lo\u1ea1i n\u01b0\u1edbc bao nhi\u00eau kg ? Bi\u1ebft kh\u1ed1i l\u01b0\u1ee3ng ri\u00eang c\u1ee7a dung d\u1ecbch n\u01b0\u1edbc mu\u1ed1i 5% l\u00e0 1, 8kg \/ l C\u00e2u 4. \u0110\u1ed1i v\u1edbi ng\u01b0\u1eddi \u00c1 \u0110\u00f4ng, \u00c2m l\u1ecbch lu\u00f4n gi\u1eef vai tr\u00f2 quan tr\u1ecdng gi\u00fap ch\u00fang ta x\u00e1c \u0111\u1ecbnh c\u00e1c d\u1ecbp l\u1ec5 T\u1ebft trong n\u0103m. V\u00e0 c\u0169ng nh\u01b0 D\u01b0\u01a1ng l\u1ecbch, \u00c2m l\u1ecbch c\u0169ng s\u1ebd c\u00f3 n\u0103m nhu\u1eadn. \u0110\u1ec3 bi\u1ebft \u0111\u01b0\u1ee3c n\u0103m \u00c2m l\u1ecbch c\u00f3 nhu\u1eadn hay kh\u00f4ng, ta l\u1ea5y n\u0103m D\u01b0\u01a1ng l\u1ecbch t\u01b0\u01a1ng \u1ee9ng chia cho 19 . N\u1ebfu s\u1ed1 d\u01b0 c\u1ee7a ph\u00e9p chia n\u00e0y l\u00e0 0; 3; 6; 9; 11; 14; 17 th\u00ec n\u0103m \u0111\u00f3 s\u1ebd l\u00e0 n\u0103m \u00c2m l\u1ecbch nhu\u1eadn. a) H\u00e3y t\u00ednh xem n\u0103m 2020 c\u00f3 ph\u1ea3i n\u0103m \u00c2m l\u1ecbch nhu\u1eadn hay kh\u00f4ng? V\u00ec sao? b) Bi\u1ebft r\u1eb1ng m\u1ed9t n\u0103m d\u01b0\u01a1ng l\u1ecbch \u0111\u01b0\u1ee3c g\u1ecdi l\u00e0 nhu\u1eadn n\u1ebfu n\u0103m D\u01b0\u01a1ng l\u1ecbch \u0111\u00f3 chia h\u1ebft cho 4 . B\u00e1c N\u0103m sinh ra v\u00e0o cu\u1ed1i th\u1ebf k\u1ec9 20 , b\u00e1c N\u0103m ch\u01b0a qu\u00e1 50 tu\u1ed5i . H\u00e3y t\u00ednh xem b\u00e1c N\u0103m sinh ra n\u0103m bao nhi\u00eau bi\u1ebft r\u1eb1ng n\u0103m sinh c\u1ee7a b\u00e1c l\u00e0 m\u1ed9t n\u0103m v\u1eeba nhu\u1eadn \u00c2m l\u1ecbch, v\u1eeba nhu\u1eadn D\u01b0\u01a1ng l\u1ecbch. C\u00e2u 5. Gi\u00e1 c\u01b0\u1edbc g\u1ecdi qu\u1ed1c t\u1ebf c\u1ee7a m\u1ed9t c\u00f4ng ty X trong d\u1ecbp khuy\u1ebfn m\u00e3i m\u1eebng ng\u00e0y th\u00e0nh l\u1eadp c\u00f4ng ty \u0111\u01b0\u1ee3c cho b\u1edfi b\u1ea3ng sau: Th\u1eddi gian Gi\u00e1 c\u01b0\u1edbc (VN\u0110) 5 ph\u00fat \u0111\u1ea7u 6000 T\u1eeb ph\u00fat th\u1ee9 6 10 5800 T\u1eeb ph\u00fat th\u1ee9 11 20 5200 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 T\u1eeb ph\u00fat th\u1ee9 21 30 5000 Tr\u00ean 30 ph\u00fat 4500 a) B\u00e1c Lan g\u1ecdi cho ng\u01b0\u1eddi th\u00e2n \u1edf n\u01b0\u1edbc ngo\u00e0i trong th\u1eddi gian 24 ph\u00fat th\u00ec s\u1ed1 ti\u1ec1n b\u00e1c Lan tr\u1ea3 l\u00e0 bao nhi\u00eau? b) M\u1ed9t ng\u01b0\u1eddi \u0111\u00e3 tr\u1ea3 197 000 \u0111\u1ed3ng \u0111\u1ec3 g\u1ecdi \u0111i\u1ec7n cho ng\u01b0\u1eddi th\u00e2n \u1edf n\u01b0\u1edbc ngo\u00e0i . T\u00ednh th\u1eddi gian ng\u01b0\u1eddi \u0111\u00f3 \u0111\u00e3 g\u1ecdi \u0111i\u1ec7n cho ng\u01b0\u1eddi th\u00e2n. C\u00e2u 6. C\u1ea7u V\u00e0m C\u1ed1ng b\u1eafc ngang s\u00f4ng H\u1eadu n\u1ed1i hai t\u1ec9nh C\u1ea7n Th\u01a1 v\u00e0 \u0110\u1ed3ng Th\u00e1p thi\u1ebft k\u1ebf theo ki\u1ec3u d\u00e2y v\u0103ng nh\u01b0 h\u00ecnh v\u1ebd. Chi\u1ec1u cao t\u1eeb s\u00e0n \u0111\u1ebfn \u0111\u1ec9nh AB 120m , d\u00e2y v\u0103ng AC 258 m , chi\u1ec1u d\u00e0i s\u00e0n c\u1ea7u t\u1eeb B \u0111\u1ebfn C l\u00e0 218m H\u1ecfi g\u00f3c nghi\u00eang c\u1ee7a s\u00e0n c\u1ea7u BC so v\u1edbi m\u1eb7t s\u00e0n n\u1eb1m ngang (gi\u1ea3 thi\u1ebft xem nh\u01b0 tr\u1ee5 \u0111\u1ee1 AB th\u1eb3ng \u0111\u1ee9ng (L\u00e0m tr\u00f2n \u0111\u1ebfn ph\u00fat) C\u00e2u 7. \u0110\u1ec3 l\u00e0m m\u0169 sinh nh\u1eadt h\u00ecnh n\u00f3n t\u1eeb mi\u1ebfng gi\u1ea5y h\u00ecnh tr\u00f2n b\u00e1n k\u00ednh 20cm . B\u1ea1n An c\u1eaft b\u1ecf ph\u1ea7n qu\u1ea1t tr\u00f2n AOB v\u1edbi AOB 60 . Sau \u0111\u00f3 d\u00e1n ph\u1ea7n h\u00ecnh qu\u1ea1t l\u1edbn c\u00f2n l\u1ea1i sao cho A B \u0111\u1ec3 l\u00e0m c\u00e1i m\u0169. a) T\u00ednh \u0111\u1ed9 d\u00e0i cung l\u1edbn AB b) H\u1ecfi th\u1ec3 t\u00edch c\u00e1i n\u00f3n l\u00e0 bao nhi\u00eau? C\u00e2u 8. T\u1eeb \u0111i\u1ec3m M n\u1eb1m ngo\u00e0i \u0111\u01b0\u1eddng tr\u00f2n O , v\u1ebd hai ti\u1ebfp tuy\u1ebfn MP,MQ ( P,Q l\u00e0 hai ti\u1ebfp \u0111i\u1ec3m). T\u1eeb \u0111i\u1ec3m N tr\u00ean cung nh\u1ecf PQ ta v\u1ebd ti\u1ebfp tuy\u1ebfn v\u1edbi \u0111\u01b0\u1eddng tr\u00f2n O , ti\u1ebfp tuy\u1ebfn n\u00e0y c\u1eaft MP,MQ l\u1ea7n l\u01b0\u1ee3t t\u1ea1iE,F a) Ch\u1ee9ng minh: P MEF 2MP . b) Ch\u1ee9ng minh: EOF OMP 90 . c) H\u1ea1 EH OF,FK OE . Ch\u1ee9ng minh: NO l\u00e0 ph\u00e2n gi\u00e1c c\u1ee7a HNK . d) Ch\u1ee9ng minh: 4 \u0111i\u1ec3m P,Q,H,K th\u1eb3ng h\u00e0ng. ----H\u1ebeT--- T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N 2","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH H\u01af\u1edaNG D\u1eaaN GI\u1ea2I C\u00e2u 1. (1,5 \u0111i\u1ec3m) Cho (P) : y = \u22122x2 v\u00e0 \u0111\u01b0\u1eddng th\u1eb3ng (d) : y = \u2212x \u2212 1 . a) V\u1ebd \u0111\u1ed3 th\u1ecb (P) v\u00e0 (d) tr\u00ean c\u00f9ng h\u1ec7 tr\u1ee5c t\u1ecda \u0111\u1ed9. b) T\u00ecm t\u1ecda \u0111\u1ed9 giao \u0111i\u1ec3m c\u1ee7a (P) v\u00e0 (d) b\u1eb1ng ph\u00e9p 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 1 \u22121 \u22122 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 1 \uf0db 2x2 \u2212 x \u22121 = 0 \uf0e9x = 1 \uf0ea \uf0db \uf0ea\uf0ea\uf0ebx = \u2212 1 2 Thay x = 1 v\u00e0o y = \u22122x2 , ta \u0111\u01b0\u1ee3c: y = \u22122.12 = \u22122 . Thay x=\u22121 v\u00e0o y = \u22122x2 , ta \u0111\u01b0\u1ee3c: y = \u2212 \uf0e6 \u2212 1 \uf0f62 =\u22121 . 2 \uf0e7\uf0e8 2 \uf0f7\uf0f8 4 V\u1eady (1; \u22122) , \uf0e6 \u2212 1 ; \u2212 1\uf0f6 l\u00e0 hai giao \u0111i\u1ec3m c\u1ea7n t\u00ecm. \uf0e7 2 \uf0f7 \uf0e8 4 \uf0f8 C\u00e2u 2. Cho ph\u01b0\u01a1ng tr\u00ecnh \u2212 2x2 + 2x + 3 = 0 c\u00f3 2 nghi\u1ec7m ph\u00e2n bi\u1ec7t 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 = x2 + 1 + x1 + 1 1 \u2212 x1 1 \u2212 x2 L\u1eddi gi\u1ea3i ( )V\u00ec \uf044 = b2 \u2212 4ac = 22 \u2212 4. \u2212 2 .3 = 4 + 12 2 \uf03e 0 T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N 3","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH 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 2 = 1 \uf0ed = x1 a \u22122 2 Theo \u0111\u1ecbnh l\u00ed Vi-et, ta c\u00f3: \uf0ef\uf0ef\uf0eeP 3 3 c = \u22122 =\u2212 2 .x2 = a Ta c\u00f3: A = x2 + 1 + x1 + 1 1 \u2212 x1 1 \u2212 x2 A = (x2 + 1)(1 \u2212 x2 ) + (x1 + 1)(1 \u2212 x1 ) (1 \u2212 x1 )(1 \u2212 x2 ) (1\u2212 x2 )(1 \u2212 x1 ) A = x2 \u2212 x22 + 1 \u2212 x2 + x1 \u2212 x12 + 1 \u2212 x1 1 \u2212 x2 \u2212 x1 + x1x2 ( ) ( )( )A = 2 \u2212 x12 + x22 2 \u2212 S2 \u2212 2P = 1 \u2212 x1 + x2 + x1x2 1\u2212S + P 2 \u2212 \uf0e9\uf0ea\uf0e6\uf0e7 1 \uf0f62 \uf0e6 3 \uf0f6\uf0f7\uf0f9\uf0fa \uf0f7 \u2212 2.\uf0e7 \u2212 A = \uf0ea\uf0eb\uf0e8 2 \uf0f8 \uf0e8 2 \uf0f8\uf0fa\uf0fb = 3 1 \uf0e6 3 \uf0f6 2 1\u2212 + \uf0e7 \u2212 \uf0f7 2 \uf0e8 2\uf0f8 C\u00e2u 3. \u1ede B\u1ea1c Li\u00eau ng\u00e0nh n\u00f4ng nghi\u1ec7p khuy\u1ebfn kh\u00edch b\u00e0 con n\u00f4ng d\u00e2n ph\u01b0\u01a1ng th\u1ee9c nu\u00f4i tr\u1ed3ng : \u201ccon t\u00f4m, c\u00e2y l\u00faa\u201d, c\u1ea3i t\u1ea1o \u0111\u1ed3ng l\u00faa n\u0103ng su\u1ea5t th\u1ea5p th\u00e0nh c\u00e1c h\u1ed3 nu\u00f4i t\u00f4m n\u01b0\u1edbc m\u1eb7n. Dung d\u1ecbch n\u01b0\u1edbc mu\u1ed1i nu\u00f4i t\u00f4m c\u00f3 n\u1ed3ng \u0111\u1ed9 5% . Nh\u01b0ng n\u01a1i \u0111\u00e2y ch\u1ec9 c\u00f3 n\u01b0\u1edbc bi\u1ec3n n\u1ed3ng \u0111\u1ed9 10% v\u00e0 n\u01b0\u1edbc l\u1ee3 n\u1ed3ng \u0111\u1ed9 mu\u1ed1i 1% . \u0110\u1ec3 \u0111\u1ed5 \u0111\u1ea7y h\u1ed3 nu\u00f4i t\u00f4m c\u00f3 dung t\u00edch 1000 l\u00edt ph\u1ea3i c\u1ea7n b\u01a1m v\u00e0o h\u1ed3 m\u1ed7i lo\u1ea1i n\u01b0\u1edbc bao nhi\u00eau kg ? Bi\u1ebft kh\u1ed1i l\u01b0\u1ee3ng ri\u00eang c\u1ee7a dung d\u1ecbch n\u01b0\u1edbc mu\u1ed1i 5% l\u00e0 1, 8kg \/ l L\u1eddi gi\u1ea3i 1800 kg S\u1ed1 kg n\u01b0\u1edbc mu\u1ed1i c\u1ea7n b\u01a1m v\u00e0o h\u1ed3 nu\u00f4i t\u00f4m l\u00e0: 1, 8 . 1000 G\u1ecdi x kg l\u00e0 s\u1ed1 kg n\u01b0\u1edbc bi\u1ec3n x 0 ,y kg l\u00e0 s\u1ed1 kg n\u01b0\u1edbc l\u1ee3 y 0 . V\u00ec s\u1ed1 kg n\u01b0\u1edbc c\u1ea7n b\u01a1m v\u00e0o h\u1ed3 l\u00e0 1800 kg n\u00ean: x y 1800 (1) V\u00ec dung d\u1ecbch n\u01b0\u1edbc n\u01b0\u1edbc c\u1ea7n b\u01a1m v\u00e0o c\u00f3 n\u1ed3ng \u0111\u1ed9 5% n\u00ean: 10%x 1%y 5%.1800 0,1x 0, 01y 90 (2) T\u1eeb (1) v\u00e0 (2) ta c\u00f3 h\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh: \uf0ecx + y = 1800 = 90 \uf0db \uf0ecx = 800 \uf0ed\uf0ee0,1a + 0,01y \uf0ed\uf0eey . = 1000 V\u1eady: ph\u1ea3i b\u01a1m v\u00e0o h\u1ed3 800 kg n\u01b0\u1edbc bi\u1ec3n v\u00e0 1000 kg n\u01b0\u1edbc l\u1ee3. C\u00e2u 4. \u0110\u1ed1i v\u1edbi ng\u01b0\u1eddi \u00c1 \u0110\u00f4ng, \u00c2m l\u1ecbch lu\u00f4n gi\u1eef vai tr\u00f2 quan tr\u1ecdng gi\u00fap ch\u00fang ta x\u00e1c \u0111\u1ecbnh c\u00e1c d\u1ecbp l\u1ec5 T\u1ebft trong n\u0103m. V\u00e0 c\u0169ng nh\u01b0 D\u01b0\u01a1ng l\u1ecbch, \u00c2m l\u1ecbch c\u0169ng s\u1ebd c\u00f3 n\u0103m nhu\u1eadn. \u0110\u1ec3 bi\u1ebft 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 \u0111\u01b0\u1ee3c n\u0103m \u00c2m l\u1ecbch c\u00f3 nhu\u1eadn hay kh\u00f4ng, ta l\u1ea5y n\u0103m D\u01b0\u01a1ng l\u1ecbch t\u01b0\u01a1ng \u1ee9ng chia cho19 . N\u1ebfu s\u1ed1 d\u01b0 c\u1ee7a ph\u00e9p chia n\u00e0y l\u00e0 0; 3; 6; 9; 11; 14; 17 th\u00ec n\u0103m \u0111\u00f3 s\u1ebd l\u00e0 n\u0103m \u00c2m l\u1ecbch nhu\u1eadn. a) H\u00e3y t\u00ednh xem n\u0103m 2020 c\u00f3 ph\u1ea3i n\u0103m \u00c2m l\u1ecbch nhu\u1eadn hay kh\u00f4ng? V\u00ec sao? b) Bi\u1ebft r\u1eb1ng m\u1ed9t n\u0103m d\u01b0\u01a1ng l\u1ecbch \u0111\u01b0\u1ee3c g\u1ecdi l\u00e0 nhu\u1eadn n\u1ebfu n\u0103m D\u01b0\u01a1ng l\u1ecbch \u0111\u00f3 chia h\u1ebft cho 4 B\u00e1c N\u0103m sinh ra v\u00e0o cu\u1ed1i th\u1ebf k\u1ec9 20 , b\u00e1c N\u0103m ch\u01b0a qu\u00e1 50 tu\u1ed5i . H\u00e3y t\u00ednh xem b\u00e1c N\u0103m sinh ra n\u0103m bao nhi\u00eau bi\u1ebft r\u1eb1ng n\u0103m sinh c\u1ee7a b\u00e1c l\u00e0 m\u1ed9t n\u0103m v\u1eeba nhu\u1eadn \u00c2m l\u1ecbch, v\u1eeba nhu\u1eadn D\u01b0\u01a1ng l\u1ecbch. L\u1eddi gi\u1ea3i a) V\u00ec 2020 : 19 d\u01b0 6 n\u00ean n\u0103m 2020 l\u00e0 n\u0103m nhu\u1eadn \u00c2m l\u1ecbch. b) V\u00ec b\u00e1c N\u0103m ch\u01b0a qu\u00e1 50 tu\u1ed5i v\u00e0 sinh v\u00e0o cu\u1ed1i th\u1ebf k\u1ec9 20 n\u00ean ta t\u00ednh t\u1eeb n\u0103m 1973 \u0111\u1ebfn n\u0103m 1999 c\u00f3 c\u00e1c n\u0103m nhu\u1eadn D\u01b0\u01a1ng l\u1ecbch (chia h\u1ebft cho 4 ) l\u00e0: 1976; 1980; 1984; 1988; 1992; 1996 . Trong c\u00e1c n\u0103m tr\u00ean ch\u1ec9 c\u00f3 n\u0103m 1976 chia 3 d\u01b0 C\u00e2u 5. 0 l\u00e0 n\u0103m nhu\u1eadn \u00c2m l\u1ecbch. V\u1eady b\u00e1c N\u0103m sinh n\u0103m 1976. Gi\u00e1 c\u01b0\u1edbc g\u1ecdi qu\u1ed1c t\u1ebf c\u1ee7a m\u1ed9t c\u00f4ng ty X trong d\u1ecbp khuy\u1ebfn m\u00e3i m\u1eebng ng\u00e0y th\u00e0nh l\u1eadp c\u00f4ng ty \u0111\u01b0\u1ee3c cho b\u1edfi b\u1ea3ng sau: Th\u1eddi gian Gi\u00e1 c\u01b0\u1edbc (VN\u0110) 5 ph\u00fat \u0111\u1ea7u 6000 T\u1eeb ph\u00fat th\u1ee9 6 10 5800 T\u1eeb ph\u00fat th\u1ee9 11 20 5200 T\u1eeb ph\u00fat th\u1ee9 21 30 5000 Tr\u00ean 30 ph\u00fat 4500 a) B\u00e1c Lan g\u1ecdi cho ng\u01b0\u1eddi th\u00e2n \u1edf n\u01b0\u1edbc ngo\u00e0i trong th\u1eddi gian 24 ph\u00fat th\u00ec s\u1ed1 ti\u1ec1n b\u00e1c Lan tr\u1ea3 l\u00e0 bao nhi\u00eau? b) M\u1ed9t ng\u01b0\u1eddi \u0111\u00e3 tr\u1ea3 197 000 \u0111\u1ed3ng \u0111\u1ec3 g\u1ecdi \u0111i\u1ec7n cho ng\u01b0\u1eddi th\u00e2n \u1edf n\u01b0\u1edbc ngo\u00e0i . T\u00ednh th\u1eddi gian ng\u01b0\u1eddi \u0111\u00f3 \u0111\u00e3 g\u1ecdi \u0111i\u1ec7n cho ng\u01b0\u1eddi th\u00e2n. L\u1eddi gi\u1ea3i 5 a) S\u1ed1 ti\u1ec1n B\u00e1c Lan ph\u1ea3i tr\u1ea3 khi g\u1ecdi 24 ph\u00fat l\u00e0: 5.6000 5.5800 10.5200 4.5000 131000 (\u0111\u1ed3ng) b) S\u1ed1 ti\u1ec1n ph\u1ea3i tr\u1ea3 cho cu\u1ed9c g\u1ecdi t\u1eeb l\u00fac b\u1eaft \u0111\u1ea7u \u0111\u1ebfn ph\u00fat th\u1ee9 30 l\u00e0: 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 5.6000 5.5800 10.5200 10.5000 161000 (\u0111\u1ed3ng) Th\u1eddi gian g\u1ecdi v\u01b0\u1ee3t qu\u00e1 30 ph\u00fat l\u00e0: 197000 161000 : 4500 8 (ph\u00fat) V\u1eady th\u1eddi gian g\u1ecdi \u0111i\u1ec7n cho ng\u01b0\u1eddi th\u00e2n l\u00e0: 30 8 38 (ph\u00fat) C\u00e2u 6. C\u1ea7u V\u00e0m C\u1ed1ng b\u1eafc ngang s\u00f4ng H\u1eadu n\u1ed1i hai t\u1ec9nh C\u1ea7n Th\u01a1 v\u00e0 \u0110\u1ed3ng Th\u00e1p thi\u1ebft k\u1ebf theo ki\u1ec3u d\u00e2y v\u0103ng nh\u01b0 h\u00ecnh v\u1ebd. Chi\u1ec1u cao t\u1eeb s\u00e0n \u0111\u1ebfn \u0111\u1ec9nh AB 120m , d\u00e2y v\u0103ngAC 258 m , chi\u1ec1u d\u00e0i s\u00e0n c\u1ea7u t\u1eeb B \u0111\u1ebfn C l\u00e0 218m H\u1ecfi g\u00f3c nghi\u00eang c\u1ee7a s\u00e0n c\u1ea7u BC so v\u1edbi m\u1eb7t s\u00e0n n\u1eb1m ngang (gi\u1ea3 thi\u1ebft xem nh\u01b0 tr\u1ee5 \u0111\u1ee1 AB th\u1eb3ng \u0111\u1ee9ng (L\u00e0m tr\u00f2n \u0111\u1ebfn ph\u00fat) L\u1eddi gi\u1ea3i A G\u1ecdiCH x 0 x 258 Ta c\u00f3: BH 2182 x 2 (Pitago) B 218 m ?C H x Suy ra: AH 120 BH 120 2182 x 2 Trong AHC c\u00f3: AC 2 AH 2 HC 2 (Pitago) 2 2582 120 2182 x 2 x 2 x 217,14m cos BCH HC 217,14 BCH 5 5 ' BC 218 V\u1eady g\u00f3c t\u1ea1o b\u1edfi s\u00e0n c\u1ea7u v\u00e0 s\u00e0n n\u1eb1m ngang l\u00e0 5 5 ' C\u00e2u 7. \u0110\u1ec3 l\u00e0m m\u0169 sinh nh\u1eadt h\u00ecnh n\u00f3n t\u1eeb mi\u1ebfng gi\u1ea5y h\u00ecnh tr\u00f2n b\u00e1n k\u00ednh 20cm . B\u1ea1n An c\u1eaft b\u1ecf ph\u1ea7n qu\u1ea1t tr\u00f2n AOB v\u1edbi AOB 60 . Sau \u0111\u00f3 d\u00e1n ph\u1ea7n h\u00ecnh qu\u1ea1t l\u1edbn c\u00f2n l\u1ea1i sao cho A B \u0111\u1ec3 l\u00e0m c\u00e1i m\u0169. a) T\u00ednh \u0111\u1ed9 d\u00e0i cung l\u1edbn AB b) H\u1ecfi th\u1ec3 t\u00edch c\u00e1i n\u00f3n l\u00e0 bao nhi\u00eau? 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 L\u1eddi gi\u1ea3i AO 60\u00b0 B O 20cm H A\u2261B a) S\u1ed1 \u0111o cung l\u1edbn AB l\u00e0: 3600 600 300 . \u0110\u1ed9 d\u00e0i cung l\u1edbn AB l\u00e0: l R.n .20.300 100 104, 72cm . 180 180 3 b) B\u00e1n k\u00ednh \u0111\u00e1y l\u00e0: R C 100 :2 50 cm . 2 3 3 \u0110\u01b0\u1eddng cao c\u1ee7a h\u00ecnh n\u00f3n l\u00e0: OH 202 50 2 10 11 cm . 3 3 Th\u1ec3 t\u00edch c\u00e1i n\u00f3n l\u00e0: V 1 R2h 1 50 2 . 10 11 3215, 89cm3 . 3 33 3 C\u00e2u 8. T\u1eeb \u0111i\u1ec3m M n\u1eb1m ngo\u00e0i \u0111\u01b0\u1eddng tr\u00f2n O , v\u1ebd hai ti\u1ebfp tuy\u1ebfn MP,MQ ( P,Q l\u00e0 hai ti\u1ebfp \u0111i\u1ec3m). T\u1eeb \u0111i\u1ec3m N tr\u00ean cung nh\u1ecf PQ ta v\u1ebd ti\u1ebfp tuy\u1ebfn v\u1edbi \u0111\u01b0\u1eddng tr\u00f2n O , ti\u1ebfp tuy\u1ebfn n\u00e0y c\u1eaft MP,MQ l\u1ea7n l\u01b0\u1ee3t t\u1ea1i E,F 2MP a) Ch\u1ee9ng minh: P MEF b) Ch\u1ee9ng minh: EOF OMP 90 c) H\u1ea1 EH OF,FK OE . Ch\u1ee9ng minh: NO l\u00e0 ph\u00e2n gi\u00e1c c\u1ee7a HNK d) Ch\u1ee9ng minh: b\u1ed1n \u0111i\u1ec3m P, Q, H, K th\u1eb3ng h\u00e0ng 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 P E NK I MO H F Q a) Ch\u1ee9ng minh: P MEF 2MP Ta c\u00f3: EN EP ( t\u00ednh ch\u1ea5t hai ti\u1ebfp tuy\u1ebfn c\u1eaft nhau) FN FQ ( t\u00ednh ch\u1ea5t hai ti\u1ebfp tuy\u1ebfn c\u1eaft nhau) Suy ra: EN FN EP EQ EF Do \u0111\u00f3: P MEF ME MF EF ME MF EP EQ MP MQ M\u00e0 MP MQ ( t\u00ednh ch\u1ea5t hai ti\u1ebfp tuy\u1ebfn c\u1eaft nhau) V\u1eady: P MEF 2MP b) Ch\u1ee9ng minh: EOF OMP 90 Ta c\u00f3: EON EOP PON 2 (T\u00ednh ch\u1ea5t hai ti\u1ebfp tuy\u1ebfn c\u1eaft nhau) FON FOQ QON 2 (T\u00ednh ch\u1ea5t hai ti\u1ebfp tuy\u1ebfn c\u1eaft nhau) PMO QMO PMQ 2 (T\u00ednh ch\u1ea5t hai ti\u1ebfp tuy\u1ebfn c\u1eaft nhau) Do \u0111\u00f3: EOF EON FON PON QON POQ 2 2 2 EOF OMP POQ PMQ 180 90 (T\u1ee9 gi\u00e1c MPOQ n\u1ed9i ti\u1ebfp) 2 22 T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N 8","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH c) Ch\u1ee9ng minh: NO l\u00e0 ph\u00e2n gi\u00e1c c\u1ee7a HNK . Ta c\u00f3: T\u1ee9 gi\u00e1c NKOF n\u1ed9i ti\u1ebfp ENK EOF (g\u00f3c ngo\u00e0i b\u1eb1ng g\u00f3c \u0111\u1ed1i trong) T\u1ee9 gi\u00e1c HNEO n\u1ed9i ti\u1ebfp FNH EOF (g\u00f3c ngo\u00e0i b\u1eb1ng g\u00f3c \u0111\u1ed1i trong) Do \u0111\u00f3: ENK HNF . 1 M\u00e0 ON EF ONE ONF 900 2 T\u1eeb 1 v\u00e0 2 ta c\u00f3: ONK ONH V\u1eady NO l\u00e0 ph\u00e2n gi\u00e1c c\u1ee7a HNK . d) Ch\u1ee9ng minh: b\u1ed1n \u0111i\u1ec3m P, Q, H, K th\u1eb3ng h\u00e0ng T\u1ee9 gi\u00e1c EFHK n\u1ed9i ti\u1ebfp KHO FEO T\u1ee9 gi\u00e1c ENHO n\u1ed9i ti\u1ebfp NHF FEO Suy ra: KHO NHF 3 Ta l\u1ea1i c\u00f3: FHN FHQ c.c.c QHF NHF 4 T\u1eeb 3 v\u00e0 4 ta c\u00f3: KHO QHF . Do \u0111\u00f3: Q; H; K th\u1eb3ng h\u00e0ng. Ch\u1ee9ng minh t\u01b0\u01a1ng t\u1ef1: P; K; H th\u1eb3ng h\u00e0ng. V\u1eady b\u1ed1n \u0111i\u1ec3m P; Q; H; K th\u1eb3ng h\u00e0ng. ----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 10 NA\u00caM HO\u00cfC: 2023 - 2024 M\u00d4N: TO\u00c1N 9 \u0110\u1ec0 THAM KH\u1ea2O M\u00c3 \u0110\u1ec0: Qu\u1eadn 10 - 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). Trong m\u1eb7t ph\u1eb3ng t\u1ecda \u0111\u1ed9 Oxy cho Parabol (P) : y = x2 v\u00e0 \u0111\u01b0\u1eddng th\u1eb3ng (d) : y = 2x + 3 . 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 8x \u2212 3 = 0 c\u00f3 hai nghi\u1ec7m x1 , x2 T\u00ednh gi\u00e1 tr\u1ecb A = x12 + x22 ; B = 2x1 + 2x2 \u2212 x12 x22 x2 x1 C\u00e2u 3. (1 \u0111i\u1ec3m). S\u1ed1 c\u00e2n n\u1eb7ng l\u00fd t\u01b0\u1edfng \u1ee9ng v\u1edbi chi\u1ec1u cao \u0111\u01b0\u1ee3c t\u00ednh theo c\u00f4ng th\u1ee9c: M = T \u2212 100 \u2212 T \u2212 150 N Trong \u0111\u00f3: M l\u00e0 c\u00e2n n\u1eb7ng t\u00ednh theo kg T l\u00e0 chi\u1ec1u cao t\u00ednh theo cm N = 4 n\u1ebfu l\u00e0 nam N = 2 n\u1ebfu l\u00e0 n\u1eef a) N\u1ebfu b\u1ea1n n\u1eef cao 1,58m . H\u1ecfi c\u00e2n n\u1eb7ng l\u00fd t\u01b0\u1edfng c\u1ee7a b\u1ea1n \u0111\u00f3 l\u00e0 bao nhi\u00eau? b) Gi\u1ea3 s\u1eed m\u1ed9t b\u1ea1n nam n\u1eb7ng 65kg . H\u1ecfi c\u00e2n n\u1eb7ng l\u00fd t\u01b0\u1edfng c\u1ee7a b\u1ea1n \u0111\u00f3 l\u00e0 bao nhi\u00eau? C\u00e2u 4. (1 \u0111i\u1ec3m). M\u1ed9t c\u00e2y kem c\u00f3 ph\u1ea7n b\u00e1nh h\u00ecnh n\u00f3n, ng\u01b0\u1eddi ta \u0111\u1ef1ng \u0111\u1ea7y kem trong ph\u1ea7n b\u00e1nh v\u00e0 th\u00eam m\u1ed9t n\u1eeda h\u00ecnh c\u1ea7u kem ph\u00eda tr\u00ean (xem h\u00ecnh). \u0110\u01b0\u1eddng k\u00ednh c\u1ee7a h\u00ecnh tr\u00f2n \u0111\u00e1y (ph\u00eda b\u00ean trong b\u00e1nh h\u00ecnh n\u00f3n) l\u00e0 4cm v\u00e0 \u0111\u1ed9 d\u00e0i \u0111\u01b0\u1eddng sinh b\u00ean trong h\u00ecnh n\u00f3n l\u00e0 8cm . T\u00ednh th\u1ec3 t\u00edch c\u1ee7a ph\u1ea7n kem. Cho bi\u1ebft: Th\u1ec3 t\u00edch h\u00ecnh n\u00f3n: V = 1 \uf070 R2h 3 (V\u1edbi R : b\u00e1n k\u00ednh \u0111\u01b0\u1eddng tr\u00f2n \u0111\u00e1y; h : chi\u1ec1u cao h\u00ecnh n\u00f3n) Th\u1ec3 t\u00edch h\u00ecnh c\u1ea7u: V = 4 \uf070 R3 3 (V\u1edbi R : b\u00e1n k\u00ednh h\u00ecnh c\u1ea7u) 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. (0,75 \u0111i\u1ec3m). M\u1ed9t ng\u01b0\u1eddi mua hai lo\u1ea1i m\u1eb7t h\u00e0ng A v\u00e0 B . N\u1ebfu t\u0103ng gi\u00e1 m\u1eb7t h\u00e0ng A th\u00eam 10% v\u00e0 m\u1eb7t h\u00e0ng B th\u00eam 20% th\u00ec ng\u01b0\u1eddi \u0111\u00f3 ph\u1ea3i tr\u1ea3 232 ngh\u00ecn \u0111\u1ed3ng. Nh\u01b0ng n\u1ebfu gi\u1ea3m gi\u00e1 c\u1ea3 hai m\u1eb7t h\u00e0ng 10% th\u00ec ng\u01b0\u1eddi \u0111\u00f3 ph\u1ea3i tr\u1ea3 t\u1ea5t c\u1ea3 180 ngh\u00ecn \u0111\u1ed3ng. T\u00ednh gi\u00e1 ti\u1ec1n m\u1ed7i lo\u1ea1i l\u00fac \u0111\u1ea7u? C\u00e2u 6. (1 \u0111i\u1ec3m). C\u00e0ng l\u00ean cao kh\u00f4ng kh\u00ed c\u00e0ng lo\u00e3ng n\u00ean \u00e1p su\u1ea5t kh\u00ed quy\u1ec3n c\u00e0ng gi\u1ea3m. V\u00ed d\u1ee5 c\u00e1c khu v\u1ef1c \u1edf th\u00e0nh ph\u1ed1 H\u1ed3 Ch\u00ed Minh \u0111\u1ec1u c\u00f3 \u0111\u1ed9 cao s\u00e1t m\u1ef1c n\u01b0\u1edbc bi\u1ec3n n\u00ean c\u00f3 \u00e1p su\u1ea5t kh\u00ed quy\u1ec3n l\u00e0 p = 760 mmHg ; c\u00f2n \u1edf th\u00e0nh ph\u1ed1 Puebla, Mexico c\u00f3 \u0111\u1ed9 cao h 2200 m so v\u1edbi m\u1ef1c n\u01b0\u1edbc bi\u1ec3n th\u00ec \u00e1p su\u1ea5t kh\u00ed quy\u1ec3n l\u00e0 p = 550, 4 mmHg . V\u1edbi nh\u1eefng \u0111\u1ed9 cao kh\u00f4ng l\u1edbn l\u1eafm th\u00ec ta c\u00f3 c\u00f4ng th\u1ee9c t\u00ednh \u00e1p su\u1ea5t kh\u00ed quy\u1ec3n t\u01b0\u01a1ng \u1ee9ng v\u1edbi \u0111\u1ed9 cao so v\u1edbi m\u1ef1c n\u01b0\u1edbc bi\u1ec3n l\u00e0 m\u1ed9t h\u00e0m s\u1ed1 b\u1eadc nh\u1ea5t p = ah + b(a \uf0b9 0) c\u00f3 \u0111\u1ed3 th\u1ecb nh\u01b0 h\u00ecnh v\u1ebd. a) X\u00e1c \u0111\u1ecbnh h\u1ec7 s\u1ed1 a v\u00e0 b . b) H\u1ecfi cao nguy\u00ean L\u00e2m \u0110\u1ed3ng c\u00f3 \u0111\u1ed9 cao 650 m so v\u1edbi m\u1ef1c n\u01b0\u1edbc bi\u1ec3n th\u00ec c\u00f3 \u00e1p su\u1ea5t kh\u00ed quy\u1ec3n l\u00e0 bao nhi\u00eau mmHg ? (l\u00e0m tr\u00f2n \u0111\u1ebfn ch\u1eef s\u1ed1 th\u1eadp ph\u00e2n th\u1ee9 nh\u1ea5t). C\u00e2u 7. (0,75 \u0111i\u1ec3m). \u0110\u1ec3 h\u00f2a chung v\u1edbi kh\u00f4ng kh\u00ed World Cup, \u1edf m\u1ed9t th\u00e0nh ph\u1ed1 t\u1ed5 ch\u1ee9c gi\u1ea3i b\u00f3ng \u0111\u00e1 l\u1ee9a tu\u1ed5i THCS bao g\u1ed3m 32 \u0111\u1ed9i tham gia chia th\u00e0nh 8 b\u1ea3ng. \u1ede v\u00f2ng b\u1ea3ng, 2 \u0111\u1ed9i c\u00f3 th\u1ee9 h\u1ea1ng cao nh\u1ea5t s\u1ebd \u0111\u01b0\u1ee3c \u0111i ti\u1ebfp v\u00e0o v\u00f2ng trong (v\u00f2ng lo\u1ea1i tr\u1ef1c ti\u1ebfp). Th\u1eafng \u0111\u01b0\u1ee3c 3 \u0111i\u1ec3m, h\u00f2a \u0111\u01b0\u1ee3c 1 \u0111i\u1ec3m, thua 0 \u0111i\u1ec3m. N\u1ebfu hai \u0111\u1ed9i c\u00f9ng \u0111i\u1ec3m s\u1ebd so hi\u1ec7u s\u1ed1 b\u00e0n th\u1eafng \u2013 thua. \u1ede b\u1ea3ng A , \u0111\u1ed9i D c\u1ee7a b\u1ea1n An n\u1eb1m trong b\u1ea3ng h\u1ea1t gi\u1ed1ng sau 2 l\u01b0\u1ee3t \u0111\u1ea5u s\u1ed1 h\u1ea1ng nh\u01b0 sau : \u2022 \u0110\u1ed9i A: 4 \u0111i\u1ec3m \u2022 \u0110\u1ed9i B: 2 \u0111i\u1ec3m \u2022 \u0110\u1ed9i C: 2 \u0111i\u1ec3m \u2022 \u0110\u1ed9i D: 1 \u0111i\u1ec3m \u1ede l\u01b0\u1ee3t \u0111\u1ea5u di\u1ec5n ra song song 2 tr\u1eadn A \u2212 C v\u00e0 B \u2212 D . C\u00e1c em h\u00e3y t\u00ednh x\u00e1c su\u1ea5t v\u00e0o v\u00f2ng trong c\u1ee7a \u0111\u1ed9i D bi\u1ebft r\u1eb1ng \u0111\u1ed9i D lu\u00f4n c\u00f3 hi\u1ec7u s\u1ed1 b\u00e0n th\u1eafng th\u1ea5p nh\u1ea5t ? X\u00e1c su\u1ea5t = (s\u1ed1 kh\u1ea3 n\u0103ng v\u00e0o v\u00f2ng trong): (s\u1ed1 kh\u1ea3 n\u0103ng x\u1ea3y ra). 100% 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 ( )C\u00e2u 8. (3 \u0111i\u1ec3m) Cho \u0111\u01b0\u1eddng tr\u00f2n O; R . T\u1eeb \u0111i\u1ec3m A b\u00ean ngo\u00e0i \u0111\u01b0\u1eddng tr\u00f2n k\u1ebb c\u00e1c ti\u1ebfp tuy\u1ebfn AB , AC ( )( B,C l\u00e0 ti\u1ebfp \u0111i\u1ec3m c\u1ee7a O ) v\u00e0 c\u00e1t tuy\u1ebfn ADE kh\u00f4ng qua t\u00e2m ( D n\u1eb1m gi\u1eefa A v\u00e0 E , AE c\u1eaft \u0111o\u1ea1n th\u1eb3ng OB ). G\u1ecdi I l\u00e0 trung \u0111i\u1ec3m c\u1ee7a DE . a) Ch\u1ee9ng minh 5 \u0111i\u1ec3m A,B,I,O,C c\u00f9ng thu\u1ed9c 1 \u0111\u01b0\u1eddng tr\u00f2n b) BC c\u1eaft AE t\u1ea1i K . Ch\u1ee9ng minh AB2 = AK.AI c) T\u1eeb D v\u1ebd DJ \/\/ AB ( J thu\u1ed9c BC ). Ch\u1ee9ng minh IJ \/\/ EB . ----H\u1ebeT--- T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N 3","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH H\u01af\u1edaNG D\u1eaaN GI\u1ea2I C\u00e2u 1. (1,5 \u0111i\u1ec3m). Trong m\u1eb7t ph\u1eb3ng t\u1ecda \u0111\u1ed9 Oxy cho Parabol (P) : y = x2 v\u00e0 \u0111\u01b0\u1eddng th\u1eb3ng (d) : y = 2x + 3 . a) V\u1ebd \u0111\u1ed3 th\u1ecb (P) v\u00e0 (d) tr\u00ean c\u00f9ng h\u1ec7 tr\u1ee5c t\u1ecda \u0111\u1ed9. b) T\u00ecm t\u1ecda \u0111\u1ed9 giao \u0111i\u1ec3m c\u1ee7a (P) v\u00e0 (d) b\u1eb1ng ph\u00e9p t\u00ednh. L\u1eddi gi\u1ea3i a) V\u1ebd \u0111\u1ed3 th\u1ecb (P) v\u00e0 (d) tr\u00ean c\u00f9ng h\u1ec7 tr\u1ee5c t\u1ecda \u0111\u1ed9. B\u1ea3ng gi\u00e1 tr\u1ecb: x \u22122 \u22121 0 12 y = x2 4 10 14 x0 1 y = 2x + 3 3 5 b) T\u00ecm t\u1ecda \u0111\u1ed9 giao \u0111i\u1ec3m c\u1ee7a (P) v\u00e0 (d) b\u1eb1ng ph\u00e9p t\u00ednh. Ph\u01b0\u01a1ng tr\u00ecnh ho\u00e0nh \u0111\u1ed9 giao \u0111i\u1ec3m c\u1ee7a (P) v\u00e0 (d) : x2 = 2x + 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 = 32 = 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 8x \u2212 3 = 0 c\u00f3 hai nghi\u1ec7m x1 , x2 . T\u00ednh gi\u00e1 tr\u1ecb A = x12 + x22 ; B= 2x1 + 2x2 \u2212 x12x22 x2 x1 L\u1eddi gi\u1ea3i V\u00ec \uf044 = b2 \u2212 4ac = (\u22128)2 \u2212 4.1.(\u22123) = 76 \uf03e 0 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 N\u00ean ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m ph\u00e2n bi\u1ec7t x1 , x2 . \uf0ec\uf0ef\uf0efS = x1 + x2 = \u2212b = 8 = 8 \uf0ed = x1 a1 Theo \u0111\u1ecbnh l\u00ed Vi-et, ta c\u00f3: \uf0ef\uf0ef\uf0eeP c = \u2212 3 = \u22123 a .x2 = 1 ( ) ( )Ta c\u00f3: A = x12 + x22 = x1 + x2 2 \u2212 2x1x2 = 82 \u2212 2. \u22123 = 70 ( ) ( )Ta c\u00f3: B = 2x1 + 2x2 \u2212 x12x22 2 x12 + x22 \u2212 x12x22 = 2.70 \u2212 \u22123 2 = \u2212167 x2 x1 = x1x2 \u22123 3 C\u00e2u 3. (1 \u0111i\u1ec3m). S\u1ed1 c\u00e2n n\u1eb7ng l\u00fd t\u01b0\u1edfng \u1ee9ng v\u1edbi chi\u1ec1u cao \u0111\u01b0\u1ee3c t\u00ednh theo c\u00f4ng th\u1ee9c: M = T \u2212 100 \u2212 T \u2212 150 N Trong \u0111\u00f3: M l\u00e0 c\u00e2n n\u1eb7ng t\u00ednh theo kg T l\u00e0 chi\u1ec1u cao t\u00ednh theo cm N = 4 n\u1ebfu l\u00e0 nam N = 2 n\u1ebfu l\u00e0 n\u1eef a) N\u1ebfu b\u1ea1n n\u1eef cao 1,58m . H\u1ecfi c\u00e2n n\u1eb7ng l\u00fd t\u01b0\u1edfng c\u1ee7a b\u1ea1n \u0111\u00f3 l\u00e0 bao nhi\u00eau? b) Gi\u1ea3 s\u1eed m\u1ed9t b\u1ea1n nam n\u1eb7ng 65kg . H\u1ecfi c\u00e2n n\u1eb7ng l\u00fd t\u01b0\u1edfng c\u1ee7a b\u1ea1n \u0111\u00f3 l\u00e0 bao nhi\u00eau? L\u1eddi gi\u1ea3i a) N\u1ebfu b\u1ea1n n\u1eef cao 1,58m . H\u1ecfi c\u00e2n n\u1eb7ng l\u00fd t\u01b0\u1edfng c\u1ee7a b\u1ea1n \u0111\u00f3 l\u00e0 bao nhi\u00eau? Thay T = 158,N = 2 v\u00e0o c\u00f4ng th\u1ee9c, ta c\u00f3 M = 158 \u2212 100 \u2212 158 \u2212 150 = 54kg 2 b) Gi\u1ea3 s\u1eed m\u1ed9t b\u1ea1n nam n\u1eb7ng 65kg . H\u1ecfi c\u00e2n n\u1eb7ng l\u00fd t\u01b0\u1edfng c\u1ee7a b\u1ea1n \u0111\u00f3 l\u00e0 bao nhi\u00eau? Thay M = 65,N = 4 v\u00e0o c\u00f4ng th\u1ee9c, ta c\u00f3 65 = T \u2212 100 \u2212 T \u2212 150 4 \uf0db 65 = 4T \u2212 400 \u2212 T + 150 4 \uf0db 260 = 3T \u2212 250 \uf0db T = 170cm C\u00e2u 4. (1 \u0111i\u1ec3m). M\u1ed9t c\u00e2y kem c\u00f3 ph\u1ea7n b\u00e1nh h\u00ecnh n\u00f3n, ng\u01b0\u1eddi ta \u0111\u1ef1ng \u0111\u1ea7y kem trong ph\u1ea7n b\u00e1nh v\u00e0 th\u00eam m\u1ed9t n\u1eeda h\u00ecnh c\u1ea7u kem ph\u00eda tr\u00ean (xem h\u00ecnh). \u0110\u01b0\u1eddng k\u00ednh c\u1ee7a h\u00ecnh tr\u00f2n \u0111\u00e1y (ph\u00eda b\u00ean trong b\u00e1nh h\u00ecnh n\u00f3n) l\u00e0 4cm v\u00e0 \u0111\u1ed9 d\u00e0i \u0111\u01b0\u1eddng sinh b\u00ean trong h\u00ecnh n\u00f3n l\u00e0 8cm . T\u00ednh th\u1ec3 t\u00edch c\u1ee7a ph\u1ea7n kem. 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 Cho bi\u1ebft: Th\u1ec3 t\u00edch h\u00ecnh n\u00f3n: V = 1 \uf070 R2h 3 (V\u1edbi R : b\u00e1n k\u00ednh \u0111\u01b0\u1eddng tr\u00f2n \u0111\u00e1y; h : chi\u1ec1u cao h\u00ecnh n\u00f3n) Th\u1ec3 t\u00edch h\u00ecnh c\u1ea7u: V = 4 \uf070 R3 3 (V\u1edbi R : b\u00e1n k\u00ednh h\u00ecnh c\u1ea7u) L\u1eddi gi\u1ea3i Chi\u1ec1u cao h\u00ecnh n\u00f3n: h = l2 \u2212 R2 = 82 \u2212 42 = 4 3cm Th\u1ec3 t\u00edch ph\u1ea7n kem trong b\u00e1nh: V1 = 1 \uf070 R2h = 1\uf070 42.4 3 = 64\uf070 3 cm3 3 3 3 Th\u1ec3 t\u00edch ph\u1ea7n kem h\u00ecnh c\u1ea7u: V2 = 4 \uf070 R3 = 4 \uf070 43 = 256\uf070 cm3 3 3 3 Th\u1ec3 t\u00edch ph\u1ea7n kem: V = V1 + V2 = 256\uf070 + 64\uf070 3 \uf0bb 384cm3 3 3 C\u00e2u 5. (0,75 \u0111i\u1ec3m). M\u1ed9t ng\u01b0\u1eddi mua hai lo\u1ea1i m\u1eb7t h\u00e0ng A v\u00e0 B . N\u1ebfu t\u0103ng gi\u00e1 m\u1eb7t h\u00e0ng A th\u00eam 10% v\u00e0 m\u1eb7t h\u00e0ng B th\u00eam 20% th\u00ec ng\u01b0\u1eddi \u0111\u00f3 ph\u1ea3i tr\u1ea3 232 ngh\u00ecn \u0111\u1ed3ng. Nh\u01b0ng n\u1ebfu gi\u1ea3m gi\u00e1 c\u1ea3 hai m\u1eb7t h\u00e0ng 10% th\u00ec ng\u01b0\u1eddi \u0111\u00f3 ph\u1ea3i tr\u1ea3 t\u1ea5t c\u1ea3 180 ngh\u00ecn \u0111\u1ed3ng. T\u00ednh gi\u00e1 ti\u1ec1n m\u1ed7i lo\u1ea1i l\u00fac \u0111\u1ea7u? L\u1eddi gi\u1ea3i 0 G\u1ecdi x (ngh\u00ecn \u0111\u1ed3ng) l\u00e0 gi\u00e1 ti\u1ec1n m\u1eb7t h\u00e0ng A x G\u1ecdi y (ngh\u00ecn \u0111\u1ed3ng) l\u00e0 gi\u00e1 ti\u1ec1n m\u1eb7t h\u00e0ng B y 0 N\u1ebfu t\u0103ng gi\u00e1 A th\u00eam 10% v\u00e0 B th\u00eam 20% th\u00ec ng\u01b0\u1eddi \u0111\u00f3 ph\u1ea3i tr\u1ea3 232 ngh\u00ecn \u0111\u1ed3ng \uf0de 1,1x + 1,2y = 232 (1) N\u1ebfu gi\u1ea3m gi\u00e1 c\u1ea3 hai m\u1eb7t h\u00e0ng 10% th\u00ec ng\u01b0\u1eddi \u0111\u00f3 ph\u1ea3i tr\u1ea3 t\u1ea5t c\u1ea3 180 ngh\u00ecn \u0111\u1ed3ng \uf0de 0,9x + 0,9y = 180 (2) T\u1eeb (1) v\u00e0 (2) \uf0de Ta c\u00f3 h\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh \uf0ec1,1x + 1,2y = 232 \uf0db \uf0ec\uf0efx = 80(n) \uf0ed\uf0ee0,9x + 0.9y = 180 \uf0ed = 120(n) \uf0ef\uf0eey V\u1eady Gi\u00e1 ti\u1ec1n m\u1eb7t h\u00e0ng A l\u00e0 80 ngh\u00ecn \u0111\u1ed3ng Gi\u00e1 ti\u1ec1n m\u1eb7t h\u00e0ng B l\u00e0 120 ngh\u00ecn \u0111\u1ed3ng T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N 6","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH C\u00e2u 6. (1 \u0111i\u1ec3m). C\u00e0ng l\u00ean cao kh\u00f4ng kh\u00ed c\u00e0ng lo\u00e3ng n\u00ean \u00e1p su\u1ea5t kh\u00ed quy\u1ec3n c\u00e0ng gi\u1ea3m. V\u00ed d\u1ee5 c\u00e1c khu v\u1ef1c \u1edf th\u00e0nh ph\u1ed1 H\u1ed3 Ch\u00ed Minh \u0111\u1ec1u c\u00f3 \u0111\u1ed9 cao s\u00e1t m\u1ef1c n\u01b0\u1edbc bi\u1ec3n n\u00ean c\u00f3 \u00e1p su\u1ea5t kh\u00ed quy\u1ec3n l\u00e0 p = 760 mmHg ; c\u00f2n \u1edf th\u00e0nh ph\u1ed1 Puebla, Mexico c\u00f3 \u0111\u1ed9 cao h 2200 m so v\u1edbi m\u1ef1c n\u01b0\u1edbc bi\u1ec3n th\u00ec \u00e1p su\u1ea5t kh\u00ed quy\u1ec3n l\u00e0 p = 550, 4 mmHg . V\u1edbi nh\u1eefng \u0111\u1ed9 cao kh\u00f4ng l\u1edbn l\u1eafm th\u00ec ta c\u00f3 c\u00f4ng th\u1ee9c t\u00ednh \u00e1p su\u1ea5t kh\u00ed quy\u1ec3n t\u01b0\u01a1ng \u1ee9ng v\u1edbi \u0111\u1ed9 cao so v\u1edbi m\u1ef1c n\u01b0\u1edbc bi\u1ec3n l\u00e0 m\u1ed9t h\u00e0m s\u1ed1 b\u1eadc nh\u1ea5t p = ah + b(a \uf0b9 0) c\u00f3 \u0111\u1ed3 th\u1ecb nh\u01b0 h\u00ecnh v\u1ebd. a) X\u00e1c \u0111\u1ecbnh h\u1ec7 s\u1ed1 a v\u00e0 b . b) H\u1ecfi cao nguy\u00ean L\u00e2m \u0110\u1ed3ng c\u00f3 \u0111\u1ed9 cao 650 m so v\u1edbi m\u1ef1c n\u01b0\u1edbc bi\u1ec3n th\u00ec c\u00f3 \u00e1p su\u1ea5t kh\u00ed quy\u1ec3n l\u00e0 bao nhi\u00eau mmHg ? (l\u00e0m tr\u00f2n \u0111\u1ebfn ch\u1eef s\u1ed1 th\u1eadp ph\u00e2n th\u1ee9 nh\u1ea5t). L\u1eddi gi\u1ea3i a) X\u00e1c \u0111\u1ecbnh c\u00e1c h\u1ec7 s\u1ed1 a v\u00e0 b . Theo \u0111\u1ec1 b\u00e0i, ta c\u00f3: \uf0ech = 0 \uf0de 760 = 0.a + b . (1) V\u1edbi \uf0ed\uf0eep = 760 V\u1edbi \uf0ech = 2200 \uf0de 550,4 = 2200.a + b . (2) \uf0ed = 550 , 4 \uf0ee p ( ) ( )T\u1eeb1 v\u00e0 2 ta c\u00f3 h\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh: \uf0ec0a + b = 760 \uf0db \uf0ec\uf0efa = \u2212 131 . \uf0ed\uf0ee2200a + b = 550,4 \uf0ed = 1375 \uf0ef\uf0eeb 760 V\u1eady: a = \u2212 131 , b = 760 v\u00e0 p = \u2212 131 h + 760 . 1375 1375 b) H\u1ecfi cao nguy\u00ean L\u00e2m \u0110\u1ed3ng c\u00f3 \u0111\u1ed9 cao 650 m so v\u1edbi m\u1ef1c n\u01b0\u1edbc bi\u1ec3n th\u00ec c\u00f3 \u00e1p su\u1ea5t kh\u00ed quy\u1ec3n l\u00e0 bao nhi\u00eau mmHg ? (l\u00e0m tr\u00f2n \u0111\u1ebfn ch\u1eef s\u1ed1 th\u1eadp ph\u00e2n th\u1ee9 nh\u1ea5t). \u00c1p su\u1ea5t t\u1ea1i cao nguy\u00ean L\u00e2m \u0110\u1ed3ng c\u00f3 chi\u1ec1u cao 650m : p = \u2212 131 .650 + 760 \uf0bb 698,1mmHg 1375 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 C\u00e2u 7. (0,75 \u0111i\u1ec3m). \u0110\u1ec3 h\u00f2a chung v\u1edbi kh\u00f4ng kh\u00ed World Cup, \u1edf m\u1ed9t th\u00e0nh ph\u1ed1 t\u1ed5 ch\u1ee9c gi\u1ea3i b\u00f3ng \u0111\u00e1 l\u1ee9a tu\u1ed5i THCS bao g\u1ed3m 32 \u0111\u1ed9i tham gia chia th\u00e0nh 8 b\u1ea3ng. \u1ede v\u00f2ng b\u1ea3ng, 2 \u0111\u1ed9i c\u00f3 th\u1ee9 h\u1ea1ng cao nh\u1ea5t s\u1ebd \u0111\u01b0\u1ee3c \u0111i ti\u1ebfp v\u00e0o v\u00f2ng trong (v\u00f2ng lo\u1ea1i tr\u1ef1c ti\u1ebfp). Th\u1eafng \u0111\u01b0\u1ee3c 3 \u0111i\u1ec3m, h\u00f2a \u0111\u01b0\u1ee3c 1 \u0111i\u1ec3m, thua 0 \u0111i\u1ec3m. N\u1ebfu hai \u0111\u1ed9i c\u00f9ng \u0111i\u1ec3m s\u1ebd so hi\u1ec7u s\u1ed1 b\u00e0n th\u1eafng \u2013 thua. \u1ede b\u1ea3ng A , \u0111\u1ed9i D c\u1ee7a b\u1ea1n An n\u1eb1m trong b\u1ea3ng h\u1ea1t gi\u1ed1ng sau 2 l\u01b0\u1ee3t \u0111\u1ea5u s\u1ed1 h\u1ea1ng nh\u01b0 sau : \u2022 \u0110\u1ed9i A: 4 \u0111i\u1ec3m \u2022 \u0110\u1ed9i B: 2 \u0111i\u1ec3m \u2022 \u0110\u1ed9i C: 2 \u0111i\u1ec3m \u2022 \u0110\u1ed9i D: 1 \u0111i\u1ec3m \u1ede l\u01b0\u1ee3t \u0111\u1ea5u di\u1ec5n ra song song 2 tr\u1eadn A-C v\u00e0 B-D. C\u00e1c em h\u00e3y t\u00ednh x\u00e1c su\u1ea5t v\u00e0o v\u00f2ng trong c\u1ee7a \u0111\u1ed9i D bi\u1ebft r\u1eb1ng \u0111\u1ed9i D lu\u00f4n c\u00f3 hi\u1ec7u s\u1ed1 b\u00e0n th\u1eafng th\u1ea5p nh\u1ea5t ? X\u00e1c su\u1ea5t = (s\u1ed1 kh\u1ea3 n\u0103ng v\u00e0o v\u00f2ng trong): (s\u1ed1 kh\u1ea3 n\u0103ng x\u1ea3y ra). 100% L\u1eddi gi\u1ea3i S\u1ed1 kh\u1ea3 n\u0103ng x\u1ea3y ra l\u00e0 9 S\u1ed1 kh\u1ea3 n\u0103ng Ph\u01b0\u1ee3ng Ho\u00e0ng v\u00e0o l\u00e0 2 TH1: A th\u1eafng C v\u00e0 B thua D : D v\u00e0o. TH2: A h\u00f2a C v\u00e0 B thua D : D v\u00e0o. V\u1eady x\u00e1c su\u1ea5t \u0111\u1ec3 \u0111\u1ed9i D \u0111\u01b0\u1ee3c v\u00e0o v\u00f2ng trong l\u00e0 2 .100% \uf0bb 22,2%. 9 C\u00e2u 8. (3 \u0111i\u1ec3m) Cho \u0111\u01b0\u1eddng tr\u00f2n (O; R) . T\u1eeb \u0111i\u1ec3m A b\u00ean ngo\u00e0i \u0111\u01b0\u1eddng tr\u00f2n k\u1ebb c\u00e1c ti\u1ebfp tuy\u1ebfn AB , AC ( B,C l\u00e0 ti\u1ebfp \u0111i\u1ec3m c\u1ee7a (O) ) v\u00e0 c\u00e1t tuy\u1ebfn ADE kh\u00f4ng qua t\u00e2m ( D n\u1eb1m gi\u1eefa A v\u00e0 E , AE c\u1eaft \u0111o\u1ea1n th\u1eb3ng OB ). G\u1ecdi I l\u00e0 trung \u0111i\u1ec3m c\u1ee7a DE . a) Ch\u1ee9ng minh 5 \u0111i\u1ec3m A,B,I,O,C c\u00f9ng thu\u1ed9c 1 \u0111\u01b0\u1eddng tr\u00f2n b) BC c\u1eaft AE t\u1ea1i K . Ch\u1ee9ng minh AB2 = AK \uf0d7 AI c) T\u1eeb D v\u1ebd DJ \/\/ AB ( J thu\u1ed9c BC ). Ch\u1ee9ng minh IJ \/\/ EB L\u1eddi gi\u1ea3i T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N 8","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH a) Ta c\u00f3 OI \u22a5 DE (\u0111\u01b0\u1eddng k\u00ednh vu\u00f4ng g\u00f3c v\u1edbi d\u00e2y cung t\u1ea1i trung \u0111i\u1ec3m d\u00e2y cung). \uf0de AIO = 90\uf0b0 \uf0de A, I ,O n\u1ed9i ti\u1ebfp \u0111\u01b0\u1eddng tr\u00f2n \u0111\u01b0\u1eddng k\u00ednh AO . Ta c\u00f3: ABO = ACO (= 90\uf0b0) \uf0de A, B,C,O n\u1ed9i ti\u1ebfp \u0111\u01b0\u1eddng tr\u00f2n \u0111\u01b0\u1eddng k\u00ednh AO . Suy ra: A, B, I ,O,C n\u1ed9i ti\u1ebfp \u0111\u01b0\u1eddng tr\u00f2n \u0111\u01b0\u1eddng k\u00ednh AO . b) Ta c\u00f3: BIA = BCA (c\u00f9ng ch\u1eafn cung AB \u0111\u01b0\u1eddng tr\u00f2n \u0111\u01b0\u1eddng k\u00ednh AO ) \uf0de BIA = CBA X\u00e9t \uf044ABK v\u00e0 \uf044AIB : \uf0ec\uf0ef AIB = ABK \uf0ed \uf0ef\uf0eeBAI (chung ) \uf0de \uf044ABK \uf044AIB ( g \u2212 g ) \uf0de AB = AK AI AB \uf0de AB2 = AK.AI c) Ta c\u00f3: CJD = CBA( DJ \/ \/ AB) M\u00e0 CBA = CIA (c\u00f9ng ch\u1eafn cung AC \u0111\u01b0\u1eddng tr\u00f2n \u0111\u01b0\u1eddng k\u00ednh AO ) Suy ra CJD = CIA \uf0de CDJI n\u1ed9i ti\u1ebfp \u0111\u01b0\u1eddng tr\u00f2n \uf0de CDE = CJI (c\u00f9ng ch\u1eafn cung CD ) M\u00e0 CDE = CBE (c\u00f9ng ch\u1eafn cung CE \u0111\u01b0\u1eddng tr\u00f2n t\u00e2m AO ) 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 Suy ra CJI = CBE \uf0de IJ \/ \/BE ( 2 g\u00f3c \u0111\u1ed3ng v\u1ecb b\u1eb1ng nhau) ----H\u1ebeT--- T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N 10","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH S\u1ede GD & \u0110T TP.H\u1ed2 CH\u00cd MINH \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 PH\u00d2NG GD & \u0110T QU\u1eacN 10 N\u0102M H\u1eccC: 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\u00e3 \u0111\u1ec1: Qu\u1eadn 10 - 4 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 = \u22122x2 v\u00e0 \u0111\u01b0\u1eddng th\u1eb3ng (d) : y = x \u2212 3 . 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 mx + m \u2212 1 = 0 ( m l\u00e0 tham s\u1ed1). T\u00ecm c\u00e1c gi\u00e1 tr\u1ecb c\u1ee7a m \u0111\u1ec3 ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m ph\u00e2n bi\u1ec7t x1 , x2 th\u1ecfa m\u00e3n x12 + x22 = x1 + x2 . C\u00e2u 3. (1 \u0111i\u1ec3m). M\u1ed9t nh\u00f3m b\u1ea1n h\u1ecdc sinh th\u1ef1c h\u00e0nh m\u00f4n c\u00f4ng ngh\u1ec7. C\u00f4 gi\u00e1o giao cho nh\u00f3m quan s\u00e1t v\u00e0 ghi l\u1ea1i chi\u1ec1u cao c\u1ee7a c\u00e2y m\u1ed7i tu\u1ea7n. Ban \u0111\u1ea7u c\u00f4 \u0111\u01b0a cho nh\u00f3m m\u00f4t lo\u1ea1i c\u00e2y non c\u00f3 chi\u1ec1u cao 2,56 cm . Sau hai tu\u1ea7n quan s\u00e1t th\u00ec chi\u1ec1u cao c\u1ee7a c\u00e2y t\u0103ng th\u00eam 1, 28cm . G\u1ecdi h (cm) l\u00e0 chi\u1ec1u cao c\u1ee7a c\u00e2y sau t (tu\u1ea7n) quan s\u00e1t li\u00ean h\u1ec7 b\u1eb1ng h\u00e0m s\u1ed1 h = at + b . a) X\u00e1c \u0111\u1ecbnh h\u1ec7 s\u1ed1 c\u1ee7a a,b ? b) H\u1ecfi sau \u00edt nh\u1ea5t bao nhi\u00eau ng\u00e0y k\u1ec3 t\u1eeb ng\u00e0y b\u1eaft \u0111\u1ea7u quan s\u00e1t th\u00ec c\u00e2y s\u1ebd \u0111at chi\u1ec1u cao 6, 76 cm ? C\u00e2u 4. (0,75 \u0111i\u1ec3m). S\u1ea3n l\u01b0\u1ee3ng c\u00e0 ph\u00ea xu\u1ea5t kh\u1ea9u c\u1ee7a Vi\u1ec7t Nam h\u00e0ng n\u0103m \u0111\u01b0\u1ee3c x\u00e1c \u0111\u1ecbnh theo h\u00e0m s\u1ed1 T =100n + 900 . V\u1edbi T l\u00e0 s\u1ea3n l\u01b0\u1ee3ng (\u0111\u01a1n v\u1ecb: ngh\u00ecn t\u1ea5n) v\u00e0 n l\u00e0 s\u1ed1 n\u0103m k\u1ec3 t\u1eeb n\u0103m 2005 . a) H\u00e3y t\u00ednh s\u1ea3n l\u01b0\u1ee3ng c\u00e0 ph\u00ea xu\u1ea5t kh\u1ea9u n\u0103m 2007 ? b) Theo h\u00e0m s\u1ed1 tr\u00ean th\u00ec s\u1ea3n l\u01b0\u1ee3ng c\u00e0 ph\u00ea xu\u1ea5t kh\u1ea9u \u0111\u1ea1t 1800 ngh\u00ecn t\u1ea5n v\u00e0o n\u0103m n\u00e0o? C\u00e2u 5. (1 \u0111i\u1ec3m). Nh\u00e2n d\u1ecbp l\u1ec5 30 \/ 04 , si\u00eau th\u1ecb \u0111i\u1ec7n m\u00e1y Nguy\u1ec5n Kim \u0111\u00e3 gi\u1ea3m gi\u00e1 nhi\u1ec1u m\u1eb7t h\u00e0ng \u0111\u1ec3 k\u00edch c\u1ea7u mua s\u1eafm. Gi\u00e1 ni\u00eam y\u1ebft t\u1ed5ng s\u1ed1 ti\u1ec1n m\u1ed9t tivi v\u00e0 m\u1ed9t m\u00e1y gi\u1eb7t l\u00e0 25, 4 tri\u1ec7u \u0111\u1ed3ng. Trong \u0111\u1ee3t n\u00e0y gi\u00e1 m\u1ed9t tivi gi\u1ea3m 40% , gi\u00e1 m\u1ed9t m\u00e1y gi\u1eb7t gi\u1ea3m 25% , n\u00ean b\u00e1c Hai mua m\u1ed9t tivi v\u00e0 m\u1ed9t m\u00e1y gi\u1eb7t v\u1edbi t\u1ed5ng s\u1ed1 ti\u1ec1n l\u00e0 16, 7 tri\u1ec7u \u0111\u1ed3ng. H\u1ecfi gi\u00e1 m\u1ed9t chi\u1ebfc tivi, m\u1ed9t chi\u1ebfc m\u00e1y gi\u1eb7t khi ch\u01b0a gi\u1ea3m gi\u00e1 l\u00e0 bao nhi\u00eau tri\u1ec7u \u0111\u1ed3ng? (l\u00e0m tr\u00f2n k\u1ebft qu\u1ea3 \u0111\u1ebfn ch\u1eef s\u1ed1 th\u1eadp ph\u00e2n th\u1ee9 nh\u1ea5t). C\u00e2u 6. (1 \u0111i\u1ec3m). M\u1ed9t h\u1ed9p s\u1eefa l\u1edbn h\u00ecnh h\u1ed9p ch\u1eef nh\u1eadt c\u00f3 di\u1ec7n t\u00edch \u0111\u00e1y l\u00e0 20 dm2 v\u00e0 chi\u1ec1u cao l\u00e0 3dm . Ng\u01b0\u1eddi ta r\u00f3t h\u1ebft s\u1eefa trong h\u1ed9p ra nh\u1eefng chai s\u1eefa nh\u1ecf m\u1ed7i chai c\u00f3 th\u1ec3 t\u00edch l\u00e0 0, 35 dm3 \u0111\u01b0\u1ee3c t\u1ea5t c\u1ea3 72 chai. H\u1ecfi l\u01b0\u1ee3ng s\u1eefa c\u00f3 trong h\u1ed9p chi\u1ebfm bao nhi\u00eau ph\u1ea7n tr\u0103m th\u1ec3 t\u00edch c\u1ee7a h\u1ed9p s\u1eefa? C\u00e2u 7. (1 \u0111i\u1ec3m). Ng\u01b0\u1eddi ta ph\u00e1t hi\u1ec7n ra r\u1eb1ng, g\u00f3c \u0111\u1ec3 n\u00e9m m\u1ed9t h\u00f2n \u0111\u00e1 \u0111i \u0111\u01b0\u1ee3c xa nh\u1ea5t tr\u00ean m\u1eb7t n\u01b0\u1edbc l\u00e0 20 \u0111\u1ed9. M\u1ed9t ng\u01b0\u1eddi cao 1, 7 m n\u00e9m m\u1ed9t h\u00f2n \u0111\u00e1 theo g\u00f3c 20 \u0111\u1ed9 xu\u1ed1ng m\u1eb7t h\u1ed3. H\u1ecfi kho\u1ea3ng c\u00e1ch t\u1eeb v\u1ecb tr\u00ed ng\u01b0\u1eddi \u0111\u00f3 \u0111\u1ebfn v\u1ecb tr\u00ed vi\u00ean \u0111\u00e1 ch\u1ea1m m\u1eb7t h\u1ed3 l\u00e0 bao xa. Bi\u1ebft v\u1ecb tr\u00ed h\u00f2n \u0111\u00e1 ngang t\u1ea7m \u0111\u1ea7u khi ng\u01b0\u1eddi \u0111\u00f3 n\u00e9m \u0111i. (L\u00e0m tr\u00f2n l\u1ea5y 1 ch\u1eef s\u1ed1 th\u1eadp ph\u00e2n). 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 8. (3 \u0111i\u1ec3m) Cho \u0111\u01b0\u1eddng tr\u00f2n v\u00e0 \u0111i\u1ec3m n\u1eb1m ngo\u00e0i \u0111\u01b0\u1eddng tr\u00f2n (O; R) . V\u1ebd hai ti\u1ebfp tuy\u1ebfn AB, AC c\u1ee7a (O) ( B,C l\u00e0 hai ti\u1ebfp \u0111i\u1ec3m). V\u1ebd c\u00e1t tuy\u1ebfn ADE c\u1ee7a (O) ( D, E thu\u1ed9c (O) ); D n\u1eb1m gi\u1eefa A v\u00e0 E ; tia AD n\u1eb1m gi\u1eefa hai tia AB v\u00e0 AO . a) Ch\u1ee9ng minh: AB2 = AD.AE . b) G\u1ecdi H l\u00e0 giao \u0111i\u1ec3m c\u1ee7a AO v\u00e0 BC . Ch\u1ee9ng minh t\u1ee9 gi\u00e1c DEOH n\u1ed9i ti\u1ebfp. c) \u0110\u01b0\u1eddng th\u1eb3ng AO c\u1eaft \u0111\u01b0\u1eddng tr\u00f2n (O) t\u1ea1i M v\u00e0 N ( M n\u1eb1m gi\u1eefa A v\u00e0 O ). Ch\u1ee9ng minh EH.AD = MH.AN . ----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 = \u22122x2 v\u00e0 \u0111\u01b0\u1eddng th\u1eb3ng (d) : y = x \u2212 3 . a) V\u1ebd \u0111\u1ed3 th\u1ecb (P) v\u00e0 (d) tr\u00ean c\u00f9ng h\u1ec7 tr\u1ee5c t\u1ecda \u0111\u1ed9. b) T\u00ecm t\u1ecda \u0111\u1ed9 giao \u0111i\u1ec3m c\u1ee7a (P) v\u00e0 (d) b\u1eb1ng ph\u00e9p t\u00ednh. L\u1eddi gi\u1ea3i a) V\u1ebd \u0111\u1ed3 th\u1ecb (P) v\u00e0 (d) tr\u00ean c\u00f9ng h\u1ec7 tr\u1ee5c t\u1ecda \u0111\u1ed9. BGT: x \u22122 \u22121 0 1 2 y = \u22122x2 \u22128 \u22122 0 \u22122 \u22128 x 01 y = x\u22123 \u22123 \u22122 b) T\u00ecm t\u1ecda \u0111\u1ed9 giao \u0111i\u00ea\u0309m c\u1ee7a (P) v\u00e0 (d) b\u1eb1ng ph\u00e9p t\u00ednh. Ph\u01b0\u01a1ng tr\u00ecnh ho\u00e0nh \u0111\u1ed9 giao \u0111i\u00ea\u0309m c\u1ee7a (P) v\u00e0 (d) : \u22122x2 = x \u2212 3 \uf0db 2x2 + x \u2212 3 = 0 \uf0e9x = 1 \uf0ea \uf0db \uf0ea\uf0ea\uf0ebx = \u2212 3 2 Thay x = 1 v\u00e0o (P) : y = \u22122x2 , ta \u0111\u01b0\u1ee3c: y = \u22122.12 = \u22122 . ( )Thay x = \u2212 3 v\u00e0o P : y = \u22122x2 , ta \u0111\u01b0\u1ee3c: y = \u22122.\uf0e6\uf0e7\uf0e8 \u2212 3 \uf0f62 = \u22129 . 2 \uf0f7\uf0f8 2 2 V\u00e2\u0323y (1; \u22122) , \uf0e6 \u22122; \u2212 9 \uf0f6 l\u00e0 hai giao \u0111i\u00ea\u0309m c\u1ea7n t\u00ecm. \uf0e7\uf0e8 2 \uf0f7\uf0f8 C\u00e2u 2. (1 \u0111i\u1ec3m). Cho ph\u01b0\u01a1ng tr\u00ecnh x2 \u2212 mx + m \u2212 1 = 0 (m l\u00e0 tham s\u1ed1). T\u00ecm c\u00e1c gi\u00e1 tr\u1ecb c\u1ee7a m \u0111\u1ec3 ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m ph\u00e2n bi\u1ec7t x1 ,x2 th\u1ecfa m\u00e3n x12 + x22 = x1 + x2 . L\u1eddi gi\u1ea3i x2 \u2212 mx + m \u2212 1 = 0 a 1;b m;c m 1 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 = (\u2212m)2 \u2212 4(m \u2212 1) = m2 \u2212 4m + 4 = (m \u2212 2)2 \uf0b3 0 N\u00ean \u0111\u00ea\u0309 ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m ph\u00e2n bi\u1ec7t x1 , x2 th\u00ec (m \u2212 2)2 \uf0b9 0 \uf0de m \u2212 2 \uf0b9 0 \uf0de m \uf0b9 2 V\u00e2\u0323y m \uf0b9 2 ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m ph\u00e2n bi\u1ec7t x1 ,x2 . Theo \u0111\u1ecbnh l\u00ed Vi-et, ta c\u00f3: \uf0ec\uf0ef\uf0efS = x1 + x2 = \u2212b = m \uf0ed = x1 a \uf0ef\uf0ef\uf0eeP .x2 = c = m\u22121 a Ta c\u00f3: x12 + x22 = x1 + x2 ( )\uf0de x1 + x2 2 \u2212 2x1x2 = x1 + x2 \uf0de m2 \u2212 2(m \u2212 1) = m \uf0de m2 \u2212 2m + 2 \u2212 m = 0 \uf0de m2 \u2212 3m + 2 = 0 \uf0de \uf0e9m = 1(n) \uf0ea\uf0ebm = 2(l) V\u00e2\u0323y m = 1tho\u0309a \u0111i\u00ea\u0300u ki\u1ec7n \u0111\u00ea\u0300 b\u00e0i. C\u00e2u 3. (1 \u0111i\u1ec3m). M\u1ed9t nh\u00f3m b\u1ea1n h\u1ecdc sinh th\u1ef1c h\u00e0nh m\u00f4n c\u00f4ng ngh\u1ec7. C\u00f4 gi\u00e1o giao cho nh\u00f3m quan s\u00e1t v\u00e0 ghi l\u1ea1i chi\u1ec1u cao c\u1ee7a c\u00e2y m\u1ed7i tu\u1ea7n. Ban \u0111\u1ea7u c\u00f4 \u0111\u01b0a cho nh\u00f3m m\u00f4t lo\u1ea1i c\u00e2y non c\u00f3 chi\u1ec1u cao 2,56cm. Sau hai tu\u1ea7n quan s\u00e1t th\u00ec chi\u1ec1u cao c\u1ee7a c\u00e2y t\u0103ng th\u00eam 1,28 cm. G\u1ecdi h (cm) l\u00e0 chi\u1ec1u cao c\u1ee7a c\u00e2y sau t (tu\u1ea7n) quan s\u00e1t li\u00ean h\u1ec7 b\u1eb1ng h\u00e0m s\u1ed1 h = at + b. a) X\u00e1c \u0111\u1ecbnh h\u1ec7 s\u1ed1 c\u1ee7a a,b? b) H\u1ecfi sau \u00edt nh\u1ea5t bao nhi\u00eau ng\u00e0y k\u1ec3 t\u1eeb ng\u00e0y b\u1eaft \u0111\u1ea7u quan s\u00e1t th\u00ec c\u00e2y s\u1ebd \u0111at chi\u1ec1u cao 6,76cm L\u1eddi gi\u1ea3i a) Xa\u0301c \u0111\u1ecbnh h\u1ec7 s\u1ed1 a , b . T\u1ea1i \uf0ect = 0 \uf0de 0a +b = 2 , 56 (1) . \uf0ed\uf0eeh = 2,56 T\u1ea1i \uf0ect = 2 + 1,28 = 3,84 \uf0de 2a + b = 3,84 (2). \uf0ed\uf0eeh = 2,56 T\u1eeb (1) v\u00e0 (2) ta c\u00f3 h\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh: \uf0ec0a + b = 2 , 56 \uf0db \uf0eca = 0,64 . \uf0ed\uf0ee2a + b = 3,84 \uf0ed\uf0eeb = 2 , 56 V\u00e2\u0323y \uf0eca = 0,64 v\u00e0 h = 0,64t + 2,56 . \uf0ed\uf0eeb = 2,56 b) Ho\u0309i sau bao nhi\u00eau ng\u00e0y k\u00ea\u0309 t\u1eeb ng\u00e0y b\u1eaft \u0111\u1ea7u quan s\u00e1t c\u00e2y s\u1ebd \u0111\u1ea1t \u0111\u01b0\u1ee3c chi\u00ea\u0300u cao 6,7 cm . 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 \u0110\u00ea\u0309 c\u00e2y \u0111\u1ea1t \u0111\u01b0\u1ee3c chi\u00ea\u0300u cao h = 6,7 cm , ta \u0111\u01b0\u1ee3c 6,7 = 0,64t + 2,56 \uf0de t \uf0bb 6,47 tu\u1ea7n V\u00e2\u0323y sau t = 6,47 tu\u1ea7n \uf0bb 45,29 ng\u00e0y th\u00ec c\u00e2y \u0111\u1ea1t \u0111\u01b0\u1ee3c chi\u00ea\u0300u cao 6,7cm . C\u00e2u 4. (0,75 \u0111i\u1ec3m). S\u1ea3n l\u01b0\u1ee3ng c\u00e0 ph\u00ea xu\u1ea5t kh\u1ea9u c\u1ee7a Vi\u1ec7t Nam h\u00e0ng n\u0103m \u0111\u01b0\u1ee3c x\u00e1c \u0111\u1ecbnh theo h\u00e0m s\u1ed1 T =100n + 900 . V\u1edbi T l\u00e0 s\u1ea3n l\u01b0\u1ee3ng (\u0111\u01a1n v\u1ecb: ngh\u00ecn t\u1ea5n) v\u00e0 n l\u00e0 s\u1ed1 n\u0103m k\u1ec3 t\u1eeb n\u0103m 2005. a. H\u00e3y t\u00ednh s\u1ea3n l\u01b0\u1ee3ng c\u00e0 ph\u00ea xu\u1ea5t kh\u1ea9u n\u0103m 2007? b. Theo h\u00e0m s\u1ed1 tr\u00ean th\u00ec s\u1ea3n l\u01b0\u1ee3ng c\u00e0 ph\u00ea xu\u1ea5t kh\u1ea9u \u0111\u1ea1t 1800 ngh\u00ecn t\u1ea5n v\u00e0o n\u0103m n\u00e0o? L\u1eddi gi\u1ea3i a) S\u1ea3n l\u01b0\u1ee3ng c\u00e0 ph\u00ea xu\u1ea5t kh\u1ea9u n\u0103m 2007 : T = 100.2 + 900 = 1100 (ngh\u00ecn t\u1ea5n) b) S\u1ea3n l\u01b0\u1ee3ng c\u00e0 ph\u00ea xu\u1ea5t kh\u1ea9u \u0111\u1ea1t 1800 ngh\u00ecn t\u1ea5n \uf0db 1800 = 100n + 900 \uf0db n = 9 . V\u00e2\u0323y s\u1ea3n l\u01b0\u1ee3ng c\u00e0 ph\u00ea xu\u1ea5t kh\u1ea9u \u0111\u1ea1t 1800 ngh\u00ecn t\u1ea5n v\u00e0o n\u0103m 2005 + 9 = 2014 . C\u00e2u 5. (1 \u0111i\u1ec3m). Nh\u00e2n d\u1ecbp l\u1ec5 30\/4, si\u00eau th\u1ecb \u0111i\u1ec7n m\u00e1y Nguy\u1ec5n Kim \u0111\u00e3 gi\u1ea3m gi\u00e1 nhi\u1ec1u m\u1eb7t h\u00e0ng \u0111\u1ec3 k\u00edch c\u1ea7u mua s\u1eafm. Gi\u00e1 ni\u00eam y\u1ebft t\u1ed5ng s\u1ed1 ti\u1ec1n m\u1ed9t tivi v\u00e0 m\u1ed9t m\u00e1y gi\u1eb7t l\u00e0 25,4 tri\u1ec7u \u0111\u1ed3ng. Trong \u0111\u1ee3t n\u00e0y gi\u00e1 m\u1ed9t tivi gi\u1ea3m 40%, gi\u00e1 m\u1ed9t m\u00e1y gi\u1eb7t gi\u1ea3m 25%, n\u00ean b\u00e1c Hai mua m\u1ed9t tivi v\u00e0 m\u1ed9t m\u00e1y gi\u1eb7t v\u1edbi t\u1ed5ng s\u1ed1 ti\u1ec1n l\u00e0 16,7 tri\u1ec7u \u0111\u1ed3ng. H\u1ecfi gi\u00e1 m\u1ed9t chi\u1ebfc tivi, m\u1ed9t chi\u1ebfc m\u00e1y gi\u1eb7t khi ch\u01b0a gi\u1ea3m gi\u00e1 l\u00e0 bao nhi\u00eau tri\u1ec7u \u0111\u1ed3ng? (l\u00e0m tr\u00f2n k\u1ebft qu\u1ea3 \u0111\u1ebfn ch\u1eef s\u1ed1 th\u1eadp ph\u00e2n th\u1ee9 nh\u1ea5t). L\u1eddi gi\u1ea3i ( )G\u1ecdi gi\u00e1 m\u1ed9t chi\u1ebfc tivi khi ch\u01b0a gi\u1ea3m gi\u00e1 l\u00e0 x (tri\u1ec7u \u0111\u1ed3ng), x \uf03e 0 Gi\u00e1 m\u1ed9t chi\u1ebfc m\u00e1y gi\u1eb7t khi ch\u01b0a gi\u1ea3m gi\u00e1 l\u00e0 25, 4 x (tri\u1ec7u \u0111\u1ed3ng) Sau khi gi\u1ea3m gi\u00e1 b\u00e1c Hai mua m\u1ed9t Tivi v\u00e0 m\u1ed9t m\u00e1y gi\u1eb7t v\u1edbi t\u1ed5ng s\u1ed1 ti\u00ea\u0300n l\u00e0 16, 7 tri\u1ec7u \u0111\u1ed3ng n\u00ean c\u00f3: x.60% 25, 4 x .75% 16, 7 x 15, 7 (nh\u00e2\u0323n) V\u00e2\u0323y gi\u00e1 m\u1ed9t chi\u1ebfc tivi khi ch\u01b0a gi\u1ea3m gi\u00e1 l\u00e0 15, 7 tri\u1ec7u \u0111\u1ed3ng Gi\u00e1 m\u1ed9t chi\u1ebfc m\u00e1y gi\u1eb7t khi ch\u01b0a gi\u1ea3m gi\u00e1 l\u00e0 25, 4 x 25, 4 15,7 9,7 tri\u1ec7u \u0111\u1ed3ng C\u00e2u 6. (1 \u0111i\u1ec3m). M\u1ed9t h\u1ed9p s\u1eefa l\u1edbn h\u00ecnh h\u1ed9p ch\u1eef nh\u1eadt c\u00f3 di\u1ec7n t\u00edch \u0111\u00e1y l\u00e0 20 dm2 v\u00e0 chi\u1ec1u cao l\u00e0 3 dm. Ng\u01b0\u1eddi ta r\u00f3t h\u1ebft s\u1eefa trong h\u1ed9p ra nh\u1eefng chai s\u1eefa nh\u1ecf m\u1ed7i chai c\u00f3 th\u1ec3 t\u00edch l\u00e0 0,35 dm3 \u0111\u01b0\u1ee3c t\u1ea5t c\u1ea3 72 chai. H\u1ecfi l\u01b0\u1ee3ng s\u1eefa c\u00f3 trong h\u1ed9p chi\u1ebfm bao nhi\u00eau ph\u1ea7n tr\u0103m th\u1ec3 t\u00edch c\u1ee7a h\u1ed9p s\u1eefa? L\u1eddi gi\u1ea3i ( )Th\u00ea\u0309 t\u00edch h\u1ed9p s\u1eefa h\u00ecnh h\u1ed9p ch\u1eef nh\u00e2\u0323t: 20.3 = 60 dm3 ( )T\u1ed5ng th\u00ea\u0309 t\u00edch 72 chai s\u1eefa: 0,35.72 = 25, 2 dm3 Ph\u1ea7n tr\u0103m th\u00ea\u0309 t\u00edch s\u1eefa c\u00f3 trong h\u1ed9p: 25, 2 .100 = 42% 5 60 T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH C\u00e2u 7. (1 \u0111i\u1ec3m). Ng\u01b0\u1eddi ta ph\u00e1t hi\u1ec7n ra r\u1eb1ng, g\u00f3c \u0111\u1ec3 n\u00e9m m\u1ed9t h\u00f2n \u0111\u00e1 \u0111i \u0111\u01b0\u1ee3c xa nh\u1ea5t tr\u00ean m\u1eb7t n\u01b0\u1edbc l\u00e0 20 \u0111\u1ed9. M\u1ed9t ng\u01b0\u1eddi cao 1,7 m n\u00e9m m\u1ed9t h\u00f2n \u0111\u00e1 theo g\u00f3c 20 \u0111\u1ed9 xu\u1ed1ng m\u1eb7t h\u1ed3. H\u1ecfi kho\u1ea3ng c\u00e1ch t\u1eeb v\u1ecb tr\u00ed ng\u01b0\u1eddi \u0111\u00f3 \u0111\u1ebfn v\u1ecb tr\u00ed vi\u00ean \u0111\u00e1 ch\u1ea1m m\u1eb7t h\u1ed3 l\u00e0 bao xa. Bi\u1ebft v\u1ecb tr\u00ed h\u00f2n \u0111\u00e1 ngang t\u1ea7m \u0111\u1ea7u khi ng\u01b0\u1eddi \u0111\u00f3 n\u00e9m \u0111i. (L\u00e0m tr\u00f2n l\u1ea5y 1 ch\u1eef s\u1ed1 th\u1eadp ph\u00e2n). L\u1eddi gi\u1ea3i G\u1ecdi: AB l\u00e0 chi\u00ea\u0300u cao c\u1ee7a ng\u01b0\u01a1\u0300i n\u00e9m h\u00f2n \u0111a\u0301 BC l\u00e0 kho\u1ea3ng c\u00e1ch t\u1eeb v\u1ecb tr\u00ed ng\u01b0\u01a1\u0300i \u0111\u00f3 \u0111\u1ebfn v\u1ecb tr\u00ed vi\u00ean \u0111a\u0301 ch\u1ea1m m\u1eb7t h\u1ed3 T\u1eeb \u0111\u00ea\u0300 b\u00e0i ta c\u00f3 h\u00ecnh v\u1ebd: D\u01b0\u0323a v\u00e0o h\u00ecnh v\u1ebd: X\u00e9t \uf044ABC vu\u00f4ng t\u1ea1i B c\u00f3: tan ACB = AB BC \uf0de tan 20o = 1,7 BC \uf0de BC = 1,7 \uf0bb 4,7 (m) tan 20o V\u00e2\u0323y kho\u1ea3ng c\u00e1ch t\u1eeb v\u1ecb tr\u00ed ng\u01b0\u01a1\u0300i \u0111\u00f3 \u0111\u1ebfn v\u1ecb tr\u00ed vi\u00ean \u0111a\u0301 ch\u1ea1m m\u1eb7t h\u1ed3 l\u00e0 4,7 m . C\u00e2u 7. (3 \u0111i\u1ec3m) Cho \u0111\u01b0\u1eddng tr\u00f2n v\u00e0 \u0111i\u1ec3m n\u1eb1m ngo\u00e0i \u0111\u01b0\u1eddng tr\u00f2n (O;R). V\u1ebd hai ti\u1ebfp tuy\u1ebfn AB, AC c\u1ee7a (O) (B,C l\u00e0 hai ti\u1ebfp \u0111i\u1ec3m). V\u1ebd c\u00e1t tuy\u1ebfn ADE c\u1ee7a (O) (D, E thu\u1ed9c (O)); D n\u1eb1m gi\u1eefa A v\u00e0 E ; tia AD n\u1eb1m gi\u1eefa hai tia AB v\u00e0 AO. a. Ch\u1ee9ng minh: AB2 = AD.AE. b. G\u1ecdi H l\u00e0 giao \u0111i\u1ec3m c\u1ee7a OA v\u00e0 BC . Ch\u1ee9ng minh t\u1ee9 gi\u00e1c DEOH n\u1ed9i ti\u1ebfp. c. \u0110\u01b0\u1eddng th\u1eb3ng AO c\u1eaft \u0111\u01b0\u1eddng tr\u00f2n (O) t\u1ea1i M v\u00e0 N ( M n\u1eb1m gi\u1eefa A v\u00e0 O). Ch\u1ee9ng minh EH.AD = MH.AN. L\u1eddi gi\u1ea3i T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N 6","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH a) Ch\u1ee9ng minh AB2 AD.AE . X\u00e9t \uf044ABD v\u00e0 ABE , ta c\u00f3: BAD v\u00e0 BAE l\u00e0 g\u00f3c chung ABD AEB 1 sdBD 2 \uf0de \uf044ABD \uf044AEB ( g \u2212 g ) AB AD AB2 AD.AE AE AB b) G\u1ecdi H l\u00e0 giao \u0111i\u1ec3m c\u1ee7a OA v\u00e0 BC . Ch\u1ee9ng minh t\u1ee9 gi\u00e1c DEOH n\u1ed9i ti\u1ebfp. Ta c\u00f3: AB2 AD.AE (cmt) AB2 AH.AO ( h\u1ec7 th\u1ee9c l\u01b0\u1ee3ng trong tam gi\u00e1c ABO vu\u00f4ng \u1edf B c\u00f3 \u0111\u01b0\u01a1\u0300ng cao BH ) AD.AE AH .AO AD AO AE AH X\u00e9t \uf044ADH v\u00e0 AOE , ta c\u00f3: DAH OAE l\u00e0 g\u00f3c chung AD AO AE AH ( ch\u1ee9ng minh tr\u00ean) \uf0de \uf044ADH \uf044AOE (c \u2212 g \u2212 c) ADH AOE (2 g\u00f3c t\u01b0\u01a1ng \u1ee9ng) X\u00e9t t\u1ee9 gi\u00e1c DEOH ta c\u00f3: ADH AOE 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 T\u1ee9 gi\u00e1c DEOH n\u1ed9i ti\u1ebfp ( c\u00f3 g\u00f3c ngo\u00e0i b\u1eb1ng g\u00f3c \u0111\u1ed1i trong kh\u00f4ng k\u00ea\u0300 v\u1edbi n\u00f3) c) \u0110\u01b0\u1eddng th\u1eb3ng AO c\u1eaft \u0111\u01b0\u1eddng tr\u00f2n O t\u1ea1i M v\u00e0 N ( M n\u1eb1m gi\u1eefa A v\u00e0 O ). Ch\u1ee9ng minh: EH.AD MH.AN Ta c\u00f3 = s\u0111 ( g\u00f3c n\u1ed9i ti\u1ebfp ch\u1eafn DM ) DOM = s\u0111 ( g\u00f3c \u1edf t\u00e2m ch\u1eafn cung DM ) DOM DEH ( 2 g\u00f3c \u1edf hai \u0111\u1ec9nh k\u00ea\u0300 c\u00f9ng nh\u00ecn 1 c\u1ea1nh DH trong t\u1ee9 gi\u00e1c DHOE n\u1ed9i ti\u1ebfp) DEM 1 DEH 2 EM l\u00e0 ph\u00e2n gi\u00e1c AEH \uf0de EH = MH (1) EA MA X\u00e9t \uf044AEM v\u00e0 AND , ta c\u00f3: A l\u00e0 g\u00f3c chung AEM AND 1 sdDM 2 \uf0de \uf044AEM \uf044AND (g \u2212 g) (sai k\u00ed hi\u1ec7u \u0111\u1ed3ng d\u1ea1ng ) \uf0de AE = AM (2) AN AD T\u1eeb (1) v\u00e0 (2) nh\u00e2n v\u1ebf theo v\u1ebf suy ra EH . AE = MH . AM \uf0db EH = MH \uf0db EH.AD = MH.AN AE AN AM AD AN AD ----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 TA\u00c2N PHU\u00d9 NA\u00caM HO\u00cfC: 2023 - 2024 \u0110\u1ec0 THAM KH\u1ea2O M\u00d4N: TO\u00c1N 9 M\u00c3 \u0110\u1ec0: Qu\u1eadn 11-1 \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. Trong m\u1eb7t ph\u1eafng t\u1ecda \u0111\u1ed9 Oxy cho Parabol (P) : y = x2 v\u00e0 \u0111\u01b0\u1eddng th\u1eb3ng (d) : y = x + 2 . a) V\u1ebd (P) v\u00e0 (d) tr\u00ean c\u00f9ng m\u1ed9t 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. Cho ph\u01b0\u01a1ng tr\u00ecnh: 2x2 \u2212 x \u2212 3 = 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 = x12 + x22 \u2212 x12x22 \u2212 2024 . C\u00e2u 3. H\u1ee3p t\u00e1c x\u00e3 A chuy\u00ean tr\u1ed3ng hoa m\u00e0u \u0111\u1ec3 b\u00e1n. Nh\u01b0ng n\u0103m nay ch\u1ecbu \u0111\u1ee3t s\u00e2u h\u1ea1i n\u00ean s\u1ed1 l\u01b0\u1ee3ng hoa m\u00e0u d\u1ef1 \u0111\u1ecbnh b\u00e1n ra \u0111\u00e3 h\u01b0 30% v\u00e0 ph\u1ea7n c\u00f2n l\u1ea1i c\u0169ng \u1ea3nh h\u01b0\u1edfng n\u00ean ch\u1ec9 b\u00e1n \u0111\u01b0\u1ee3c v\u1edbi gi\u00e1 b\u00e1n b\u1eb1ng 3 gi\u00e1 b\u00e1n d\u1ef1 \u0111\u1ecbnh l\u00fac \u0111\u1ea7u. N\u1ebfu b\u00e1n h\u1ebft ph\u1ea7n c\u00f2n l\u1ea1i n\u00e0y v\u1edbi gi\u00e1 nh\u01b0 tr\u00ean th\u00ec s\u1ed1 ti\u1ec1n 4 s\u1ebd \u00edt h\u01a1n 152 tri\u1ec7u \u0111\u1ed3ng so v\u1edbi d\u1ef1 t\u00ednh l\u00fac \u0111\u1ea7u. H\u1ecfi n\u1ebfu kh\u00f4ng b\u1ecb h\u01b0 h\u1ea1i v\u00e0 kh\u00f4ng gi\u1ea3m gi\u00e1 th\u00ec theo d\u1ef1 t\u00ednh, h\u1ee3p t\u00e1c x\u00e3 n\u00e0y s\u1ebd thu v\u1ec1 bao nhi\u00eau ti\u1ec1n t\u1eeb hoa m\u00e0u? C\u00e2u 4. M\u1ed9t v\u00e9 xem phim c\u00f3 m\u1ee9c gi\u00e1 l\u00e0 60000 \u0111\u1ed3ng. Trong d\u1ecbp khuy\u1ebfn m\u00e3i cu\u1ed1i n\u0103m 2019, s\u1ed1 l\u01b0\u1ee3ng ng\u01b0\u1eddi xem phim t\u0103ng th\u00eam 45% n\u00ean t\u1ed5ng doanh thu c\u0169ng t\u0103ng 8,75% . H\u1ecfi r\u1ea1p phim \u0111\u00e3 gi\u1ea3m gi\u00e1 m\u1ed7i v\u00e9 bao nhi\u00eau ph\u1ea7n tr\u0103m so v\u1edbi gi\u00e1 ban \u0111\u1ea7u? C\u00e2u 5. C\u00f3 2 \u0111\u1ed9i c\u00f4ng nh\u00e2n c\u00f9ng l\u00e0m 1 c\u00f4ng tr\u00ecnh. L\u1ea7n th\u1ee9 nh\u1ea5t: \u0111\u1ed9i 1 l\u00e0m trong 6 ng\u00e0y, \u0111\u1ed9i 2 l\u00e0m trong 4 ng\u00e0y th\u00ec xong c\u00f4ng tr\u00ecnh. L\u1ea7n th\u1ee9 hai: \u0111\u1ed9i 1 l\u00e0m trong 4 ng\u00e0y, \u0111\u1ed9i 2 l\u00e0m trong 8 ng\u00e0y th\u00ec xong c\u00f4ng tr\u00ecnh. H\u1ecfi n\u1ebfu l\u00e0m m\u1ed9t m\u00ecnh th\u00ec m\u1ed7i \u0111\u1ed9i c\u00f4ng nh\u00e2n l\u00e0m trong bao l\u00e2u ho\u00e0n th\u00e0nh c\u00f4ng tr\u00ecnh? 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. M\u1ed9t chi\u1ebfc b\u00e0n h\u00ecnh tr\u00f2n \u0111\u01b0\u1ee3c gh\u00e9p b\u1edfi hai n\u1eeda h\u00ecnh tr\u00f2n \u0111\u01b0\u1eddng k\u00ednh 1,2 m . Ng\u01b0\u1eddi ta mu\u1ed1n n\u1edbi r\u1ed9ng m\u1eb7t b\u00e0n b\u1eb1ng c\u00e1ch gh\u00e9p th\u00eam v\u00e0o gi\u1eefa m\u1ed9t m\u1eb7t h\u00ecnh ch\u1eef nh\u1eadt c\u00f3 1 k\u00edch th\u01b0\u1edbc l\u00e0 1,2 m nh\u01b0 h\u00ecnh v\u1ebd d\u01b0\u1edbi. a) K\u00edch th\u01b0\u1edbc kia c\u1ee7a h\u00ecnh ch\u1eef nh\u1eadt ph\u1ea3i l\u00e0 bao nhi\u00eau \u0111\u1ec3 di\u1ec7n t\u00edch m\u1eb7t b\u00e0n t\u0103ng g\u1ea5p ba sau khi n\u1edbi? b) K\u00edch th\u01b0\u1edbc kia c\u1ee7a h\u00ecnh ch\u1eef nh\u1eadt ph\u1ea3i l\u00e0 bao nhi\u00eau \u0111\u1ec3 chu vi m\u1eb7t b\u00e0n t\u0103ng g\u1ea5p \u0111\u00f4i sau khi n\u1edbi? C\u00e2u 7. M\u1ed9t kh\u1ed1i g\u1ed7 h\u00ecnh tr\u1ee5 c\u00f3 b\u00e1n k\u00ednh \u0111\u00e1y l\u00e0 3 cm , chi\u1ec1u cao 4 cm \u0111\u01b0\u1ee3c \u0111\u1eb7t \u0111\u1ee9ng tr\u00ean m\u1eb7t b\u00e0n. M\u1ed9t ph\u1ea7n c\u1ee7a kh\u1ed1i g\u1ed7 b\u1ecb c\u1eaft r\u1eddi theo c\u00e1c b\u00e1n k\u00ednh OA,OB v\u00e0 theo chi\u1ec1u d\u00e0i th\u1eb3ng \u0111\u1ee9ng t\u1eeb tr\u00ean xu\u1ed1ng d\u01b0\u1edbi v\u1edbi AOB = 30\uf0b0 nh\u01b0 h\u00ecnh v\u1ebd b\u00ean d\u01b0\u1edbi: a) T\u00ednh th\u1ec3 t\u00edch c\u1ee7a kh\u1ed1i g\u1ed7 c\u00f2n l\u1ea1i sau khi b\u1ecb c\u1eaft r\u1eddi. b) Di\u1ec7n t\u00edch to\u00e0n ph\u1ea7n c\u1ee7a kh\u1ed1i g\u1ed7 c\u00f2n l\u1ea1i sau khi \u0111\u00e3 b\u1ecb c\u1eaft. C\u00e2u 8. Cho \uf044ABC n\u1ed9i ti\u1ebfp trong \u0111\u01b0\u1eddng tr\u00f2n (O; R) . Ba \u0111\u01b0\u1eddng cao AD,BE,CF c\u1eaft nhau t\u1ea1i H . a) Ch\u1ee9ng minh c\u00e1c t\u1ee9 gi\u00e1c AEHF,BCEF n\u1ed9i ti\u1ebfp. b) K\u1ebb \u0111\u01b0\u1eddng k\u00ednh AK c\u1ee7a (O) . Ch\u1ee9ng minh AB.AC = 2R.AD . c) G\u1ecdi M l\u00e0 trung \u0111i\u1ec3m c\u1ee7a BC,I l\u00e0 giao \u0111i\u1ec3m EF v\u00e0 BC . Ch\u1ee9ng minh t\u1ee9 gi\u00e1c EFDM n\u1ed9i ti\u1ebfp v\u00e0 IB.IC = ID.IM . ----H\u1ebeT--- T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N 2","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH H\u01af\u1edaNG D\u1eaaN GI\u1ea2I C\u00e2u 1. Trong m\u1eb7t ph\u1eafng t\u1ecda \u0111\u1ed9 Oxy cho Parabol ( P) : y = x2 v\u00e0 \u0111\u01b0\u1eddng th\u1eb3ng (d ) : y = x + 2 . a) V\u1ebd ( P) v\u00e0 (d ) tr\u00ean c\u00f9ng m\u1ed9t 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) ( P) : y = x2 x \u22122 \u22121 0 1 2 y = x2 4 1 0 1 4 (d): y = x + 2 x \u22122 \u22121 0 1 2 y = x+2 0 1 2 3 4 b) T\u1ecda \u0111\u1ed9 giao \u0111i\u1ec3m ( P) v\u00e0 (d ) . Ph\u01b0\u01a1ng tr\u00ecnh ho\u00e0nh \u0111\u1ed9 giao \u0111i\u1ec3m ( P) v\u00e0 (d ) 3 x2 = x + 2 \uf0db x2 \u2212 x \u2212 2 = 0 \uf0db ( x \u2212 2)( x +1) = 0 T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH \uf0db \uf0e9x \u22122=0 \uf0ea\uf0eb x +1= 0 Th\u1ebf x \u2212 2 v\u00e0o ( P) y = x2 \uf0db y = 22 = 4 V\u1eady t\u1ecda \u0111\u1ed9 giao \u0111i\u1ec3m ( P) v\u00e0 (d ) l\u00e0 (2; 4) C\u00e2u 2. Cho ph\u01b0\u01a1ng tr\u00ecnh: 2x2 \u2212 x \u2212 3 = 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 = x12 + x22 \u2212 x12 x22 \u2212 2024. L\u1eddi gi\u1ea3i 2x2 \u2212 x \u2212 3 = 0 (a = 2;b = \u22121;c = \u22123) \uf044 = b2 \u2212 4ac \uf044 = (\u22121)2 \u2212 4.2(\u22123) \uf044 = 1 + 24 \uf044 = 25 \uf03e 0 V\u00ec \uf044 \uf03e 0 n\u00ean ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 2 nghi\u1ec7m ph\u00e2n bi\u1ec7t \u00c1p d\u1ee5ng \u0111\u1ecbnh l\u00fd Vi-et ta c\u00f3: \uf0ec x1 + x2 = \u2212 b = 1 \uf0ef\uf0ef a 2 \uf0ed \uf0ef c 3 \uf0ef\uf0ee x1 x2 = a = \u2212 2 Ta c\u00f3: A = x2 + x22 \u2212 x12 x12 \u2212 2024 A = \uf0e9\uf0eb( x1 + x2 )2 \u2212 2 x1 x2 \uf0f9 \u2212 ( x1x2 )2 \u2212 2014 \uf0fb A = \uf0e9\uf0ea\uf0ea\uf0eb\uf0e6\uf0e7\uf0e8 1 \uf0f62 .2.\uf0e6\uf0e7\uf0e8 \u2212 3 \uf0f6\uf0f7\uf0f8\uf0f9\uf0fa\uf0fa\uf0fb \u2212 \uf0e6 \u2212 3 \uf0f62 \u2212 2024 2 \uf0f7\uf0f8 2 \uf0e7\uf0e8 2 \uf0f7\uf0f8 A = \uf0e9 \u22121 + 3\uf0f6\uf0f8\uf0f7 \u2212 9 \u2212 2024 \uf0ea\uf0eb 4 4 A = \u22122023 4 V\u1eady gi\u00e1 tr\u1ecb c\u1ee7a bi\u1ec3u th\u1ee9c A l\u00e0 \u22122023 T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH C\u00e2u 3. H\u1ee3p t\u00e1c x\u00e3 A chuy\u00ean tr\u1ed3ng hoa m\u00e0u \u0111\u1ec3 b\u00e1n. Nh\u01b0ng n\u0103m nay ch\u1ecbu \u0111\u1ee3t s\u00e2u h\u1ea1i n\u00ean s\u1ed1 l\u01b0\u1ee3ng hoa m\u00e0u d\u1ef1 \u0111\u1ecbnh b\u00e1n ra \u0111\u00e3 h\u01b0 30% v\u00e0 ph\u1ea7n c\u00f2n l\u1ea1i c\u0169ng \u1ea3nh h\u01b0\u1edfng n\u00ean ch\u1ec9 b\u00e1n \u0111\u01b0\u1ee3c v\u1edbi gi\u00e1 b\u00e1n b\u1eb1ng 3 gi\u00e1 b\u00e1n d\u1ef1 \u0111\u1ecbnh l\u00fac \u0111\u1ea7u. N\u1ebfu b\u00e1n h\u1ebft ph\u1ea7n c\u00f2n l\u1ea1i n\u00e0y v\u1edbi gi\u00e1 nh\u01b0 tr\u00ean th\u00ec s\u1ed1 ti\u1ec1n s\u1ebd \u00edt h\u01a1n 152 4 tri\u1ec7u \u0111\u1ed3ng so v\u1edbi d\u1ef1 t\u00ednh l\u00fac \u0111\u1ea7u. H\u1ecfi n\u1ebfu kh\u00f4ng b\u1ecb h\u01b0 h\u1ea1i v\u00e0 kh\u00f4ng gi\u1ea3m gi\u00e1 th\u00ec theo d\u1ef1 t\u00ednh, h\u1ee3p t\u00e1c x\u00e3 n\u00e0y s\u1ebd thu v\u1ec1 bao nhi\u00eau ti\u1ec1n t\u1eeb hoa m\u00e0u? L\u1eddi gi\u1ea3i G\u1ecdi x l\u00e0 s\u1ed1 hoa d\u1ef1 \u0111\u1ecbnh b\u00e1n v\u00e0 y l\u00e0 gi\u00e1 t\u00ecnh d\u1ef1 \u0111\u1ecbnh b\u00e1n ( x \uf0ce * , y \uf03e 0 ) S\u1ed1 ti\u1ec1n theo d\u1ef1 t\u00ednh x\u00e3 A nh\u1eadn \u0111\u01b0\u1ee3c khi b\u00e1n h\u1ebft hoa l\u00e0: xy (\u0111) S\u1ed1 l\u01b0\u1ee3ng hoa c\u00f2n l\u1ea1i sau khi h\u01b0 30% : x(100% \u2212 30%) = 0.7x (hoa) Gi\u00e1 b\u00e1n c\u1ee7a hoa l\u00fac sau: 3 y 4 N\u1ebfu b\u00e1n h\u1ebft ph\u1ea7n c\u00f2n l\u1ea1i v\u1edbi gi\u00e1 nh\u01b0 tr\u00ean th\u00ec \u00edt h\u01a1n 152 tri\u1ec7u \u0111\u1ed3ng so v\u1edbi d\u1ef1 t\u00ednh l\u00fac \u0111\u1ea7u: 0, 7x. 3 y = xy \u2212152 (tri\u1ec7u) 4 \uf0db 21 xy \u2212 xy = 152 (tri\u1ec7u) 40 \uf0db \u2212 19 xy = \u2212152 (tri\u1ec7u) 40 \uf0db xy = 320 (tri\u1ec7u) C\u00e2u 4. V\u1eady n\u1ebfu kh\u00f4ng b\u1ecb h\u01b0 h\u1ea1i v\u00e0 kh\u00f4ng gi\u1ea3m gi\u00e1 tr\u1ecb th\u00ec theo d\u1ef1 t\u00ednh s\u1ebd thu v\u1ec1 320 tri\u1ec7u \u0111\u1ed3ng. M\u1ed9t v\u00e9 xem phim c\u00f3 m\u1ee9c gi\u00e1 l\u00e0 60000 \u0111\u1ed3ng. Trong d\u1ecbp khuy\u1ebfn m\u00e3i cu\u1ed1i n\u0103m 2019, s\u1ed1 l\u01b0\u1ee3ng ng\u01b0\u1eddi xem phim t\u0103ng th\u00eam 45% n\u00ean t\u1ed5ng doanh thu c\u0169ng t\u0103ng 8, 75% . H\u1ecfi r\u1ea1p phim \u0111\u00e3 gi\u1ea3m gi\u00e1 m\u1ed7i v\u00e9 bao nhi\u00eau ph\u1ea7n tr\u0103m so v\u1edbi gi\u00e1 ban \u0111\u1ea7u? 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 G\u1ecdi x l\u00e0 s\u1ed1 l\u01b0\u1ee3ng ng\u01b0\u1eddi xem tr\u01b0\u1edbc khi khuy\u1ebfn m\u00e3i (ng\u01b0\u1eddi, x \uf0ce * ) Doanh thu l\u00fac \u0111\u00f3 : 60000x (\u0111) V\u00ec khi c\u00f3 \u0111\u1ee3t khuy\u1ec5n m\u00e3i s\u1ed1 l\u01b0\u1ee3ng ng\u01b0\u1eddi xem t\u0103ng th\u1ebfm 45% x(100% + 45%) = 1, 45x Doanh thu \u0111\u1ee3t khuy\u1ebfn m\u00e3: 60000x(100% + 8, 75%) = 65250x (\u0111) Gi\u00e1 v\u00e9 l\u00fac sau: 65250x :1, 45x = 45000 (\u0111) G\u1ecdi y l\u00e0 % gi\u1ea3m gi\u00e1 cho m\u1ed7i v\u00e9 xem phim (%, y \uf03e 0 ) Theo \u0111\u1ec1 b\u00e0i ta c\u00f3: 60000(100% \u2212 y%) = 45000 \uf0db 100% \u2212 y% = 0, 75 \uf0db y% = 0, 25 \uf0db y = 0, 25% V\u1eady ph\u1ea7n tr\u0103m gi\u1ea3m gi\u00e1 cho m\u1ed7i v\u00e9 l\u00e0 25% C\u00e2u 5. C\u00f3 2 \u0111\u1ed9i c\u00f4ng nh\u00e2n c\u00f9ng l\u00e0m 1 c\u00f4ng tr\u00ecnh. L\u1ea7n th\u1ee9 nh\u1ea5t: \u0111\u1ed9i 1 l\u00e0m trong 6 ng\u00e0y, \u0111\u1ed9i 2 l\u00e0m trong 4 ng\u00e0y th\u00ec xong c\u00f4ng tr\u00ecnh. L\u1ea7n th\u1ee9 hai: \u0111\u1ed9i 1 l\u00e0m trong 4 ng\u00e0y, \u0111\u1ed9i 2 l\u00e0m trong 8 ng\u00e0y th\u00ec xong c\u00f4ng tr\u00ecnh. H\u1ecfi n\u1ebfu l\u00e0m m\u1ed9t m\u00ecnh th\u00ec m\u1ed7i \u0111\u1ed9i c\u00f4ng nh\u00e2n l\u00e0m trong bao l\u00e2u ho\u00e0n th\u00e0nh c\u00f4ng tr\u00ecnh? L\u1eddi gi\u1ea3i G\u1ecdi x l\u00e0 th\u1eddi gian \u0111\u1ed9i 1 l\u00e0m m\u1ed9t m\u00ecnh xong c\u00f4ng vi\u1ec7c ( x \uf03e 0 ) G\u1ecdi y l\u00e0 th\u1eddi gian \u0111\u1ed9i 2 l\u00e0m m\u1ed9t m\u00ecnh xong c\u00f4ng vi\u1ec7c ( y \uf03e 0 ) V\u00ec l\u1ea7n th\u1ee9 nh\u1ea5t \u0111\u1ed9i 1 l\u00e0m 6 ng\u00e0y , \u0111\u1ed9i 2 l\u00e0m 4 ng\u00e0y th\u00ec xong c\u00f4ng vi\u1ec7c n\u00ean: 6 + 4 = 1 (1) xy V\u00ec l\u1ea7n th\u1ee9 nh\u1ea5t \u0111\u1ed9i 2 l\u00e0m 4 ng\u00e0y , \u0111\u1ed9i 2 l\u00e0m 8 ng\u00e0y th\u00ec xong c\u00f4ng vi\u1ec7c n\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 4 + 8 = 1 (2) xy \uf0ec6 + 4 = 1 \uf0ef\uf0ef x + y = 1 T\u1eeb (1), (2) \uf0db \uf0ed 8 \uf0ef 4 y \uf0ef\uf0ee x \u0110\u1eb7t a = 1 ; b = 1 xy \uf0db \uf0ec6a + 4b =1 \uf0ed\uf0ee4a + 8b =1 \uf0db \uf0ec\uf0ef\uf0efa = 1 \u2212 4b \uf0ed \uf0ef 4 (1 6 8b = 1 \uf0ef\uf0ee \u2212 4b) + 6 \uf0db \uf0ec\uf0ef\uf0efa = 1 \u2212 4b \uf0ed 6 \uf0ef 4 48b = 6 \u221216b + 6 6 \uf0ef\uf0ee 6 \uf0ec\uf0efa = 1 \u2212 4b \uf0ed 6 \uf0db \uf0ef\uf0ee4 \u221216b + 48b = 6 \uf0ec\uf0ef3 = 1 \u2212 4b \uf0ed 6 \uf0db \uf0ef\uf0ee32b = 2 \uf0db \uf0ec\uf0ef\uf0efa = 1 \u2212 4b \uf0db \uf0ec\uf0ef\uf0efa = 1 \uf0ed = 6 \uf0ed = 8 \uf0ef\uf0ef\uf0eeb \uf0ef\uf0ef\uf0eeb 1 1 16 16 \uf0ec\uf0ef\uf0efa = 1 = 1 \uf0ec1 = 1 \uf0ed = x = 8 \uf0ef\uf0ef x = 8 \uf0db \uf0ef\uf0ef\uf0eeb 1 1 \uf0db \uf0ed 1 \uf0db \uf0ecx = 8 y 16 \uf0ef 16 \uf0ed = 16 1 \uf0ee y \uf0ef\uf0ee y V\u1eady n\u1ebfu hai \u0111\u1ed9i l\u00e0m m\u1ed9t m\u00ecnh th\u00ec \u0111\u1ed9i 1 h\u1ebft 8 ng\u00e0y, \u0111\u1ed9i 2 h\u1ebft 16 ng\u00e0y l\u00e0 xong c\u00f4ng vi\u1ec7c. 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 C\u00e2u 6. M\u1ed9t chi\u1ebfc b\u00e0n h\u00ecnh tr\u00f2n \u0111\u01b0\u1ee3c gh\u00e9p b\u1edfi hai n\u1eeda h\u00ecnh tr\u00f2n \u0111\u01b0\u1eddng k\u00ednh 1, 2m . Ng\u01b0\u1eddi ta mu\u1ed1n n\u1edbi r\u1ed9ng m\u1eb7t b\u00e0n b\u1eb1ng c\u00e1ch gh\u00e9p th\u00eam v\u00e0o gi\u1eefa m\u1ed9t m\u1eb7t h\u00ecnh ch\u1eef nh\u1eadt c\u00f3 1 k\u00edch th\u01b0\u1edbc l\u00e0 1, 2m nh\u01b0 h\u00ecnh v\u1ebd d\u01b0\u1edbi. L\u1eddi gi\u1ea3i a) Di\u1ec7n t\u00edch h\u00ecnh tr\u00f2n l\u00fac \u0111\u1ea7u (1, 2)2 = .3,14 = 1,1304 ( )S m2 2 Di\u1ec7n t\u00edch l\u00fac sau g\u1ea5p 3 l\u1ea7n: ( )1,1304.3 = 3,3912 m2 V\u1eady \u0111\u1ec3 di\u1ec7n t\u00edch c\u1ee7a b\u00e0n t\u0103ng g\u1ea5p 3 l\u1ea7n sau khi n\u1edbi th\u00ec k\u00edch th\u01b0\u1edbc c\u1ee7a h\u00ecnh ch\u1eef nh\u1eadt 1,884m b) Chu vi h\u00ecnh tr\u00f2n ban \u0111\u1ea7u 3,14.1, 2 = 3, 768(m) Chu vi l\u00fac sau t\u0103ng g\u1ea5p 2 l\u1ea7n: 3,768.2 = 7,536(m) K\u00edch th\u01b0\u1edbc h\u00ecnh ch\u1eef nh\u1eadt: (7,536 \u2212 3,768) : 2 = 1,884(m) C\u00e2u 7. V\u1eady \u0111\u1ec3 chu vi m\u1eb7t b\u00e0n t\u0103ng g\u1ea5p \u0111\u00f4i sau khi n\u1edbi th\u00ec k\u00edch th\u01b0\u1edbc c\u1ee7a h\u00ecnh ch\u1eef nh\u1eadt l\u00e0 1,884m M\u1ed9t kh\u1ed1i g\u1ed7 h\u00ecnh tr\u1ee5 c\u00f3 b\u00e1n k\u00ednh \u0111\u00e1y l\u00e0 3cm , chi\u1ec1u cao 4cm \u0111\u01b0\u1ee3c \u0111\u1eb7t \u0111\u1ee9ng tr\u00ean m\u1eb7t b\u00e0n. M\u1ed9t ph\u1ea7n c\u1ee7a kh\u1ed1i g\u1ed7 b\u1ecb c\u1eaft r\u1eddi theo c\u00e1c b\u00e1n k\u00ednh OA,OB v\u00e0 theo chi\u1ec1u d\u00e0i th\u1eb3ng \u0111\u1ee9ng t\u1eeb tr\u00ean xu\u1ed1ng d\u01b0\u1edbi v\u1edbi AOB = 30\uf0b0 nh\u01b0 h\u00ecnh v\u1ebd b\u00ean d\u01b0\u1edbi: 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 a) T\u00ednh th\u1ec3 t\u00edch c\u1ee7a kh\u1ed1i g\u1ed7 c\u00f2n l\u1ea1i sau khi b\u1ecb c\u1eaft r\u1eddi. b) Di\u1ec7n t\u00edch to\u00e0n ph\u1ea7n c\u1ee7a kh\u1ed1i g\u1ed7 c\u00f2n l\u1ea1i sau khi \u0111\u00e3 b\u1ecb c\u1eaft. L\u1eddi gi\u1ea3i a) Th\u1ec3 t\u00edch ph\u1ea7n b\u1ecb c\u1eaft l\u00e0 \uf070 .32.30 \uf0d7 4 = 3\uf070 360 ( )V1 = Squat \uf0d7h = cm3 Th\u1ec3 t\u00edch ph\u1ea7n c\u00f2n l\u1ea1i l\u00e0: ( )V2 = V \u2212V1 = \uf070.32.4 \u2212 3\uf070 = 33\uf070 cm3 b) Di\u1ec7n t\u00edch ph\u1ea7n c\u00f2n l\u1ea1i c\u1ee7a hai \u0111\u1ea7u: \uf0e6 \uf070 .9.30 \uf0f6 33 \uf0e7\uf0e8 360 \uf0f7\uf0f8 2 ( )2 \uf070 .9 \u2212 = \uf070 cm2 Di\u1ec7n t\u00edch xung quanh: ( )2\uf070 Rh \u2212 \uf070 R.30 \uf0d7 h + 2Rh = 22\uf070 + 24 cm2 180 Di\u1ec7n t\u00edch to\u00e0n ph\u1ea7n l\u00e0: C\u00e2u 8. ( )33 \uf070 + 22\uf070 + 24 = 38 1 \uf070 + 24 cm2 22 Cho \uf044ABC n\u1ed9i ti\u1ebfp trong \u0111\u01b0\u1eddng tr\u00f2n (O; R) . Ba \u0111\u01b0\u1eddng cao AD, BE,CF c\u1eaft nhau t\u1ea1i H . a) Ch\u1ee9ng minh c\u00e1c t\u1ee9 gi\u00e1c AEHF, BCEF n\u1ed9i ti\u1ebfp. b) K\u1ebb \u0111\u01b0\u1eddng k\u00ednh AK c\u1ee7a (O) . Ch\u1ee9ng minh AB.AC = 2R.AD . c) G\u1ecdi M l\u00e0 trung \u0111i\u1ec3m c\u1ee7a BC, I l\u00e0 giao \u0111i\u1ec3m EF v\u00e0 BC . Ch\u1ee9ng minh t\u1ee9 gi\u00e1c EFDM n\u1ed9i ti\u1ebfp v\u00e0 IB.IC = ID.IM . T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N 9"]