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 L\u1eddi gi\u1ea3i a) X\u00e9t tam t\u1ee9 gi\u00e1c AEHF ta c\u00f3 \uf0ec\uf0ef AFH = 90\uf0b0(CF la\u00f8 \u00f1\u00f6\u00f4\u00f8ng cao) \uf0ed \uf0ef\uf0ee AEH = 90\uf0b0 (BE la\u00f8 \u00f1\u00f6\u00f4\u00f8ng cao) m\u00e0 AFH + AEH = 180\uf0b0 \uf0de T\u1ee9 gi\u00e1c AEHF n\u1ed9i ti\u1ebfp (2 g\u00f3c \u0111\u1ed1i b\u00f9 nhau) X\u00e9t t\u1ee9 gi\u00e1c BCEF ta c\u00f3 BFC = BEC = 90\uf0b0 \uf0de T\u1ee9 gi\u00e1c BCEF n\u1ed9i ti\u1ebfp (t\u1ee9 gi\u00e1c c\u00f3 2 \u0111\u1ec9nh li\u00ean ti\u1ebfp c\u00f9ng nh\u00ecn c\u1ea1nh d\u01b0\u1edbi hai g\u00f3c b\u1eb1ng nhau) b) V\u00ec ACK l\u00e0 g\u00f3c n\u1ed9i ti\u1ebfp ch\u1eafn n\u1eeda cung ch\u1eafn n\u1eeda \u0111\u01b0\u1eddng tr\u00f2n (O) , \u0111\u01b0\u1eddng k\u00ednh AB \uf0de ACK = 90\uf0b0 X\u00e9t \uf044ADB v\u00e0 \uf044ACK \uf0ec ABD = AKC \uf0e6 = 1 AC \uf0f6 \uf0ef\uf0ef \uf0e7\uf0e8 2 \uf0f8\uf0f7 Ta c\u00f3: \uf0ed \uf0ef ADB = ACK (= 90\uf0b0) \uf0ef\uf0ee \uf0de \uf044ADB \u223d \uf044AKC (g.g) \uf0de AB = AD (ts\u0111d) AK AC \uf0de AB.AC = AK.AD M\u00e0 AK = 2R ( AK l\u00e0 \u0111\u01b0\u1eddng k\u00ednh) \uf0de AB.AC = 2R.AD 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 c) X\u00e9t t\u1ee9 gi\u00e1c BFEC ta c\u00f3: BFC = CEB (= 90\uf0b0) M\u00e0 hai g\u00f3c c\u00f9ng nh\u00ecn c\u1ea1nh BC \uf0de T\u1ee9 gi\u00e1c BFEC n\u1ed9i ti\u1ebfp (t\u1ee9 gi\u00e1c c\u00f3 2 \u0111\u1ec9nh li\u00ean ti\u1ebfp c\u00f9ng nh\u00ecn c\u1ea1nh d\u01b0\u1edbi hai g\u00f3c b\u1eb1ng nhau) \uf0de EFC = CEB (1) X\u00e9t t\u1ee9 gi\u00e1c BFED ta c\u00f3 \uf0ec\uf0efBFH = 90\uf0b0 \uf0ed \uf0ef\uf0eeBDH = 90\uf0b0 M\u00e0 BFH + BDH = 180\uf0b0 \uf0de t\u1ee9 gi\u00e1c BFED n\u1ed9i ti\u1ebfp (2 g\u00f3c \u0111\u1ed1i b\u00f9 nhau) \uf0de DFC = EBC (2) T\u1eeb (1), (2) \uf0de EFC = DFC \uf0de FC l\u00e0 tia ph\u00e2n gi\u00e1c g\u00f3c DFE \uf0de DFC = 2.EFC M\u00e0 EMC = 2.EFC \uf0de DFE = EMC X\u00e9t t\u1ee9 gi\u00e1c EFDM Ta c\u00f3 DFE = EMC n\u1ed9i ti\u1ebfp (t\u1ee9 gi\u00e1c c\u00f3 g\u00f3c ngo\u00e0i b\u1eb1ng g\u00f3c \u0111\u1ed1i trong) X\u00e9t \uf044IFD v\u00e0 \uf044IME ta c\u00f3: \uf0ec\uf0efEIM chung \uf0ed \uf0ef\uf0eeIDF = IEM (ch\u00f6\u00f9ng minh tre\u00e2n) \uf0de \uf044IFD \u223d \uf044IME (g.g) \uf0de ID = IF (ts\u0111d) IE IM \uf0de ID.IM = IE.IF (3) Ta c\u00f3: \uf0ec\uf0efBFE + CFE = 180\uf0b0( BCEF no\u00e4i tie\u00e1p) \uf0ed + IFB = 180\uf0b0(Ke\u00e0 bu\u00f8) \uf0ef\uf0eeBFE \uf0de BCE = IFB 11 X\u00e9t \uf044IBF v\u00e0 \uf044IEC c\u00f3 T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH \uf0ec\uf0efFIB: chung \uf0ed \uf0ef\uf0eeIFB = ICE \uf0de \uf044IBF \u223d \uf044IEC (g.g) \uf0de IF = IB (ts\u0111d) IC IE \uf0de IC.IB = IF.IE (4) T\u1eeb (3) v\u00e0 (4) \uf0de IC.IB = ID.IM ----H\u1ebeT--- T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N 12","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH S\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 QU\u1eacN 11 NA\u00caM HO\u00cfC: 2023 - 2024 \u0110\u1ec0 THAM KH\u1ea2O M\u00d4N: TO\u00c1N 9 M\u00c3 \u0110\u1ec0: Qu\u1eadn 11 - 2 \u0110\u00ea thi g\u1ed3m 8 c\u00e2u ho\u0309i t\u01b0\u0323 lu\u00e2\u0323n. Th\u01a1\u0300i gian: 120 phu\u0301t (kh\u00f4ng k\u00ea\u0309 th\u01a1\u0300i gian pha\u0301t \u0111\u00ea\u0300) C\u00e2u 1. (1,5 \u0111i\u1ec3m). Cho (P) : y = \u2212x2 v\u00e0 \u0111\u01b0\u1eddng th\u1eb3ng (d) : y = x \u2212 2 . 42 a) V\u1ebd \u0111\u1ed3 th\u1ecb (P) v\u00e0 (d) tr\u00ean c\u00f9ng h\u1ec7 tr\u1ee5c t\u1ecda \u0111\u1ed9. b) T\u00ecm t\u1ecda \u0111\u1ed9 giao \u0111i\u1ec3m c\u1ee7a (P) v\u00e0 (d) b\u1eb1ng ph\u00e9p t\u00ednh. C\u00e2u 2. (1 \u0111i\u1ec3m). Cho ph\u01b0\u01a1ng tr\u00ecnh x2 \u2212 3x \u2212 5 = 0 c\u00f3 2 nghi\u1ec7m l\u00e0 x1 , x2 . Kh\u00f4ng gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh, h\u00e3y t\u00ednh gi\u00e1 tr\u1ecb c\u1ee7a bi\u1ec3u th\u1ee9c A = 2x1 + 2x2 x2 x1 L\u01b0u \u00fd: T\u1eeb b\u00e0i n\u00e0y, c\u00e1c s\u1ed1 li\u1ec7u t\u00ednh to\u00e1n v\u1ec1 \u0111\u1ed9 d\u00e0i khi l\u00e0m tr\u00f2n (n\u1ebfu c\u00f3) l\u1ea5y \u0111\u1ebfn m\u1ed9t ch\u1eef s\u1ed1 th\u1eadp ph\u00e2n, s\u1ed1 \u0111o g\u00f3c l\u00e0m tr\u00f2n \u0111\u1ebfn ph\u00fat. C\u00e2u 3. (1 \u0111i\u1ec3m). Quy t\u1eafc sau \u0111\u00e2y cho ta c\u00e1ch t\u00ednh ng\u00e0y cu\u1ed1i c\u00f9ng c\u1ee7a th\u00e1ng hai trong n\u0103m 20ab l\u00e0 th\u1ee9 m\u1ea5y ? - L\u1ea5y ab chia 12 \u0111\u01b0\u1ee3c th\u01b0\u01a1ng l\u00e0 x d\u01b0 l\u00e0 y - L\u1ea5y y chia 4 \u0111\u01b0\u1ee3c th\u01b0\u01a1ng l\u00e0 z - T\u00ednh M x y z - L\u1ea5y M chia 7 \u0111\u01b0\u1ee3c d\u01b0 r N\u1ebfu r 0 \u0111\u00f3 l\u00e0 th\u1ee9 3 N\u1ebfu r 1 \u0111\u00f3 l\u00e0 th\u1ee9 4 N\u1ebfu r 5 \u0111\u00f3 l\u00e0 ch\u1ee7 nh\u1eadt N\u1ebfu r 6 \u0111\u00f3 l\u00e0 th\u1ee9 hai Em h\u00e3y d\u00f9ng quy t\u1eafc tr\u00ean t\u00ednh xem ng\u00e0y cu\u1ed1i c\u00f9ng c\u1ee7a th\u00e1ng hai trong n\u0103m 2024 l\u00e0 th\u1ee9 m\u1ea5y ? T\u1eeb \u0111\u00f3 cho bi\u1ebft ng\u00e0y 29 \/ 01 \/ 2024 l\u00e0 th\u1ee9 m\u1ea5y ? C\u00e2u 4. (0,75 \u0111i\u1ec3m). Ti\u1ec1n v\u1ed1n v\u00e0 l\u00e3i b\u00e1n h\u00e0ng c\u1ee7a m\u1ed9t c\u1eeda h\u00e0ng kinh doanh 6 th\u00e1ng \u0111\u1ea7u n\u0103m \u0111\u01b0\u1ee3c bi\u1ec3u th\u1ecb b\u1eb1ng m\u1ed9t do\u1ea1n th\u1eb3ng, v\u1edbi v\u1ed1n ban \u0111\u1ea7u l\u00e0 : 15 tri\u1ec7u \u0111\u1ed3ng ( h\u00ecnh v\u1ebd ). a) H\u00e3y x\u00e1c \u0111\u1ecbnh h\u1ec7 s\u1ed1 a v\u00e0 b bi\u1ebft ph\u01b0\u01a1ng tr\u00ecnh \u0111\u1ed3 th\u1ecb tr\u00ean l\u00e0 m\u1ed9t \u0111\u01b0\u1eddng th\u1eb3ng c\u00f3 d\u1ea1ng y ax b a 0 v\u1edbi y l\u00e0 s\u1ed1 ti\u1ec1n v\u1ed1n v\u00e0 l\u00e3i b\u00e1n h\u00e0ng; x l\u00e0 s\u1ed1 th\u00e1ng b\u00e1n h\u00e0ng. 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. b) T\u00ednh s\u1ed1 ti\u1ec1n v\u1ed1n v\u00e0 l\u00e3i \u1edf th\u00e1ng th\u1ee9 t\u01b0 ? (1 \u0111i\u1ec3m). B\u1ea1n Mai mua 38 c\u00e1i b\u00e1nh cho l\u1edbp li\u00ean hoan. T\u1ea1i c\u1eeda h\u00e0ng b\u00e1nh A gi\u00e1 b\u00e1nh Mai mu\u1ed1n mua l\u00e0 16 000 \u0111\u1ed3ng\/ 1 c\u00e1i. C\u1eeda h\u00e0ng b\u00e1nh A \u0111ang c\u00f3 ch\u01b0\u01a1ng tr\u00ecnh khuy\u1ebfn m\u00e3i, n\u1ebfu mua h\u01a1n 10 c\u00e1i s\u1ebd \u0111\u01b0\u1ee3c gi\u1ea3m gi\u00e1 9% tr\u00ean t\u1ed5ng s\u1ed1 ti\u1ec1n mua b\u00e1nh a) N\u1ebfu b\u1ea1n Mai mua 38 c\u00e1i b\u00e1nh \u1edf c\u1eeda h\u00e0ng A th\u00ec ph\u1ea3i tr\u1ea3 bao nhi\u00eau ti\u1ec1n ? b) T\u1ea1i c\u1eeda h\u00e0ng B b\u00e1n c\u00f9ng lo\u1ea1i b\u00e1nh n\u00f3i tr\u00ean ( ch\u1ea5t l\u01b0\u1ee3ng nh\u01b0 nhau) \u0111\u1ed3ng gi\u00e1 16 000 \u0111\u1ed3ng\/ 1 c\u00e1i, nh\u01b0ng n\u1ebfu mua ba c\u00e1i th\u00ec ch\u1ec9 tr\u1ea3 43000 \u0111\u1ed3ng. B\u1ea1n Mai n\u00ean mua b\u00e1nh \u1edf c\u1eeda h\u00e0ng n\u00e0o \u0111\u1ec3 c\u00f3 l\u1ee3i h\u01a1n. C\u00e2u 6. (1 \u0111i\u1ec3m). H\u00f4m nay, ba b\u1ea1n Tu\u1ea5n nh\u1edd b\u1ea1n thay d\u00f9m n\u01b0\u1edbc trong b\u1ec3 c\u00e1 l\u00f2ng b\u1ec3 l\u00e0 h\u00ecnh h\u1ed9p ch\u1eef nh\u1eadt, \u0111\u00e1y c\u00f3 k\u00edch th\u01b0\u1edbc 30cm 60cm , chi\u1ec1u cao 40cm . N\u01b0\u1edbc hi\u1ec7n nay ch\u1ee9a trong b\u1ec3 c\u00e1ch m\u00e9p tr\u00ean c\u1ee7a b\u1ec3 l\u00e0 5cm , ba c\u1ee7a Tu\u1ea5n d\u1eb7n d\u00f9ng \u1ed1ng h\u00fat d\u1eabn n\u01b0\u1edbc v\u00e0o c\u00e1c s\u00f4 tr\u00e1nh n\u01b0\u1edbc tr\u00e0n n\u01b0\u1edbc ra s\u00e0n nh\u00e0, k\u1ec3 c\u1ea3 h\u00fat ch\u1ea5t b\u1ea9n d\u01b0\u1edbi \u0111\u00e1y b\u1ec3, ch\u1eeba l\u1ea1i n\u01b0\u1edbc c\u0169 n\u1eeda b\u1ec3, sau \u0111\u00f3 \u0111\u1ed5 n\u01b0\u1edbc m\u1edbi v\u00e0o n\u1ebfu th\u1ef1c hi\u1ec7n \u0111ung y\u00eau c\u1ea7u tr\u00ean, b\u1ea1n Tu\u1ea5n \u0111\u00e3 r\u00fat n\u01b0\u1edbc t\u1eeb b\u1ec3 c\u00e1 ra ngo\u00e0i bao nhi\u00eau l\u00edt n\u01b0\u1edbc ( Cho bi\u1ebft c\u00f4ng th\u1ee9c t\u00ednh th\u1ec3 t\u00edch h\u00ecnh h\u1ed9p ch\u1eef nh\u1eadt l\u00e0 V S.h , trong \u0111\u00f3 S l\u00e0 di\u1ec7n t\u00edch \u0111\u00e1y,h l\u00e0 chi\u1ec1u cai h\u00ecnh h\u1ed9p v\u00e0 th\u1ec3 t\u00edch c\u00e1, rong, s\u1ecfi l\u00f3t d\u01b0\u1edbi \u0111\u00e1y b\u1ec3 chi\u1ebfm kho\u1ea3ng 10% th\u1ec3 t\u00edch b\u1ec3, th\u1ec3 t\u00edch c\u00e1c ch\u1ea5t b\u1ea9n kh\u00f4ng \u0111\u00e1ng k\u1ec3, 1dm3 n\u01b0\u1edbc 1 l\u00edt n\u01b0\u1edbc ( l\u00e0m tr\u00f2n m\u1ed9t ch\u1eef s\u1ed1 th\u1eadp ph\u00e2n) C\u00e2u 7. (1 \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- thua. \u1ede b\u1ea3ng A , \u0111\u1ed9i Ph\u01b0\u1ee3ng Ho\u00e0ng 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: 1. \u0110\u1ed9i B\u00e1o \u0110en: 4 \u0111i\u1ec3m 2. \u0110\u1ed9i Th\u1ecf Tr\u1eafng: 2 \u0111i\u1ec3m 3. \u0110\u1ed9i S\u01b0 T\u1eed: 2 \u0111i\u1ec3m 4. \u0110\u1ed9i Ph\u01b0\u1ee3ng Ho\u00e0ng 1 \u0111i\u1ec3m \u1ede l\u01b0\u1ee3t \u0111\u1ea5u di\u1ec5n ra song song 2 tr\u1eadn B\u00e1o \u0110en- S\u01b0 T\u1eed v\u00e0 Th\u1ecf Tr\u1eafng-Ph\u01b0\u1ee3ng Ho\u00e0ng. C\u00e1c em h\u00e3y t\u00ednh x\u00e1c su\u1ea5t v\u00e0o v\u00f2ng trong c\u1ee7a \u0111\u1ed9i Ph\u01b0\u1ee3ng Ho\u00e0ng bi\u1ebft r\u1eb1ng \u0111\u1ed9i Ph\u01b0\u1ee3ng Ho\u00e0ng 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 C\u00e2u 8. ra).100% (3 \u0111i\u1ec3m) Cho \u0111\u01b0\u1eddng tr\u00f2n t\u00e2m O c\u00f3 \u0111\u01b0\u1eddng k\u00ednh AB 2R . G\u1ecdi I l\u00e0 trung \u0111i\u1ec3m c\u1ee7a \u0111o\u1ea1n th\u1eb3ng OA v\u00e0 E l\u00e0 \u0111i\u1ec3m thu\u1ed9c \u0111\u01b0\u1eddng tr\u00f2n t\u00e2m O ( E kh\u00f4ng tr\u00f9ng v\u1edbi A v\u00e0 B ). G\u1ecdi Ax v\u00e0 By l\u00e0 c\u00e1c ti\u1ebfp tuy\u1ebfn t\u1ea1i A v\u00e0 B c\u1ee7a \u0111\u01b0\u01a1ng tr\u00f2n O ( Ax , By c\u00f9ng thu\u1ed9c m\u1ed9t n\u1eeda m\u1eb7t ph\u1eb3ng b\u1edd AB c\u00f3 ch\u1ee9a \u0111i\u1ec3m E ). Qua \u0111i\u1ec3m E k\u1ebb \u0111\u01b0\u1eddng th\u1eb3ng d vu\u00f4ng g\u00f3c v\u1edbi EI c\u1eaft Ax v\u00e0 By l\u1ea7n l\u01b0\u1ee3t t\u1ea1i M v\u00e0 N . 1. Ch\u1ee9ng minh t\u1ee9 gi\u00e1c AMEI n\u1ed9i ti\u1ebfp. 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 2. Ch\u1ee9ng minh ENI EBI v\u00e0 AE.IN BE.IM. 3. G\u1ecdi P l\u00e0 giao \u0111i\u1ec3m c\u1ee7a AE v\u00e0 MI ; Q l\u00e0 giao \u0111i\u1ec3m c\u1ee7a BE v\u00e0 NI . Ch\u1ee9ng minh hai \u0111\u01b0\u1eddng th\u1eb3ng PQ v\u00e0 BN vu\u00f4ng g\u00f3c v\u1edbi nhau. 4. G\u1ecdi F l\u00e0 \u0111i\u1ec3m ch\u00ednh gi\u1eefa cung AB kh\u00f4ng ch\u1ee9a \u0111i\u1ec3m E c\u1ee7a \u0111\u01b0\u1eddng tr\u00f2n O . T\u00ednh di\u1ec7n t\u00edch tam gi\u00e1c OMN theo R khi ba \u0111i\u1ec3m E,I,F th\u1eb3ng h\u00e0ng. ----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). Cho (P) : y = \u2212x2 v\u00e0 \u0111\u01b0\u1eddng th\u1eb3ng (d) : y = x \u2212 2 . 42 a) V\u1ebd \u0111\u1ed3 th\u1ecb (P) v\u00e0 (d) tr\u00ean c\u00f9ng h\u1ec7 tr\u1ee5c t\u1ecda \u0111\u1ed9. b) T\u00ecm t\u1ecda \u0111\u1ed9 giao \u0111i\u1ec3m c\u1ee7a (P) v\u00e0 (d) b\u1eb1ng ph\u00e9p t\u00ednh. L\u1eddi gi\u1ea3i (1 \u0111i\u1ec3m) a) V\u1ebd \u0111\u1ed3 th\u1ecb (P) v\u00e0 (d) tr\u00ean c\u00f9ng h\u1ec7 tr\u1ee5c t\u1ecda \u0111\u1ed9. BGT: x \u22124 \u22122 0 2 4 y = \u2212 x2 \u22124 \u22121 0 \u22121 \u22124 4 x 02 y= x\u22122 \u22122 \u22121 2 b) T\u00ecm t\u1ecda \u0111\u1ed9 giao \u0111i\u1ec3m c\u1ee7a (P) v\u00e0 (d) b\u1eb1ng ph\u00e9p t\u00ednh. Ph\u01b0\u01a1ng tr\u00ecnh ho\u00e0nh \u0111\u1ed9 giao \u0111i\u1ec3m c\u1ee7a (P) v\u00e0 (d) : \u2212 x2 = x \u2212 2 42 \uf0db \u2212 1 x2 \u2212 1 x + 2 = 0 42 \uf0db \uf0e9x = 2 \uf0ea = \u22124 \uf0eb x Thay x = 2 v\u00e0o y = 1x\u22122 , ta \u0111\u01b0\u1ee3c: y = 1 .2 \u2212 2 = \u22121 . 2 2 Thay x = \u22124 v\u00e0o y = 1 x \u2212 2 , ta \u0111\u01b0\u1ee3c: y = 1 .(\u22124) \u2212 2 = \u22124 . 22 V\u1eady (2; \u22121) , (\u22124; \u2212 4) l\u00e0 hai giao \u0111i\u1ec3m c\u1ea7n t\u00ecm. C\u00e2u 2. (1 \u0111i\u1ec3m) Cho ph\u01b0\u01a1ng tr\u00ecnh x2 \u2212 3x \u2212 5 = 0 c\u00f3 2 nghi\u1ec7m l\u00e0 x1 , x2 . Kh\u00f4ng gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh, h\u00e3y t\u00ednh gi\u00e1 tr\u1ecb c\u1ee7a bi\u1ec3u th\u1ee9c A = 2x1 + 2x2 x2 x1 L\u1eddi gi\u1ea3i V\u00ec \uf044 = b2 \u2212 4ac = (\u22123)2 \u2212 4.(1).(\u22125) = 29 \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 = 3 =3 \uf0ed = x1 a 1 \u22125 Theo \u0111\u1ecbnh l\u00ed Vi-et, ta c\u00f3: \uf0ef\uf0ef\uf0eeP c \u22125 a = 1 = .x2 = Ta c\u00f3: A = 2x1 + 2x2 x2 x1 ( ) ( ) ( ( ))A = 2x12 + 2x22 \uf0e9 2 \uf0f9 x12 + x22 2 \uf0ea\uf0eb x1 + x2 \u2212 2x1x2 \uf0fa\uf0fb 32 \u2212 2. 2 2 \u22125 = \u221238 = = = x1x2 x1x2 x1x2 \u22125 5 L\u01b0u \u00fd: T\u1eeb b\u00e0i n\u00e0y, c\u00e1c s\u1ed1 li\u1ec7u t\u00ednh to\u00e1n v\u1ec1 \u0111\u1ed9 d\u00e0i khi l\u00e0m tr\u00f2n (n\u1ebfu c\u00f3) l\u1ea5y \u0111\u1ebfn m\u1ed9t ch\u1eef s\u1ed1 th\u1eadp ph\u00e2n, s\u1ed1 \u0111o g\u00f3c l\u00e0m tr\u00f2n \u0111\u1ebfn ph\u00fat. C\u00e2u 3. (1 \u0111i\u1ec3m) Quy t\u1eafc sau \u0111\u00e2y cho ta c\u00e1ch t\u00ednh ng\u00e0y cu\u1ed1i c\u00f9ng c\u1ee7a th\u00e1ng hai trong n\u0103m 20ab l\u00e0 th\u1ee9 m\u1ea5y ? - L\u1ea5y ab chia 12 \u0111\u01b0\u1ee3c th\u01b0\u01a1ng l\u00e0 x d\u01b0 l\u00e0 y - L\u1ea5y y chia 4 \u0111\u01b0\u1ee3c th\u01b0\u01a1ng l\u00e0 z - T\u00ednh M x y z - L\u1ea5y M chia 7 \u0111\u01b0\u1ee3c d\u01b0 r N\u1ebfu r 0 \u0111\u00f3 l\u00e0 th\u1ee9 3 N\u1ebfu r 1 \u0111\u00f3 l\u00e0 th\u1ee9 4 \u2026\u2026 N\u1ebfu r 5 \u0111\u00f3 l\u00e0 ch\u1ee7 nh\u1eadt N\u1ebfu r 6 \u0111\u00f3 l\u00e0 th\u1ee9 hai Em h\u00e3y d\u00f9ng quy t\u1eafc tr\u00ean t\u00ednh xem ng\u00e0y cu\u1ed1i c\u00f9ng c\u1ee7a th\u00e1ng hai trong n\u0103m 2024 l\u00e0 th\u1ee9 m\u1ea5y ? T\u1eeb \u0111\u00f3 cho bi\u1ebft ng\u00e0y 29 \/ 01 \/ 2024 l\u00e0 th\u1ee9 m\u1ea5y ? L\u1eddi gi\u1ea3i Ta c\u00f3 : 2024 c\u00f3 24 chia 12 \u0111\u01b0\u1ee3c th\u01b0\u01a1ng l\u00e0 2 d\u01b0 l\u00e0 0 L\u1ea5y 0 chia 4 \u0111\u01b0\u1ee3c th\u01b0\u01a1ng l\u00e0 0 T\u1eeb \u0111\u00f3 M 2 0 0 2 L\u1ea5y M chia 7 d\u01b0 r 2 . T\u1eeb \u0111\u00f3 ta c\u00f3 ng\u00e0y cu\u1ed1i c\u00f9ng c\u1ee7a th\u00e1ng hai trong n\u0103m 2024 t\u00ednh theo quy t\u1eafc l\u00e0 th\u1ee9 5 . T\u1eeb 29 \/ 01 \/ 2024 \u0111\u1ebfn 28 \/ 02 \/ 2024 l\u00e0 31 ng\u00e0y. Ta c\u00f3 : 31 7.4 3 . N\u00ean th\u1ee9 5 l\u00e0 28 \/ 02 \/ 2024 l\u00f9i l\u1ea1i n\u00ean c\u00f3 29 \/ 01 \/ 2024 l\u00e0 th\u1ee9 3 T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N 5","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH C\u00e2u 4. (0,75 \u0111i\u1ec3m).Ti\u1ec1n v\u1ed1n v\u00e0 l\u00e3i b\u00e1n h\u00e0ng c\u1ee7a m\u1ed9t c\u1eeda h\u00e0ng kinh doanh 6 th\u00e1ng \u0111\u1ea7u n\u0103m \u0111\u01b0\u1ee3c bi\u1ec3u th\u1ecb b\u1eb1ng m\u1ed9t do\u1ea1n th\u1eb3ng, v\u1edbi v\u1ed1n ban \u0111\u1ea7u l\u00e0 : 15 tri\u1ec7u \u0111\u1ed3ng ( h\u00ecnh v\u1ebd ). a) H\u00e3y x\u00e1c \u0111\u1ecbnh h\u1ec7 s\u1ed1 a v\u00e0 b bi\u1ebft ph\u01b0\u01a1ng tr\u00ecnh \u0111\u1ed3 th\u1ecb tr\u00ean l\u00e0 m\u1ed9t \u0111\u01b0\u1eddng th\u1eb3ng c\u00f3 d\u1ea1ng y ax b a 0 v\u1edbi y l\u00e0 s\u1ed1 ti\u1ec1n v\u1ed1n v\u00e0 l\u00e3i b\u00e1n h\u00e0ng; x l\u00e0 s\u1ed1 th\u00e1ng b\u00e1n h\u00e0ng. b) T\u00ednh s\u1ed1 ti\u1ec1n v\u1ed1n v\u00e0 l\u00e3i \u1edf th\u00e1ng th\u1ee9 t\u01b0 ? L\u1eddi gi\u1ea3i a) Ta c\u00f3 y ax b V\u1edbi x 0 v\u00e0 y 15 th\u00ec 15 a.0 b b 15 V\u1edbi x 5 v\u00e0 y 25 th\u00ec 25 a.5 b 25 a.5 15 a 2 T\u1eeb \u0111\u00f3 ta c\u00f3 : y 2x 15 b) Ta c\u00f3: y 2x 15 V\u1edbi x 4 th\u00ec y 2.4 15 23 tri\u1ec7u \u0111\u1ed3ng. C\u00e2u 5. (1 \u0111i\u1ec3m) B\u1ea1n Mai mua 38 c\u00e1i b\u00e1nh cho l\u1edbp li\u00ean hoan. T\u1ea1i c\u1eeda h\u00e0ng b\u00e1nh A gi\u00e1 b\u00e1nh Mai mu\u1ed1n mua l\u00e0 16 000 \u0111\u1ed3ng\/ 1 c\u00e1i. C\u1eeda h\u00e0ng b\u00e1nh A \u0111ang c\u00f3 ch\u01b0\u01a1ng tr\u00ecnh khuy\u1ebfn m\u00e3i, n\u1ebfu mua h\u01a1n 10 c\u00e1i s\u1ebd \u0111\u01b0\u1ee3c gi\u1ea3m gi\u00e1 9% tr\u00ean t\u1ed5ng s\u1ed1 ti\u1ec1n mua b\u00e1nh a) N\u1ebfu b\u1ea1n Mai mua 38 c\u00e1i b\u00e1nh \u1edf c\u1eeda h\u00e0ng A th\u00ec ph\u1ea3i tr\u1ea3 bao nhi\u00eau ti\u1ec1n ? b) T\u1ea1i c\u1eeda h\u00e0ng B, b\u00e1n c\u00f9ng lo\u1ea1i b\u00e1nh n\u00f3i tr\u00ean ( ch\u1ea5t l\u01b0\u1ee3ng nh\u01b0 nhau) \u0111\u1ed3ng gi\u00e1 16 000 \u0111\u1ed3ng\/ 1 c\u00e1i, nh\u01b0ng n\u1ebfu mua ba c\u00e1i th\u00ec ch\u1ec9 tr\u1ea3 43000 \u0111\u1ed3ng. B\u1ea1n Mai n\u00ean mua b\u00e1nh \u1edf c\u1eeda h\u00e0ng n\u00e0o \u0111\u1ec3 c\u00f3 l\u1ee3i h\u01a1n. L\u1eddi gi\u1ea3i a) Do b\u1ea1n Mai mua 38 c\u00e1i b\u00e1nh \u1edf c\u1eeda h\u00e0ng A n\u00ean \u0111\u01b0\u1ee3c c\u1eeda h\u00e0ng b\u00e1nh gi\u1ea3m 9% tr\u00ean t\u1ed5ng s\u1ed1 ti\u1ec1n mua b\u00e1nh. V\u1eady s\u1ed1 ti\u1ec1n b\u1ea1n Mai ph\u1ea3i tr\u1ea3 l\u00e0: 16 000. 100% 9% .38 553 280 \u0111\u1ed3ng. b) T\u1ea1i c\u1eeda h\u00e0ng B , n\u1ebfu mua 3 c\u00e1i b\u00e1nh ch\u1ec9 ph\u1ea3i tr\u1ea3 43 000 \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 38 c\u00e1i b\u00e1nh \u0111\u01b0\u1ee3c chia th\u00e0nh: 38 12.3 2 . \u0110\u1ec3 mua 38 c\u00e1i b\u00e1nh, s\u1ed1 ti\u1ec1n b\u1ea1n Mai c\u1ea7n tr\u1ea3 l\u00e0: 12.43 000 2.16 000 548 000 \u0111\u1ed3ng. Do 548 000 553280 n\u00ean b\u1ea1n Mai mua b\u00e1nh \u1edf c\u1eeda h\u00e0ng B th\u00ec tr\u1ea3 ti\u1ec1n \u00edt h\u01a1n. C\u00e2u 6. (1 \u0111i\u1ec3m) H\u00f4m nay, ba b\u1ea1n Tu\u1ea5n nh\u1edd b\u1ea1n thay d\u00f9m n\u01b0\u1edbc trong b\u1ec3 c\u00e1 l\u00f2ng b\u1ec3 l\u00e0 h\u00ecnh h\u1ed9p ch\u1eef nh\u1eadt, \u0111\u00e1y c\u00f3 k\u00edch th\u01b0\u1edbc 30cm 60cm , chi\u1ec1u cao 40cm . N\u01b0\u1edbc hi\u1ec7n nay ch\u1ee9a trong b\u1ec3 c\u00e1ch m\u00e9p tr\u00ean c\u1ee7a b\u1ec3 l\u00e0 5cm , ba c\u1ee7a Tu\u1ea5n d\u1eb7n d\u00f9ng \u1ed1ng h\u00fat d\u1eabn n\u01b0\u1edbc v\u00e0o c\u00e1c s\u00f4 tr\u00e1nh n\u01b0\u1edbc tr\u00e0n n\u01b0\u1edbc ra s\u00e0n nh\u00e0, k\u1ec3 c\u1ea3 h\u00fat ch\u1ea5t b\u1ea9n d\u01b0\u1edbi \u0111\u00e1y b\u1ec3, ch\u1eeba l\u1ea1i n\u01b0\u1edbc c\u0169 n\u1eeda b\u1ec3, sau \u0111\u00f3 \u0111\u1ed5 n\u01b0\u1edbc m\u1edbi v\u00e0o n\u1ebfu th\u1ef1c hi\u1ec7n \u0111\u00fang y\u00eau c\u1ea7u tr\u00ean, b\u1ea1n Tu\u1ea5n \u0111\u00e3 r\u00fat n\u01b0\u1edbc t\u1eeb b\u1ec3 c\u00e1 ra ngo\u00e0i bao nhi\u00eau l\u00edt n\u01b0\u1edbc ( Cho bi\u1ebft c\u00f4ng th\u1ee9c t\u00ednh th\u1ec3 t\u00edch h\u00ecnh h\u1ed9p ch\u1eef nh\u1eadt l\u00e0 V S.h , trong \u0111\u00f3 S l\u00e0 di\u1ec7n t\u00edch \u0111\u00e1y,h l\u00e0 chi\u1ec1u cao h\u00ecnh h\u1ed9p v\u00e0 th\u1ec3 t\u00edch c\u00e1, rong, s\u1ecfi l\u00f3t d\u01b0\u1edbi \u0111\u00e1y b\u1ec3 chi\u1ebfm kho\u1ea3ng 10% th\u1ec3 t\u00edch b\u1ec3, th\u1ec3 t\u00edch c\u00e1c ch\u1ea5t b\u1ea9n kh\u00f4ng \u0111\u00e1ng k\u1ec3, 1dm3 n\u01b0\u1edbc 1 l\u00edt n\u01b0\u1edbc ( l\u00e0m tr\u00f2n m\u1ed9t ch\u1eef s\u1ed1 th\u1eadp ph\u00e2n) L\u1eddi gi\u1ea3i Ta c\u00f3: B\u1ec3 c\u00e1 l\u00f2ng b\u1ec3 l\u00e0 h\u00ecnh h\u1ed9p ch\u1eef nh\u1eadt, \u0111\u00e1y c\u00f3 k\u00edch th\u01b0\u1edbc 30cm 60cm , chi\u1ec1u cao 40cm . N\u01b0\u1edbc hi\u1ec7n nay ch\u1ee9a trong b\u1ec3 c\u00e1ch m\u00e9p tr\u00ean c\u1ee7a b\u1ec3 l\u00e0 5cm n\u00ean ta c\u00f3 chi\u1ec1u cao n\u01b0\u1edbc l\u00e0 35cm . Th\u1ec3 t\u00edch h\u1ed3 c\u00e1 khi \u0111\u00f3 l\u00e0 V S.h 30.60.35 63000 cm3 63 dm3 . Th\u1ec3 t\u00edch n\u01b0\u1edbc trong h\u1ed3 c\u00e1 khi \u0111\u00f3 l\u00e0 Vnuoc 90%.V 56, 7 dm3 Ba Tu\u1ea5n d\u1eb7n ch\u1eeba l\u1ea1i n\u01b0\u1edbc c\u0169 n\u1eeda b\u1ec3 n\u00ean l\u01b0\u1ee3ng n\u01b0\u1edbc b\u1ecb r\u00fat ra kh\u1ecfi b\u1ec3 l\u00e0 56, 7 28, 35 28, 4 l 2 C\u00e2u 7. (1 \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\u1ed9 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- thua. \u1ede b\u1ea3ng A, \u0111\u1ed9i Ph\u01b0\u1ee3ng Ho\u00e0ng 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: 1. \u0110\u1ed9i B\u00e1o \u0110en: 4 \u0111i\u1ec3m 2. \u0110\u1ed9i Th\u1ecf Tr\u1eafng: 2 \u0111i\u1ec3m 3. \u0110\u1ed9i S\u01b0 T\u1eed: 2 \u0111i\u1ec3m 4. \u0110\u1ed9i Ph\u01b0\u1ee3ng Ho\u00e0ng 1 \u0111i\u1ec3m 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 \u1ede l\u01b0\u1ee3t \u0111\u1ea5u di\u1ec5n ra song song 2 tr\u1eadn B\u00e1o \u0110en- S\u01b0 T\u1eed v\u00e0 Th\u1ecf Tr\u1eafng-Ph\u01b0\u1ee3ng Ho\u00e0ng. C\u00e1c em h\u00e3y t\u00ednh x\u00e1c su\u1ea5t v\u00e0o v\u00f2ng trong c\u1ee7a \u0111\u1ed9i Ph\u01b0\u1ee3ng Ho\u00e0ng bi\u1ebft r\u1eb1ng \u0111\u1ed9i Ph\u01b0\u1ee3ng Ho\u00e0ng 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\u00e0 v\u00f2ng trong): ( s\u1ed1 kh\u1ea3 n\u0103ng x\u1ea3y ra).100% Qu\u1eadn 7-01 L\u1eddi gi\u1ea3i S\u1ed1 kh\u1ea3 n\u0103ng x\u1ea3y ra l\u00e0 9 = 3.3 ( Tr\u1eadn B\u00e1o \u0110en- S\u01b0 T\u1eed c\u00f3 3 kh\u1ea3 n\u0103ng v\u00e0 tr\u1eadn Th\u1ecf Tr\u1eafng- Ph\u01b0\u1ee3ng Ho\u00e0ng c\u00f3 3 kh\u1ea3 n\u0103ng) S\u1ed1 kh\u1ea3 n\u0103ng Ph\u01b0\u1ee3ng Ho\u00e0ng v\u00e0o l\u00e0 2 TH: B\u00e1o \u0110en th\u1eafng S\u01b0 T\u1eed v\u00e0 Th\u1ecf Tr\u1eafng thua Ph\u01b0\u1ee3ng Ho\u00e0ng: Ph\u01b0\u1ee3ng Ho\u00e0ng v\u00e0o. TH: B\u00e1o \u0110en h\u00f2a S\u01b0 T\u1eed v\u00e0 Th\u1ecf Tr\u1eafng thua Ph\u01b0\u1ee3ng Ho\u00e0ng: Ph\u01b0\u1ee3ng Ho\u00e0ng v\u00e0o. V\u1eady x\u00e1c su\u1ea5t \u0111\u1ec3 Ph\u01b0\u1ee3ng Ho\u00e0ng \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 t\u00e2m O c\u00f3 \u0111\u01b0\u1eddng k\u00ednh AB 2R . G\u1ecdi I l\u00e0 trung \u0111i\u1ec3m c\u1ee7a \u0111o\u1ea1n th\u1eb3ng OA v\u00e0 E l\u00e0 \u0111i\u1ec3m thu\u1ed9c \u0111\u01b0\u1eddng tr\u00f2n t\u00e2m O ( E kh\u00f4ng tr\u00f9ng v\u1edbi A v\u00e0 B ). G\u1ecdi Ax v\u00e0 By l\u00e0 c\u00e1c ti\u1ebfp tuy\u1ebfn t\u1ea1i A v\u00e0 B c\u1ee7a \u0111\u01b0\u01a1ng tr\u00f2n O ( Ax , By c\u00f9ng thu\u1ed9c m\u1ed9t n\u1eeda m\u1eb7t ph\u1eb3ng b\u1edd AB c\u00f3 ch\u1ee9a \u0111i\u1ec3m E ). Qua \u0111i\u1ec3m E k\u1ebb \u0111\u01b0\u1eddng th\u1eb3ng d vu\u00f4ng g\u00f3c v\u1edbi EI c\u1eaft Ax v\u00e0 By l\u1ea7n l\u01b0\u1ee3t t\u1ea1i M v\u00e0 N . 1. Ch\u1ee9ng minh t\u1ee9 gi\u00e1c AMEI n\u1ed9i ti\u1ebfp. 2. Ch\u1ee9ng minh ENI EBI v\u00e0 AE.IN BE.IM. 3. G\u1ecdi P l\u00e0 giao \u0111i\u1ec3m c\u1ee7a AE v\u00e0 MI ; Q l\u00e0 giao \u0111i\u1ec3m c\u1ee7a BE v\u00e0 NI . Ch\u1ee9ng minh hai \u0111\u01b0\u1eddng th\u1eb3ng PQ v\u00e0 BN vu\u00f4ng g\u00f3c v\u1edbi nhau. 4. G\u1ecdi F l\u00e0 \u0111i\u1ec3m ch\u00ednh gi\u1eefa cung AB kh\u00f4ng ch\u1ee9a \u0111i\u1ec3m E c\u1ee7a \u0111\u01b0\u1eddng tr\u00f2n O . T\u00ednh di\u1ec7n t\u00edch tam gi\u00e1c OMN theo R khi ba \u0111i\u1ec3m E,I,F 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 8","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH 1) Ch\u1ee9ng minh t\u1ee9 gi\u00e1c AMEI n\u1ed9i ti\u1ebfp. X\u00e9t t\u1ee9 gi\u00e1c AMEI , c\u00f3: \uf0ef\uf0ecMAI = 90\uf0b0( AM \u22a5 AB) \uf0ed \uf0ee\uf0efMEI = 90\uf0b0 ( EM \u22a5 EI ) \uf0de MAI + MEI = 180\uf0b0 \uf0de T\u1ee9 gi\u00e1c AMEI n\u1ed9i ti\u1ebfp v\u00ec c\u00f3 hai g\u00f3c \u0111\u1ed1i b\u00f9 nhau. 2) Ch\u1ee9ng minh ENI EBI v\u00e0 AE.IN BE .IM . X\u00e9t t\u1ee9 gi\u00e1c BNEI , c\u00f3: \uf0ef\uf0ecNBI = 90\uf0b0(BN \u22a5 AB) \uf0ed \uf0ee\uf0efNEI = 90\uf0b0 ( EN \u22a5 EI ) \uf0de NBI + NEI = 180\uf0b0 \uf0de T\u1ee9 gi\u00e1c BNEI n\u1ed9i ti\u1ebfp v\u00ec c\u00f3 hai g\u00f3c \u0111\u1ed1i b\u00f9 nhau. ENI EBI ( 2 g\u00f3c c\u00f3 \u0111\u1ec9nh k\u1ec1 nhau c\u00f9ng nh\u00ecn 1 c\u1ea1nh trong t\u1ee9 gi\u00e1c n\u1ed9i ti\u1ebfp) X\u00e9t ABE v\u00e0 IMN , c\u00f3: ABE MNI EBI ENI BAE IMN ( 2 g\u00f3c c\u00f3 \u0111\u1ec9nh k\u1ec1 nhau c\u00f9ng nh\u00ecn c\u1ea1nh EI trong t\u1ee9 gi\u00e1c AMEI n\u1ed9i ti\u1ebfp) ABE MNI g g AE BE AE.NI MI .BE MI NI 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 3) G\u1ecdi P l\u00e0 giao \u0111i\u1ec3m c\u1ee7a AE v\u00e0 MI ; Q l\u00e0 giao \u0111i\u1ec3m c\u1ee7a BE v\u00e0 NI . Ch\u1ee9ng minh hai \u0111\u01b0\u1eddng th\u1eb3ng PQ v\u00e0 BN vu\u00f4ng g\u00f3c v\u1edbi nhau. X\u00e9t O ta c\u00f3 : AEB l\u00e0 g\u00f3c n\u1ed9i ti\u1ebfp ch\u1eafn n\u1eeda \u0111\u01b0\u1eddng tr\u00f2n AEB 90 Ta c\u00f3 : MNI 90 ABE MIN AEB X\u00e9t t\u1ee9 gi\u00e1c EPIQ , c\u00f3: ( )\uf0ec\uf0efPEQ = 90\uf0b0 AEB = 90 ( )\uf0ed \uf0ef\uf0eePIQ = 90\uf0b0 MIN = 90 \uf0de PEQ + PIQ = 180\uf0b0 \uf0de T\u1ee9 gi\u00e1c EPIQ n\u1ed9i ti\u1ebfp v\u00ec c\u00f3 hai g\u00f3c \u0111\u1ed1i b\u00f9 nhau. EQP EIP ( v\u00ec 2 g\u00f3c \u1edf 2 \u0111\u1ec9nh k\u1ec1 c\u00f9ng nh\u00ecn 1 c\u1ea1nh trong t\u1ee9 gi\u00e1c n\u1ed9i ti\u1ebfp ) Ta c\u00f3: EQP EIP cmt EAM EIM ( v\u00ec 2 g\u00f3c \u1edf 2 \u0111\u1ec9nh k\u1ec1 c\u00f9ng nh\u00ecn EM trong t\u1ee9 gi\u00e1c AMEI n\u1ed9i ti\u1ebfp ) EAM EBA ( v\u00ec g\u00f3c t\u1ea1o b\u1edfi ti\u1ebfp tuy\u1ebfn v\u00e0 d\u00e2y cung EA v\u00e0 g\u00f3c n\u1ed9i ti\u1ebfp ch\u1eafn cung EA ) EQP EBA M\u00e0 2 g\u00f3c EQP v\u00e0 EBA \u1edf n\u1eb1m v\u1ecb tr\u00ed \u0111\u1ed3ng v\u1ecb N\u00ean PQ \/ \/BA M\u1eb7t kh\u00e1c AB BN Suy ra PQ BN 4) G\u1ecdi F l\u00e0 \u0111i\u1ec3m ch\u00ednh gi\u1eefa cung AB kh\u00f4ng ch\u1ee9a \u0111i\u1ec3m E c\u1ee7a \u0111\u01b0\u1eddng tr\u00f2n O . T\u00ednh di\u1ec7n t\u00edch tam gi\u00e1c OMN theo R khi ba \u0111i\u1ec3m E,I,F th\u1eb3ng h\u00e0ng. Khi E,I,F th\u1eb3ng h\u00e0ng th\u00ec AF BF AEI IEB 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 M\u00e0 AEI IEB AEB 90 N\u00ean AEI IEB 45 Ta c\u00f3 : AEI IEB 45 AEI AMI ( v\u00ec 2 g\u00f3c \u1edf 2 \u0111\u1ec9nh k\u1ec1 c\u00f9ng nh\u00ecn AI trong t\u1ee9 gi\u00e1c AMEI n\u1ed9i ti\u1ebfp ) BEI BNI ( v\u00ec 2 g\u00f3c \u1edf 2 \u0111\u1ec9nh k\u1ec1 c\u00f9ng nh\u00ecn BI trong t\u1ee9 gi\u00e1c BNEI n\u1ed9i ti\u1ebfp ) AEI AMI BEI BNI 45 Ta c\u00f3 : MAI vu\u00f4ng \u1edf A v\u00e0 AMI 45 BNI vu\u00f4ng \u1edf B v\u00e0 BNI 45 MAI vu\u00f4ng c\u00e2n \u1edf A v\u00e0 BNI vu\u00f4ng c\u00e2n \u1edf B MA AI R NB BI 2 3R 2 Ta c\u00f3 : S OMN SMABN SMAI SNBI AB. MA NB MA.IA IB.BN 2 22 2R. R 3R 22 R . R 3R . 3R 2 2 2 2 2 2 2 2R2 R2 9R2 3R2 8 84 ---H\u1ebeT--- T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N 11","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH S\u1ede GD $ \u0110T TP H\u1ed2 CH\u00cd MINH \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 PH\u00d2NG GI\u00c1O D\u1ee4C V\u00c0 \u0110\u00c0O T\u1ea0O QU\u1eacN 11 N\u0102M H\u1eccC : 2023 - 2024 \u0110\u1ec0 THAM KH\u1ea2O M\u00d4N: TO\u00c1N 9 M\u00c3 \u0110\u1ec0: Qu\u1eadn 11 - 3 \u0110\u00ea thi g\u1ed3m 8 c\u00e2u ho\u0309i t\u01b0\u0323 lu\u00e2\u0323n. Th\u01a1\u0300i gian: 120 phu\u0301t (kh\u00f4ng k\u00ea\u0309 th\u01a1\u0300i gian pha\u0301t \u0111\u00ea\u0300) C\u00e2u 1. (1,5 \u0111i\u1ec3m). Cho parabol (P) : y = 1 x2 v\u00e0 \u0111\u01b0\u1eddng th\u1eb3ng (d ) : y = \u2212x + 2 . 42 a) V\u1ebd \u0111\u1ed3 th\u1ecb (P) v\u00e0 (d) tr\u00ean c\u00f9ng h\u1ec7 tr\u1ee5c t\u1ecda \u0111\u1ed9. b) T\u00ecm t\u1ecda \u0111\u1ed9 giao \u0111i\u1ec3m c\u1ee7a (P) v\u00e0 (d) b\u1eb1ng ph\u00e9p to\u00e1n. C\u00e2u 2. (1 \u0111i\u1ec3m). Cho ph\u01b0\u01a1ng tr\u00ecnh 4x2 + 3x \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 = (x1 \u2212 2)(x2 \u2212 2) C\u00e2u 3. (1 \u0111i\u1ec3m). M\u1ed9t c\u1eeda h\u00e0ng s\u00e1ch c\u0169 c\u00f3 m\u1ed9t ch\u00ednh s\u00e1ch nh\u01b0 sau: N\u1ebfu kh\u00e1ch h\u00e0ng \u0111\u0103ng k\u00fd l\u00e0m h\u1ed9i vi\u00ean c\u1ee7a c\u1eeda h\u00e0ng s\u00e1ch th\u00ec m\u1ed7i n\u0103m ph\u1ea3i \u0111\u00f3ng ph\u00ed th\u00e0nh vi\u00ean l\u00e0 50000 \u0111\u1ed3ng\/n\u0103m. Bi\u1ebft r\u1eb1ng, l\u00e0 h\u1ed9i vi\u00ean khi thu\u00ea 2 cu\u1ed1n s\u00e1ch th\u00ec ph\u1ea3i tr\u1ea3 60000 \u0111\u1ed3ng (\u0110\u00e3 t\u00ednh ph\u00ed th\u00e0nh vi\u00ean). G\u1ecdi s (\u0111\u1ed3ng), l\u00e0 t\u1ed5ng s\u1ed1 ti\u1ec1n m\u1ed7i h\u00e0nh kh\u00e1ch l\u00e0 h\u1ed9i vi\u00ean ph\u1ea3i tr\u1ea3 trong m\u1ed7i n\u0103m v\u00e0 t l\u00e0 s\u1ed1 cu\u1ed1n s\u00e1ch m\u00e0 kh\u00e1ch h\u00e0ng thu\u00ea, bi\u1ebft s l\u00e0 h\u00e0m s\u1ed1 b\u1eadc nh\u1ea5t c\u00f3 d\u1ea1ng: s = at + b . a) T\u00ecm h\u1ec7 s\u1ed1 a v\u00e0 b . b) N\u1ebfu kh\u00e1ch h\u00e0ng kh\u00f4ng ph\u1ea3i l\u00e0 h\u1ed9i vi\u00ean th\u00ec s\u1ebd thu\u00ea s\u00e1ch v\u1edbi gi\u00e1 10000 \u0111\u1ed3ng\/ cu\u1ed1n s\u00e1ch . Nam l\u00e0 h\u1ed9i vi\u00ean c\u1ee7a c\u1eeda h\u00e0ng s\u00e1ch, n\u0103m ngo\u00e1i Nam \u0111\u00e3 ph\u1ea3i tr\u1ea3 cho c\u1eeda h\u00e0ng s\u00e1ch t\u1ed5ng c\u1ed9ng 90000 \u0111\u1ed3ng. H\u1ecfi n\u1ebfu Nam kh\u00f4ng ph\u1ea3i l\u00e0 h\u1ed9i vi\u00ean c\u1ee7a c\u1eeda h\u00e0ng s\u00e1ch th\u00ec s\u1ed1 ti\u1ec1n ph\u1ea3i tr\u1ea3 l\u00e0 bao nhi\u00eau? C\u00e2u 4. (1 \u0111i\u1ec3m). M\u1ed9t c\u1eeda h\u00e0ng \u0111i\u1ec7n m\u00e1y nh\u1eadp v\u1ec1 m\u1ed9t l\u00f4 h\u00e0ng g\u1ed3m 100 chi\u1ebfc \u0111i\u1ec7n tho\u1ea1i di \u0111\u1ed9ng v\u00e0 b\u00e1n v\u1edbi gi\u00e1 ni\u00eam y\u1ebft l\u00e0 8500000 \u0111\u1ed3ng. a) Ng\u01b0\u1eddi ch\u1ee7 c\u1eeda h\u00e0ng cho bi\u1ebft m\u1ed7i \u0111i\u1ec7n tho\u1ea1i di \u0111\u1ed9ng b\u00e1n ra v\u1edbi gi\u00e1 tr\u00ean \u0111em l\u1ea1i l\u1ee3i nhu\u1eadn 70% so v\u1edbi gi\u00e1 nh\u1eadp v\u00e0o. H\u00e3y cho bi\u1ebft gi\u00e1 nh\u1eadp v\u00e0o c\u1ee7a l\u00f4 h\u00e0ng tr\u00ean. b) Sau khi b\u00e1n \u0111\u01b0\u1ee3c 60 chi\u1ebfc \u0111i\u1ec7n tho\u1ea1i di \u0111\u1ed9ng th\u00ec ng\u01b0\u1eddi ch\u1ee7 gi\u1ea3m gi\u00e1 20% v\u00e0 b\u00e1n \u0111\u01b0\u1ee3c s\u1ed1 \u0111i\u1ec7n tho\u1ea1i c\u00f2n l\u1ea1i. H\u00e3y t\u00ednh t\u1ec9 l\u1ec7 ph\u1ea7n tr\u0103m l\u1ee3i nhu\u1eadn m\u00e0 c\u1eeda h\u00e0ng \u0111\u1ea1t \u0111\u01b0\u1ee3c c\u1ee7a l\u00f4 h\u00e0ng tr\u00ean. C\u00e2u 5. (0,75 \u0111i\u1ec3m). \u0110\u1ec3 h\u01b0\u1edfng \u1ee9ng phong tr\u00e0o l\u00e0m l\u1ed3ng \u0111\u00e8n t\u1eb7ng c\u00e1c em nh\u1ecf v\u00f9ng s\u00e2u trong d\u1ecbp Trung Thu, l\u1edbp 9A g\u1ed3m 46 b\u1ea1n ti\u1ebfn h\u00e0nh l\u00e0m v\u00e0o c\u00e1c ng\u00e0y cu\u1ed1i tu\u1ea7n. C\u00f4 gi\u00e1o \u0111\u1eb7t ch\u1ec9 ti\u00eau: m\u1ed9t b\u1ea1n nam s\u1ebd l\u00e0m 2 c\u00e1i l\u1ed3ng \u0111\u00e8n c\u00f2n b\u1ea1n n\u1eef th\u00ec m\u1ed9t b\u1ea1n l\u00e0m ra 3 c\u00e1i. Sau 2 th\u00e1ng, c\u1ea3 l\u1edbp l\u00e0m t\u1ed5ng c\u1ed9ng \u0111\u01b0\u1ee3c 118 c\u00e1i. H\u1ecfi l\u1edbp \u0111\u00f3 c\u00f3 bao nhi\u00eau b\u1ea1n h\u1ecdc sinh nam v\u00e0 bao nhi\u00eau h\u1ecdc sinh n\u1eef? 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. (0,75 \u0111i\u1ec3m). T\u1eeb n\u00f3c m\u1ed9t cao \u1ed1c cao 50m ng\u01b0\u1eddi ta nh\u00ecn th\u1ea5y ch\u00e2n v\u00e0 \u0111\u1ec9nh m\u1ed9t c\u1ed9t \u0103ng- ten v\u1edbi c\u00e1c g\u00f3c h\u1ea1 v\u00e0 n\u00e2ng l\u1ea7n l\u01b0\u1ee3t l\u00e0 62\uf0b0 v\u00e0 34\uf0b0 . T\u00ednh chi\u1ec1u cao c\u1ee7a c\u1ed9t \u0103ng- ten. C\u00e2u 7. (1 \u0111i\u1ec3m). M\u1ed9t chi\u1ebfc n\u00f3n \u00f4ng gi\u00e0 Noel th\u01b0\u1eddng g\u1ed3m c\u00f3 ba ph\u1ea7n: H\u00ecnh tr\u1ee5 \u0111\u1ec3 l\u00e0m \u0111\u1ebf n\u00f3n, ph\u1ea7n m\u0169 ch\u00ednh l\u00e0 h\u00ecnh n\u00f3n, tr\u00ean \u0111\u1ec9nh n\u00f3n l\u00e0 qu\u1ea3 b\u00f3ng tr\u1eafng c\u00f3 h\u00ecnh c\u1ea7u v\u00e0 c\u00f3 c\u00e1c k\u00edch th\u01b0\u1edbc t\u01b0\u01a1ng \u1ee9ng nh\u01b0 h\u00ecnh v\u1ebd. T\u00ednh t\u1ed5ng di\u1ec7n t\u00edch ph\u1ea7n v\u1ea3i \u0111\u1ec3 may n\u00f3n, bi\u1ebft r\u1eb1ng chi\u1ec1u cao c\u1ee7a \u0111\u1ebf n\u00f3n b\u1eb1ng \u0111\u01b0\u1eddng k\u00ednh c\u1ee7a qu\u1ea3 b\u00f3ng (k\u1ebft qu\u1ea3 l\u00e0m tr\u00f2n \u0111\u1ebfn h\u00e0ng \u0111\u01a1n v\u1ecb). C\u00e2u 8. (3 \u0111i\u1ec3m) T\u1eeb \u0111i\u1ec3m A \u1edf ngo\u00e0i \u0111\u01b0\u1eddng tr\u00f2n (O; R) v\u1ebd hai ti\u1ebfp tuy\u1ebfn AB v\u00e0 AC v\u00e0 m\u1ed9t c\u00e1t tuy\u1ebfn ADE kh\u00f4ng \u0111i qua t\u00e2m O ( B,C t\u1ea1i l\u00e0 c\u00e1c ti\u1ebfp \u0111i\u1ec3m v\u00e0 AD \uf03c AE ) a) Ch\u1ee9ng minh t\u1ee9 gi\u00e1c ABOC n\u1ed9i ti\u1ebfp, x\u00e1c \u0111\u1ecbnh t\u00e2m v\u00e0 b\u00e1n k\u00ednh c\u1ee7a \u0111\u01b0\u1eddng tr\u00f2n \u0111\u00f3?. b) G\u1ecdi H l\u00e0 giao \u0111i\u1ec3m c\u1ee7a OA v\u1edbi BC . Ch\u1ee9ng minh AH.AO = AD.AE = AB2 . c) G\u1ecdi I l\u00e0 trung \u0111i\u1ec3m c\u1ee7a DE . Qua B v\u1ebd d\u00e2y BK \/ \/DE . Ch\u1ee9ng minh ba \u0111i\u1ec3m K,I,C 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 = 1 x2 v\u00e0 \u0111\u01b0\u1eddng th\u1eb3ng (d ) : y = \u2212x + 2 . 42 a) V\u1ebd \u0111\u1ed3 th\u1ecb (P) v\u00e0 (d) tr\u00ean c\u00f9ng h\u1ec7 tr\u1ee5c t\u1ecda \u0111\u1ed9. b) T\u00ecm t\u1ecda \u0111\u1ed9 giao \u0111i\u1ec3m c\u1ee7a (P) v\u00e0 (d) b\u1eb1ng ph\u00e9p t\u00ednh. L\u1eddi gi\u1ea3i a) V\u1ebd \u0111\u1ed3 th\u1ecb (P) v\u00e0 (d) tr\u00ean c\u00f9ng h\u1ec7 tr\u1ee5c t\u1ecda \u0111\u1ed9. BGT: x \u22124 \u22122 0 2 4 y = 1 x2 4 1 0 1 4 4 x \u22124 0 y = \u2212x + 2 4 2 2 b) T\u00ecm t\u1ecda \u0111\u1ed9 giao \u0111i\u1ec3m c\u1ee7a (P) v\u00e0 (d) b\u1eb1ng ph\u00e9p t\u00ednh. Ph\u01b0\u01a1ng tr\u00ecnh ho\u00e0nh \u0111\u1ed9 giao \u0111i\u1ec3m c\u1ee7a (P) v\u00e0 (d) : 1 x2 = \u2212x + 2 42 \uf0db 1 x2 + x \u2212 2 = 0 42 \uf0db \uf0e9x = 2 \uf0ea\uf0ebx = \u22124 Thay x = \u22124 v\u00e0o y = 1 x2 , ta \u0111\u01b0\u1ee3c: y = 1 .42 = 4 . 44 Thay x = 2 v\u00e0o y = 1 x2 , ta \u0111\u01b0\u1ee3c: y = 1 .22 = 1 . 44 V\u1eady (\u22124; 4) , (2; 1) l\u00e0 hai giao \u0111i\u1ec3m c\u1ea7n t\u00ecm. T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N 3","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH C\u00e2u 2. (1 \u0111i\u1ec3m) Cho ph\u01b0\u01a1ng tr\u00ecnh 4x2 + 3x \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 = (x1 \u2212 2)(x2 \u2212 2) L\u1eddi gi\u1ea3i V\u00ec \uf044 = b2 \u2212 4ac = 32 \u2212 4.4.(\u22121) = 25 \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 = \u22123 \uf0ed = x1 a 4 Theo \u0111\u1ecbnh l\u00ed Vi-et, ta c\u00f3: \uf0ef\uf0ef\uf0eeP \u22121 c = 4 .x2 = a Ta c\u00f3: A = (x1 \u2212 2)(x2 \u2212 2) A = x1x2 \u2212 2x1 \u2212 2x2 + 4 A = x1x2 \u2212 2(x1 + x2 ) + 4 A = \u22121 \u2212 2.\uf0e6\uf0e7 \u22123 \uf0f6 + 4 4 \uf0e8 4 \uf0f7 \uf0f8 A = 21 4 C\u00e2u 3. (1 \u0111i\u1ec3m) M\u1ed9t c\u1eeda h\u00e0ng s\u00e1ch c\u0169 c\u00f3 m\u1ed9t ch\u00ednh s\u00e1ch nh\u01b0 sau: : N\u1ebfu kh\u00e1ch h\u00e0ng \u0111\u0103ng k\u00fd l\u00e0m h\u1ed9i vi\u00ean c\u1ee7a c\u1eeda h\u00e0ng s\u00e1ch th\u00ec m\u1ed7i n\u0103m ph\u1ea3i \u0111\u00f3ng ph\u00ed th\u00e0nh vi\u00ean l\u00e0 50000 \u0111\u1ed3ng\/n\u0103m. Bi\u1ebft r\u1eb1ng, l\u00e0 h\u1ed9i vi\u00ean khi thu\u00ea 2 cu\u1ed1n s\u00e1ch th\u00ec ph\u1ea3i tr\u1ea3 60000 \u0111\u1ed3ng (\u0110\u00e3 t\u00ednh ph\u00ed th\u00e0nh vi\u00ean). G\u1ecdi s (\u0111\u1ed3ng), l\u00e0 t\u1ed5ng s\u1ed1 ti\u1ec1n m\u1ed7i h\u00e0nh kh\u00e1ch l\u00e0 h\u1ed9i vi\u00ean ph\u1ea3i tr\u1ea3 trong m\u1ed7i n\u0103m v\u00e0 t l\u00e0 s\u1ed1 cu\u1ed1n s\u00e1ch m\u00e0 kh\u00e1ch h\u00e0ng thu\u00ea, bi\u1ebft s l\u00e0 h\u00e0m s\u1ed1 b\u1eadc nh\u1ea5t c\u00f3 d\u1ea1ng: s = at + b a) T\u00ecm h\u1ec7 s\u1ed1 a v\u00e0 b b) N\u1ebfu kh\u00e1ch h\u00e0ng kh\u00f4ng ph\u1ea3i l\u00e0 h\u1ed9i vi\u00ean th\u00ec s\u1ebd thu\u00ea s\u00e1ch v\u1edbi gi\u00e1 10000 \u0111\u1ed3ng\/ cu\u1ed1n s\u00e1ch . Nam l\u00e0 h\u1ed9i vi\u00ean c\u1ee7a c\u01b0\u1ea3 h\u00e0ng s\u00e1ch, n\u0103m ngo\u00e1i Nam \u0111\u00e3 ph\u1ea3i tr\u1ea3 cho c\u1eeda h\u00e0ng s\u00e1ch t\u1ed5ng c\u1ed9ng 90000 \u0111\u1ed3ng. H\u1ecfi n\u1ebfu Nam kh\u00f4ng ph\u1ea3i l\u00e0 h\u1ed9i vi\u00ean c\u1ee7a c\u1eeda h\u00e0ng s\u00e1ch th\u00ec s\u1ed1 ti\u1ec1n ph\u1ea3i tr\u1ea3 l\u00e0 bao nhi\u00eau? 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: \uf0ect = 0 \uf0de 50000 = 0.a + b . (1) V\u1edbi \uf0ed\uf0ees = 50000 V\u1edbi \uf0ect = 2 \uf0de 60000 = 2.a + b . (2) \uf0ed\uf0ees = 60000 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 T\u1eeb (1) v\u00e0 (2) ta c\u00f3 h\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh: \uf0ec0a + b = 50000 \uf0db \uf0eca = 5000 . \uf0ee\uf0ed2a + b = 60000 \uf0ee\uf0edb = 50000 V\u1eady: a = 5000 , b = 50000 v\u00e0 s = 5000t + 50000 . b) N\u1ebfu kh\u00e1ch h\u00e0ng kh\u00f4ng ph\u1ea3i l\u00e0 h\u1ed9i vi\u00ean th\u00ec s\u1ebd thu\u00ea s\u00e1ch v\u1edbi gi\u00e1 10000 \u0111\u1ed3ng\/ cu\u1ed1n s\u00e1ch . Nam l\u00e0 h\u1ed9i vi\u00ean c\u1ee7a c\u01b0\u1ea3 h\u00e0ng s\u00e1ch, n\u0103m ngo\u00e1i Nam \u0111\u00e3 ph\u1ea3i tr\u1ea3 cho c\u1eeda h\u00e0ng s\u00e1ch t\u1ed5ng c\u1ed9ng 90000 \u0111\u1ed3ng. H\u1ecfi n\u1ebfu Nam kh\u00f4ng ph\u1ea3i l\u00e0 h\u1ed9i vi\u00ean c\u1ee7a c\u1eeda h\u00e0ng s\u00e1ch th\u00ec s\u1ed1 ti\u1ec1n ph\u1ea3i tr\u1ea3 l\u00e0 bao nhi\u00eau? Thay s = 90000 v\u00e0o s = 5000t + 50000 ta c\u00f3: 90000 = 5000t + 50000 \uf0dbt=8 V\u1eady Nam ph\u1ea3i tr\u1ea3 s\u1ed1 ti\u1ec1n thu\u00ea 8 cu\u1ed1n s\u00e1ch khi kh\u00f4ng ph\u1ea3i l\u00e0 h\u1ed9i vi\u00ean l\u00e0 : 8.10000 = 80000 \u0111\u1ed3ng. C\u00e2u 4. (1 \u0111i\u1ec3m). M\u1ed9t c\u1eeda h\u00e0ng \u0111i\u1ec7n m\u00e1y nh\u1eadp v\u1ec1 m\u1ed9t l\u00f4 h\u00e0ng g\u1ed3m 100 chi\u1ebfc \u0111i\u1ec7n tho\u1ea1i di \u0111\u1ed9ng v\u00e0 b\u00e1n v\u1edbi gi\u00e1 ni\u00eam y\u1ebft l\u00e0 8500000 \u0111\u1ed3ng. a) Ng\u01b0\u1eddi ch\u1ee7 c\u1eeda h\u00e0ng cho bi\u1ebft m\u1ed7i \u0111i\u1ec7n tho\u1ea1i di \u0111\u1ed9ng b\u00e1n ra v\u1edbi gi\u00e1 tr\u00ean \u0111em l\u1ea1i l\u1ee3i nhu\u1eadn 70% so v\u1edbi gi\u00e1 nh\u1eadp v\u00e0o. H\u00e3y cho bi\u1ebft gi\u00e1 nh\u1eadp v\u00e0o c\u1ee7a l\u00f4 h\u00e0ng tr\u00ean. b) Sau khi b\u00e1n \u0111\u01b0\u1ee3c 60 chi\u1ebfc \u0111i\u1ec7n tho\u1ea1i di \u0111\u1ed9ng th\u00ec ng\u01b0\u1eddi ch\u1ee7 gi\u1ea3m gi\u00e1 20% v\u00e0 b\u00e1n \u0111\u01b0\u1ee3c s\u1ed1 \u0111i\u1ec7n tho\u1ea1i c\u00f2n l\u1ea1i. H\u00e3y t\u00ednh t\u1ec9 l\u1ec7 ph\u1ea7n tr\u0103m l\u1ee3i nhu\u1eadn m\u00e0 c\u1eeda h\u00e0ng \u0111\u1ea1t \u0111\u01b0\u1ee3c c\u1ee7a l\u00f4 h\u00e0ng tr\u00ean. L\u1eddi gi\u1ea3i Gi\u00e1 nh\u1eadp v\u00e0o c\u1ee7a m\u1ed9t chi\u1ebfc \u0111i\u1ec7n tho\u1ea1i l\u00e0: 8 500000 : (100% + 70%) = 5000000 (\u0111\u1ed3ng) Gi\u00e1 nh\u1eadp v\u00e0o c\u1ee7a l\u00f4 h\u00e0ng l\u00e0: 5000000.100 = 500000000 (\u0111\u1ed3ng) b) S\u1ed1 ti\u1ec1n thu v\u1ec1 khi b\u00e1n h\u1ebft 100 chi\u1ebfc \u0111i\u1ec7n tho\u1ea1i l\u00e0: 60.8 500000 + 40.8 500000.80% =782000000 (\u0111\u1ed3ng) L\u1ee3i nhu\u1eadn thu \u0111\u01b0\u1ee3c t\u1eeb vi\u1ec7c b\u00e1n 100 chi\u1ebfc \u0111i\u1ec7n tho\u1ea1i l\u00e0: 782000000 \u2212 500000000 = 282000000 (\u0111\u1ed3ng) T\u1ec9 l\u1ec7 ph\u1ea7n tr\u0103m l\u1ee3i nhu\u1eadn m\u00e0 c\u1eeda h\u00e0ng \u0111\u1ea1t \u0111\u01b0\u1ee3c c\u1ee7a l\u00f4 h\u00e0ng tr\u00ean l\u00e0: 282000000 : 500000000.100% = 56,4% C\u00e2u 5. (0,75 \u0111i\u1ec3m) \u0110\u1ec3 h\u01b0\u1edfng \u1ee9ng phong tr\u00e0o l\u00e0m l\u1ed3ng \u0111\u00e8n t\u1eb7ng c\u00e1c em nh\u1ecf v\u00f9ng s\u00e2u trong d\u1ecbp Trung Thu, l\u1edbp 9A g\u1ed3m 46 b\u1ea1n ti\u1ebfn h\u00e0nh l\u00e0m v\u00e0o c\u00e1c ng\u00e0y cu\u1ed1i tu\u1ea7n. C\u00f4 gi\u00e1o \u0111\u1eb7t ch\u1ec9 ti\u00eau: m\u1ed9t b\u1ea1n nam s\u1ebd l\u00e0m 2 c\u00e1i l\u1ed3ng \u0111\u00e8n c\u00f2n b\u1ea1n n\u1eef th\u00ec m\u1ed9t b\u1ea1n l\u00e0m ra 3 c\u00e1i. Sau 2 th\u00e1ng, c\u1ea3 l\u1edbp 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\u00e0m t\u1ed5ng c\u1ed9ng \u0111\u01b0\u1ee3c 118 c\u00e1i. H\u1ecfi l\u1edbp \u0111\u00f3 c\u00f3 bao nhi\u00eau b\u1ea1n h\u1ecdc sinh nam v\u00e0 bao nhi\u00eau h\u1ecdc sinh n\u1eef? L\u1eddi gi\u1ea3i G\u1ecdi x (h\u1ecdc sinh) l\u00e0 s\u1ed1 h\u1ecdc sinh Nam c\u1ee7a l\u1edbp ( x \uf0ce \uf02a ) G\u1ecdi y (h\u1ecdc sinh) l\u00e0 s\u1ed1 h\u1ecdc sinh N\u1eef c\u1ee7a l\u1edbp ( y \uf0ce \uf02a ) ( )Do l\u1edbp c\u00f3 46 h\u1ecdc sinh n\u00ean ta c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh: x + y = 46 1 Do m\u1ed9t b\u1ea1n nam s\u1ebd l\u00e0m 2 c\u00e1i l\u1ed3ng \u0111\u00e8n c\u00f2n b\u1ea1n n\u1eef th\u00ec m\u1ed9t b\u1ea1n l\u00e0m ra 3 c\u00e1i m\u00e0 c\u1ea3 l\u1edbp l\u00e0m ( )t\u1ed5ng c\u1ed9ng \u0111\u01b0\u1ee3c 118 c\u00e1i n\u00ean ta c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh: 2x + 3y = 118 2 ( ) ( )T\u1eeb 1 v\u00e0 2 ta c\u00f3 h\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh: \uf0ecx + y = 46 \uf0db \uf0ecx = 20 (nh\u1eadn) \uf0ed\uf0ee2x + 3y = 118 \uf0ed\uf0eey = 26 V\u1eady l\u1edbp c\u00f3 20 h\u1ecdc sinh Nam v\u00e0 26 h\u1ecdc sinh n\u1eef. C\u00e2u 6. (0,75 \u0111i\u1ec3m) T\u1eeb n\u00f3c m\u1ed9t cao \u1ed1c cao 50m ng\u01b0\u1eddi ta nh\u00ecn th\u1ea5y ch\u00e2n v\u00e0 \u0111\u1ec9nh m\u1ed9t c\u1ed9t \u0103ng- ten v\u1edbi c\u00e1c g\u00f3c h\u1ea1 v\u00e0 n\u00e2ng l\u1ea7n l\u01b0\u1ee3t l\u00e0 62\uf0b0 v\u00e0 34\uf0b0 . T\u00ednh chi\u1ec1u cao c\u1ee7a c\u1ed9t \u0103ng- ten L\u1eddi gi\u1ea3i Ta c\u00f3 AB = CD = 50m X\u00e9t BDC vu\u00f4ng t\u1ea1i D , ta c\u00f3: Tan DBC DC DB (tslg) Tan 62 50 DB BD 26, 6m X\u00e9t BDE vu\u00f4ng t\u1ea1i D , ta c\u00f3: 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 Tan DBE DE DB (tslg) Tan 340 DE 26, 6 DE 17, 9 m Do \u0111\u00f3 CE CD DE 50 17, 9 67, 9 m V\u1eady chi\u1ec1u cao c\u1ee7a c\u1ed9t \u0103ng- ten l\u00e0 kho\u1ea3ng 67, 9 m . C\u00e2u 7. (1 \u0111i\u1ec3m) M\u1ed9t chi\u1ebfc n\u00f3n \u00f4ng gi\u00e0 Noel th\u01b0\u1eddng g\u1ed3m c\u00f3 ba ph\u1ea7n: H\u00ecnh tr\u1ee5 \u0111\u1ec3 l\u00e0m \u0111\u1ebf n\u00f3n, ph\u1ea7n m\u0169 ch\u00ednh l\u00e0 h\u00ecnh n\u00f3n, tr\u00ean \u0111\u1ec9nh n\u00f3n l\u00e0 qu\u1ea3 b\u00f3ng tr\u1eafng c\u00f3 h\u00ecnh c\u1ea7u v\u00e0 c\u00f3 c\u00e1c k\u00edch th\u01b0\u1edbc t\u01b0\u01a1ng \u1ee9ng nh\u01b0 h\u00ecnh v\u1ebd. T\u00ednh t\u1ed5ng di\u1ec7n t\u00edch ph\u1ea7n v\u1ea3i \u0111\u1ec3 may n\u00f3n, bi\u1ebft r\u1eb1ng chi\u1ec1u cao c\u1ee7a \u0111\u1ebf n\u00f3n b\u1eb1ng \u0111\u01b0\u1eddng k\u00ednh c\u1ee7a qu\u1ea3 b\u00f3ng. (k\u1ebft qu\u1ea3 l\u00e0m tr\u00f2n \u0111\u1ebfn h\u00e0ng \u0111\u01a1n v\u1ecb). L\u1eddi gi\u1ea3i Di\u1ec7n t\u00edch v\u1ea3i \u0111\u1ec3 l\u00e0m qu\u1ea3 b\u00f3ng tr\u1eafng c\u00f3 h\u00ecnh c\u1ea7u l\u00e0: 4\uf070 R2 = 4\uf070 .\uf0e6\uf0e7\uf0e8 7 \uf0f62 = 49\uf070 cm2 2 \uf0f7\uf0f8 Di\u1ec7n t\u00edch v\u1ea3i \u0111\u1ec3 l\u00e0m \u0111\u1ebf n\u00f3n h\u00ecnh tr\u1ee5 l\u00e0: 2\uf070 rh = 2\uf070 .\uf0e6\uf0e7\uf0e8 32 \uf0f62 .7 = 3584\uf070 cm2 2 \uf0f7\uf0f8 \u0110\u1ed9 d\u00e0i c\u1ea1nh OH l\u00e0: 45 \u2212 7 \u2212 7 = 34,5 cm 2 \u0110\u1ed9 d\u00e0i c\u1ea1nh OA l\u00e0: OH 2 + HA2 = 34,52 +162 \uf0bb 38 cm Di\u1ec7n t\u00edch ph\u1ea7n v\u1ea3i \u0111\u1ec3 l\u00e0m ph\u1ea7n m\u0169 h\u00ecnh n\u00f3n l\u00e0: 2\uf070 rl = 2\uf070 .\uf0e6\uf0e7\uf0e8 32 \uf0f6 .38 = 1216\uf070 2 \uf0f7\uf0f8 T\u1ed5ng di\u1ec7n t\u00edch ph\u1ea7n v\u1ea3i \u0111\u1ec3 may n\u00f3n: 49\uf070 + 3584\uf070 +1216\uf070 \uf0bb15234 cm2 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 8. (3 \u0111i\u1ec3m) T\u1eeb \u0111i\u1ec3m A \u1edf ngo\u00e0i \u0111\u01b0\u1eddng tr\u00f2n (O; R) v\u1ebd hai ti\u1ebfp tuy\u1ebfn AB v\u00e0 AC v\u00e0 m\u1ed9t c\u00e1t tuy\u1ebfn ADE kh\u00f4ng \u0111i qua t\u00e2m O ( B,C t\u1ea1i l\u00e0 c\u00e1c ti\u1ebfp \u0111i\u1ec3m v\u00e0 AD \uf03c AE ) a) Ch\u1ee9ng minh t\u1ee9 gi\u00e1c ABOC n\u1ed9i ti\u1ebfp, x\u00e1c \u0111\u1ecbnh t\u00e2m v\u00e0 b\u00e1n k\u00ednh c\u1ee7a \u0111\u01b0\u1eddng tr\u00f2n \u0111\u00f3?. b) G\u1ecdi H l\u00e0 giao \u0111i\u1ec3m c\u1ee7a OA v\u1edbi BC . Ch\u1ee9ng minh AH.AO = AD.AE = AB2 . c) G\u1ecdi I l\u00e0 trung \u0111i\u1ec3m c\u1ee7a DE . Qua B v\u1ebd d\u00e2y BK \/ \/DE . Ch\u1ee9ng minh ba \u0111i\u1ec3m K,I,C th\u1eb3ng h\u00e0ng . L\u1eddi gi\u1ea3i a) Ch\u1ee9ng minh t\u1ee9 gi\u00e1c ABOC n\u1ed9i ti\u1ebfp, x\u00e1c \u0111\u1ecbnh t\u00e2m v\u00e0 b\u00e1n k\u00ednh c\u1ee7a \u0111\u01b0\u1eddng tr\u00f2n \u0111\u00f3? 8 X\u00e9t t\u1ee9 gi\u00e1c ABOC , c\u00f3: ABO = ACO = 90\uf0b0 ( AB , AC l\u00e0 hai ti\u1ebfp tuy\u1ebfn c\u1ee7a (O,R) \uf0de ABO + ACO = 180\uf0b0 \uf0de T\u1ee9 gi\u00e1c ABOC n\u1ed9i ti\u1ebfp \u0111\u01b0\u1eddng tr\u00f2n \u0111\u01b0\u1eddng k\u00ednh OA v\u00ec c\u00f3 hai g\u00f3c \u0111\u1ed1i b\u00f9 nhau. T\u00e2m c\u1ee7a \u0111\u01b0\u1eddng tr\u00f2n l\u00e0 trung \u0111i\u1ec3m c\u1ee7a OA v\u00e0 b\u00e1n k\u00ednh b\u1eb1ng OA 2 b) G\u1ecdi H l\u00e0 giao \u0111i\u1ec3m c\u1ee7a OA v\u1edbi BC . Ch\u1ee9ng minh AH.AO = AD.AE = AB2 . X\u00e9t \uf044ABD v\u00e0 \uf044AEB c\u00f3: BAE l\u00e0 g\u00f3c chung ABD = AEB ( 2 gnt c\u00f9ng ch\u1eafn BD ) \uf0de \uf044ABD \uf044AEB (g \u2013 g). \uf0de AD = AB (TS\u0110D). AB AE T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH \uf0de AD.AE = AB2 (1) Ta c\u00f3: AB = AC (t\u00ednh ch\u1ea5t hai ti\u1ebfp tuy\u1ebfn c\u1eaft nhau) OA = OB = R \uf0de OA l\u00e0 \u0111\u01b0\u1eddng trung tr\u1ef1c c\u1ee7a BC . \uf0de OA \u22a5 BC t\u1ea1i H X\u00e9t \uf044ABO vu\u00f4ng t\u1ea1i B , c\u00f3 BH l\u00e0 \u0111\u01b0\u1eddng cao: AH.AO = AB2 (HTL) (2) T\u1eeb (1) , (2) suy ra: AH.AO = AD.AE = AB2 c) G\u1ecdi I l\u00e0 trung \u0111i\u1ec3m c\u1ee7a DE . Qua B v\u1ebd d\u00e2y BK \/ \/DE . Ch\u1ee9ng minh ba \u0111i\u1ec3m K,I,C th\u1eb3ng h\u00e0ng . Ta c\u00f3 BK \/ \/DE(gt) \uf0de EKBD l\u00e0 h\u00ecnh thang M\u00e0 EKBD l\u00e0 t\u1ee9 gi\u00e1c n\u1ed9i ti\u1ebfp (do E,K,B,D \uf0ce(O; R) ) \uf0de EKBD l\u00e0 h\u00ecnh thang c\u00e2n D\u1ec5 d\u00e0ng Cm: \uf044KIE = \uf044BID(c.g.c) \uf0de IK = IB \uf0de \uf044KIB c\u00e2n t\u1ea1i I \uf0de IKB = IBK Ta c\u00f3: BIA = IBK (SLT do BK \/ \/ED) (3) L\u1ea1i c\u00f3 CKB = BCA\uf0e6\uf0e7 = 1 sd BC \uf0f6 (4) \uf0e8 2 \uf0f7 \uf0f8 D\u1ec5 d\u00e0ng ch\u1ee9ng minh 5 \u0111i\u1ec3m A,B,C,O,I c\u00f9ng thu\u1ed9c m\u1ed9t \u0111\u01b0\u1eddng tr\u00f2n \uf0de BCA = BIA (5) T\u1eeb (3) , (4) (5) suy ra: IBK = CKB M\u00e0 IKB = IBK (ch\u1ee9ng minh tr\u00ean) \uf0de IKB = CKB \uf0de K,I,C 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 12 NA\u00caM HO\u00cfC: 2023 - 2024 \u0110\u1ec0 THAM KH\u1ea2O M\u00d4N: TO\u00c1N 9 M\u00c3 \u0110\u1ec0: Qu\u1eadn 12 - 1 \u0110\u00ea thi g\u1ed3m 8 c\u00e2u ho\u0309i t\u01b0\u0323 lu\u00e2\u0323n. Th\u01a1\u0300i gian: 120 phu\u0301t (kh\u00f4ng k\u00ea\u0309 th\u01a1\u0300i gian pha\u0301t \u0111\u00ea\u0300) C\u00e2u 1. (1,5 \u0111i\u1ec3m). Cho (P) : y = x2 v\u00e0 \u0111\u01b0\u1eddng th\u1eb3ng (D) : y = x + 1 . 22 a) V\u1ebd \u0111\u1ed3 th\u1ecb (P) v\u00e0 (D) tr\u00ean c\u00f9ng h\u1ec7 tr\u1ee5c t\u1ecda \u0111\u1ed9. b) T\u00ecm t\u1ecda \u0111\u1ed9 giao \u0111i\u1ec3m c\u1ee7a (P) v\u00e0 (D) b\u1eb1ng ph\u00e9p t\u00ednh. C\u00e2u 2. (1,0 \u0111i\u1ec3m). Cho ph\u01b0\u01a1ng tr\u00ecnh 3x2 + 2x \u2212 4 = 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= 3x12 + 5x1x2 + 3x22 4x13x2 + 4x1x23 C\u00e2u 3. (1,0 \u0111i\u1ec3m). Gi\u00e1 b\u00e1n m\u1ed9t c\u00e1i b\u00e1nh \u1edf hai c\u1eeda h\u00e0ng A v\u00e0 B \u0111\u1ec1u l\u00e0 15000 \u0111\u1ed3ng, nh\u01b0ng m\u1ed7i c\u1eeda h\u00e0ng c\u00f3 h\u00ecnh th\u1ee9c khuy\u1ebfn m\u00e3i kh\u00e1c nhau. - C\u1eeda h\u00e0ng A : n\u1ebfu kh\u00e1ch h\u00e0ng mua b\u1ed1n c\u00e1i b\u00e1nh tr\u1edf l\u00ean th\u00ec ba c\u00e1i b\u00e1nh \u0111\u1ea7u ti\u00ean gi\u00e1 m\u1ed7i c\u00e1i b\u00e1nh v\u1eabn l\u00e0 15000 \u0111\u1ed3ng nh\u01b0ng t\u1eeb c\u00e1i b\u00e1nh th\u1ee9 t\u01b0 tr\u1edf \u0111i kh\u00e1ch h\u00e0ng ch\u1ec9 ph\u1ea3i tr\u1ea3 80 % gi\u00e1 \u0111ang b\u00e1n. - C\u1eeda h\u00e0ng B : n\u1ebfu kh\u00e1ch h\u00e0ng mua 3 c\u00e1i b\u00e1nh th\u00ec \u0111\u01b0\u1ee3c t\u1eb7ng m\u1ed9t c\u00e1i b\u00e1nh mi\u1ec5n ph\u00ed. M\u1ed9t nh\u00f3m b\u1ea1n h\u1ecdc sinh mua 15 c\u00e1i b\u00e1nh th\u00ec ch\u1ecdn c\u1eeda h\u00e0ng n\u00e0o c\u00f3 l\u1ee3i h\u01a1n v\u00e0 l\u1ee3i h\u01a1n bao nhi\u00eau? C\u00e2u 4. (0,75 \u0111i\u1ec3m). Ba l\u00e0ng A, B, C n\u1eb1m tr\u00ean c\u00f9ng m\u1ed9t qu\u1ed1c l\u1ed9, l\u00e0ng B n\u1eb1m gi\u1eefa l\u00e0ng A v\u00e0 l\u00e0ng C, l\u00e0ng A c\u00e1ch l\u00e0ng B 5 km. M\u1ed9t ng\u01b0\u1eddi \u0111i b\u1ed9 theo h\u01b0\u1edbng t\u1eeb l\u00e0ng B \u0111\u1ebfn l\u00e0ng C m\u1ed7i gi\u1edd c\u00e1ch l\u00e0ng A th\u00eam 4 km. Bi\u1ebft r\u1eb1ng m\u1ed1i li\u00ean h\u1ec7 gi\u1eefa kho\u1ea3ng c\u00e1ch t\u1eeb l\u00e0ng A \u0111\u1ebfn ng\u01b0\u1eddi \u0111i b\u1ed9 y km v\u00e0 th\u1eddi gian \u0111i b\u1ed9 c\u1ee7a ng\u01b0\u1eddi \u0111\u00f3 l\u00e0 x (gi\u1edd) l\u00e0 m\u1ed9t h\u00e0m s\u1ed1 b\u1eadc nh\u1ea5t c\u00f3 d\u1ea1ng y ax b. a) X\u00e1c \u0111\u1ecbnh c\u00e1c h\u1ec7 s\u1ed1 a v\u00e0 b. b) N\u1ebfu ng\u01b0\u1eddi \u0111i b\u1ed9 c\u00e1ch l\u00e0ng A 7 km th\u00ec ng\u01b0\u1eddi \u1ea5y ph\u1ea3i \u0111i b\u1ed9 trong bao nhi\u00eau ph\u00fat? C\u00e2u 5. (0,75 \u0111i\u1ec3m). M\u1ed9t ng\u01b0\u1eddi mang cam \u0111i \u0111\u1ed5i l\u1ea5y t\u00e1o v\u00e0 l\u00ea. C\u1ee9 9 qu\u1ea3 cam th\u00ec \u0111\u1ed5i \u0111\u01b0\u1ee3c 2 qu\u1ea3 t\u00e1o v\u00e0 1 qu\u1ea3 l\u00ea, 5 qu\u1ea3 t\u00e1o th\u00ec \u0111\u1ed5i \u0111\u01b0\u1ee3c 2 qu\u1ea3 l\u00ea. N\u1ebfu ng\u01b0\u1eddi \u0111\u00f3 \u0111\u1ed5i h\u1ebft s\u1ed1 cam mang \u0111i th\u00ec \u0111\u01b0\u1ee3c 17 qu\u1ea3 t\u00e1o v\u00e0 13 qu\u1ea3 l\u00ea. H\u1ecfi ng\u01b0\u1eddi \u0111\u00f3 mang \u0111i bao nhi\u00eau qu\u1ea3 cam? C\u00e2u 6. (1,0 \u0111i\u1ec3m). M\u1ed9t xe ch\u1edf x\u0103ng d\u1ea7u, b\u00ean tr\u00ean c\u00f3 m\u1ed9t b\u1ed3n ch\u1ee9a h\u00ecnh tr\u1ee5 d\u00e0i 2, 6 m v\u00e0 \u0111\u01b0\u1eddng k\u00ednh \u0111\u00e1y l\u00e0 1, 4 m. Theo ti\u00eau chu\u1ea9n an to\u00e0n th\u00ec b\u1ed3n ch\u1ec9 ch\u1ee9a \u0111\u01b0\u1ee3c t\u1ed1i \u0111a 80 % th\u1ec3 t\u00edch khi xe di chuy\u1ec3n tr\u00ean \u0111\u01b0\u1eddng. 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 3,14 ). a) M\u1ed7i chuy\u1ebfn xe c\u00f3 th\u1ec3 ch\u1edf nhi\u1ec1u nh\u1ea5t bao nhi\u00eau l\u00edt nhi\u00ean li\u1ec7u? (cho b) Tr\u00ean \u0111\u01b0\u1eddng v\u1eadn chuy\u1ec3n, xe ch\u1edf x\u0103ng d\u1ea7u tr\u00ean ph\u1ea3i \u0111i qua 1 c\u00e2y c\u1ea7u c\u00f3 t\u1ea3i tr\u1ecdng 5 t\u1ea5n. Bi\u1ebft xe khi ch\u01b0a ch\u1edf h\u00e0ng n\u1eb7ng 3 t\u1ea5n. H\u1ecfi n\u1ebfu mu\u1ed1n \u0111i qua c\u00e2y c\u1ea7u \u0111\u00f3 th\u00ec xe ch\u1edf t\u1ed1i \u0111a bao nhi\u00eau l\u00edt x\u0103ng? Bi\u1ebft kh\u1ed1i l\u01b0\u1ee3ng ri\u00eang c\u1ee7a x\u0103ng l\u00e0 0, 713 kg\/l\u00edt (C\u00e1c k\u1ebft qu\u1ea3 l\u00e0m tr\u00f2n \u0111\u1ebfn h\u00e0ng \u0111\u01a1n v\u1ecb). C\u00e2u 7. (1 \u0111i\u1ec3m). \u0110i\u1ec3m trung b\u00ecnh c\u1ee7a 100 h\u1ecdc sinh trong hai l\u1edbp 9A v\u00e0 9B l\u00e0 7, 608. T\u00ednh \u0111i\u1ec3m trung b\u00ecnh c\u1ee7a c\u00e1c h\u1ecdc sinh m\u1ed7i l\u1edbp, bi\u1ebft r\u1eb1ng s\u1ed1 h\u1ecdc sinh c\u1ee7a l\u1edbp 9A h\u01a1n l\u1edbp 9B 2 h\u1ecdc sinh v\u00e0 \u0111i\u1ec3m trung b\u00ecnh c\u1ee7a h\u1ecdc sinh l\u1edbp 9B b\u1eb1ng 9 \u0111i\u1ec3m trung b\u00ecnh c\u1ee7a h\u1ecdc sinh l\u1edbp 9A. 10 C\u00e2u 8. (3 \u0111i\u1ec3m) T\u1eeb \u0111i\u1ec3m A n\u1eb1m ngo\u00e0i \u0111\u01b0\u1eddng tr\u00f2n (O; R) k\u1ebb hai ti\u1ebfp tuy\u1ebfn AB v\u00e0 AC . V\u1ebd c\u00e1t tuy\u1ebfn AMN v\u1edbi O ( M n\u1eb1m gi\u1eefa A v\u00e0 N ; \u0111i\u1ec3m O n\u1eb1m ngo\u00e0i BAN ). G\u1ecdi H l\u00e0 giao \u0111i\u1ec3m c\u1ee7a OA v\u00e0 CB. a) Ch\u1ee9ng minh t\u1ee9 gi\u00e1c OBAC n\u1ed9i ti\u1ebfp. b) Ch\u1ee9ng minh AB2 AM.AN OA2 R2. c) Ti\u1ebfp tuy\u1ebfn t\u1ea1i M v\u00e0 N c\u1ee7a O c\u1eaft nhau t\u1ea1i S. Ch\u1ee9ng minh t\u1ee9 gi\u00e1c OHMN n\u1ed9i ti\u1ebfp v\u00e0 ba \u0111i\u1ec3m S, B, C th\u1eb3ng h\u00e0ng. ----H\u1ebeT--- T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N 2","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH H\u01af\u1edaNG D\u1eaaN GI\u1ea2I C\u00e2u 1. (1,5 \u0111i\u1ec3m) Cho (P) : y = x2 v\u00e0 \u0111\u01b0\u1eddng th\u1eb3ng (D) : y = x + 1 . 22 a) V\u1ebd \u0111\u1ed3 th\u1ecb (P) v\u00e0 (D) tr\u00ean c\u00f9ng h\u1ec7 tr\u1ee5c t\u1ecda \u0111\u1ed9. b) T\u00ecm t\u1ecda \u0111\u1ed9 giao \u0111i\u1ec3m c\u1ee7a (P) v\u00e0 (D) b\u1eb1ng ph\u00e9p t\u00ednh. L\u1eddi gi\u1ea3i a) V\u1ebd \u0111\u1ed3 th\u1ecb (P) v\u00e0 (D) tr\u00ean c\u00f9ng h\u1ec7 tr\u1ee5c t\u1ecda \u0111\u1ed9. BGT: x \u22124 \u22122 0 2 4 y = x2 82028 2 x 24 y = x +1 2 3 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) : x2 = x + 1 22 \uf0db x2 \u2212 x \u22121 = 0 22 \uf0db \uf0e9x = 2 \uf0ea\uf0ebx = \u22121 Thay x = 2 v\u00e0o y = x2 , ta \u0111\u01b0\u1ee3c: y = 22 = 2. 22 Thay x = \u22121 v\u00e0o y = x2 , ta \u0111\u01b0\u1ee3c: y = (\u22121)2 = 1 . 2 22 V\u1eady (2;2) , \uf0e6 \u22121; 1 \uf0f6 l\u00e0 hai giao \u0111i\u1ec3m c\u1ea7n t\u00ecm. \uf0e7 2 \uf0f7 \uf0e8 \uf0f8 C\u00e2u 2. (1 \u0111i\u1ec3m) Cho ph\u01b0\u01a1ng tr\u00ecnh 3x2 + 2x \u2212 4 = 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= 3x12 + 5x1x2 + 3x22 . 4x13x2 + 4x1x23 T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N 3","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH L\u1eddi gi\u1ea3i V\u00ec a.c = 3.(\u22124) = \u221212 \uf03c 0 N\u00ean ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m ph\u00e2n bi\u1ec7t tr\u00e1i d\u1ea5u x1 ,x2 . \uf0ec\uf0ef\uf0efS = x1 + x2 = \u2212b = \u2212 2 \uf0ed = x1 a 3 Theo \u0111\u1ecbnh l\u00ed Vi-et, ta c\u00f3: \uf0ef\uf0ef\uf0eeP =\u22124 c .x2 = a 3 Ta c\u00f3: A= 3x12 + 5x1x2 + 3x22 4x13x2 + 4x1x23 ( )3 A= x12 + x22 + 5x1x2 ( )4x1x2 x12 + x22 ( )3 S2 \u2212 2P + 5P ( )A = 4P S2 \u2212 2P A = 3S2 \u2212 P 4S2P \u2212 8P2 3.\uf0e6\uf0e7\uf0e8 \u22122 \uf0f62 \u2212 \uf0e6 \u2212 4 \uf0f6 \u22129 . 3 \uf0f7\uf0f8 \uf0e7\uf0e8 3 \uf0f7\uf0f8 56 A= = \uf0f62 \uf0f62 4 \uf0e6 \u22122 \uf0f7\uf0f8 \uf0e6 \u2212 4 \uf0f6 \u2212 8 \uf0e6 \u2212 4 \uf0f7\uf0f8 \uf0e7\uf0e8 3 \uf0e7\uf0e8 3 \uf0f7\uf0f8 \uf0e7\uf0e8 3 C\u00e2u 3. (1,0 \u0111i\u1ec3m). Gi\u00e1 b\u00e1n m\u1ed9t c\u00e1i b\u00e1nh \u1edf hai c\u1eeda h\u00e0ng A v\u00e0 B \u0111\u1ec1u l\u00e0 15000 \u0111\u1ed3ng, nh\u01b0ng m\u1ed7i c\u1eeda h\u00e0ng c\u00f3 h\u00ecnh th\u1ee9c khuy\u1ebfn m\u00e3i kh\u00e1c nhau. - C\u1eeda h\u00e0ng A : n\u1ebfu kh\u00e1ch h\u00e0ng mua b\u1ed1n c\u00e1i b\u00e1nh tr\u1edf l\u00ean th\u00ec ba c\u00e1i b\u00e1nh \u0111\u1ea7u ti\u00ean gi\u00e1 m\u1ed7i c\u00e1i b\u00e1nh v\u1eabn l\u00e0 15000 \u0111\u1ed3ng nh\u01b0ng t\u1eeb c\u00e1i b\u00e1nh th\u1ee9 t\u01b0 tr\u1edf \u0111i kh\u00e1ch h\u00e0ng ch\u1ec9 ph\u1ea3i tr\u1ea3 80 % gi\u00e1 \u0111ang b\u00e1n. - C\u1eeda h\u00e0ng B : n\u1ebfu kh\u00e1ch h\u00e0ng mua 3 c\u00e1i b\u00e1nh th\u00ec \u0111\u01b0\u1ee3c t\u1eb7ng m\u1ed9t c\u00e1i b\u00e1nh mi\u1ec5n ph\u00ed. M\u1ed9t nh\u00f3m b\u1ea1n h\u1ecdc sinh mua 15 c\u00e1i b\u00e1nh th\u00ec ch\u1ecdn c\u1eeda h\u00e0ng n\u00e0o c\u00f3 l\u1ee3i h\u01a1n v\u00e0 l\u1ee3i h\u01a1n bao nhi\u00eau? L\u1eddi gi\u1ea3i S\u1ed1 ti\u1ec1n ph\u1ea3i tr\u1ea3 khi mua \u1edf c\u1eeda h\u00e0ng A 15000.3 15000.80%.12 189000 (\u0111\u1ed3ng) Ta c\u00f3 15 : 4 3 d\u01b0 3 \u0111\u01b0\u1ee3c t\u1eb7ng 3 c\u00e1i b\u00e1nh S\u1ed1 ti\u1ec1n ph\u1ea3i tr\u1ea3 khi mua \u1edf c\u1eeda h\u00e0ng B 15000.12 180000 (\u0111\u1ed3ng) V\u1eady mua \u1edf c\u1eeda h\u00e0ng B l\u1ee3i h\u01a1n v\u00e0 l\u1ee3i h\u01a1n 9000 \u0111\u1ed3ng. T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N 4","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH C\u00e2u 4. (0,75 \u0111i\u1ec3m). Ba l\u00e0ng A, B, C n\u1eb1m tr\u00ean c\u00f9ng m\u1ed9t qu\u1ed1c l\u1ed9, l\u00e0ng B n\u1eb1m gi\u1eefa l\u00e0ng A v\u00e0 l\u00e0ng C, l\u00e0ng A c\u00e1ch l\u00e0ng B 5 km. M\u1ed9t ng\u01b0\u1eddi \u0111i b\u1ed9 theo h\u01b0\u1edbng t\u1eeb l\u00e0ng B \u0111\u1ebfn l\u00e0ng C m\u1ed7i gi\u1edd c\u00e1ch l\u00e0ng A th\u00eam 4 km. Bi\u1ebft r\u1eb1ng m\u1ed1i li\u00ean h\u1ec7 gi\u1eefa kho\u1ea3ng c\u00e1ch t\u1eeb l\u00e0ng A \u0111\u1ebfn ng\u01b0\u1eddi \u0111i b\u1ed9 y km v\u00e0 th\u1eddi gian \u0111i b\u1ed9 c\u1ee7a ng\u01b0\u1eddi \u0111\u00f3 l\u00e0 x (gi\u1edd) l\u00e0 m\u1ed9t h\u00e0m s\u1ed1 b\u1eadc nh\u1ea5t c\u00f3 d\u1ea1ng y ax b. a. X\u00e1c \u0111\u1ecbnh c\u00e1c h\u1ec7 s\u1ed1 a v\u00e0 b. b. N\u1ebfu ng\u01b0\u1eddi \u0111i b\u1ed9 c\u00e1ch l\u00e0ng A 7 km th\u00ec ng\u01b0\u1eddi \u1ea5y ph\u1ea3i \u0111i b\u1ed9 trong bao nhi\u00eau ph\u00fat? L\u1eddi gi\u1ea3i a. X\u00e1c \u0111\u1ecbnh c\u00e1c h\u1ec7 s\u1ed1 a v\u00e0 b. Kho\u1ea3ng c\u00e1ch t\u1eeb l\u00e0ng A \u0111\u1ebfn ng\u01b0\u1eddi \u0111i b\u1ed9 = V\u1eadn t\u1ed1c x th\u1eddi gian + kho\u1ea3ng c\u00e1ch t\u1eeb l\u00e0ng A c\u00e1ch l\u00e0ng B y 4x 5 V\u1eady a = 4, b = 5. b. N\u1ebfu ng\u01b0\u1eddi \u0111i b\u1ed9 c\u00e1ch l\u00e0ng A 7 km th\u00ec ng\u01b0\u1eddi \u1ea5y ph\u1ea3i \u0111i b\u1ed9 trong bao nhi\u00eau ph\u00fat? Th\u1ebf y 7 v\u00e0o y 4x 5 7 4x 5 x 0, 5 V\u1eady ng\u01b0\u1eddi \u0111\u00f3 ph\u1ea3i \u0111i b\u1ed9 trong 30 ph\u00fat. C\u00e2u 5. (0,75 \u0111i\u1ec3m). M\u1ed9t ng\u01b0\u1eddi mang cam \u0111i \u0111\u1ed5i l\u1ea5y t\u00e1o v\u00e0 l\u00ea. C\u1ee9 9 qu\u1ea3 cam th\u00ec \u0111\u1ed5i \u0111\u01b0\u1ee3c 2 qu\u1ea3 t\u00e1o v\u00e0 1 qu\u1ea3 l\u00ea, 5 qu\u1ea3 t\u00e1o th\u00ec \u0111\u1ed5i \u0111\u01b0\u1ee3c 2 qu\u1ea3 l\u00ea. N\u1ebfu ng\u01b0\u1eddi \u0111\u00f3 \u0111\u1ed5i h\u1ebft s\u1ed1 cam mang \u0111i th\u00ec \u0111\u01b0\u1ee3c 17 qu\u1ea3 t\u00e1o v\u00e0 13 qu\u1ea3 l\u00ea. H\u1ecfi ng\u01b0\u1eddi \u0111\u00f3 mang \u0111i bao nhi\u00eau qu\u1ea3 cam? L\u1eddi gi\u1ea3i C\u00e1ch 1: C\u1ee9 9 qu\u1ea3 cam th\u00ec \u0111\u1ed5i \u0111\u01b0\u1ee3c 2 qu\u1ea3 t\u00e1o v\u00e0 1 qu\u1ea3 l\u00ea 18 qu\u1ea3 cam th\u00ec \u0111\u1ed5i \u0111\u01b0\u1ee3c 4 qu\u1ea3 t\u00e1o v\u00e0 2 qu\u1ea3 l\u00ea M\u00e0 2 qu\u1ea3 l\u00ea th\u00ec \u0111\u1ed5i \u0111\u01b0\u1ee3c 5 qu\u1ea3 t\u00e1o N\u00ean 18 qu\u1ea3 cam th\u00ec \u0111\u1ed5i \u0111\u01b0\u1ee3c 9 qu\u1ea3 t\u00e1o 2 qu\u1ea3 cam th\u00ec \u0111\u1ed5i \u0111\u01b0\u1ee3c 1 qu\u1ea3 t\u00e1o M\u00e0 5 qu\u1ea3 t\u00e1o th\u00ec \u0111\u1ed5i \u0111\u01b0\u1ee3c 2 qu\u1ea3 l\u00ea 10 qu\u1ea3 cam th\u00ec \u0111\u1ed5i \u0111\u01b0\u1ee3c 2 qu\u1ea3 l\u00ea 5 qu\u1ea3 cam th\u00ec \u0111\u1ed5i \u0111\u01b0\u1ee3c 1 qu\u1ea3 l\u00ea V\u1eady s\u1ed1 qu\u1ea3 cam ng\u01b0\u1eddi \u0111\u00f3 mang \u0111i 17.2 13.5 99 (qu\u1ea3) C\u00e1ch 2: G\u1ecdi s\u1ed1 qu\u1ea3 cam \u0111\u1ed5i \u0111\u01b0\u1ee3c 1 qu\u1ea3 t\u00e1o l\u00e0 x (qu\u1ea3) 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 G\u1ecdi s\u1ed1 qu\u1ea3 cam \u0111\u1ed5i \u0111\u01b0\u1ee3c 1 qu\u1ea3 l\u00ea l\u00e0 y (qu\u1ea3) (\u0110K: x,y * ) C\u1ee9 9 qu\u1ea3 cam th\u00ec \u0111\u1ed5i \u0111\u01b0\u1ee3c 2 qu\u1ea3 t\u00e1o v\u00e0 1 qu\u1ea3 l\u00ea, n\u00ean ta c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh: 2x 1y 9 1 5 qu\u1ea3 t\u00e1o th\u00ec \u0111\u1ed5i \u0111\u01b0\u1ee3c 2 qu\u1ea3 l\u00ea, n\u00ean ta c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh: 5x 2y 5x 2y 0 1 2x y 9 x2 T\u1eeb 1 v\u00e0 2 ta c\u00f3 h\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh: 5x 2y 0 y 5 (nh\u1eadn). V\u1eady s\u1ed1 qu\u1ea3 cam ng\u01b0\u1eddi \u0111\u00f3 mang \u0111i 17.2 13.5 99 (qu\u1ea3). C\u00e2u 6. (1 \u0111i\u1ec3m) M\u1ed9t xe ch\u1edf x\u0103ng d\u1ea7u, b\u00ean tr\u00ean c\u00f3 m\u1ed9t b\u1ed3n ch\u1ee9a h\u00ecnh tr\u1ee5 d\u00e0i 2, 6 m v\u00e0 \u0111\u01b0\u1eddng k\u00ednh \u0111\u00e1y l\u00e0 1, 4 m. Theo ti\u00eau chu\u1ea9n an to\u00e0n th\u00ec b\u1ed3n ch\u1ec9 ch\u1ee9a \u0111\u01b0\u1ee3c t\u1ed1i \u0111a 80 % th\u1ec3 t\u00edch khi xe di chuy\u1ec3n tr\u00ean \u0111\u01b0\u1eddng. 3,14 ). a. M\u1ed7i chuy\u1ebfn xe c\u00f3 th\u1ec3 ch\u1edf nhi\u1ec1u nh\u1ea5t bao nhi\u00eau l\u00edt nhi\u00ean li\u1ec7u? (cho b. Tr\u00ean \u0111\u01b0\u1eddng v\u1eadn chuy\u1ec3n, xe ch\u1edf x\u0103ng d\u1ea7u tr\u00ean ph\u1ea3i \u0111i qua 1 c\u00e2y c\u1ea7u c\u00f3 t\u1ea3i tr\u1ecdng 5 t\u1ea5n. Bi\u1ebft xe khi ch\u01b0a ch\u1edf h\u00e0ng n\u1eb7ng 3 t\u1ea5n. H\u1ecfi n\u1ebfu mu\u1ed1n \u0111i qua c\u00e2y c\u1ea7u \u0111\u00f3 th\u00ec xe ch\u1edf t\u1ed1i \u0111a bao nhi\u00eau l\u00edt x\u0103ng? Bi\u1ebft kh\u1ed1i l\u01b0\u1ee3ng ri\u00eang c\u1ee7a x\u0103ng l\u00e0 0, 713 kg\/l\u00edt (C\u00e1c k\u1ebft qu\u1ea3 l\u00e0m tr\u00f2n \u0111\u1ebfn h\u00e0ng \u0111\u01a1n v\u1ecb). L\u1eddi gi\u1ea3i 3,14 ). a) M\u1ed7i chuy\u1ebfn xe c\u00f3 th\u1ec3 ch\u1edf nhi\u1ec1u nh\u1ea5t bao nhi\u00eau l\u00edt nhi\u00ean li\u1ec7u? (cho B\u00e1n k\u00ednh \u0111\u00e1y: 1, 4 : 2 0, 7 m S\u1ed1 l\u00edt nhi\u00ean li\u1ec7u xe c\u00f3 th\u1ec3 ch\u1edf nhi\u1ec1u nh\u1ea5t: 3,14.2, 6.0, 72.80% 3,2 m3 3200 (l\u00edt) b) Tr\u00ean \u0111\u01b0\u1eddng v\u1eadn chuy\u1ec3n, xe ch\u1edf x\u0103ng d\u1ea7u tr\u00ean ph\u1ea3i \u0111i qua 1 c\u00e2y c\u1ea7u c\u00f3 t\u1ea3i tr\u1ecdng 5 t\u1ea5n. Bi\u1ebft xe khi ch\u01b0a ch\u1edf h\u00e0ng n\u1eb7ng 3 t\u1ea5n. H\u1ecfi n\u1ebfu mu\u1ed1n \u0111i qua c\u00e2y c\u1ea7u \u0111\u00f3 th\u00ec xe ch\u1edf t\u1ed1i \u0111a bao nhi\u00eau l\u00edt x\u0103ng? Bi\u1ebft kh\u1ed1i l\u01b0\u1ee3ng ri\u00eang c\u1ee7a x\u0103ng l\u00e0 0, 713 kg\/l\u00edt (C\u00e1c k\u1ebft qu\u1ea3 l\u00e0m tr\u00f2n \u0111\u1ebfn h\u00e0ng \u0111\u01a1n v\u1ecb). Kh\u1ed1i l\u01b0\u1ee3ng x\u0103ng xe c\u00f3 th\u1ec3 ch\u1edf t\u1ed1i \u0111a \u0111\u1ec3 qua c\u1ea7u 5 3 2 (t\u1ea5n) 2000 kg S\u1ed1 l\u00edt x\u0103ng xe ch\u1edf t\u1ed1i \u0111a \u0111\u1ec3 qua c\u1ea7u 2000 : 0, 713 2805 (l\u00edt). T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N 6","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH C\u00e2u 7. (1 \u0111i\u1ec3m). \u0110i\u1ec3m trung b\u00ecnh c\u1ee7a 100 h\u1ecdc sinh trong hai l\u1edbp 9A v\u00e0 9B l\u00e0 7, 608. T\u00ednh \u0111i\u1ec3m trung b\u00ecnh c\u1ee7a c\u00e1c h\u1ecdc sinh m\u1ed7i l\u1edbp, bi\u1ebft r\u1eb1ng s\u1ed1 h\u1ecdc sinh c\u1ee7a l\u1edbp 9A h\u01a1n l\u1edbp 9B 2 h\u1ecdc sinh v\u00e0 \u0111i\u1ec3m trung b\u00ecnh c\u1ee7a h\u1ecdc sinh l\u1edbp 9B b\u1eb1ng 9 \u0111i\u1ec3m trung b\u00ecnh c\u1ee7a h\u1ecdc sinh l\u1edbp 9A. 10 L\u1eddi gi\u1ea3i S\u1ed1 h\u1ecdc sinh l\u1edbp 9A : 100 2 : 2 51 (h\u1ecdc sinh) S\u1ed1 h\u1ecdc sinh l\u1edbp 9B : 100 2 : 2 49 (h\u1ecdc sinh) G\u1ecdi \u0111i\u1ec3m trung b\u00ecnh c\u1ee7a l\u1edbp 9A v\u00e0 9B l\u1ea7n l\u01b0\u1ee3t l\u00e0 x,y (\u0110K x,y 0 ) \u0110i\u1ec3m trung b\u00ecnh c\u1ee7a h\u1ecdc sinh l\u1edbp 9B b\u1eb1ng 9 \u0111i\u1ec3m trung b\u00ecnh c\u1ee7a h\u1ecdc sinh l\u1edbp 9A. , n\u00ean ta 10 c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh: y 9 x 9 x y 01 10 10 \u0110i\u1ec3m trung b\u00ecnh c\u1ee7a 100 h\u1ecdc sinh trong hai l\u1edbp 9A v\u00e0 9B l\u00e0 7, 608. , n\u00ean ta c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh: 51x 49y 100.7, 608 760, 8 2 T\u1eeb 1 v\u00e0 2 ta c\u00f3 h\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh: 9 x y 0 x8 10 y 7,2 (nh\u1eadn). 51x 49y 760, 8 V\u1eady \u0111i\u1ec3m trung b\u00ecnh c\u1ee7a l\u1edbp 9A v\u00e0 9B l\u1ea7n l\u01b0\u1ee3t l\u00e0 8 v\u00e0 7,2. C\u00e2u 8. (3 \u0111i\u1ec3m) T\u1eeb \u0111i\u1ec3m A n\u1eb1m ngo\u00e0i \u0111\u01b0\u1eddng tr\u00f2n (O; R) k\u1ebb hai ti\u1ebfp tuy\u1ebfn AB v\u00e0 AC . V\u1ebd c\u00e1t tuy\u1ebfn AMN v\u1edbi O ( M n\u1eb1m gi\u1eefa A v\u00e0 N ; \u0111i\u1ec3m O n\u1eb1m ngo\u00e0i BAN ). G\u1ecdi H l\u00e0 giao \u0111i\u1ec3m c\u1ee7a OA v\u00e0 CB. a) Ch\u1ee9ng minh t\u1ee9 gi\u00e1c OBAC n\u1ed9i ti\u1ebfp. b) Ch\u1ee9ng minh AB2 AM.AN OA2 R2. c) Ti\u1ebfp tuy\u1ebfn t\u1ea1i M v\u00e0 N c\u1ee7a O c\u1eaft nhau t\u1ea1i S. Ch\u1ee9ng minh t\u1ee9 gi\u00e1c OHMN n\u1ed9i ti\u1ebfp v\u00e0 ba \u0111i\u1ec3m S, B, C 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 a) Ch\u1ee9ng minh t\u1ee9 gi\u00e1c OBAC n\u1ed9i ti\u1ebfp. X\u00e9t t\u1ee9 gi\u00e1c OBAC ta c\u00f3 OBA OCA 90 g.t OBA OCA 180 T\u1ee9 gi\u00e1c OBAC n\u1ed9i ti\u1ebfp (t\u1ed5ng hai g\u00f3c \u0111\u1ed1i b\u1eb1ng 180 ) b) Ch\u1ee9ng minh AB2 AM.AN OA2 R2. X\u00e9t ABM v\u00e0 ANB ta c\u00f3 BAN chung v\u00e0 ABM ANB (c\u00f9ng ch\u1eafn BM ) ABM \u223d ANB g.g AB AM AB2 AM.AN . AN AB X\u00e9t ABO vu\u00f4ng t\u1ea1i B , ta c\u00f3: OA2 OB2 AB2 AB2 OA2 R2 . c) Ti\u1ebfp tuy\u1ebfn t\u1ea1i M v\u00e0 N c\u1ee7a O c\u1eaft nhau t\u1ea1i S. Ch\u1ee9ng minh t\u1ee9 gi\u00e1c OHMN n\u1ed9i ti\u1ebfp v\u00e0 ba \u0111i\u1ec3m S, B, C th\u1eb3ng h\u00e0ng. Ta c\u00f3 OB OC R . V\u00e0 AB AC (t\u00ednh ch\u1ea5t hai ti\u1ebfp tuy\u1ebfn c\u1eaft nhau) OA l\u00e0 \u0111\u01b0\u1eddng trung tr\u1ef1c c\u1ee7a BC . OA BC t\u1ea1i H . X\u00e9t ABO vu\u00f4ng t\u1ea1i B c\u00f3 \u0111\u01b0\u1eddng cao BH , ta c\u00f3: AB2 AH.AO M\u00e0 AB2 AM.AN c.m.t AM.AN AH .AO AM AO AH AN . X\u00e9t AHM v\u00e0 ANO ta c\u00f3 OAN chung v\u00e0 AM AO c.m.t AH AN AHM \u223d ANO c.g.c AHM ANO . 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 1 T\u1ee9 gi\u00e1c OHMN n\u1ed9i ti\u1ebfp (g\u00f3c ngo\u00e0i b\u1eb1ng g\u00f3c \u0111\u1ed1i trong) Ch\u1ee9ng minh t\u01b0\u01a1ng t\u1ef1 c\u00e2u a 2 T\u1ee9 gi\u00e1c OMSN n\u1ed9i ti\u1ebfp \u0111\u01b0\u1eddng tr\u00f2n \u0111\u01b0\u1eddng k\u00ednh OS T\u1eeb 1 v\u00e0 2 5 \u0111i\u1ec3m O,H,M,S,N c\u00f9ng thu\u1ed9c \u0111\u01b0\u1eddng tr\u00f2n \u0111\u01b0\u1eddng k\u00ednh OS SHO 900 (g\u00f3c n\u1ed9i ti\u1ebfp ch\u1eafn n\u1eeda \u0111\u01b0\u1eddng tr\u00f2n) SH OA t\u1ea1i H M\u00e0 BC OA t\u1ea1i H (c.m.t) ba \u0111i\u1ec3m S, B, C 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 12 NA\u00caM HO\u00cfC: 2023 - 2024 \u0110\u1ec0 THAM KH\u1ea2O M\u00d4N: TO\u00c1N 9 M\u00c3 \u0110\u1ec0: Qu\u1eadn 12 - 2 \u0110\u00ea thi g\u1ed3m 8 c\u00e2u ho\u0309i t\u01b0\u0323 lu\u00e2\u0323n. Th\u01a1\u0300i gian: 120 phu\u0301t (kh\u00f4ng k\u00ea\u0309 th\u01a1\u0300i gian pha\u0301t \u0111\u00ea\u0300) C\u00e2u 1. (1,5 \u0111i\u1ec3m). Cho (P) : y = x2 v\u00e0 \u0111\u01b0\u1eddng th\u1eb3ng (d) : y = \u22124 + 3x . 2 a) V\u1ebd \u0111\u1ed3 th\u1ecb (P) v\u00e0 (d) tr\u00ean c\u00f9ng m\u1eb7t ph\u1eb3ng t\u1ecda \u0111\u1ed9. b) T\u00ecm t\u1ecda \u0111\u1ed9 giao \u0111i\u1ec3m c\u1ee7a (P) v\u00e0 (d) b\u1eb1ng ph\u00e9p t\u00ednh. C\u00e2u 2. (1 \u0111i\u1ec3m). Cho ph\u01b0\u01a1ng tr\u00ecnh 3x2 + 5x \u2212 6 = 0 c\u00f3 2 nghi\u1ec7m l\u00e0 x1 , x2 . Kh\u00f4ng gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh, h\u00e3y t\u00ednh gi\u00e1 tr\u1ecb c\u1ee7a bi\u1ec3u th\u1ee9c A = x12 + x22 . x2 \u2212 1 x1 \u2212 1 C\u00e2u 3. (1 \u0111i\u1ec3m). Hi\u1ec7n nay, c\u00e1c em h\u1ecdc sinh trung h\u1ecdc ng\u00e0y c\u00e0ng quan t\u00e2m h\u01a1n v\u1ec1, c\u00e2n n\u1eb7ng v\u00e0 chi\u1ec1u cao c\u1ee7a b\u1ea3n th\u00e2n. C\u00e1c b\u1ea1n nam th\u01b0\u1eddng ch\u00fa tr\u1ecdng \u0111\u1ebfn chi\u1ec1u cao c\u00f2n c\u00e1c b\u1ea1n n\u1eef l\u1ea1i r\u1ea5t hay \u0111\u1ec3 \u00fd \u0111\u1ebfn c\u00e2n n\u1eb7ng. C\u00e2n n\u1eb7ng l\u00fd t\u01b0\u1edfng \u1ee9ng v\u1edbi chi\u1ec1u cao c\u1ee7a m\u1ed9t ng\u01b0\u1eddi \u0111\u01b0\u1ee3c t\u00ednh theo c\u00f4ng th\u1ee9c sau : w = h \u2212 100 \u2212 h \u2212 150 (1) c Trong \u0111\u00f3 : w l\u00e0 c\u00e2n n\u1eb7ng (t\u00ednh theo kg ), h l\u00e0 chi\u1ec1u cao (t\u00ednh theo cm ) v\u00e0 c = 4 n\u1ebfu ng\u01b0\u1eddi \u0111\u00f3 l\u00e0 nam v\u00e0 c = 2 n\u1ebfu ng\u01b0\u1eddi \u0111\u00f3 l\u00e0 n\u1eef. C\u00f4ng th\u1ee9c n\u00e0y cho ph\u00e9p t\u00ednh \u0111\u01b0\u1ee3c c\u00e2n n\u1eb7ng l\u00fd t\u01b0\u1edfng c\u1ee7a m\u1ed9t ng\u01b0\u1eddi \u0111\u1ec3 c\u00f3 m\u1ed9t c\u01a1 th\u1ec3 c\u00e2n \u0111\u1ed1i, \u01b0a nh\u00ecn khi bi\u1ebft chi\u1ec1u cao c\u1ee7a ng\u01b0\u1eddi \u0111\u00f3. a) M\u1ed9t b\u1ea1n n\u1eef c\u00f3 chi\u1ec1u cao l\u00e0 1,52m th\u00ec c\u00e2n n\u1eb7ng l\u00fd t\u01b0\u1edfng c\u1ee7a b\u1ea1n \u1ea5y l\u00e0 bao nhi\u00eau kg ? b) N\u1ebfu m\u1ed9t b\u1ea1n nam c\u00f3 c\u00e2n n\u1eb7ng l\u00e0 56 kg th\u00ec chi\u1ec1u cao c\u1ee7a b\u1ea1n nam n\u00e0y c\u1ea7n ph\u1ea3i l\u00e0 bao nhi\u00eau \u0111\u1ec3 56 kg l\u00e0 c\u00e2n n\u1eb7ng l\u00fd t\u01b0\u1edfng cho b\u1ea1n \u1ea5y ? C\u00e2u 4. (0,75 \u0111i\u1ec3m). L\u1edbp 9A c\u00f3 30 h\u1ecdc sinh, m\u1ed7i b\u1ea1n d\u1ef1 \u0111\u1ecbnh \u0111\u00f3ng m\u1ed7i th\u00e1ng 70000 \u0111\u1ed3ng \u0111\u1ec3 mua qu\u00e0 t\u1eb7ng c\u00e1c tr\u1ebb em \u1edf m\u00e1i \u1ea5m t\u00ecnh th\u01b0\u01a1ng v\u00e0 sau 3 th\u00e1ng s\u1ebd \u0111\u1ee7 ti\u1ec1n \u0111\u1ec3 t\u1eb7ng m\u1ed7i em \u1edf m\u00e1i \u1ea5m 3 g\u00f3i qu\u00e0 (gi\u00e1 tr\u1ecb m\u1ed7i g\u00f3i qu\u00e0 l\u00e0 nh\u01b0 nhau). Khi c\u00e1c h\u1ecdc sinh \u0111\u00e3 \u0111\u00f3ng \u0111\u1ee7 ti\u1ec1n th\u00ec m\u00e1i \u1ea5m nh\u1eadn ch\u0103m s\u00f3c th\u00eam 9 em v\u00e0 gi\u00e1 ti\u1ec1n m\u1ed7i g\u00f3i qu\u00e0 t\u0103ng th\u00eam 5% n\u00ean ch\u1ec9 t\u1eb7ng \u0111\u01b0\u1ee3c m\u1ed7i em 2 g\u00f3i qu\u00e0. H\u1ecfi l\u00fac \u0111\u1ea7u c\u00f3 bao nhi\u00eau em \u1edf m\u00e1i \u1ea5m \u0111\u01b0\u1ee3c t\u1eb7ng qu\u00e0? C\u00e2u 5. (1 \u0111i\u1ec3m). \u0110\u1ec3 l\u1eadp \u0111\u1ed9i tuy\u1ec3n n\u0103ng khi\u1ebfu v\u1ec1 b\u00f3ng r\u1ed5 c\u1ee7a tr\u01b0\u1eddng th\u1ea7y th\u1ec3 d\u1ee5c \u0111\u01b0a ra quy \u0111\u1ecbnh tuy\u1ec3n ch\u1ecdn nh\u01b0 sau: m\u1ed7i b\u1ea1n d\u1ef1 tuy\u1ec3n s\u1ebd \u0111\u01b0\u1ee3c n\u00e9m 10 qu\u1ea3 b\u00f3ng v\u00e0o r\u1ed5, qu\u1ea3 b\u00f3ng v\u00e0o r\u1ed5 \u0111\u01b0\u1ee3c c\u1ed9ng 4 \u0111i\u1ec3m; qu\u1ea3 b\u00f3ng n\u00e9m ra ngo\u00e0i th\u00ec b\u1ecb tr\u1eeb 2 \u0111i\u1ec3m. N\u1ebfu b\u1ea1n n\u00e0o c\u00f3 s\u1ed1 \u0111i\u1ec3m t\u1eeb 22 \u0111i\u1ec3m tr\u1edf l\u00ean th\u00ec s\u1ebd \u0111\u01b0\u1ee3c ch\u1ecdn v\u00e0o \u0111\u1ed9i tuy\u1ec3n. H\u1ecfi m\u1ed9t h\u1ecdc sinh mu\u1ed1n \u0111\u01b0\u1ee3c ch\u1ecdn v\u00e0o \u0111\u1ed9i tuy\u1ec3n th\u00ec ph\u1ea3i n\u00e9m \u00edt nh\u1ea5t bao nhi\u00eau qu\u1ea3 v\u00e0o r\u1ed5? C\u00e2u 6. (1 \u0111i\u1ec3m). B\u1ea3ng c\u01b0\u1edbc ph\u00ed d\u1ecbch v\u1ee5 Mobiphone \u00e1p d\u1ee5ng cho thu\u00ea bao tr\u1ea3 tr\u01b0\u1edbc, c\u01b0\u1edbc g\u1ecdi li\u00ean m\u1ea1ng trong n\u01b0\u1edbc (\u0111\u00e3 bao g\u1ed3m VAT) quy \u0111\u1ecbnh r\u1eb1ng : n\u1ebfu g\u1ecdi trong 5 gi\u00e2y \u0111\u1ea7u th\u00ec t\u00ednh c\u01b0\u1edbc 200 \u0111\u1ed3ng\/ 5 gi\u00e2y \u0111\u1ea7u, c\u00f2n k\u1ec3 t\u1eeb sau gi\u00e2y th\u1ee9 5 tr\u1edf \u0111i, h\u1ecd t\u00ednh th\u00eam 28 \u0111\u1ed3ng cho m\u1ed7i gi\u00e2y. T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N 1","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH a) G\u1ecdi m l\u00e0 s\u1ed1 ti\u1ec1n ph\u1ea3i tr\u1ea3 (t\u00ednh b\u1eb1ng \u0111\u1ed3ng) v\u00e0 n l\u00e0 th\u1eddi gian g\u1ecdi nhi\u1ec1u h\u01a1n 5 gi\u00e2y ( t \uf03e 5 ). H\u00e3y l\u1eadp c\u00f4ng th\u1ee9c bi\u1ec3u th\u1ecb m theo n ? b) H\u1ecfi b\u1ea1n Anh g\u1ecdi trong bao l\u00e2u m\u00e0 b\u1ea1n tr\u1ea3 3420 \u0111\u1ed3ng ? C\u00e2u 7. (0,75 \u0111i\u1ec3m): N\u01b0\u1edbc gi\u1ea3i kh\u00e1t th\u01b0\u1eddng \u0111\u1ef1ng trong lon nh\u00f4m v\u00e0 c\u1ee1 lon ph\u1ed5 bi\u1ebfn tr\u00ean th\u1ebf gi\u1edbi th\u01b0\u1eddng ch\u1ee9a \u0111\u01b0\u1ee3c kho\u1ea3ng 335 ml ch\u1ea5t l\u1ecfng, \u0111\u01b0\u1ee3c thi\u1ebft k\u1ebf h\u00ecnh tr\u1ee5 v\u1edbi chi\u1ec1u cao g\u1ea7n g\u1ea5p \u0111\u00f4i \u0111\u01b0\u1eddng k\u00ednh \u0111\u00e1y (cao 12cm , \u0111\u01b0\u1eddng k\u00ednh \u0111\u00e1y 6,5cm ). Nh\u01b0ng hi\u1ec7n nay c\u00e1c nh\u00e0 s\u1ea3n xu\u1ea5t c\u00f3 xu h\u01b0\u1edbng t\u1ea1o ra nh\u1eefng lon nh\u00f4m v\u1edbi ki\u1ec3u d\u00e1ng thon cao d\u00e0i. Tuy chi ph\u00ed s\u1ea3n xu\u1ea5t c\u1ee7a nh\u1eefng chi\u1ebfc lon n\u00e0y t\u1ed1n k\u00e9m h\u01a1n, do n\u00f3 c\u00f3 di\u1ec7n t\u00edch m\u1eb7t ngo\u00e0i (di\u1ec7n t\u00edch to\u00e0n ph\u1ea7n c\u1ee7a h\u00ecnh tr\u1ee5) l\u1edbn h\u01a1n, nh\u01b0ng n\u00f3 l\u1ea1i d\u1ec5 \u0111\u00e1nh l\u1eeba th\u1ecb gi\u00e1c v\u00e0 \u0111\u01b0\u1ee3c ng\u01b0\u1eddi ti\u00eau d\u00f9ng \u01b0a chu\u1ed9ng h\u01a1n. a) M\u1ed9t lon n\u01b0\u1edbc ng\u1ecdt cao 14,2 cm , \u0111\u01b0\u1eddng k\u00ednh \u0111\u00e1y 5,9cm . H\u1ecfi lon n\u01b0\u1edbc ng\u1ecdt cao n\u00e0y c\u00f3 th\u1ec3 ch\u1ee9a \u0111\u01b0\u1ee3c h\u1ebft l\u01b0\u1ee3ng n\u01b0\u1edbc ng\u1ecdt c\u1ee7a m\u1ed9t lon c\u1ee1 ph\u1ed5 bi\u1ebfn kh\u00f4ng ? V\u00ec sao ? Bi\u1ebft th\u1ec3 t\u00edch h\u00ecnh tr\u1ee5 V = \uf070 R2h v\u1edbi \uf070 \uf0bb 3,14 (k\u1ebft qu\u1ea3 l\u00e0m tr\u00f2n \u0111\u1ebfn ch\u1eef s\u1ed1 th\u1eadp ph\u00e2n th\u1ee9 hai). b) H\u1ecfi chi ph\u00ed s\u1ea3n xu\u1ea5t lon n\u01b0\u1edbc ng\u1ecdt cao \u1edf c\u00e2u a, t\u0103ng bao nhi\u00eau ph\u1ea7n tr\u0103m so v\u1edbi s\u1ea3n xu\u1ea5t lon n\u01b0\u1edbc ng\u1ecdt c\u1ee1 ph\u1ed5 bi\u1ebfn ? Bi\u1ebft Sxq = 2\uf070 R.h v\u00e0 Stp = Sxq + 2.S\u00f1 . C\u00e2u 8. (3 \u0111i\u1ec3m): T\u1eeb \u0111i\u1ec3m A n\u1eb1m ngo\u00e0i \u0111\u01b0\u1eddng tr\u00f2n (O; R) sao cho OA \uf03e 2R ; v\u1ebd hai ti\u1ebfp tuy\u1ebfn AB , AC ( C , B l\u00e0 hai ti\u1ebfp \u0111i\u1ec3m). G\u1ecdi K l\u00e0 trung \u0111i\u1ec3m c\u1ee7a AB ; CK c\u1eaft (O) t\u1ea1i N ; tia AN c\u1eaft (O) t\u1ea1i M . a) Ch\u1ee9ng minh: OA \u22a5 BC t\u1ea1i H v\u00e0 BK2 = KN.KC . b) Ch\u1ee9ng minh: MC \/ \/ AB . c) Ch\u1ee9ng minh: T\u1ee9 gi\u00e1c BHNK n\u1ed9i ti\u1ebfp v\u00e0 tia NB l\u00e0 tia ph\u00e2n gi\u00e1c c\u1ee7a MNK . ----H\u1ebeT--- T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N 2","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH H\u01af\u1edaNG D\u1eaaN GI\u1ea2I C\u00e2u 1. (1,5 \u0111i\u1ec3m). Cho (P) : y = x2 v\u00e0 \u0111\u01b0\u1eddng th\u1eb3ng (d) : y = \u22124 + 3x . 2 a) V\u1ebd \u0111\u1ed3 th\u1ecb (P) v\u00e0 (d) tr\u00ean c\u00f9ng m\u1eb7t ph\u1eb3ng t\u1ecda \u0111\u1ed9. b) T\u00ecm t\u1ecda \u0111\u1ed9 giao \u0111i\u1ec3m c\u1ee7a (P) v\u00e0 (d) b\u1eb1ng ph\u00e9p t\u00ednh. L\u1eddi gi\u1ea3i a) V\u1ebd \u0111\u1ed3 th\u1ecb (P) v\u00e0 (d) tr\u00ean c\u00f9ng h\u1ec7 tr\u1ee5c t\u1ecda \u0111\u1ed9. a) BGT: x \u22124 \u22122 0 2 4 y = x2 82028 2 x 2 4 y = \u22124 + 3x 2 8 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 = 3x \u2212 4 2 \uf0db x2 \u2212 3x + 4 = 0 2 \uf0db \uf0e9x = 4 \uf0ea\uf0ebx = 2 Thay x = 4 v\u00e0o (P) : y = x2 , ta \u0111\u01b0\u1ee3c: y = 42 = 8 . 22 Thay x = 2 v\u00e0o (P) : y = x2 , ta \u0111\u01b0\u1ee3c: y = 22 = 2 . 22 V\u1eady (4; 8) , (2; 2) l\u00e0 hai giao \u0111i\u1ec3m c\u1ea7n t\u00ecm. T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N 3","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH C\u00e2u 2. (1 \u0111i\u1ec3m) Cho ph\u01b0\u01a1ng tr\u00ecnh 3x2 + 5x \u2212 6 = 0 c\u00f3 2 nghi\u1ec7m l\u00e0 x1 , x2 . Kh\u00f4ng gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh, h\u00e3y t\u00ednh gi\u00e1 tr\u1ecb c\u1ee7a bi\u1ec3u th\u1ee9c A = x12 + x22 x2 \u2212 1 x1 \u2212 1 L\u1eddi gi\u1ea3i V\u00ec \uf044 = b2 \u2212 4ac = (5)2 \u2212 4.3.(\u22126) = 97 \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 = x1 a 3 Theo \u0111\u1ecbnh l\u00ed Vi-et, ta c\u00f3: \uf0ef\uf0ef\uf0eeP = \u22122 c .x2 = a Ta c\u00f3: A = x12 + x22 x2 \u2212 1 x1 \u2212 1 A = x12 + x22 x2 \u2212 1 x1 \u2212 1 A = ( ) (x12 . x1 \u2212 1 + x22. x2 \u2212 1) (x2 \u2212 1)(x1 \u2212 1) A = x13 \u2212 x12 + x23 \u2212 x22 x1.x2 \u2212 x1 \u2212 x2 + 1 ( )A = x13 + x23 \u2212 x12 \u2212 x22 x1.x2 \u2212 x1 + x2 + 1 ( )( ) ( )A = x1 + x2 x12 \u2212 x1.x2 + x22 \u2212 x12 + x22 ( )x1.x2 \u2212 x1 + x2 + 1 ( )( ) ( )A = x1 + x2 x12 + x22 \u2212 x1.x2 \u2212 x12 + x22 ( )x1.x2 \u2212 x1 + x2 + 1 ( ) ( )( ) ( ) ( )x1 + x2 x1 + x2 2 \u2212 2x1.x2 \u2212 x1.x2 \u2212 x1 + x2 2 \u2212 2x1.x2 ( )A = x1.x2 \u2212 x1 + x2 + 1 \uf0e6 \u22125 \uf0f6 \uf0e6 \uf0e6 \u22125 \uf0f62 3.(\u22122) \uf0f6 \uf0e6 \uf0e6 \u22125 \uf0f62 2.(\u22122) \uf0f6 \uf0e7\uf0e8 3 \uf0f7\uf0f8 \uf0e7 \uf0e7\uf0e8 3 \uf0f7\uf0f8 \u2212 \uf0f7 \u2212 \uf0e7 \uf0e7\uf0e8 3 \uf0f7\uf0f8 \u2212 \uf0f7 \uf0e7\uf0e8 \uf0f7\uf0f8 \uf0e7\uf0e8 \uf0f7\uf0f8 A = = \u2212289 (\u22122) \u2212 \uf0e6 \u22125 \uf0f6 + 1 9 \uf0e7\uf0e8 3 \uf0f7\uf0f8 T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N 4","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH C\u00e2u 3. (1 \u0111i\u1ec3m). Hi\u1ec7n nay, c\u00e1c em h\u1ecdc sinh trung h\u1ecdc ng\u00e0y c\u00e0ng quan t\u00e2m h\u01a1n v\u1ec1, c\u00e2n n\u1eb7ng v\u00e0 chi\u1ec1u cao c\u1ee7a b\u1ea3n th\u00e2n. C\u00e1c b\u1ea1n nam th\u01b0\u1eddng ch\u00fa tr\u1ecdng \u0111\u1ebfn chi\u1ec1u cao c\u00f2n c\u00e1c b\u1ea1n n\u1eef l\u1ea1i r\u1ea5t hay \u0111\u1ec3 \u00fd \u0111\u1ebfn c\u00e2n n\u1eb7ng. C\u00e2n n\u1eb7ng l\u00fd t\u01b0\u1edfng \u1ee9ng v\u1edbi chi\u1ec1u cao c\u1ee7a m\u1ed9t ng\u01b0\u1eddi \u0111\u01b0\u1ee3c t\u00ednh theo c\u00f4ng th\u1ee9c sau : w = h \u2212 100 \u2212 h \u2212 150 (1) c Trong \u0111\u00f3 : w l\u00e0 c\u00e2n n\u1eb7ng (t\u00ednh theo kg ), h l\u00e0 chi\u1ec1u cao (t\u00ednh theo cm ) v\u00e0 c = 4 n\u1ebfu ng\u01b0\u1eddi \u0111\u00f3 l\u00e0 nam v\u00e0 c = 2 n\u1ebfu ng\u01b0\u1eddi \u0111\u00f3 l\u00e0 n\u1eef. C\u00f4ng th\u1ee9c n\u00e0y cho ph\u00e9p t\u00ednh \u0111\u01b0\u1ee3c c\u00e2n n\u1eb7ng l\u00fd t\u01b0\u1edfng c\u1ee7a m\u1ed9t ng\u01b0\u1eddi \u0111\u1ec3 c\u00f3 m\u1ed9t c\u01a1 th\u1ec3 c\u00e2n \u0111\u1ed1i, \u01b0a nh\u00ecn khi bi\u1ebft chi\u1ec1u cao c\u1ee7a ng\u01b0\u1eddi \u0111\u00f3. a) M\u1ed9t b\u1ea1n n\u1eef c\u00f3 chi\u1ec1u cao l\u00e0 1,52m th\u00ec c\u00e2n n\u1eb7ng l\u00fd t\u01b0\u1edfng c\u1ee7a b\u1ea1n \u1ea5y l\u00e0 bao nhi\u00eau kg ? b) N\u1ebfu m\u1ed9t b\u1ea1n nam c\u00f3 c\u00e2n n\u1eb7ng l\u00e0 56 kg th\u00ec chi\u1ec1u cao c\u1ee7a b\u1ea1n nam n\u00e0y c\u1ea7n ph\u1ea3i l\u00e0 bao nhi\u00eau \u0111\u1ec3 56 kg l\u00e0 c\u00e2n n\u1eb7ng l\u00fd t\u01b0\u1edfng cho b\u1ea1n \u1ea5y ? L\u1eddi gi\u1ea3i a) Ta c\u00f3 c\u00f4ng th\u1ee9c w = h \u2212 100 \u2212 h \u2212 150 (1) c M\u1ed9t b\u1ea1n n\u1eef c\u00f3 chi\u1ec1u cao l\u00e0 1,52m th\u00ec c = 2 v\u00e0 h = 152 . Thay c = 2 v\u00e0 h = 152 v\u00e0o c\u00f4ng th\u1ee9c (1) , ta \u0111\u01b0\u1ee3c w = 152 \u2212 100 \u2212 152 \u2212 150 = 51 . 2 B\u1ea1n n\u1eef c\u00f3 chi\u1ec1u cao l\u00e0 1,52m th\u00ec c\u00e2n n\u1eb7ng l\u00fd t\u01b0\u1edfng c\u1ee7a b\u1ea1n \u1ea5y l\u00e0 51kg . b) B\u1ea1n nam c\u00f3 c\u00e2n n\u1eb7ng l\u00e0 56 kg th\u00ec c = 4 v\u00e0 w = 56 Thay c = 4 v\u00e0 w = 56 v\u00e0o c\u00f4ng th\u1ee9c (1) Ta \u0111\u01b0\u1ee3c 56 = h \u2212 100 \u2212 h \u2212 150 4 \uf0db h \u2212 100 \u2212 h \u2212 150 = 56 4 \uf0db 4h \u2212 400 \u2212 (h \u2212 150) = 224 44 \uf0db 4h \u2212 400 \u2212 (h \u2212 150) = 224 \uf0db 4h \u2212 400 \u2212 h + 150 = 224 \uf0db 4h \u2212 h = 224 + 400 \u2212 150 \uf0db 3h = 474 \uf0db h = 158 V\u1eady b\u1ea1n nam c\u00f3 c\u00e2n n\u1eb7ng l\u00e0 56 kg th\u00ec chi\u1ec1u cao c\u1ee7a b\u1ea1n nam n\u00e0y l\u00e0 1,58m \u0111\u1ec3 56 kg l\u00e0 c\u00e2n n\u1eb7ng l\u00fd t\u01b0\u1edfng cho b\u1ea1n \u1ea5y T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N 5","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH C\u00e2u 4. (0,75 \u0111i\u1ec3m). L\u1edbp 9A c\u00f3 30 h\u1ecdc sinh, m\u1ed7i b\u1ea1n d\u1ef1 \u0111\u1ecbnh \u0111\u00f3ng m\u1ed7i th\u00e1ng 70 000 \u0111\u1ed3ng \u0111\u1ec3 mua qu\u00e0 t\u1eb7ng c\u00e1c tr\u1ebb em \u1edf m\u00e1i \u1ea5m t\u00ecnh th\u01b0\u01a1ng v\u00e0 sau 3 th\u00e1ng s\u1ebd \u0111\u1ee7 ti\u1ec1n \u0111\u1ec3 t\u1eb7ng m\u1ed7i em \u1edf m\u00e1i \u1ea5m 3 g\u00f3i qu\u00e0 (gi\u00e1 tr\u1ecb m\u1ed7i g\u00f3i qu\u00e0 l\u00e0 nh\u01b0 nhau). Khi c\u00e1c h\u1ecdc sinh \u0111\u00e3 \u0111\u00f3ng \u0111\u1ee7 ti\u1ec1n th\u00ec m\u00e1i \u1ea5m nh\u1eadn ch\u0103m s\u00f3c th\u00eam 9 em v\u00e0 gi\u00e1 ti\u1ec1n m\u1ed7i g\u00f3i qu\u00e0 t\u0103ng th\u00eam 5% n\u00ean ch\u1ec9 t\u1eb7ng \u0111\u01b0\u1ee3c m\u1ed7i em 2 g\u00f3i qu\u00e0. H\u1ecfi l\u00fac \u0111\u1ea7u c\u00f3 bao nhi\u00eau em \u1edf m\u00e1i \u1ea5m \u0111\u01b0\u1ee3c t\u1eb7ng qu\u00e0? L\u1eddi gi\u1ea3i ( )G\u1ecdi x (ng\u01b0\u1eddi) l\u00e0 s\u1ed1 em \u1edf m\u00e1i \u1ea5m \u0111\u01b0\u1ee3c t\u1eb7ng qu\u00e0 l\u00fac \u0111\u1ea7u x \uf0ce * . T\u1ed5ng ti\u1ec1n 30 b\u1ea1n d\u1ef1 \u0111\u1ecbnh \u0111\u00f3ng trong 3 th\u00e1ng l\u00e0 : 70 000 . 30 . 3 = 6 300 000 (\u0111\u1ed3ng). S\u1ed1 ph\u1ea7n qu\u00e0 c\u00e1c em \u0111\u01b0\u1ee3c nh\u1eadn ban \u0111\u1ea7u l\u00e0 3x (ph\u1ea7n qu\u00e0). Gi\u00e1 tr\u1ecb 1 ph\u1ea7n qu\u00e0 ban \u0111\u1ea7u l\u00e0 6 300000 (\u0111\u1ed3ng). 3x S\u1ed1 em \u1edf m\u00e1i \u1ea5m \u0111\u01b0\u1ee3c t\u1eb7ng qu\u00e0 l\u00fac sau l\u00e0 x + 9 (em). S\u1ed1 ph\u1ea7n qu\u00e0 c\u00e1c em \u0111\u01b0\u1ee3c nh\u1eadn 2.(x + 9) (ph\u1ea7n qu\u00e0). Gi\u00e1 tr\u1ecb 1 ph\u1ea7n qu\u00e0 l\u00fac sau l\u00e0 6 300000 (\u0111\u1ed3ng). 2(x + 9) V\u00ec gi\u00e1 ti\u1ec1n m\u1ed7i g\u00f3i qu\u00e0 t\u0103ng th\u00eam 5% gi\u00e1 tr\u1ecb c\u1ee7a c\u00e1c ph\u1ea7n qu\u00e0 l\u00e0 nh\u01b0 nhau , N\u00ean ta c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh: 6 300 000 .(1 + 5%) = 6 300000 3x 2(x + 9) \uf0db 1 .(1 + 5%) = 1 9) 3x 2(x + \uf0db 0 , 35 = 1 9) x 2(x + \uf0de x = 0,35.2(x + 9) \uf0de x = 0,7x + 6,3 \uf0de 0,3x = 6,3 \uf0de x = 21(nh\u1eadn) V\u1eady c\u00f3 21 em \u1edf m\u00e1i \u1ea5m \u0111\u01b0\u1ee3c t\u1eb7ng qu\u00e0 l\u00fac \u0111\u1ea7u. 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 5. (1 \u0111i\u1ec3m). \u0110\u1ec3 l\u1eadp \u0111\u1ed9i tuy\u1ec3n n\u0103ng khi\u1ebfu v\u1ec1 b\u00f3ng r\u1ed5 c\u1ee7a tr\u01b0\u1eddng th\u1ea7y th\u1ec3 d\u1ee5c \u0111\u01b0a ra quy \u0111\u1ecbnh tuy\u1ec3n ch\u1ecdn nh\u01b0 sau: m\u1ed7i b\u1ea1n d\u1ef1 tuy\u1ec3n s\u1ebd \u0111\u01b0\u1ee3c n\u00e9m 10 qu\u1ea3 b\u00f3ng v\u00e0o r\u1ed5, qu\u1ea3 b\u00f3ng v\u00e0o r\u1ed5 \u0111\u01b0\u1ee3c c\u1ed9ng 4 \u0111i\u1ec3m; qu\u1ea3 b\u00f3ng n\u00e9m ra ngo\u00e0i th\u00ec b\u1ecb tr\u1eeb 2 \u0111i\u1ec3m. N\u1ebfu b\u1ea1n n\u00e0o c\u00f3 s\u1ed1 \u0111i\u1ec3m t\u1eeb 22 \u0111i\u1ec3m tr\u1edf l\u00ean th\u00ec s\u1ebd \u0111\u01b0\u1ee3c ch\u1ecdn v\u00e0o \u0111\u1ed9i tuy\u1ec3n. H\u1ecfi m\u1ed9t h\u1ecdc sinh mu\u1ed1n \u0111\u01b0\u1ee3c ch\u1ecdn v\u00e0o \u0111\u1ed9i tuy\u1ec3n th\u00ec ph\u1ea3i n\u00e9m \u00edt nh\u1ea5t bao nhi\u00eau qu\u1ea3 v\u00e0o r\u1ed5? L\u1eddi gi\u1ea3i C\u00e1ch 1: ( )G\u1ecdi x (qu\u1ea3 b\u00f3ng) l\u00e0 s\u1ed1 qu\u1ea3 b\u00f3ng m\u1ed9t h\u1ecdc sinh n\u00e9m v\u00e0o r\u1ed5 x \uf0ce * , x \uf0a3 10 . ( )G\u1ecdi y (qu\u1ea3 b\u00f3ng) l\u00e0 s\u1ed1 qu\u1ea3 b\u00f3ng m\u1ed9t h\u1ecdc sinh n\u00e9m ra ngo\u00e0i y \uf0ce * , y \uf0a3 10 . V\u00ec m\u1ed7i b\u1ea1n d\u1ef1 tuy\u1ec3n s\u1ebd \u0111\u01b0\u1ee3c n\u00e9m 10 qu\u1ea3 b\u00f3ng v\u00e0o r\u1ed5 N\u00ean x + y = 10 (1) . V\u00ec qu\u1ea3 b\u00f3ng v\u00e0o r\u1ed5 \u0111\u01b0\u1ee3c c\u1ed9ng 4 \u0111i\u1ec3m, qu\u1ea3 b\u00f3ng n\u00e9m ra ngo\u00e0i b\u1ecb tr\u1eeb 2 \u0111i\u1ec3m v\u00e0 t\u1ed5ng \u0111i\u1ec3m l\u00e0 22 \u0111i\u1ec3m, n\u00ean 4x \u2212 2y = 22 (2) . T\u1eeb (1) v\u00e0 (2) ta c\u00f3 h\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh : \uf0ecx + y = 10 22 \uf0db \uf0ec\uf0efx = 7 (n) \uf0ed\uf0ee4x \u2212 2y = \uf0ed = 3 (n) \uf0ef\uf0eey V\u1eady m\u1ed7i h\u1ecdc sinh c\u1ea7n n\u00e9m v\u00e0o r\u1ed7 \u00edt nh\u1ea5t l\u00e0 7 qu\u1ea3. C\u00e1ch 2: ( )G\u1ecdi x (qu\u1ea3 b\u00f3ng) l\u00e0 s\u1ed1 qu\u1ea3 b\u00f3ng m\u1ed9t h\u1ecdc sinh n\u00e9m v\u00e0o r\u1ed5 x \uf0ce * , x \uf0a3 10 Suy ra: 10 \u2212 x l\u00e0 s\u1ed1 qu\u1ea3 b\u00f3ng m\u1ed9t h\u1ecdc sinh n\u00e9m ra ngo\u00e0i. V\u00ec qu\u1ea3 b\u00f3ng v\u00e0o r\u1ed5 \u0111\u01b0\u1ee3c c\u1ed9ng 4 \u0111i\u1ec3m; qu\u1ea3 b\u00f3ng n\u00e9m ra ngo\u00e0i th\u00ec b\u1ecb tr\u1eeb 2 \u0111i\u1ec3m v\u00e0 s\u1ed1 \u0111i\u1ec3m t\u1eeb 22 \u0111i\u1ec3m tr\u1edf l\u00ean. N\u00ean ta c\u00f3 b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh : 4x \u2212 2(10 \u2212 x) \uf0b3 22 \uf0db 4x \u2212 20 + 2x \uf0b3 22 \uf0db 6x \uf0b3 42 \uf0dbx\uf0b37 V\u1eady m\u1ed7i h\u1ecdc sinh c\u1ea7n n\u00e9m v\u00e0o r\u1ed7 \u00edt nh\u1ea5t l\u00e0 7 qu\u1ea3. 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. (1 \u0111i\u1ec3m). B\u1ea3ng c\u01b0\u1edbc ph\u00ed d\u1ecbch v\u1ee5 Mobiphone \u00e1p d\u1ee5ng cho thu\u00ea bao tr\u1ea3 tr\u01b0\u1edbc, c\u01b0\u1edbc g\u1ecdi li\u00ean m\u1ea1ng trong n\u01b0\u1edbc (\u0111\u00e3 bao g\u1ed3m VAT) quy \u0111\u1ecbnh r\u1eb1ng : n\u1ebfu g\u1ecdi trong 5 gi\u00e2y \u0111\u1ea7u th\u00ec t\u00ednh c\u01b0\u1edbc 200 \u0111\u1ed3ng\/ 5 gi\u00e2y \u0111\u1ea7u, c\u00f2n k\u1ec3 t\u1eeb sau gi\u00e2y th\u1ee9 5 tr\u1edf \u0111i, h\u1ecd t\u00ednh th\u00eam 28 \u0111\u1ed3ng cho m\u1ed7i gi\u00e2y. a) G\u1ecdi m l\u00e0 s\u1ed1 ti\u1ec1n ph\u1ea3i tr\u1ea3 (t\u00ednh b\u1eb1ng \u0111\u1ed3ng) v\u00e0 n l\u00e0 th\u1eddi gian g\u1ecdi nhi\u1ec1u h\u01a1n 5 gi\u00e2y (t \uf03e 5). H\u00e3y l\u1eadp c\u00f4ng th\u1ee9c bi\u1ec3u th\u1ecb m theo n ? b) H\u1ecfi b\u1ea1n Anh g\u1ecdi trong bao l\u00e2u m\u00e0 b\u1ea1n tr\u1ea3 3420 \u0111\u1ed3ng ? L\u1eddi gi\u1ea3i a) C\u00f4ng th\u1ee9c bi\u1ec3u th\u1ecb m theo n c\u00f3 d\u1ea1ng m = a.n + b . Theo \u0111\u1ec1 b\u00e0i, ta c\u00f3: V\u1edbi \uf0ec\uf0efn = 6 5).28 = 228 \uf0de 228 = 6.a + b . (1) \uf0ed\uf0ef\uf0eem = 200 + (6 \u2212 V\u1edbi \uf0ec\uf0efn = 7 \u2212 5).28 = 256 \uf0de 256 = 7.a + b . (2) \uf0ed\uf0ef\uf0eem = 200 + (7 T\u1eeb (1) v\u00e0 (2) ta c\u00f3 h\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh: \uf0ec6a + b = 228 \uf0db \uf0eca = 28 . \uf0ed\uf0ee7a + b = 256 \uf0ed\uf0eeb = 60 V\u1eady: a = 28 , b = 60 v\u00e0 m = 28n + 60 . b) H\u1ecfi b\u1ea1n Anh g\u1ecdi trong bao l\u00e2u m\u00e0 b\u1ea1n tr\u1ea3 3420 \u0111\u1ed3ng ? Thay m = 3420 v\u00e0o bi\u1ec3u th\u1ee9c m = 28n + 60 Ta \u0111\u01b0\u1ee3c : 3420 = 28n + 60 \uf0db n = 120 . V\u1eady b\u1ea1n Anh g\u1ecdi 120 gi\u00e2y, b\u1ea1n tr\u1ea3 3420 \u0111\u1ed3ng. T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N 8","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH C\u00e2u 7. (0,75 \u0111i\u1ec3m): N\u01b0\u1edbc gi\u1ea3i kh\u00e1t th\u01b0\u1eddng \u0111\u1ef1ng trong lon nh\u00f4m v\u00e0 c\u1ee1 lon ph\u1ed5 bi\u1ebfn tr\u00ean th\u1ebf gi\u1edbi th\u01b0\u1eddng ch\u1ee9a \u0111\u01b0\u1ee3c kho\u1ea3ng 335 ml ch\u1ea5t l\u1ecfng, \u0111\u01b0\u1ee3c thi\u1ebft k\u1ebf h\u00ecnh tr\u1ee5 v\u1edbi chi\u1ec1u cao g\u1ea7n g\u1ea5p \u0111\u00f4i \u0111\u01b0\u1eddng k\u00ednh \u0111\u00e1y (cao 12cm , \u0111\u01b0\u1eddng k\u00ednh \u0111\u00e1y 6,5cm ). Nh\u01b0ng hi\u1ec7n nay c\u00e1c nh\u00e0 s\u1ea3n xu\u1ea5t c\u00f3 xu h\u01b0\u1edbng t\u1ea1o ra nh\u1eefng lon nh\u00f4m v\u1edbi ki\u1ec3u d\u00e1ng thon cao d\u00e0i. Tuy chi ph\u00ed s\u1ea3n xu\u1ea5t c\u1ee7a nh\u1eefng chi\u1ebfc lon n\u00e0y t\u1ed1n k\u00e9m h\u01a1n, do n\u00f3 c\u00f3 di\u1ec7n t\u00edch m\u1eb7t ngo\u00e0i (di\u1ec7n t\u00edch to\u00e0n ph\u1ea7n c\u1ee7a h\u00ecnh tr\u1ee5) l\u1edbn h\u01a1n, nh\u01b0ng n\u00f3 l\u1ea1i d\u1ec5 \u0111\u00e1nh l\u1eeba th\u1ecb gi\u00e1c v\u00e0 \u0111\u01b0\u1ee3c ng\u01b0\u1eddi ti\u00eau d\u00f9ng \u01b0a chu\u1ed9ng h\u01a1n. c) M\u1ed9t lon n\u01b0\u1edbc ng\u1ecdt cao 14,2 cm , \u0111\u01b0\u1eddng k\u00ednh \u0111\u00e1y 5,9cm . H\u1ecfi lon n\u01b0\u1edbc ng\u1ecdt cao n\u00e0y c\u00f3 th\u1ec3 ch\u1ee9a \u0111\u01b0\u1ee3c h\u1ebft l\u01b0\u1ee3ng n\u01b0\u1edbc ng\u1ecdt c\u1ee7a m\u1ed9t lon c\u1ee1 ph\u1ed5 bi\u1ebfn kh\u00f4ng ? V\u00ec sao ? Bi\u1ebft th\u1ec3 t\u00edch h\u00ecnh tr\u1ee5 V = \uf070 R2h v\u1edbi \uf070 \uf0bb 3,14 (k\u1ebft qu\u1ea3 l\u00e0m tr\u00f2n \u0111\u1ebfn ch\u1eef s\u1ed1 th\u1eadp ph\u00e2n th\u1ee9 hai) d) H\u1ecfi chi ph\u00ed s\u1ea3n xu\u1ea5t lon n\u01b0\u1edbc ng\u1ecdt cao \u1edf c\u00e2u a, t\u0103ng bao nhi\u00eau ph\u1ea7n tr\u0103m so v\u1edbi s\u1ea3n xu\u1ea5t lon n\u01b0\u1edbc ng\u1ecdt c\u1ee1 ph\u1ed5 bi\u1ebfn ? Bi\u1ebft Sxq = 2\uf070 R.h v\u00e0 Stp = Sxq + 2.S\u00f1 L\u1eddi gi\u1ea3i a) B\u00e1n k\u00ednh \u0111\u00e1y lon cao : 5,9 : 2 = 2,95(cm) . ( )( )Th\u1ec3 t\u00edch lon cao : V = \uf070 R2h = 3,14. 2,95 2 .14,2 = 388,03 cm3 = 388,03ml . V\u00ec 388,03ml \uf03e 335ml N\u00ean Lon n\u01b0\u1edbc ng\u1ecdt cao n\u00e0y c\u00f3 th\u1ec3 ch\u1ee9a \u0111\u01b0\u1ee3c h\u1ebft l\u01b0\u1ee3ng n\u01b0\u1edbc ng\u1ecdt c\u1ee7a m\u1ed9t lon c\u1ee1 ph\u1ed5 bi\u1ebfn. b) Di\u1ec7n t\u00edch to\u00e0n ph\u1ea7n lon ph\u1ed5 bi\u1ebfn ( ) ( ) ( )Stp = Sxq + 2.S\u00f1 = 2\uf070 R.h + 2.\uf070 R2 = 2.3,14. 3,25 .12 + 2.3,14. 3,25 2 = 311,25 cm3 . Di\u1ec7n t\u00edch to\u00e0n ph\u1ea7n lon cao ( ) ( ) ( )Stp = Sxq + 2.S\u00f1 = 2\uf070 R.h + 2.\uf070 R2 = 2.3,14. 2,95 .14,2 + 2.3,14. 2.95 2 = 317,72 cm3 . % Nguy\u00ean li\u1ec7u \u0111\u1ec3 s\u1ea3n xu\u1ea5t lon n\u01b0\u1edbc ng\u1ecdt cao t\u0103ng so v\u1edbi s\u1ea3n xu\u1ea5t lon n\u01b0\u1edbc ng\u1ecdt c\u1ee1 ph\u1ed5 bi\u1ebfn l\u00e0: 317,72 \u2212 311,25 \uf0bb 0,0208 = 2,08% . 311,25 V\u1eady chi ph\u00ed s\u1ea3n xu\u1ea5t lon n\u01b0\u1edbc ng\u1ecdt cao \u1edf c\u00e2u a, t\u0103ng 2,08%so v\u1edbi s\u1ea3n xu\u1ea5t lon n\u01b0\u1edbc ng\u1ecdt c\u1ee1 ph\u1ed5 bi\u1ebfn. 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 C\u00e2u 8. (3 \u0111i\u1ec3m): T\u1eeb \u0111i\u1ec3m A n\u1eb1m ngo\u00e0i \u0111\u01b0\u1eddng tr\u00f2n (O; R) sao cho OA \uf03e 2R ; v\u1ebd hai ti\u1ebfp tuy\u1ebfn AB , AC ( C , B l\u00e0 hai ti\u1ebfp \u0111i\u1ec3m). G\u1ecdi K l\u00e0 trung \u0111i\u1ec3m c\u1ee7a AB ; CK c\u1eaft (O) t\u1ea1i N ; tia AN c\u1eaft (O) t\u1ea1i M . c) Ch\u1ee9ng minh: OA \u22a5 BC t\u1ea1i H v\u00e0 BK2 = KN.KC . d) Ch\u1ee9ng minh: MC \/\/ AB . e) Ch\u1ee9ng minh: T\u1ee9 gi\u00e1c BHNK n\u1ed9i ti\u1ebfp v\u00e0 tia NB l\u00e0 tia ph\u00e2n gi\u00e1c c\u1ee7a MNK . L\u1eddi gi\u1ea3i a) Ch\u1ee9ng minh: OA \u22a5 BC t\u1ea1i H v\u00e0 BK2 = KN.KC . Taco\u00f9 : ( )\uf0ec\uf0efAB = AC AC, AB la\u00f8 tie\u00e1ptuye\u00e1ncu\u00fba(O) ( )\uf0ed \uf0ef\uf0eeOB = OC OA,OB la\u00f8 ba\u00f9n k\u00ednhcu\u00fba(O) \uf0de OA la\u00f8 \u00f1\u00f6\u00f4\u00f8ng trung tr\u00f6\u00efc cu\u00fba BC \uf0de OA \u22a5 BC ta\u00efi H Xe\u00f9t\uf044BKN va\u00f8 \uf044CKB, taco\u00f9 \uf0ec\uf0efK la\u00f8 go\u00f9c chung ( )\uf0ed \uf0ef\uf0eeKBN = KCB cu\u00f8ng cha\u00e9n cung BN Va\u00e4y\uf044BKN \u223d\uf044CKB(g \u2212 g) \uf0de BK = CK (t\u00e6 so\u00e1 \u00f1o\u00e0ng da\u00efng) KN BK \uf0de BK 2 = KN.KC 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 b) Ch\u1ee9ng minh: MC \/ \/ AB Ta co\u00f9 : ( )\uf0ec\uf0efBK 2 = KN.KC cm ca\u00e2u a ( )\uf0ed \uf0ef\uf0eeKB = KA K la\u00f8 trung \u00f1ie\u00e5m AB \uf0de AK 2 = KN.KC \uf0de KA = KC KN KA Xe\u00f9t\uf044AKN va\u00f8 \uf044CKA, taco\u00f9 \uf0ec\uf0efK la\u00f8 go\u00f9c chung \uf0ed KA KC \uf0ef\uf0ee KN = KA ( cmt ) Va\u00e4y\uf044AKN \u223d\uf044CKA(g \u2212 g) ( )\uf0de KAN = ACK hai go\u00f9c t\u00f6\u00f4ng \u00f6\u00f9ng La\u00efico\u00f9 : CMN = ACK (cu\u00f8ng cha\u00e9n cung CN ) \uf0de KAN = CMN \uf0de MC \/ \/ AB c) Ch\u1ee9ng minh: T\u1ee9 gi\u00e1c BHNK n\u1ed9i ti\u1ebfp v\u00e0 tia NB l\u00e0 tia ph\u00e2n gi\u00e1c c\u1ee7a MNK Cm : BHNK nt Xe\u00f9t \uf044ABC,taco\u00f9: \uf0ec\uf0efHB = HC (cm ca\u00e2u a) ( )\uf0ed \uf0ef\uf0eeKB = KA K la\u00f8 trung \u00f1ie\u00e5m AB \uf0de HK la\u00f8 \u00f1\u00f6\u00f4\u00f8ng trung b\u00ecnh \uf044ABC \uf0de HK \/ \/ AC Ta co\u00f9 : \uf0ec BNK = KBC ( \uf044BKN \u223d \uf044CKB ) \uf0ef ( )\uf0ef\uf0edKBC = ACB cu\u00f8ng cha\u00e9n cung BC \uf0ef \uf0ef\uf0eeACB = BHK (HK \/ \/ AC ) \uf0de BNK = BHK \uf0de BHNK no\u00e4i tie\u00e1p ( )t\u00f6\u00f9gia\u00f9c co\u00f9hai \u00f1\u00e6nh ke\u00e0 nhaucu\u00f8ng nh\u00ecn mo\u00e4t ca\u00efnh d\u00f6\u00f4\u00f9i hai go\u00f9c ba\u00e8ng nhau CM : NB la\u00f8 tia pha\u00e2n gia\u00f9c MNK Ta co\u00f9 : \uf0ec BNK = KBC ( \uf044BKN \u223d \uf044CKB ) \uf0ef \uf0ef\uf0edKBC = BCM ( AB \/ \/ MC ) ( )\uf0ef \uf0efBCM = BNM cu\u00f8ng cha\u00e9n cung BM \uf0ee \uf0de BNK = BNM \uf0de NB la\u00f8 tia pha\u00e2n gia\u00f9c MNK T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N 11","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH S\u00d4\u00db GD&\u00d1T TP HO\u00c0 CH\u00cd MINH \u00d1E\u00c0 THAM KHA\u00dbO TUYE\u00c5N SINH 10 PHO\u00d8NG G\u00d1&\u00d1T QUA\u00c4N 12 NA\u00caM HO\u00cfC: 2021 - 2022 \u0110\u1ec0 THAM KH\u1ea2O M\u00d4N: TO\u00c1N 9 M\u00c3 \u0110\u1ec0: Qu\u1eadn 12 - 3 \u0110\u00ea thi g\u1ed3m 8 c\u00e2u ho\u0309i t\u01b0\u0323 lu\u00e2\u0323n. Th\u01a1\u0300i gian: 120 phu\u0301t (kh\u00f4ng k\u00ea\u0309 th\u01a1\u0300i gian pha\u0301t \u0111\u00ea\u0300) C\u00e2u 1. (1,5 \u0111i\u1ec3m). Cho (P) : y = \u2212x2 v\u00e0 \u0111\u01b0\u1eddng th\u1eb3ng (d) : y = x \u2212 4 . 2 a) V\u1ebd (P) v\u00e0 (d) tr\u00ean c\u00f9ng m\u1eb7t ph\u1eb3ng t\u1ecda \u0111\u1ed9. b) T\u00ecm t\u1ecda \u0111\u1ed9 giao \u0111i\u1ec3m c\u1ee7a (P) v\u00e0 (d) b\u1eb1ng ph\u00e9p to\u00e1n. C\u00e2u 2. (1 \u0111i\u1ec3m). Cho ph\u01b0\u01a1ng tr\u00ecnh 2x2 \u2212 5x \u2212 7 = 0 . G\u1ecdi x1 , x2 l\u00e0 hai nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh. Kh\u00f4ng gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh, h\u00e3y t\u00ednh gi\u00e1 tr\u1ecb c\u1ee7a bi\u1ec3u th\u1ee9c M = x12 + x22 x2 x1 C\u00e2u 3. (1 \u0111i\u1ec3m). M\u1ed9t b\u00ecnh x\u0103ng c\u1ee7a m\u1ed9t \u00f4 t\u00f4 ch\u1ee9a \u0111\u01b0\u1ee3c 50 l\u00edt. Bi\u1ebft r\u1eb1ng trung b\u00ecnh c\u1ee9 \u0111i 9 km th\u00ec h\u1ebft 1 l\u00edt x\u0103ng.Bi\u1ebft r\u1eb1ng m\u1ed1i li\u00ean h\u1ec7 gi\u1eefa y (l\u00edt x\u0103ng) c\u00f2n l\u1ea1i trong b\u00ecnh v\u00e0 qu\u00e3ng \u0111\u01b0\u1eddng x (km) l\u00e0 m\u1ed9t h\u00e0m s\u1ed1 b\u1eadc nh\u1ea5t y = ax + b . a) H\u00e3y vi\u1ebft h\u00e3y x\u00e1c \u0111\u1ecbnh h\u1ec7 s\u1ed1 a v\u00e0 b . b) Khi x\u0103ng trong b\u00ecnh c\u00f2n l\u1ea1i 4 l\u00edt x\u0103ng th\u00ec \u00f4 t\u00f4 d\u1eebng l\u1ea1i \u0111\u1ed5 x\u0103ng . H\u1ecfi khi \u0111\u00f3 \u00f4 t\u00f4 \u0111i \u0111\u01b0\u1ee3c qu\u00e3ng \u0111\u01b0\u1eddng l\u00e0 bao nhi\u00eau? C\u00e2u 4. (0,75 \u0111i\u1ec3m). Nh\u00e2n d\u1ecbp trung thu m\u1ed9t c\u1eeda h\u00e0ng b\u00e1n b\u00e1nh k\u1eb9o \u0111\u01b0a ra h\u00ecnh th\u1ee9c khuy\u1ebfn m\u00e3i cho m\u1ed9t lo\u1ea1i b\u00e1nh A \u0111ang c\u00f3 gi\u00e1 b\u00e1n l\u00e0 120 000 \u0111\/h\u1ed9p nh\u01b0 sau: H\u00ecnh th\u1ee9c khuy\u1ebfn m\u00e3i 1 : Mua 3 h\u1ed9p \u0111\u1ea7u gi\u00e1 120000 \u0111\/h\u1ed9p , t\u1eeb h\u1ed9p th\u1ee9 t\u01b0 tr\u1edf \u0111i m\u1ed7i h\u1ed9p gi\u1ea3m 30% H\u00ecnh th\u1ee9c khuy\u1ebfn m\u00e3i 2 : Mua 3 t\u1eb7ng 1 B\u1ea1n Lan c\u1ea7n mua gi\u00fap cho m\u1eb9 9 h\u1ed9p b\u00e1nh A \u0111\u1ec3 l\u00e0m qu\u00e0. Em h\u00e3y t\u00ednh gi\u00fap b\u1ea1n Lan n\u00ean ch\u1ecdn h\u00ecnh th\u1ee9c khuy\u1ebfn m\u00e3i n\u00e0o th\u00ec c\u00f3 l\u1ee3i h\u01a1n?( Tr\u1ea3 ti\u1ec1n \u00edt h\u01a1n) C\u00e2u 5. (1 \u0111i\u1ec3m). Nh\u1eefng n\u0103m d\u01b0\u01a1ng l\u1ecbch n\u00e0o chia h\u1ebft cho 4 th\u00ec n\u0103m \u0111\u00f3 l\u00e0 n\u0103m nhu\u1eadn. V\u00ed d\u1ee5: 2016 chia h\u1ebft cho 4 n\u00ean n\u0103m 2016 l\u00e0 n\u0103m nhu\u1eadn. Ngo\u00e0i ra, v\u1edbi nh\u1eefng n\u0103m tr\u00f2n th\u1ebf k\u1ef7 (nh\u1eefng n\u0103m c\u00f3 2 s\u1ed1 cu\u1ed1i l\u00e0 s\u1ed1 0 ) th\u00ec c\u00e1c b\u1ea1n l\u1ea5y s\u1ed1 n\u0103m chia cho 400 , n\u1ebfu chia h\u1ebft th\u00ec n\u0103m \u0111\u00f3 l\u00e0 n\u0103m c\u00f3 nhu\u1eadn (ho\u1eb7c 2 s\u1ed1 \u0111\u1ea7u trong n\u0103m chia h\u1ebft cho 4 ). a) Em h\u00e3y cho bi\u1ebft n\u0103m 2023 c\u00f3 ph\u1ea3i l\u00e0 n\u0103m nhu\u1eadn d\u01b0\u01a1ng l\u1ecbch kh\u00f4ng ? Gi\u1ea3i th\u00edch? b) Ban An sinh nh\u1eadt l\u1ea7n th\u1ee9 14 v\u00e0o th\u1ee9 b\u1ea3y ng\u00e0y 8 th\u00e1ng t\u01b0 n\u0103m 2023 . H\u1ecfi sinh nh\u1eadt l\u1ea7n th\u1ee9 15 c\u1ee7a b\u1ea1n v\u00e0o th\u1ee9 m\u1ea5y? C\u00e2u 6. (1 \u0111i\u1ec3m M\u1ed9t ng\u01b0\u1eddi \u0111o \u0111\u1ea1c chi\u1ec1u cao c\u1ee7a m\u1ed9t ng\u1ecdn n\u00fai t\u1ea1i hai \u0111i\u1ec3m A,B tr\u00ean m\u1eb7t \u0111\u1ea5t. K\u1ebft qu\u1ea3 \u0111\u01b0\u1ee3c cho trong h\u00ecnh v\u1ebd. T\u00ednh chi\u1ec1u cao c\u1ee7a ng\u1ecdn n\u00fai.( L\u00e0m tr\u00f2n \u0111\u1ebfn 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 7. (1 \u0111i\u1ec3m). M\u1ed9t th\u00f9ng \u0111\u1ef1ng n\u01b0\u1edbc c\u00f3 d\u1ea1ng h\u00ecnh tr\u1ee5 chi\u1ec1u cao l\u00e0 35 cm \u0111\u01b0\u1eddng k\u00ednh \u0111\u00e1y 30 cm . a) T\u00ednh th\u1ec3 t\u00edch c\u1ee7a th\u00f9ng. b) Ng\u01b0\u1eddi ta s\u1eed d\u1ee5ng th\u00f9ng tr\u00ean \u0111\u1ec3 m\u00fac n\u01b0\u1edbc \u0111\u1ed5 v\u00e0o m\u1ed9t b\u1ec3 ch\u1ee9a c\u00f3 dung t\u00edch 1m3 . H\u1ecfi c\u1ea7n ph\u1ea3i \u0111\u1ed5 \u00edt nh\u1ea5t bao nhi\u00eau th\u00f9ng th\u00ec \u0111\u1ea7y b\u1ec3 ch\u1ee9a ? Bi\u1ebft r\u1eb1ng m\u1ed7i l\u1ea7n x\u00e1ch ng\u01b0\u1eddi ta ch\u1ec9 \u0111\u1ed5 \u0111\u1ea7y 90% th\u00f9ng \u0111\u1ec3 n\u01b0\u1edbc kh\u00f4ng \u0111\u1ed5 ra ngo\u00e0i. ( )C\u00e2u 8. (3 \u0111i\u1ec3m) Cho tam gi\u00e1c ABC nh\u1ecdn (AB \uf03c AC) n\u1ed9i ti\u1ebfp O . C\u00e1c \u0111\u01b0\u1eddng cao AD,BE,CF c\u1eaft nhau t\u1ea1i H . Tia EF c\u1eaft tia CB t\u1ea1i K . a) Ch\u1ee9ng minh t\u1ee9 gi\u00e1c BFEC n\u1ed9i ti\u1ebfp v\u00e0 KF.KE = KB.KC . b) \u0110\u01b0\u1eddng th\u1eb3ng KA c\u1eaft (O) t\u1ea1i M . Ch\u1ee9ng minh t\u1ee9 gi\u00e1c AEFM n\u1ed9i ti\u1ebfp. c) G\u1ecdi N l\u00e0 trung \u0111i\u1ec3m c\u1ee7a BC . Ch\u1ee9ng minh MN,BE,CF \u0111\u1ed3ng quy( c\u00f9ng \u0111i qua m\u1ed9t \u0111i\u1ec3m ). ----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 = \u2212x2 v\u00e0 \u0111\u01b0\u1eddng th\u1eb3ng (d) : y = x \u2212 4 . 2 a) V\u1ebd \u0111\u1ed3 th\u1ecb (P) v\u00e0 (d) tr\u00ean c\u00f9ng h\u1ec7 tr\u1ee5c t\u1ecda \u0111\u1ed9. b) T\u00ecm t\u1ecda \u0111\u1ed9 giao \u0111i\u1ec3m c\u1ee7a (P) v\u00e0 (d) b\u1eb1ng ph\u00e9p t\u00ednh. L\u1eddi gi\u1ea3i a) V\u1ebd \u0111\u1ed3 th\u1ecb (P) v\u00e0 (d) tr\u00ean c\u00f9ng h\u1ec7 tr\u1ee5c t\u1ecda \u0111\u1ed9. BGT: x \u22124 \u22122 0 2 4 y = \u2212x2 \u22128 \u22122 0 \u22122 \u22128 2 x 01 y = x\u22124 \u22124 \u22123 b) T\u00ecm t\u1ecda \u0111\u1ed9 giao \u0111i\u1ec3m c\u1ee7a (P) v\u00e0 (d) b\u1eb1ng ph\u00e9p t\u00ednh. Ph\u01b0\u01a1ng tr\u00ecnh ho\u00e0nh \u0111\u1ed9 giao \u0111i\u1ec3m c\u1ee7a (P) v\u00e0 (d) : \u2212x2 = x \u2212 4 2 \uf0db x2 + 2x \u2212 8 = 0 \uf0db \uf0e9x = 2 \uf0ea\uf0ebx = \u22124 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 Thay x = 2 v\u00e0o y = \u2212x2 , ta \u0111\u01b0\u1ee3c: y = \u221222 = \u22122 . 22 Thay x = \u22124 v\u00e0o y = \u2212x2 , ta \u0111\u01b0\u1ee3c: y = \u2212(\u22124)2 = \u22128 . 22 V\u1eady (2; \u2212 2) , (\u22124; \u2212 8) l\u00e0 hai giao \u0111i\u1ec3m c\u1ea7n t\u00ecm. C\u00e2u 2. (1 \u0111i\u1ec3m) Cho ph\u01b0\u01a1ng tr\u00ecnh 2x2 \u2212 5x \u2212 7 = 0 c\u00f3 2 nghi\u1ec7m l\u00e0 x1 ,x2 . Kh\u00f4ng gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh, h\u00e3y t\u00ednh gi\u00e1 tr\u1ecb c\u1ee7a bi\u1ec3u th\u1ee9c M = x12 + x22 . x2 x1 L\u1eddi gi\u1ea3i V\u00ec \uf044 = b2 \u2212 4ac = (\u22125)2 \u2212 4.2.(\u22127) = 81 \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 = 5 \uf0ed = x1 a2 Theo \u0111\u1ecbnh l\u00ed Vi-et, ta c\u00f3: \uf0ef\uf0ef\uf0eeP c = \u22127 a .x2 = 2 Ta c\u00f3: M = x12 + x22 x2 x1 ( )( )M = x13 + x23 = x1 + x2 x12 + x22 \u2212 x1x2 x2 x1 x2 x1 ( )S S2 \u2212 3P = \u2212 335 = P 28 C\u00e2u 3. (1 \u0111i\u1ec3m) M\u1ed9t b\u00ecnh x\u0103ng c\u1ee7a m\u1ed9t \u00f4 t\u00f4 ch\u1ee9a \u0111\u01b0\u1ee3c 50 l\u00edt. Bi\u1ebft r\u1eb1ng trung b\u00ecnh c\u1ee9 \u0111i 9 km th\u00ec h\u1ebft 1 l\u00edt x\u0103ng.Bi\u1ebft r\u1eb1ng m\u1ed1i li\u00ean h\u1ec7 gi\u1eefa y (l\u00edt x\u0103ng) c\u00f2n l\u1ea1i trong b\u00ecnh v\u00e0 qu\u00e3ng \u0111\u01b0\u1eddng x (km) l\u00e0 m\u1ed9t h\u00e0m s\u1ed1 b\u1eadc nh\u1ea5t y = ax + b . a)H\u00e3y vi\u1ebft h\u00e3y x\u00e1c \u0111\u1ecbnh h\u1ec7 s\u1ed1 a v\u00e0b . b)Khi x\u0103ng trong b\u00ecnh c\u00f2n l\u1ea1i 4 l\u00edt x\u0103ng th\u00ec \u00f4 t\u00f4 d\u1eebng l\u1ea1i \u0111\u1ed5 x\u0103ng . H\u1ecfi khi \u0111\u00f3 \u00f4 t\u00f4 \u0111i \u0111\u01b0\u1ee3c qu\u00e3ng \u0111\u01b0\u1eddng l\u00e0 bao nhi\u00eau?. 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: V\u1edbi \uf0ecx =0 \uf0de 50 = 0.a + b . (1) \uf0ed\uf0eey = 50 T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N 4","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH V\u1edbi \uf0ecx = 9 \u22121 = 49 \uf0de 49 = 9.a + b . (2) \uf0ed\uf0eep = 50 T\u1eeb (1) v\u00e0 (2) ta c\u00f3 h\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh: \uf0ec0a + b = 50 \uf0db \uf0ec\uf0efa = \u22121 \uf0ed\uf0ee9a + b = 49 \uf0ed = 9 \uf0ef\uf0eeb 50 V\u1eady: a = \u2212 1 , b = 50 v\u00e0 y = \u2212 1 x + 50 . 99 b) Khi x\u0103ng trong b\u00ecnh c\u00f2n l\u1ea1i 4 l\u00edt x\u0103ng th\u00ec \u00f4 t\u00f4 d\u1eebng l\u1ea1i \u0111\u1ed5 x\u0103ng . H\u1ecfi khi \u0111\u00f3 \u00f4 t\u00f4 \u0111i \u0111\u01b0\u1ee3c qu\u00e3ng \u0111\u01b0\u1eddng l\u00e0 bao nhi\u00eau? Khi x\u0103ng trong b\u00ecnh c\u00f2n l\u1ea1i 4 l\u00edt x\u0103ng : 4 = \u2212 1 .x + 50 \uf0de x = 414 (km). 9 V\u1eady khi \u0111\u00f3 \u00f4 t\u00f4 \u0111i \u0111\u01b0\u1ee3c qu\u00e3ng \u0111\u01b0\u1eddng l\u00e0 x = 414 (km). C\u00e2u 4. (0,75 \u0111i\u1ec3m). ). Nh\u00e2n d\u1ecbp trung thu m\u1ed9t c\u1eeda h\u00e0ng b\u00e1n b\u00e1nh k\u1eb9o \u0111\u01b0a ra h\u00ecnh th\u1ee9c khuy\u1ebfn m\u00e3i cho m\u1ed9t lo\u1ea1i b\u00e1nh A \u0111ang c\u00f3 gi\u00e1 b\u00e1n l\u00e0 120 000\u0111\/h\u1ed9p nh\u01b0 sau: H\u00ecnh th\u1ee9c khuy\u1ebfn m\u00e3i 1 : Mua 3 h\u1ed9p \u0111\u1ea7u gi\u00e1 120000 \u0111\/h\u1ed9p , t\u1eeb h\u1ed9p th\u1ee9 t\u01b0 tr\u1edf \u0111i m\u1ed7i h\u1ed9p gi\u1ea3m 30% H\u00ecnh th\u1ee9c khuy\u1ebfn m\u00e3i 2 : Mua 3 t\u1eb7ng 1 B\u1ea1n Lan c\u1ea7n mua gi\u00fap cho m\u1eb9 9 h\u1ed9p b\u00e1nh A \u0111\u1ec3 l\u00e0m qu\u00e0. Em h\u00e3y t\u00ednh gi\u00fap b\u1ea1n Lan n\u00ean ch\u1ecdn h\u00ecnh th\u1ee9c khuy\u1ebfn m\u00e3i n\u00e0o th\u00ec c\u00f3 l\u1ee3i h\u01a1n?( Tr\u1ea3 ti\u1ec1n \u00edt h\u01a1n) L\u1eddi gi\u1ea3i H\u00ecnh th\u1ee9c khuy\u1ebfn m\u00e3i 1 : Mua 3 h\u1ed9p \u0111\u1ea7u gi\u00e1 120000 \u0111\/h\u1ed9p , t\u1eeb h\u1ed9p th\u1ee9 t\u01b0 tr\u1edf \u0111i m\u1ed7i h\u1ed9p gi\u1ea3m 30% n\u00ean s\u1ed1 ti\u1ec1n ph\u1ea3i tr\u1ea3 l\u00e0 : 3.120000 (9 3).120000.(1 30%) 864000 (\u0111\u1ed3ng) H\u00ecnh th\u1ee9c khuy\u1ebfn m\u00e3i 2 : Mua 3 t\u1eb7ng 1 n\u00ean s\u1ed1 ti\u1ec1n ph\u1ea3i tr\u1ea3 l\u00e0 : (6 1).120000 840000 (\u0111\u1ed3ng) V\u1eady Lan n\u00ean mua theo h\u00ecnh th\u1ee9c khuy\u1ebfn m\u00e3i 2 v\u00ec 840000 864000 . C\u00e2u 5. (1 \u0111i\u1ec3m) Nh\u1eefng n\u0103m d\u01b0\u01a1ng l\u1ecbch n\u00e0o chia h\u1ebft cho 4 th\u00ec n\u0103m \u0111\u00f3 l\u00e0 n\u0103m nhu\u1eadn. V\u00ed d\u1ee5: 2016 chia h\u1ebft cho 4 n\u00ean n\u0103m 2016 l\u00e0 n\u0103m nhu\u1eadn. Ngo\u00e0i ra, v\u1edbi nh\u1eefng n\u0103m tr\u00f2n th\u1ebf k\u1ef7 (nh\u1eefng n\u0103m c\u00f3 2 s\u1ed1 cu\u1ed1i l\u00e0 s\u1ed1 0 ) th\u00ec c\u00e1c b\u1ea1n l\u1ea5y s\u1ed1 n\u0103m chia cho 400, n\u1ebfu chia h\u1ebft th\u00ec n\u0103m \u0111\u00f3 l\u00e0 n\u0103m c\u00f3 nhu\u1eadn (ho\u1eb7c 2 s\u1ed1 \u0111\u1ea7u trong n\u0103m chia h\u1ebft cho 4). a) Em h\u00e3y cho bi\u1ebft n\u0103m 2023 c\u00f3 ph\u1ea3i l\u00e0 n\u0103m nhu\u1eadn d\u01b0\u01a1ng l\u1ecbch kh\u00f4ng ? Gi\u1ea3i th\u00edch? b) Ban An sinh nh\u1eadt l\u1ea7n th\u1ee9 14 v\u00e0o th\u1ee9 b\u1ea3y ng\u00e0y 8 th\u00e1ng t\u01b0 n\u0103m 2023 . H\u1ecfi sinh nh\u1eadt l\u1ea7n th\u1ee9 15 c\u1ee7a b\u1ea1n v\u00e0o th\u1ee9 m\u1ea5y? T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N 5","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH L\u1eddi gi\u1ea3i a) N\u0103m 2023 c\u00f3 2 s\u1ed1 t\u1eadn c\u00f9ng 23 kh\u00f4ng chia h\u1ebft cho 4 n\u00ean kh\u00f4ng l\u00e0 n\u0103m nhu\u1eadn d\u01b0\u01a1ng l\u1ecbch. b) Sinh nh\u1eadt l\u1ea7n th\u1ee9 15 c\u1ee7a b\u1ea1n An v\u00e0o n\u0103m 2024 c\u00f3 2 s\u1ed1 t\u1eadn c\u00f9ng l\u00e0 24 chia h\u1ebft cho 4 n\u00ean l\u00e0 n\u0103m nhu\u1eadn. V\u1eady t\u1eeb sinh nh\u1eadt th\u1ee9 14 \u0111\u1ebfn th\u1ee9 15 l\u00e0 366 ng\u00e0y. X\u00e9t 366 : 7 52 R 2 n\u00ean ng\u00e0y sinh nh\u1eadt th\u1ee9 15 l\u00e0 th\u1ee9 hai. C\u00e2u 6. (1 \u0111i\u1ec3m) M\u1ed9t ng\u01b0\u1eddi \u0111o \u0111\u1ea1c chi\u1ec1u cao c\u1ee7a m\u1ed9t ng\u1ecdn n\u00fai t\u1ea1i hai \u0111i\u1ec3m A,B tr\u00ean m\u1eb7t \u0111\u1ea5t. K\u1ebft qu\u1ea3 \u0111\u01b0\u1ee3c cho trong h\u00ecnh v\u1ebd. T\u00ednh chi\u1ec1u cao c\u1ee7a ng\u1ecdn n\u00fai.( L\u00e0m tr\u00f2n \u0111\u1ebfn 1 ch\u1eef s\u1ed1 th\u1eadp ph\u00e2n). L\u1eddi gi\u1ea3i \u00c1p d\u1ee5ng h\u1ec7 th\u1ee9c l\u01b0\u1ee3ng cho c\u00e1c tam gi\u00e1c vu\u00f4ng AHC, BHC , ta c\u00f3: AC = HC.cot (33,7\uf0b0), BC = HC.cot (41,6\uf0b0) HA \u2212 HB = 400 \uf0db h.cot (33,7\uf0b0) \u2212 h.cot (41,6\uf0b0) = 400 \uf0de Chi\u1ec1u cao ng\u1ecdn n\u00fai: h 400 1072,1(m) cot 33, 7 cot 41, 6 T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N 6","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH C\u00e2u 7. (1 \u0111i\u1ec3m) M\u1ed9t th\u00f9ng \u0111\u1ef1ng n\u01b0\u1edbc c\u00f3 d\u1ea1ng h\u00ecnh tr\u1ee5 chi\u1ec1u cao l\u00e0 35 cm \u0111\u01b0\u1eddng k\u00ednh \u0111\u00e1y 30 cm . a)T\u00ednh th\u1ec3 t\u00edch c\u1ee7a th\u00f9ng. b)Ng\u01b0\u1eddi ta s\u1eed d\u1ee5ng th\u00f9ng tr\u00ean \u0111\u1ec3 m\u00fac n\u01b0\u1edbc \u0111\u1ed5 v\u00e0o m\u1ed9t b\u1ec3 ch\u1ee9a c\u00f3 dung t\u00edch 1 m3 . H\u1ecfi c\u1ea7n ph\u1ea3i \u0111\u1ed5 \u00edt nh\u1ea5t bao nhi\u00eau th\u00f9ng th\u00ec \u0111\u1ea7y b\u1ec3 ch\u1ee9a ? Bi\u1ebft r\u1eb1ng m\u1ed7i l\u1ea7n x\u00e1ch ng\u01b0\u1eddi ta ch\u1ec9 \u0111\u1ed5 \u0111\u1ea7y 90% th\u00f9ng \u0111\u1ec3 n\u01b0\u1edbc kh\u00f4ng \u0111\u1ed5 ra ngo\u00e0i. L\u1eddi gi\u1ea3i a) B\u00e1n k\u00ednh \u0111\u00e1y h\u00ecnh tr\u1ee5 l\u00e0 R = 30 : 2 =15(cm) . ( )Th\u1ec3 t\u00edch tr\u1ee5: V = \uf070R2h = \uf070.152.35 = 7875\uf070 \uf0bb 24740 cm3 ( ) ( )b) Th\u1ec3 t\u00edch n\u01b0\u1edbc m\u1ed7i l\u1ea7n x\u00e1ch l\u00e0: 24740.90% = 22266 cm3 = 0, 022266 cm3 . S\u1ed1 th\u00f9ng \u00edt nh\u1ea5t c\u1ea7n \u0111\u1ed5 \u0111\u1ec3 \u0111\u1ea7y b\u1ec3 l\u00e0:1: 0, 022266 = 44,91 n\u00ean s\u1ed1 th\u00f9ng c\u1ea7n l\u00e0 50 th\u00f9ng. C\u00e2u 8. (3 \u0111i\u1ec3m) Cho tam gi\u00e1c ABC nh\u1ecdn (AB \uf03c AC) n\u1ed9i ti\u1ebfp (O) . C\u00e1c \u0111\u01b0\u1eddng cao AD, BE,CF c\u1eaft nhau t\u1ea1i H . Tia EF c\u1eaft tia CB t\u1ea1i K . a)Ch\u1ee9ng minh t\u1ee9 gi\u00e1c BFEC n\u1ed9i ti\u1ebfp v\u00e0 KF.KE = KB.KC . b)\u0110\u01b0\u1eddng th\u1eb3ng KA c\u1eaft (O) t\u1ea1i M . Ch\u1ee9ng minh t\u1ee9 gi\u00e1c AEFM n\u1ed9i ti\u1ebfp. c)G\u1ecdi N l\u00e0 trung \u0111i\u1ec3m c\u1ee7a BC . Ch\u1ee9ng minh MN, BE,CF \u0111\u1ed3ng quy( c\u00f9ng \u0111i qua m\u1ed9t \u0111i\u1ec3m ). T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N 7"]