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 c) Ch\u1ee9ng minh BC = AB + AC . DH DI DK X\u00e9t \uf044AID v\u00e0 \uf044CHD , ta c\u00f3: IAD = HCD ( 2 gnt (O) c\u00f9ng ch\u1eafn BD ) AID = CHD(= 90\uf0b0) \uf0de \uf044AID \uf044CHD (g.g). \uf0de AI = ID (ts\u0111d) \uf0de AI = CH (1) CH DH ID DH X\u00e9t \uf044AKD v\u00e0 \uf044BHD , ta c\u00f3: DAK = DBH ( 2 gnt (O) c\u00f9ng ch\u1eafn DC ) AKD = BHD(= 90\uf0b0) \uf0de \uf044AKD \uf044BHD (g.g). \uf0de AK = DK (ts\u0111d) \uf0de AK = BH (2) BH DH DK DH X\u00e9t \uf044BID v\u00e0 \uf044CKD , ta c\u00f3: DBI = DCK (T\u1ee9 gi\u00e1c ABDC n\u1ed9i ti\u1ebfp) DIB = DKC (= 90\uf0b0) \uf0de \uf044BID \uf044CKD (g.g). \uf0de BI = ID (ts\u0111d) \uf0de BI = CK (2) CK BK ID BK C\u1ed9ng (1) v\u00e0 (2) theo v\u1ebf, ta \u0111\u01b0\u1ee3c: AI + AK = CH + BH ID DK DH DH \uf0de AB + BI + AC \u2212 KC = BC ID DK DH \uf0de AB + BI + AC \u2212 KC = BC ID ID DK DK DH \uf0de AB + AC = BC \uf0e6 BI = KC \uf0f6 ID DK DH \uf0e7\uf0e8 ID DK \uf0f7\uf0f8 ----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 S\u1ede GD \u2013 \u0110T TP H\u1ed2 CH\u00cd MINH S\u00d4\u00db GD&\u00d1T TP HO\u00c0 CH\u00cd MINH \u00d1E\u00c0 THAM KHA\u00dbO TUYE\u00c5N SINH 10 PHO\u00d8NG G\u00d1&\u00d1T QUA\u00c4N PHU\u00d9 NHUA\u00c4N NA\u00caM HO\u00cfC: 2023 - 2024 \u0110\u1ec0 THAM KH\u1ea2O M\u00d4N: TO\u00c1N 9 M\u00c3 \u0110\u1ec0: Qu\u1eadn Ph\u00fa Nhu\u1eadn - 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 hai h\u00e0m s\u1ed1: y = x2 v\u00e0 y = x + 4 c\u00f3 \u0111\u1ed3 th\u1ecb l\u1ea7n l\u01b0\u1ee3t l\u00e0 (P) v\u00e0 (D) . 2 a) V\u1ebd (P) v\u00e0 (D) tr\u00ean c\u00f9ng m\u1ed9t m\u1eb7t ph\u1eb3ng t\u1ecda \u0111\u1ed9 Oxy . 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: \u2212x2 + 9x = 5 . G\u1ecdi x1 v\u00e0 x2 l\u00e0 hai nghi\u1ec7m n\u1ebfu c\u00f3 c\u1ee7a ph\u01b0\u01a1ng 3 tr\u00ecnh. T\u00ednh gi\u00e1 tr\u1ecb c\u1ee7a bi\u1ec3u th\u1ee9c A = x12 + x22 \u2212 x1x2 . 4 L\u01b0u \u00fd: T\u1eeb b\u00e0i n\u00e0y, c\u00e1c s\u1ed1 li\u1ec7u t\u00ednh to\u00e1n v\u1ec1 \u0111\u1ed9 d\u00e0i khi l\u00e0m tr\u00f2n (n\u1ebfu c\u00f3) l\u1ea5y \u0111\u1ebfn m\u1ed9t ch\u1eef s\u1ed1 th\u1eadp ph\u00e2n, s\u1ed1 \u0111o g\u00f3c l\u00e0m tr\u00f2n \u0111\u1ebfn ph\u00fat. C\u00e2u 3. (1 \u0111i\u1ec3m). Nh\u1eb1m gi\u00fap b\u00e0 con n\u00f4ng d\u00e2n c\u00e1c t\u1ec9nh Mi\u1ec1n Trung kh\u00f4i ph\u1ee5c s\u1ea3n su\u1ea5t n\u00f4ng nghi\u1ec7p \u1ed5n \u0111\u1ecbnh cu\u1ed9c s\u1ed1ng sau \u0111\u1ee3t b\u00e3o l\u0169, ng\u00e2n h\u00e0ng AGRIBANK cho vay v\u1ed1n \u01b0u \u0111\u00e3i v\u1edbi l\u00e3i su\u1ea5t 5% \/n\u0103m. B\u00e1c H\u00e0 \u0111\u00e3 vay 100 tri\u1ec7u \u0111\u1ed3ng l\u00e0m v\u1ed1n ch\u0103n nu\u00f4i g\u00e0 ta th\u1ea3 v\u01b0\u1eddn. B\u00e1c H\u00e0 \u0111\u00e3 nu\u00f4i \u0111\u01b0\u1ee3c hai l\u1ee9a g\u00e0 trong m\u1ed9t n\u0103m, l\u1ee9a th\u1ee9 nh\u1ea5t b\u00e1c H\u00e0 l\u00e3i \u0111\u01b0\u1ee3c 42% so v\u1edbi v\u1ed1n b\u1ecf ra. V\u00ec th\u1ea5y c\u00f4ng vi\u1ec7c ch\u0103n nu\u00f4i thu\u1eadn l\u1ee3i, b\u00e1c H\u00e0 d\u1ed3n c\u1ea3 v\u1ed1n l\u1eabn l\u00e3i c\u1ee7a \u0111\u1ee3t nu\u00f4i l\u1ee9a g\u00e0 th\u1ee9 nh\u1ea5t \u0111\u1ec3 \u0111\u1ea7u t\u01b0 v\u00e0o nu\u00f4i ti\u1ebfp l\u1ee9a g\u00e0 th\u1ee9 hai. Sau \u0111\u1ee3t nu\u00f4i th\u1ee9 hai, nh\u1edd c\u00f3 kinh nghi\u1ec7m t\u1eeb l\u1ee9a th\u1ee9 nh\u1ea5t b\u00e1c H\u00e0 \u0111\u00e3 l\u00e3i \u0111\u01b0\u1ee3c 50% so v\u1edbi v\u1ed1n b\u1ecf ra. H\u1ecfi sau m\u1ed9t n\u0103m, qua hai \u0111\u1ee3t ch\u0103n nu\u00f4i g\u00e0 ta th\u1ea3 v\u01b0\u1eddn, b\u00e1c H\u00e0 l\u00e3i \u0111\u01b0\u1ee3c bao nhi\u00eau ti\u1ec1n? C\u00e2u 4. (0,75 \u0111i\u1ec3m). C\u00f4ng ty A th\u1ef1c hi\u1ec7n m\u1ed9t cu\u1ed9c kh\u1ea3o s\u00e1t \u0111\u1ec3 t\u00ecm hi\u1ec3u v\u1ec1 m\u1ed1i li\u00ean h\u1ec7 gi\u1eefa y (s\u1ea3n ph\u1ea9m) l\u00e0 s\u1ed1 l\u01b0\u1ee3ng s\u1ea3n ph\u1ea9m b\u00e1n ra v\u1edbi x (ngh\u00ecn \u0111\u1ed3ng) l\u00e0 gi\u00e1 b\u00e1n ra c\u1ee7a m\u1ed7i s\u1ea3n ph\u1ea9m v\u00e0 nh\u1eadn th\u1ea5y r\u1eb1ng y = ax + b (v\u1edbi a,b l\u00e0 h\u1eb1ng s\u1ed1). Bi\u1ebft r\u1eb1ng: v\u1edbi gi\u00e1 b\u00e1n l\u00e0 400 ngh\u00ecn \u0111\u1ed3ng\/s\u1ea3n ph\u1ea9m th\u00ec s\u1ed1 l\u01b0\u1ee3ng s\u1ea3n ph\u1ea9m b\u00e1n ra l\u00e0 1200 s\u1ea3n ph\u1ea9m; v\u1edbi gi\u00e1 b\u00e1n l\u00e0 460 ngh\u00ecn \u0111\u1ed3ng\/s\u1ea3n ph\u1ea9m th\u00ec s\u1ed1 l\u01b0\u1ee3ng s\u1ea3n ph\u1ea9m b\u00e1n ra l\u00e0 1800 s\u1ea3n ph\u1ea9m. a) X\u00e1c \u0111\u1ecbnh c\u00e1c h\u1ec7 s\u1ed1 a v\u00e0 b . b) B\u1eb1ng ph\u00e9p t\u00ednh, h\u00e3y t\u00ednh s\u1ed1 l\u01b0\u1ee3ng s\u1ea3n ph\u1ea9m b\u00e1n ra v\u1edbi gi\u00e1 b\u00e1n l\u00e0 440 ngh\u00ecn \u0111\u1ed3ng\/s\u1ea3n ph\u00e2m? 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 S\u1ede GD \u2013 \u0110T TP H\u1ed2 CH\u00cd MINH C\u00e2u 5. (0,75 \u0111i\u1ec3m). S\u1ea3n l\u01b0\u1ee3ng c\u1ee7a m\u1ed9t x\u00ed nghi\u1ec7p trong 3 qu\u00fd c\u1ee7a n\u0103m 2020 c\u00f3 k\u1ebft qu\u1ea3 nh\u01b0 sau: Qu\u00fd Hai c\u00f3 s\u1ea3n l\u01b0\u1ee3ng \u00edt h\u01a1n 20% so v\u1edbi Qu\u00fd M\u1ed9t; s\u1ea3n l\u01b0\u1ee3ng c\u1ee7a Qu\u00fd Ba \u0111\u1ea1t nhi\u1ec1u h\u01a1n 8% so v\u1edbi Qu\u00fd M\u1ed9t. Nh\u01b0 v\u1eady, s\u1ea3n l\u01b0\u1ee3ng c\u1ee7a Qu\u00fd Ba t\u0103ng bao nhi\u00eau ph\u1ea7n tr\u0103m so v\u1edbi Qu\u00fd Hai? C\u00e2u 6. (1,5 di\u1ec3m). X\u00fac x\u1eafc hay c\u00f2n g\u1ecdi l\u00e0 x\u00ed ng\u1ea7u l\u00e0 m\u1ed9t kh\u1ed1i nh\u1ecf h\u00ecnh l\u1eadp ph\u01b0\u01a1ng \u0111\u01b0\u1ee3c \u0111\u00e1nh d\u1ea5u ch\u1ea5m tr\u00f2n v\u1edbi s\u1ed1 l\u01b0\u1ee3ng t\u1eeb m\u1ed9t \u0111\u1ebfn s\u00e1u cho c\u1ea3 s\u00e1u m\u1eb7t. Hai vi\u00ean x\u00fac x\u1eafc h\u00ecnh l\u1eadp ph\u01b0\u01a1ng \u0111\u01b0\u1ee3c l\u00e0m b\u1eb1ng g\u1ed7 c\u00f3 t\u1ed5ng di\u1ec7n t\u00edch to\u00e0n ph\u1ea7n l\u00e0 23,52 cm2 . a) T\u00ednh kh\u1ed1i l\u01b0\u1ee3ng c\u1ee7a hai vi\u00ean x\u00fac x\u1eafc? Cho bi\u1ebft: kh\u1ed1i l\u01b0\u1ee3ng m = V .D , trong \u0111\u00f3 V l\u00e0 th\u1ec3 t\u00edch v\u00e0 kh\u1ed1i l\u01b0\u1ee3ng ri\u00eang c\u1ee7a g\u1ed7 l\u00e0 D = 0,8 gam \/ cm3 . b) Ng\u01b0\u1eddi ta gieo hai vi\u00ean x\u00fac x\u1eafc tr\u00ean c\u00f9ng m\u1ed9t l\u1ea7n. H\u1ecfi x\u00e1c xu\u1ea5t \u0111\u1ec3 xu\u1ea5t hi\u1ec7n hai m\u1eb7t gi\u1ed1ng nhau l\u00e0 bao nhi\u00eau ph\u1ea7n tr\u0103m? K\u1ebft qu\u1ea3 l\u00e0m tr\u00f2n \u0111\u1ebfn ch\u1eef s\u1ed1 th\u1eadp ph\u00e2n th\u1ee9 nh\u1ea5t. ( )C\u00e2u 7. (0,5 \u0111i\u1ec3m) M\u1ed9t xe \u00f4t\u00f4 chuy\u1ec3n \u0111\u1ed9ng theo h\u00e0m s\u1ed1 S = 30.t + 4.t2 , trong \u0111\u00f3: S km2 l\u00e0 qu\u00e3ng \u0111\u01b0\u1eddng xe \u0111i \u0111\u01b0\u1ee3c; t (gi\u1edd) l\u00e0 th\u1eddi gian chuy\u1ec3n \u0111\u1ed9ng c\u1ee7a xe t\u00ednh t\u1eeb l\u00fac 7 gi\u1edd s\u00e1ng. Xem nh\u01b0 xe chuy\u1ec3n \u0111\u1ed9ng \u0111\u1ec1u tr\u00ean m\u1ed9t \u0111o\u1ea1n \u0111\u01b0\u1eddng th\u1eb3ng v\u00e0 kh\u00f4ng ngh\u1ec9. a) H\u1ecfi t\u1eeb l\u00fac 7 gi\u1edd 30 ph\u00fat \u0111\u1ebfn l\u00fac 8 gi\u1edd 15 ph\u00fat xe \u0111\u00e3 \u0111i \u0111\u01b0\u1ee3c qu\u00e3ng \u0111\u01b0\u1eddng d\u00e0i bao nhi\u00eau km ? b) \u0110\u1ebfn l\u00fac m\u1ea5y gi\u1edd th\u00ec xe \u0111i \u0111\u01b0\u1ee3c qu\u00e3ng \u0111\u01b0\u1eddng d\u00e0i 34 km (t\u00ednh t\u1eeb l\u00fac 7 gi\u1edd s\u00e1ng)? C\u00e2u 8. (3 \u0111i\u1ec3m)Cho \u0111\u01b0\u1eddng tr\u00f2n (O; R) v\u00e0 \u0111i\u1ec3m A n\u1eb1m ngo\u00e0i \u0111\u01b0\u1eddng tr\u00f2n (O) . V\u1ebd hai ti\u1ebfp tuy\u1ebfn AB , AC c\u1ee7a (O) (v\u1edbi B , C l\u00e0 ti\u1ebfp \u0111i\u1ec3m). V\u1ebd c\u00e1t tuy\u1ebfn ADE c\u1ee7a (O) (v\u1edbi D , E thu\u1ed9c (O) ; D n\u1eb1m gi\u1eefa A v\u00e0 E ; tia AD n\u1eb1m gi\u1eefa hai tia AB v\u00e0 AO ). a) Ch\u1ee9ng minh t\u1ee9 gi\u00e1c OBAC n\u1ed9i ti\u1ebfp v\u00e0 AB2 = AD.AE . b) G\u1ecdi H l\u00e0 giao \u0111i\u1ec3m c\u1ee7a OA v\u00e0 BC . Ch\u1ee9ng minh BD.CE = BE.CD v\u00e0 t\u1ee9 gi\u00e1c DEOH n\u1ed9i ti\u1ebfp. c) \u0110\u01b0\u1eddng th\u1eb3ng AO c\u1eaft \u0111\u01b0\u1eddng tr\u00f2n (O) t\u1ea1i M v\u00e0 N ( M n\u1eb1m gi\u1eefa A v\u00e0 O ). Ch\u1ee9ng minh: HC2 = HD.HE v\u00e0 EH.AD = MH.AN . ----H\u1ebeT--- T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N 2","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 S\u1ede GD \u2013 \u0110T TP H\u1ed2 CH\u00cd MINH H\u01af\u1edaNG D\u1eaaN GI\u1ea2I C\u00e2u 1. (1,5 \u0111i\u1ec3m). Cho hai h\u00e0m s\u1ed1: y = x2 v\u00e0 y = x + 4 c\u00f3 \u0111\u1ed3 th\u1ecb l\u1ea7n l\u01b0\u1ee3t l\u00e0 (P) v\u00e0 (D) . 2 a) V\u1ebd (P) v\u00e0 (D) tr\u00ean c\u00f9ng m\u1ed9t m\u1eb7t ph\u1eb3ng t\u1ecda \u0111\u1ed9 Oxy . 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 \u22122 0 y=x+4 2 4 b) T\u00ecm t\u1ecda \u0111\u1ed9 giao \u0111i\u1ec3m c\u1ee7a (P) v\u00e0 (d) b\u1eb1ng ph\u00e9p t\u00ednh. Ph\u01b0\u01a1ng tr\u00ecnh ho\u00e0nh \u0111\u1ed9 giao \u0111i\u1ec3m c\u1ee7a (P) v\u00e0 (d) : x2 = x + 4 2 \uf0db x2 \u2212 2x \u2212 8 = 0 \uf0db \uf0e9x = 4 \uf0ea\uf0ebx = \u22122 Thay x = 4 v\u00e0o y = x2 , ta \u0111\u01b0\u1ee3c: y = 4 2 = 8 . 22 Thay x = \u22122 v\u00e0o y = x2 , ta \u0111\u01b0\u1ee3c: y = (\u22122)2 = 2 . 22 V\u1eady (4; 8) , (\u22122; 2) l\u00e0 hai giao \u0111i\u1ec3m c\u1ea7n t\u00ecm. C\u00e2u 2. (1 \u0111i\u1ec3m). Cho ph\u01b0\u01a1ng tr\u00ecnh: \u2212x2 + 9x = 5 . G\u1ecdi x1 v\u00e0 x2 l\u00e0 hai nghi\u1ec7m n\u1ebfu c\u00f3 c\u1ee7a ph\u01b0\u01a1ng 3 tr\u00ecnh. T\u00ednh gi\u00e1 tr\u1ecb c\u1ee7a bi\u1ec3u th\u1ee9c A = x12 + x22 \u2212 x1x2 . 4 L\u1eddi gi\u1ea3i T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N 3","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 S\u1ede GD \u2013 \u0110T TP H\u1ed2 CH\u00cd MINH Ta c\u00f3: \u2212x2 + 9x = 5 \uf0db \u2212 1 x2 + 9x \u2212 5 = 0 33 V\u00ec \uf044 = b2 \u2212 4ac = 92 \u2212 4.\uf0e6\uf0e7 \u2212 1 \uf0f6\uf0f7.(\u22125) = 223 \uf03e0 \uf0e8 3 3 \uf0f8 N\u00ean ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m ph\u00e2n bi\u1ec7t x1 , x2 . \uf0ef\uf0ef\uf0ecS = x1 + x2 = \u2212b = 27 \uf0ed = x1 a Theo \u0111\u1ecbnh l\u00ed Vi-et, ta c\u00f3: \uf0ef\uf0ee\uf0efP c = 15 a .x2 = Ta c\u00f3: A = x12 + x22 \u2212 x1x2 4 A = S2 \u2212 2P \u2212 P = S2 \u2212 9P = 272 \u2212 9 .15 = 2781 = 3 309 . 44 4 42 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). Nh\u1eb1m gi\u00fap b\u00e0 con n\u00f4ng d\u00e2n c\u00e1c t\u1ec9nh Mi\u1ec1n Trung kh\u00f4i ph\u1ee5c s\u1ea3n su\u1ea5t n\u00f4ng nghi\u1ec7p \u1ed5n \u0111\u1ecbnh cu\u1ed9c s\u1ed1ng sau \u0111\u1ee3t b\u00e3o l\u0169, ng\u00e2n h\u00e0ng AGRIBANK cho vay v\u1ed1n \u01b0u \u0111\u00e3i v\u1edbi l\u00e3i su\u1ea5t 5% \/n\u0103m. B\u00e1c H\u00e0 \u0111\u00e3 vay 100 tri\u1ec7u \u0111\u1ed3ng l\u00e0m v\u1ed1n ch\u0103n nu\u00f4i g\u00e0 ta th\u1ea3 v\u01b0\u1eddn. B\u00e1c H\u00e0 \u0111\u00e3 nu\u00f4i \u0111\u01b0\u1ee3c hai l\u1ee9a g\u00e0 trong m\u1ed9t n\u0103m, l\u1ee9a th\u1ee9 nh\u1ea5t b\u00e1c H\u00e0 l\u00e3i \u0111\u01b0\u1ee3c 42% so v\u1edbi v\u1ed1n b\u1ecf ra. V\u00ec th\u1ea5y c\u00f4ng vi\u1ec7c ch\u0103n nu\u00f4i thu\u1eadn l\u1ee3i, b\u00e1c H\u00e0 d\u1ed3n c\u1ea3 v\u1ed1n l\u1eabn l\u00e3i c\u1ee7a \u0111\u1ee3t nu\u00f4i l\u1ee9a g\u00e0 th\u1ee9 nh\u1ea5t \u0111\u1ec3 \u0111\u1ea7u t\u01b0 v\u00e0o nu\u00f4i ti\u1ebfp l\u1ee9a g\u00e0 th\u1ee9 hai. Sau \u0111\u1ee3t nu\u00f4i th\u1ee9 hai, nh\u1edd c\u00f3 kinh nghi\u1ec7m t\u1eeb l\u1ee9a th\u1ee9 nh\u1ea5t b\u00e1c H\u00e0 \u0111\u00e3 l\u00e3i \u0111\u01b0\u1ee3c 50% so v\u1edbi v\u1ed1n b\u1ecf ra. H\u1ecfi sau m\u1ed9t n\u0103m, qua hai \u0111\u1ee3t ch\u0103n nu\u00f4i g\u00e0 ta th\u1ea3 v\u01b0\u1eddn, b\u00e1c H\u00e0 l\u00e3i \u0111\u01b0\u1ee3c bao nhi\u00eau ti\u1ec1n? L\u1eddi gi\u1ea3i T\u1ed5ng s\u1ed1 ti\u1ec1n b\u00e1c H\u00e0 thu v\u1ec1 khi b\u00e1n to\u00e0n b\u1ed9 l\u1ee9a g\u00e0 th\u1ee9 nh\u1ea5t: 100(1 + 42%) = 142 tri\u1ec7u \u0111\u1ed3ng. T\u1ed5ng s\u1ed1 ti\u1ec1n b\u00e1c H\u00e0 thu v\u1ec1 khi b\u00e1n to\u00e0n b\u1ed9 l\u1ee9a g\u00e0 th\u1ee9 hai: 142(1 + 50%) = 213 tri\u1ec7u \u0111\u1ed3ng. T\u1ed5ng s\u1ed1 ti\u1ec1n b\u00e1c H\u00e0 ph\u1ea3i tr\u1ea3 ng\u00e2n h\u00e0ng sau m\u1ed9t n\u0103m thu\u00ea ti\u1ec1n l\u00e0: 100(1 + 5%) = 105 tri\u1ec7u \u0111\u1ed3ng Sau hai \u0111\u1ee3t nu\u00f4i g\u00e0, s\u1ed1 ti\u1ec1n b\u00e1c H\u00e0 l\u00e3i l\u00e0: 213 \u2212 105 = 108 tri\u1ec7u \u0111\u1ed3ng. C\u00e2u 4. (0,75 \u0111i\u1ec3m). C\u00f4ng ty A th\u1ef1c hi\u1ec7n m\u1ed9t cu\u1ed9c kh\u1ea3o s\u00e1t \u0111\u1ec3 t\u00ecm hi\u1ec3u v\u1ec1 m\u1ed1i li\u00ean h\u1ec7 gi\u1eefa y (s\u1ea3n ph\u1ea9m) l\u00e0 s\u1ed1 l\u01b0\u1ee3ng s\u1ea3n ph\u1ea9m b\u00e1n ra v\u1edbi x (ngh\u00ecn \u0111\u1ed3ng) l\u00e0 gi\u00e1 b\u00e1n ra c\u1ee7a m\u1ed7i s\u1ea3n ph\u1ea9m v\u00e0 nh\u1eadn th\u1ea5y r\u1eb1ng y = ax + b (v\u1edbi a,b l\u00e0 h\u1eb1ng s\u1ed1). Bi\u1ebft r\u1eb1ng: v\u1edbi gi\u00e1 b\u00e1n l\u00e0 400 ngh\u00ecn \u0111\u1ed3ng\/s\u1ea3n 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 S\u1ede GD \u2013 \u0110T TP H\u1ed2 CH\u00cd MINH ph\u1ea9m th\u00ec s\u1ed1 l\u01b0\u1ee3ng s\u1ea3n ph\u1ea9m b\u00e1n ra l\u00e0 1200 s\u1ea3n ph\u1ea9m; v\u1edbi gi\u00e1 b\u00e1n l\u00e0 460 ngh\u00ecn \u0111\u1ed3ng\/s\u1ea3n ph\u1ea9m th\u00ec s\u1ed1 l\u01b0\u1ee3ng s\u1ea3n ph\u1ea9m b\u00e1n ra l\u00e0 1800 s\u1ea3n ph\u1ea9m. a) X\u00e1c \u0111\u1ecbnh c\u00e1c h\u1ec7 s\u1ed1 a v\u00e0 b . b) B\u1eb1ng ph\u00e9p t\u00ednh, h\u00e3y t\u00ednh s\u1ed1 l\u01b0\u1ee3ng s\u1ea3n ph\u1ea9m b\u00e1n ra v\u1edbi gi\u00e1 b\u00e1n l\u00e0 440 ngh\u00ecn \u0111\u1ed3ng\/s\u1ea3n ph\u00e2m? L\u1eddi gi\u1ea3i a) X\u00e1c \u0111\u1ecbnh c\u00e1c h\u1ec7 s\u1ed1 a v\u00e0 b . V\u1edbi \uf0ecx = 400 \uf0de 400a + b = 1200 (1) \uf0ed\uf0eey = 1200 V\u1edbi \uf0ecx = 460 \uf0de 460a + b = 1800 (2) \uf0ed\uf0eey = 1800 T\u1eeb (1) v\u00e0 (2) , ta c\u00f3 h\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh: \uf0ec400a + b = 1200 \uf0db \uf0eca = 10 \uf0ed\uf0ee460a + b = 1800 \uf0ee\uf0edb = \u22122800 V\u1eady a = 10 , b = \u22122800 . b) B\u1eb1ng ph\u00e9p t\u00ednh, h\u00e3y t\u00ednh s\u1ed1 l\u01b0\u1ee3ng s\u1ea3n ph\u1ea9m b\u00e1n ra v\u1edbi gi\u00e1 b\u00e1n l\u00e0 440 ngh\u00ecn \u0111\u1ed3ng\/s\u1ea3n ph\u00e2m? V\u1edbi x = 440 , ta \u0111\u01b0\u1ee3c y = 10.440 \u2212 2800 = 1600 s\u1ea3n ph\u1ea9m. V\u1eady v\u1edbi gi\u00e1 440 ng\u00ecnh \u0111\u1ed3ng th\u00ec c\u1eeda h\u00e0ng b\u00e1n \u0111\u01b0\u1ee3c 1600 s\u1ea3n ph\u1ea9m. C\u00e2u 5. (0,75 \u0111i\u1ec3m). S\u1ea3n l\u01b0\u1ee3ng c\u1ee7a m\u1ed9t x\u00ed nghi\u1ec7p trong 3 qu\u00fd c\u1ee7a n\u0103m 2020 c\u00f3 k\u1ebft qu\u1ea3 nh\u01b0 sau: Qu\u00fd Hai c\u00f3 s\u1ea3n l\u01b0\u1ee3ng \u00edt h\u01a1n 20% so v\u1edbi Qu\u00fd M\u1ed9t; s\u1ea3n l\u01b0\u1ee3ng c\u1ee7a Qu\u00fd Ba \u0111\u1ea1t nhi\u1ec1u h\u01a1n 8% so v\u1edbi Qu\u00fd M\u1ed9t. Nh\u01b0 v\u1eady, s\u1ea3n l\u01b0\u1ee3ng c\u1ee7a Qu\u00fd Ba t\u0103ng bao nhi\u00eau ph\u1ea7n tr\u0103m so v\u1edbi Qu\u00fd Hai? L\u1eddi gi\u1ea3i G\u1ecdi x l\u00e0 s\u1ea3n l\u01b0\u1ee3ng c\u1ee7a x\u00ed nghi\u1ec7p trong qu\u00fd m\u1ed9t (x \uf03e 0) . Suy ra: x(1 \u2212 20%) = 0,8x l\u00e0 s\u1ea3n l\u01b0\u1ee3ng c\u1ee7a x\u00ed nghi\u1ec7p trong qu\u00fd hai. x(1+ 8%) = 1,08x l\u00e0 s\u1ea3n l\u01b0\u1ee3ng c\u1ee7a x\u00ed nghi\u1ec7p trong qu\u00fd Ba. % t\u0103ng s\u1ea3n l\u01b0\u1ee3ng gi\u1eefa qu\u00fd Ba v\u00e0 qu\u00fd Hai l\u00e0: 1,08x \u2212 0,8x .100% = 35% . 0,8x C\u00e2u 6. (1,5 di\u1ec3m). X\u00fac x\u1eafc hay c\u00f2n g\u1ecdi l\u00e0 x\u00ed ng\u1ea7u l\u00e0 m\u1ed9t kh\u1ed1i nh\u1ecf h\u00ecnh l\u1eadp ph\u01b0\u01a1ng \u0111\u01b0\u1ee3c \u0111\u00e1nh d\u1ea5u ch\u1ea5m tr\u00f2n v\u1edbi s\u1ed1 l\u01b0\u1ee3ng t\u1eeb m\u1ed9t \u0111\u1ebfn s\u00e1u cho c\u1ea3 s\u00e1u m\u1eb7t. Hai vi\u00ean x\u00fac x\u1eafc h\u00ecnh l\u1eadp ph\u01b0\u01a1ng \u0111\u01b0\u1ee3c l\u00e0m b\u1eb1ng g\u1ed7 c\u00f3 t\u1ed5ng di\u1ec7n t\u00edch to\u00e0n ph\u1ea7n l\u00e0 23,52 cm2 . 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 S\u1ede GD \u2013 \u0110T TP H\u1ed2 CH\u00cd MINH a) T\u00ednh kh\u1ed1i l\u01b0\u1ee3ng c\u1ee7a hai vi\u00ean x\u00fac x\u1eafc? Cho bi\u1ebft: kh\u1ed1i l\u01b0\u1ee3ng m = V .D , trong \u0111\u00f3 V l\u00e0 th\u1ec3 t\u00edch v\u00e0 kh\u1ed1i l\u01b0\u1ee3ng ri\u00eang c\u1ee7a g\u1ed7 l\u00e0 D = 0,8 gam \/ cm3 . b) Ng\u01b0\u1eddi ta gieo hai vi\u00ean x\u00fac x\u1eafc tr\u00ean c\u00f9ng m\u1ed9t l\u1ea7n. H\u1ecfi x\u00e1c xu\u1ea5t \u0111\u1ec3 xu\u1ea5t hi\u1ec7n hai m\u1eb7t gi\u1ed1ng nhau l\u00e0 bao nhi\u00eau ph\u1ea7n tr\u0103m? K\u1ebft qu\u1ea3 l\u00e0m tr\u00f2n \u0111\u1ebfn ch\u1eef s\u1ed1 th\u1eadp ph\u00e2n th\u1ee9 nh\u1ea5t. L\u1eddi gi\u1ea3i a) T\u00ednh kh\u1ed1i l\u01b0\u1ee3ng c\u1ee7a hai vi\u00ean x\u00fac x\u1eafc? Cho bi\u1ebft: kh\u1ed1i l\u01b0\u1ee3ng m = V .D , trong \u0111\u00f3 V l\u00e0 th\u1ec3 t\u00edch v\u00e0 kh\u1ed1i l\u01b0\u1ee3ng ri\u00eang c\u1ee7a g\u1ed7 l\u00e0 D = 0,8 gam \/ cm3 . G\u1ecdi x(cm) l\u00e0 \u0111\u1ed9 d\u00e0i c\u1ea1nh c\u1ee7a vi\u00ean x\u00fac x\u1eafc h\u00ecnh l\u1eadp ph\u01b0\u01a1ng (x \uf03e 0) . Ta c\u00f3: Stp = 6x2 = 23, 52 \uf0db x = 7 2 cm . 5 Th\u1ec3 t\u00edch c\u1ee7a vi\u1ec5n x\u00fac x\u1eafc: V = x3 = \uf0e6 7 2 \uf0f63 = 686 2 \uf0bb 7,76cm3 . \uf0e7\uf0e7\uf0e8 5 \uf0f7\uf0f7\uf0f8 125 Kh\u1ed1i l\u01b0\u1ee3ng c\u1ee7a hai vi\u00ean x\u00fac x\u1eafc: m = 2.DV = 2.0,8.7,76 = 12,416(gam) . b) Ng\u01b0\u1eddi ta gieo hai vi\u00ean x\u00fac x\u1eafc tr\u00ean c\u00f9ng m\u1ed9t l\u1ea7n. H\u1ecfi x\u00e1c xu\u1ea5t \u0111\u1ec3 xu\u1ea5t hi\u1ec7n hai m\u1eb7t gi\u1ed1ng nhau l\u00e0 bao nhi\u00eau ph\u1ea7n tr\u0103m? K\u1ebft qu\u1ea3 l\u00e0m tr\u00f2n \u0111\u1ebfn ch\u1eef s\u1ed1 th\u1eadp ph\u00e2n th\u1ee9 nh\u1ea5t. V\u00ec m\u1ed7i con x\u00fac x\u1eafc c\u00f3 s\u00e1u m\u1eb7t kh\u00e1c nhau, nh\u01b0 v\u1eady c\u00f3 s\u00e1u tr\u01b0\u1eddng h\u1ee3p th\u1ecfa \u0111i\u1ec1u ki\u1ec7n hai m\u1eb7t xu\u1ea5t hi\u1ec7n gi\u1ed1ng nhau l\u00e0: (1;1) , (2; 2) , (3;3) , (4; 4) \u00b8 (5;5) , (6;6) . X\u00e1c xu\u1ea5t \u0111\u1ec3 gieo hai con x\u00fac x\u00fac c\u00f9ng m\u1ed9t l\u00fac c\u00f3 hai m\u1eb7t xu\u1ea5t hi\u1ec7n gi\u1ed1ng nhau l\u00e0: 6 = 1 . 6.6 6 ( )C\u00e2u 7. (0,5 \u0111i\u1ec3m) M\u1ed9t xe \u00f4t\u00f4 chuy\u1ec3n \u0111\u1ed9ng theo h\u00e0m s\u1ed1 S = 30.t + 4.t2 , trong \u0111\u00f3: S km2 l\u00e0 qu\u00e3ng \u0111\u01b0\u1eddng xe \u0111i \u0111\u01b0\u1ee3c; t (gi\u1edd) l\u00e0 th\u1eddi gian chuy\u1ec3n \u0111\u1ed9ng c\u1ee7a xe t\u00ednh t\u1eeb l\u00fac 7 gi\u1edd s\u00e1ng. Xem nh\u01b0 xe chuy\u1ec3n \u0111\u1ed9ng \u0111\u1ec1u tr\u00ean m\u1ed9t \u0111o\u1ea1n \u0111\u01b0\u1eddng th\u1eb3ng v\u00e0 kh\u00f4ng ngh\u1ec9. a) H\u1ecfi t\u1eeb l\u00fac 7 gi\u1edd 30 ph\u00fat \u0111\u1ebfn l\u00fac 8 gi\u1edd 15 ph\u00fat xe \u0111\u00e3 \u0111i \u0111\u01b0\u1ee3c qu\u00e3ng \u0111\u01b0\u1eddng d\u00e0i bao nhi\u00eau km ? b) \u0110\u1ebfn l\u00fac m\u1ea5y gi\u1edd th\u00ec xe \u0111i \u0111\u01b0\u1ee3c qu\u00e3ng \u0111\u01b0\u1eddng d\u00e0i 34 km (t\u00ednh t\u1eeb l\u00fac 7 gi\u1edd s\u00e1ng)? L\u1eddi gi\u1ea3i T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N 6","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 S\u1ede GD \u2013 \u0110T TP H\u1ed2 CH\u00cd MINH a) H\u1ecfi t\u1eeb l\u00fac 7 gi\u1edd 30 ph\u00fat \u0111\u1ebfn l\u00fac 8 gi\u1edd 15 ph\u00fat xe \u0111\u00e3 \u0111i \u0111\u01b0\u1ee3c qu\u00e3ng \u0111\u01b0\u1eddng d\u00e0i bao nhi\u00eau km ? X\u00e9t th\u1eddi \u0111i\u1ec3m g\u1ed1c l\u00e0 7 gi\u1edd s\u00e1ng T\u1ea1i th\u1eddi \u0111i\u1ec3m 7 gi\u1edd 30 ph\u00fat, ta c\u00f3 t = 30 ph\u00fat hay t = 0,5 gi\u1edd n\u00ean qu\u00e3ng \u0111\u01b0\u1eddng xe \u00f4 t\u00f4 \u0111\u00e3 \u0111i \u0111\u01b0\u1ee3c: S = 30.0,5 + 4.0,52 = 16 km . T\u1ea1i th\u1eddi \u0111i\u1ec3m 8 gi\u1edd 15 ph\u00fat, ta c\u00f3 t = 1 gi\u1edd 15 ph\u00fat hay t = 1,25 gi\u1edd n\u00ean qu\u00e3ng \u0111\u01b0\u1eddng xe \u00f4 t\u00f4 \u0111\u00e3 \u0111i \u0111\u01b0\u1ee3c: S = 30.1,25 + 4.1,252 = 43,75 km . b) \u0110\u1ebfn l\u00fac m\u1ea5y gi\u1edd th\u00ec xe \u0111i \u0111\u01b0\u1ee3c qu\u00e3ng \u0111\u01b0\u1eddng d\u00e0i 34 km (t\u00ednh t\u1eeb l\u00fac 7 gi\u1edd s\u00e1ng)? \uf0e9t = 1 \uf0ea V\u1edbi S = 34 km , ta c\u00f3: 4t2 + 30t = 34 \uf0db 4t 2 + 30t \u2212 34 = 0 \uf0db \uf0ea\uf0ea\uf0ebt = \u2212 17 . 2 V\u00ec th\u1eddi gian l\u00e0 gi\u00e1 tr\u1ecb d\u01b0\u01a1ng n\u00ean nh\u1eadn gi\u00e1 tr\u1ecb t = 1 . V\u1eady \u0111\u1ebfn l\u00fac 8 gi\u1edd s\u00e1ng th\u00ec xe \u0111\u00e3 \u0111i \u0111\u01b0\u1ee3c 34 km k\u1ec3 t\u1eeb l\u00fac 7 gi\u1edd s\u00e1ng. C\u00e2u 8. (3 \u0111i\u1ec3m) Cho \u0111\u01b0\u1eddng tr\u00f2n (O; R) v\u00e0 \u0111i\u1ec3m A n\u1eb1m ngo\u00e0i \u0111\u01b0\u1eddng tr\u00f2n (O) . V\u1ebd hai ti\u1ebfp tuy\u1ebfn AB , AC c\u1ee7a (O) (v\u1edbi B , C l\u00e0 ti\u1ebfp \u0111i\u1ec3m). V\u1ebd c\u00e1t tuy\u1ebfn ADE c\u1ee7a (O) (v\u1edbi D , E thu\u1ed9c (O) ; D n\u1eb1m gi\u1eefa A v\u00e0 E ; tia AD n\u1eb1m gi\u1eefa hai tia AB v\u00e0 AO ). a) Ch\u1ee9ng minh t\u1ee9 gi\u00e1c OBAC n\u1ed9i ti\u1ebfp v\u00e0 AB2 = AD.AE . b) G\u1ecdi H l\u00e0 giao \u0111i\u1ec3m c\u1ee7a OA v\u00e0 BC . Ch\u1ee9ng minh BD.CE = BE.CD v\u00e0 t\u1ee9 gi\u00e1c DEOH n\u1ed9i ti\u1ebfp. c) \u0110\u01b0\u1eddng th\u1eb3ng AO c\u1eaft \u0111\u01b0\u1eddng tr\u00f2n (O) t\u1ea1i M v\u00e0 N ( M n\u1eb1m gi\u1eefa A v\u00e0 O ). Ch\u1ee9ng minh: HC2 = HD.HE v\u00e0 EH.AD = MH.AN . L\u1eddi gi\u1ea3i T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N 7","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 S\u1ede GD \u2013 \u0110T TP H\u1ed2 CH\u00cd MINH a) Ch\u1ee9ng minh t\u1ee9 gi\u00e1c OBAC n\u1ed9i ti\u1ebfp v\u00e0 AB2 = AD.AE . X\u00e9t t\u1ee9 gi\u00e1c OBAC , c\u00f3: \uf0ec\uf0efABO = 90\uf0b0( AB \u22a5 OB) \uf0ed \uf0ee\uf0ef ACO = 90\uf0b0( AC \u22a5 OC ) \uf0de ABO + ACO = 180\uf0b0 \uf0de T\u1ee9 gi\u00e1c OBAC n\u1ed9i ti\u1ebfp v\u00ec c\u00f3 hai g\u00f3c \u0111\u1ed1i b\u00f9 nhau. X\u00e9t \uf044ABD v\u00e0 \uf044AEB , ta c\u00f3: ABD = AEB (g\u00f3c t\u1ea1o b\u1edfi tt v\u00e0 d\u00e2y c\u00f9ng v\u1edbi g\u00f3c n\u1ed9i ti\u1ebfp ch\u1eafn BD ) BAE chung \uf0de \uf044ABD \uf044AEB (g.g) \uf0de AB = AD (t\u1ec9 s\u1ed1 \u0111\u1ed3ng d\u1ea1n) AE AB \uf0de AB2 = AD.AE . b) G\u1ecdi H l\u00e0 giao \u0111i\u1ec3m c\u1ee7a OA v\u00e0 BC . Ch\u1ee9ng minh BD.CE = BE.CD v\u00e0 t\u1ee9 gi\u00e1c DEOH n\u1ed9i ti\u1ebfp. X\u00e9t \uf044ACD v\u00e0 \uf044AEC , ta c\u00f3: ACD = AEC (g\u00f3c t\u1ea1o b\u1edfi tt v\u00e0 d\u00e2y c\u00f9ng v\u1edbi g\u00f3c n\u1ed9i ti\u1ebfp ch\u1eafn CD ) EAC chung \uf0de \uf044ACD \uf044AEC (g.g) \uf0de AC = CD (t\u1ec9 s\u1ed1 \u0111\u1ed3ng d\u1ea1n) AE EC M\u00e0: AB = BD (\uf044ABD \uf044AEB) AE EB N\u00ean: CD = BD ( AB = AC) EC EB \uf0de CD.EB = BD.EC . Ta c\u00f3: AB = AC (t\/c 2 ti\u1ebfp tuy\u1ebfn c\u1eaft nhau t\u1ea1i A ) 8 ( )M\u00e0: OB = OC = R(O) N\u00ean: AO l\u00e0 \u0111\u01b0\u1eddng trung tr\u1ef1c c\u1ee7a BC \uf0de AO \u22a5 BC t\u1ea1i H . 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 S\u1ede GD \u2013 \u0110T TP H\u1ed2 CH\u00cd MINH X\u00e9t \uf044ABO vu\u00f4ng t\u1ea1i B c\u00f3 BH l\u00e0 \u0111\u01b0\u1eddng cao Ta c\u00f3: AB2 = AH.AO M\u00e0: AB2 = AD.AE (cmt) N\u00ean: AH.AO = AD.AE \uf0de AH = AD . AE AO X\u00e9t \uf044AHD v\u00e0 \uf044AEO , ta c\u00f3: AH = AD (cmt) AE AO EAO chung \uf0de \uf044AHD \uf044AEO (c.g.c) \uf0de AHD = AEO ( 2 g\u00f3c t\u01b0\u01a1ng \u1ee9ng) \uf0de T\u1ee9 gi\u00e1c DEOH n\u1ed9i ti\u1ebfp v\u00ec c\u00f3 g\u00f3c ngo\u00e0i b\u1eb1ng g\u00f3c trong \u0111\u1ed1i di\u1ec7n. c) \u0110\u01b0\u1eddng th\u1eb3ng AO c\u1eaft \u0111\u01b0\u1eddng tr\u00f2n (O) t\u1ea1i M v\u00e0 N ( M n\u1eb1m gi\u1eefa A v\u00e0 O ). Ch\u1ee9ng minh: HC2 = HD.HE v\u00e0 EH.AD = MH.AN . Ta c\u00f3 DEM = 1 DOM (g\u00f3c nt v\u00e0 g\u00f3c \u1edf t\u00e2m c\u00f9ng ch\u1eafn DM ) 2 M\u00e0: DEH = DOM (t\u1ee9 gi\u00e1c DEOH n\u1ed9i ti\u1ebfp) N\u00ean: DEM = 1 DEH 2 \uf0de EM l\u00e0 ph\u00e2n gi\u00e1c AEH \uf0de EH = MH (t\/c \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c trong \uf044AEH ) (1) EA MA X\u00e9t \uf044ADM v\u00e0 \uf044ANE , ta c\u00f3: ADM = ANE (t\u1ee9 gi\u00e1c DMNE n\u1ed9i ti\u1ebfp) EAN chung \uf0de \uf044ADM \uf044ANE (g.g) \uf0de AD = AM (t\u1ec9 s\u1ed1 \u0111\u1ed3ng d\u1ea1ng) Hay AE = AM (2) AN AE AN AD Nh\u1eadn (1) v\u00e0 (2) theo v\u1ebf, ta \u0111\u01b0\u1ee3c: EH . AE = MH . AM \uf0de EH = MH \uf0de EH.AD = MH.AN . EA AN MA AD AN AD ----H\u1ebeT--- 9 T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH 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 B\u00ccNH TA\u00c2N NA\u00caM HO\u00cfC: 2022 - 2023 \u0110\u1ec0 THAM KH\u1ea2O M\u00d4N: TO\u00c1N 9 M\u00c3 \u0110\u1ec0: Qu\u1eadn B\u00ecnh T\u00e2n - 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 = \u2212x2 v\u00e0 \u0111\u01b0\u1eddng th\u1eb3ng (d) : y = \u22125 x + 3 . 22 a) V\u1ebd \u0111\u1ed3 th\u1ecb (P) v\u00e0 (d) tr\u00ean c\u00f9ng h\u1ec7 tr\u1ee5c t\u1ecda \u0111\u1ed9. b) T\u00ecm t\u1ecda \u0111\u1ed9 giao \u0111i\u1ec3m c\u1ee7a (P) v\u00e0 (d) b\u1eb1ng ph\u00e9p t\u00ednh. C\u00e2u 2. (1 \u0111i\u1ec3m). Cho ph\u01b0\u01a1ng tr\u00ecnh 2x2 + x \u2212 6 = 0 . Kh\u00f4ng gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh, h\u00e3y t\u00ednh gi\u00e1 tr\u1ecb c\u1ee7a bi\u1ec3u th\u1ee9c A = x12 + x22 \u2212 5x1 \u2212 5x2 . 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). C\u00f4ng th\u1ee9c t\u00ednh di\u1ec7n t\u00edch h\u00ecnh tam gi\u00e1c khi bi\u1ebft s\u1ed1 \u0111o ba c\u1ea1nh c\u1ee7a n\u00f3 \u0111\u01b0\u1ee3c cho b\u1edfi c\u00f4ng th\u1ee9c: S p p a p b p c ,p a b c 2 Trong \u0111\u00f3 a, b, c l\u00e0 s\u1ed1 \u0111o ba c\u1ea1nh c\u1ee7a tam gi\u00e1c, p l\u00e0 n\u1eeda chu vi. Ngo\u00e0i ra, di\u1ec7n t\u00edch h\u00ecnh tam gi\u00e1c c\u00f2n \u0111\u01b0\u1ee3c cho b\u1edfi c\u00f4ng th\u1ee9c: S = abc . 4R Trong \u0111\u00f3, a, b, c l\u00e0 s\u1ed1 \u0111o ba c\u1ea1nh c\u1ee7a tam gi\u00e1c, R l\u00e0 b\u00e1n k\u00ednh c\u1ee7a \u0111\u01b0\u1eddng tr\u00f2n ngo\u1ea1i ti\u1ebfp tam gi\u00e1c. C\u00e1c nh\u00e0 kh\u1ea3o c\u1ed5 v\u1eeba ph\u00e1t hi\u1ec7n \u0111\u01b0\u1ee3c m\u1ed9t chi\u1ebfc \u0111\u0129a c\u1ed5 h\u00ecnh tr\u00f2n \u0111\u00e3 b\u1ecb b\u1ec3, bi\u1ebft s\u1ed1 \u0111o ba c\u1ea1nh AB, AC v\u00e0 BC c\u1ee7a ABC tr\u00ean \u0111\u0129a nh\u01b0 sau: a BC 6 cm ; b AC 7 cm ; c AB 5 cm. H\u00e3y x\u00e1c \u0111\u1ecbnh b\u00e1n k\u00ednh c\u1ee7a chi\u1ebfc \u0111\u0129a tr\u00ean. C\u00e2u 4. (0,75 \u0111i\u1ec3m). Anh T\u00fa l\u00e0 sinh vi\u00ean \u0111\u1ea1i h\u1ecdc. Anh d\u00e0nh m\u1ed9t s\u1ed1 bu\u1ed5i t\u1ed1i \u0111\u1ec3 \u0111i l\u00e0m th\u00eam c\u00f4ng vi\u1ec7c ph\u1ee5c v\u1ee5 t\u1ea1i m\u1ed9t qu\u00e1n n\u01b0\u1edbc, m\u1ed7i bu\u1ed5i \u0111\u01b0\u1ee3c tr\u1ea3 150 000 \u0111\u1ed3ng. Do th\u00e1ng n\u00e0y c\u00f3 d\u1ecbp T\u1ebft \u0111\u00f4ng kh\u00e1ch n\u00ean anh T\u00fa l\u00e0m t\u0103ng th\u00eam 6 bu\u1ed5i v\u1edbi ti\u1ec1n \u0111\u01b0\u1ee3c tr\u1ea3 m\u1ed7i bu\u1ed5i t\u0103ng g\u1ea5p 150% so v\u1edbi ng\u00e0y th\u01b0\u1eddng. Ngo\u00e0i ra, m\u1ed7i bu\u1ed5i l\u00e0m anh c\u00f2n \u0111\u01b0\u1ee3c h\u1ed7 tr\u1ee3 th\u00eam 20 000 \u0111\u1ed3ng ti\u1ec1n \u0103n \u0111\u01b0\u1ee3c T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N 1","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH t\u00ednh g\u1ed9p trong ti\u1ec1n l\u01b0\u01a1ng tr\u1ea3 cu\u1ed1i th\u00e1ng. H\u1ecfi th\u00e1ng v\u1eeba qua anh T\u00fa \u0111\u00e3 l\u00e0m bao nhi\u00eau bu\u1ed5i bi\u1ebft anh nh\u1eadn \u0111\u01b0\u1ee3c ti\u1ec1n l\u01b0\u01a1ng t\u1ea5t c\u1ea3 l\u00e0 3170 000 \u0111\u1ed3ng? C\u00e2u 5. (1 \u0111i\u1ec3m). C\u00e2n n\u1eb7ng trung b\u00ecnh c\u1ee7a tr\u1ebb s\u01a1 sinh \u0111\u1ee7 th\u00e1ng l\u00e0 kho\u1ea3ng 3 000 g . Tr\u1ebb 6 th\u00e1ng c\u00f3 c\u00e2n n\u1eb7ng g\u1ea5p \u0111\u00f4i, 6 th\u00e1ng ti\u1ebfp theo m\u1ed7i th\u00e1ng t\u0103ng 500 g . T\u1eeb n\u0103m th\u1ee9 hai tr\u1edf \u0111i, trung b\u00ecnh m\u1ed7i n\u0103m t\u0103ng th\u00eam 1, 5 kg . G\u1ecdi P kg l\u00e0 c\u00e2n n\u1eb7ng c\u1ee7a tr\u1ebb em d\u01b0\u1edbi 14 tu\u1ed5i; N l\u00e0 s\u1ed1 tu\u1ed5i (d\u1ef1a v\u00e0o h\u1eb1ng s\u1ed1 sinh h\u1ecdc ng\u01b0\u1eddi Vi\u1ec7t Nam n\u0103m 1975 ). a) L\u1eadp c\u00f4ng th\u1ee9c P theo N. b) D\u1ef1a v\u00e0o h\u1eb1ng s\u1ed1 sinh h\u1ecdc ng\u01b0\u1eddi Vi\u1ec7t Nam n\u0103m 1975 , m\u1ed1i li\u00ean h\u1ec7 s\u1ed1 tu\u1ed5i v\u00e0 chi\u1ec1u cao c\u1ee7a tr\u1ebb em tr\u00ean 1 tu\u1ed5i \u0111\u01b0\u1ee3c cho b\u1edfi c\u00f4ng th\u1ee9c: h 75 5 N 1 v\u1edbi h l\u00e0 chi\u1ec1u cao cm ; N l\u00e0 s\u1ed1 tu\u1ed5i. H\u1ecfi theo h\u1eb1ng s\u1ed1 sinh h\u1ecdc ng\u01b0\u1eddi Vi\u1ec7t Nam m\u1ed9t tr\u1ebb em n\u1eb7ng 16, 5 kg th\u00ec chi\u1ec1u cao t\u01b0\u01a1ng \u1ee9ng l\u00e0 bao nhi\u00eau? C\u00e2u 6. (1 \u0111i\u1ec3m). M\u1ed9t b\u1ec3 ch\u1ee9a n\u01b0\u1edbc tr\u00ean n\u00f3c m\u1ed9t t\u00f2a chung c\u01b0 c\u00f3 d\u1ea1ng h\u00ecnh c\u1ea7u, \u0111\u01b0\u1eddng k\u00ednh b\u00ean trong b\u1ec3 c\u00f3 \u0111\u1ed9 d\u00e0i l\u00e0 8 m . Th\u1ec3 t\u00edch h\u00ecnh c\u1ea7u: V 4 R3 3 a) B\u1ec3 ch\u1ee9a \u0111\u01b0\u1ee3c bao nhi\u00eau l\u00edt n\u01b0\u1edbc khi \u0111\u01b0\u1ee3c b\u01a1m \u0111\u1ea7y (l\u00e0m tr\u00f2n \u0111\u1ebfn ch\u1eef s\u1ed1 th\u00e2\u0323p ph\u00e2n th\u1ee9 nh\u1ea5t). b) T\u00f2a nh\u00e0 chung c\u01b0 c\u00f3 608 ng\u01b0\u1eddi, trung b\u00ecnh m\u1ed7i ng\u00e0y m\u1ed7i ng\u01b0\u1eddi d\u00f9ng 18, 5 l\u00edt n\u01b0\u1edbc. H\u1ecfi khi \u0111\u01b0\u1ee3c b\u01a1m \u0111\u1ea7y th\u00ec l\u01b0\u1ee3ng n\u01b0\u1edbc trong b\u1ec3 \u0111\u1ee7 dung cho c\u00e1c h\u1ed9 d\u00e2n to\u00e0n nh\u00e0 chung c\u01b0 trong bao nhi\u00eau ng\u00e0y bi\u1ebft r\u1eb1ng l\u01b0\u1ee3ng n\u01b0\u1edbc hao h\u1ee5t trong qu\u00e1 tr\u00ecnh d\u1eabn t\u1eeb b\u1ec3 ch\u1ee9a xu\u1ed1ng h\u1ed9 d\u00e2n l\u00e0 0, 5% (l\u00e0m tr\u00f2n \u0111\u1ebfn ng\u00e0y). C\u00e2u 7. (0,75 \u0111i\u1ec3m). B\u1ea1n A c\u00f3 m\u1ed9t \u1ed5 kh\u00f3a xe \u0111\u1ea1p nh\u01b0 trong h\u00ecnh. \u1ed5 kh\u00f3a c\u00f3 c\u00e1c s\u1ed1 t\u1eeb 0 \u0111\u1ebfn 9 tr\u00ean m\u1ed7i v\u00f2ng quay. Kh\u00f3a s\u1ebd k\u00eau t\u00e1ch nh\u1eb9 khi b\u1ea1n A quay l\u00ean hay quay xu\u1ed1ng m\u1ed9t s\u1ed1 tr\u00ean m\u1ed7i v\u00f2ng k\u1ec3 c\u1ea3 khi quay t\u1eeb 0 \u0111\u1ebfn 9 hay ng\u01b0\u1ee3c l\u1ea1i. Khi nh\u00ecn v\u00e0o \u1ed5 kh\u00f3a th\u00ec A th\u1ea5y c\u00f3 c\u00e1c s\u1ed1 m\u1ed7i v\u00f2ng \u0111ang \u1edf v\u1ecb tr\u00ed 9 0 4 nh\u01b0 h\u00ecnh v\u1ebd. M\u00e3 kh\u00f3a A \u0111\u00e3 c\u00e0i l\u00e0 5 8 7. 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 a) Em h\u00e3y t\u00ednh s\u1ed1 ti\u1ebfng t\u00e1ch \u00edt nh\u1ea5t khi A c\u1ea7n \u0111\u1ec3 m\u1edf \u1ed5 kh\u00f3a. b) B\u1ea1n c\u1ee7a A c\u0169ng \u0111\u00e3 m\u1edf \u0111\u01b0\u1ee3c kh\u00f3a t\u1eeb v\u1ecb tr\u00ed 9 0 4 v\u1edbi s\u1ed1 ti\u1ebfng t\u00e1ch l\u00e0 nhi\u1ec1u nh\u1ea5t. T\u00ednh s\u1ed1 ti\u1ebfng t\u00e1ch trung b\u00ecnh c\u1ea7n \u0111\u1ec3 m\u1edf \u0111\u01b0\u1ee3c \u1ed5 kh\u00f3a. Xem nh\u01b0 g\u1ea7n v\u1edbi trung b\u00ecnh c\u1ed9ng c\u1ee7a s\u1ed1 ti\u1ebfng \u00edt nh\u1ea5t v\u00e0 nhi\u1ec1u nh\u1ea5t. C\u00e2u 8. (3 \u0111i\u1ec3m) Cho \u0111i\u1ec3m A n\u1eb1m ngo\u00e0i \u0111\u01b0\u1eddng tr\u00f2n (O; R) , k\u1ebb c\u00e1c ti\u1ebfp tuy\u1ebfn AB , AC v\u1edbi (O) (l\u00e0 c\u00e1c ti\u1ebfp \u0111i\u1ec3m). V\u1ebd \u0111\u01b0\u1eddng k\u00ednh CE c\u1ee7a (O) . G\u1ecdi H l\u00e0 giao \u0111i\u1ec3m c\u1ee7a OA v\u00e0 BC . a) Ch\u1ee9ng minh t\u1ee9 gi\u00e1c ABOC n\u1ed9i ti\u1ebfp v\u00e0 BE\/\/OA . b) AE c\u1eaft (O) t\u1ea1i D (kh\u00e1c E ), BD c\u1eaft OA t\u1ea1i M . Ch\u1ee9ng minh r\u1eb1ng MAD = MBA v\u00e0 AHD = ACD. c) V\u1ebd EI vu\u00f4ng g\u1edbi v\u1edbi OA t\u1ea1i I ; v\u1ebd DK l\u00e0 \u0111\u01b0\u1eddng k\u00ednh c\u1ee7a (O) . Ch\u1ee9ng minh K ,I ,B 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 = \u22125 x + 3 . 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 \u22122 \u22121 0 1 2 y = \u2212x2 \u22122 \u22121 0 \u22121 \u22122 2 22 x 12 y = \u22125 x + 3 1 \u22122 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) : \u2212x2 = \u22125 x + 3 22 \uf0db \u2212x2 + 5 x \u2212 3 = 0 22 \uf0db \uf0e9x = 3 \uf0ea\uf0ebx = 2 Thay x = 3 v\u00e0o y = \u2212x2 , ta \u0111\u01b0\u1ee3c: y = \u221232 = \u2212 9 . 2 22 Thay x = 2 v\u00e0o y = \u2212x2 , ta \u0111\u01b0\u1ee3c: y = \u221222 = \u22122 . 22 V\u1eady \uf0e6 3; \u2212 9 \uf0f6 , (2; \u2212 2) l\u00e0 hai giao \u0111i\u1ec3m c\u1ea7n t\u00ecm. \uf0e7 2 \uf0f7 \uf0e8 \uf0f8 T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N 4","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH C\u00e2u 2. (1 \u0111i\u1ec3m) Cho ph\u01b0\u01a1ng tr\u00ecnh 2x2 + x \u2212 6 = 0 . Kh\u00f4ng gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh, h\u00e3y t\u00ednh gi\u00e1 tr\u1ecb c\u1ee7a bi\u1ec3u th\u1ee9c A = x12 + x22 \u2212 5x1 \u2212 5x2 . L\u1eddi gi\u1ea3i V\u00ec \uf044 = b2 \u2212 4ac = 12 \u2212 4.2.(\u22126) = 49 \uf03e 0 N\u00ean ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m ph\u00e2n bi\u1ec7t x1 ,x2 . \uf0ef\uf0ec\uf0efS = x1 + x2 = \u2212b =\u22121 \uf0ed = x1 a 2 Theo \u0111\u1ecbnh l\u00ed Vi-et, ta c\u00f3: \uf0ef\uf0ef\uf0eeP c a = \u22126 = \u22123 .x2 = 2 Ta c\u00f3: A = x12 + x22 \u2212 5x1 \u2212 5x2 ( ) ( )A = x12 + x22 \u2212 5 x1 + x2 A = S2 \u2212 2P \u2212 5S A = \uf0e6 \u22121 \uf0f62 \u2212 2.(\u22123) \u2212 5.\uf0e6\uf0e7 \u22121 \uf0f6 = 35 . \uf0e7 2 \uf0f7 \uf0e8 2 \uf0f7 4 \uf0e8 \uf0f8 \uf0f8 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) C\u00f4ng th\u1ee9c t\u00ednh di\u1ec7n t\u00edch h\u00ecnh tam gi\u00e1c khi bi\u1ebft s\u1ed1 \u0111o ba c\u1ea1nh c\u1ee7a n\u00f3 \u0111\u01b0\u1ee3c cho b\u1edfi c\u00f4ng th\u1ee9c: S p p a p b p c ,p a b c 2 Trong \u0111\u00f3 a, b, c l\u00e0 s\u1ed1 \u0111o ba c\u1ea1nh c\u1ee7a tam gi\u00e1c, p l\u00e0 n\u1eeda chu vi. Ngo\u00e0i ra, di\u1ec7n t\u00edch h\u00ecnh tam gi\u00e1c c\u00f2n \u0111\u01b0\u1ee3c cho b\u1edfi c\u00f4ng th\u1ee9c: S = abc . 4R Trong \u0111\u00f3, a, b, c l\u00e0 s\u1ed1 \u0111o ba c\u1ea1nh c\u1ee7a tam gi\u00e1c, R l\u00e0 b\u00e1n k\u00ednh c\u1ee7a \u0111\u01b0\u1eddng tr\u00f2n ngo\u1ea1i ti\u1ebfp tam gi\u00e1c. C\u00e1c nh\u00e0 kh\u1ea3o c\u1ed5 v\u1eeba ph\u00e1t hi\u1ec7n \u0111\u01b0\u1ee3c m\u1ed9t chi\u1ebfc \u0111\u0129a c\u1ed5 h\u00ecnh tr\u00f2n \u0111\u00e3 b\u1ecb b\u1ec3, bi\u1ebft s\u1ed1 \u0111o ba c\u1ea1nh AB, AC v\u00e0 BC c\u1ee7a ABC tr\u00ean \u0111\u0129a nh\u01b0 sau: a BC 6 cm ; b AC 7 cm ; c AB 5 cm. H\u00e3y x\u00e1c \u0111\u1ecbnh b\u00e1n k\u00ednh c\u1ee7a chi\u1ebfc \u0111\u0129a tr\u00ean. T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N 5","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH L\u1eddi gi\u1ea3i N\u1eeda chu vi ABC l\u00e0: p a b c 6 7 5 9 cm 22 Di\u1ec7n t\u00edch c\u1ee7a ABC l\u00e0: S p p a p b p c 9 9 6 9 7 9 5 6 6 cm2 B\u00e1n k\u00ednh c\u1ee7a chi\u1ebfc \u0111\u0129a l\u00e0: S = abc \uf0de R = abc = 6.7.5 = 35 6 (cm) 4R 4S 4.6 6 24 C\u00e2u 4. (0,75 \u0111i\u1ec3m). Anh T\u00fa l\u00e0 sinh vi\u00ean \u0111\u1ea1i h\u1ecdc. Anh d\u00e0nh m\u1ed9t s\u1ed1 bu\u1ed5i t\u1ed1i \u0111\u1ec3 \u0111i l\u00e0m th\u00eam c\u00f4ng vi\u1ec7c ph\u1ee5c v\u1ee5 t\u1ea1i m\u1ed9t qu\u00e1n n\u01b0\u1edbc, m\u1ed7i bu\u1ed5i \u0111\u01b0\u1ee3c tr\u1ea3 150 000 \u0111\u1ed3ng. Do th\u00e1ng n\u00e0y c\u00f3 d\u1ecbp T\u1ebft \u0111\u00f4ng kh\u00e1ch n\u00ean anh T\u00fa l\u00e0m t\u0103ng th\u00eam 6 bu\u1ed5i v\u1edbi ti\u1ec1n \u0111\u01b0\u1ee3c tr\u1ea3 m\u1ed7i bu\u1ed5i t\u0103ng g\u1ea5p 150% so v\u1edbi ng\u00e0y th\u01b0\u1eddng. Ngo\u00e0i ra, m\u1ed7i bu\u1ed5i l\u00e0m anh c\u00f2n \u0111\u01b0\u1ee3c h\u1ed7 tr\u1ee3 th\u00eam 20 000 \u0111\u1ed3ng ti\u1ec1n \u0103n \u0111\u01b0\u1ee3c t\u00ednh g\u1ed9p trong ti\u1ec1n l\u01b0\u01a1ng tr\u1ea3 cu\u1ed1i th\u00e1ng. H\u1ecfi th\u00e1ng v\u1eeba qua anh T\u00fa \u0111\u00e3 l\u00e0m bao nhi\u00eau bu\u1ed5i bi\u1ebft anh nh\u1eadn \u0111\u01b0\u1ee3c ti\u1ec1n l\u01b0\u01a1ng t\u1ea5t c\u1ea3 l\u00e0 3170 000 \u0111\u1ed3ng? L\u1eddi gi\u1ea3i G\u1ecdi x (bu\u1ed5i) l\u00e0 s\u1ed1 bu\u1ed5i anh T\u00fa \u0111\u00e3 l\u00e0m trong th\u00e1ng v\u1eeba qua. x S\u1ed1 ti\u1ec1n anh T\u00fa nh\u1eadn \u0111\u01b0\u1ee3c khi \u0111i l\u00e0m b\u00ecnh th\u01b0\u1eddng l\u00e0 150 000 x 6 (\u0111\u1ed3ng) S\u1ed1 ti\u1ec1n anh T\u00fa nh\u1eadn \u0111\u01b0\u1ee3c khi l\u00e0m t\u0103ng th\u00eam 6 bu\u1ed5i l\u00e0 6.150 000.150% 1350 000 (\u0111\u1ed3ng) S\u1ed1 ti\u1ec1n \u0103n anh T\u00fa \u0111\u01b0\u1ee3c h\u1ed7 tr\u1ee3 th\u00eam l\u00e0 20 000x (\u0111\u1ed3ng) Theo \u0111\u1ec1 b\u00e0i ta c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh: 150 000 x 6 1350 000 20 000x 3170 000 170 000x 450 000 3170 000 170 000x 2 720 000 x 16 n V\u1eady th\u00e1ng v\u1eeba quan anh T\u00fa l\u00e0 \u0111\u01b0\u1ee3c 16 bu\u1ed5i. C\u00e2u 5. (1 \u0111i\u1ec3m) C\u00e2n n\u1eb7ng trung b\u00ecnh c\u1ee7a tr\u1ebb s\u01a1 sinh \u0111\u1ee7 th\u00e1ng l\u00e0 kho\u1ea3ng 3 000 g . Tr\u1ebb 6 th\u00e1ng c\u00f3 c\u00e2n n\u1eb7ng g\u1ea5p \u0111\u00f4i, 6 th\u00e1ng ti\u1ebfp theo m\u1ed7i th\u00e1ng t\u0103ng 500 g . T\u1eeb n\u0103m th\u1ee9 hai tr\u1edf \u0111i, trung b\u00ecnh m\u1ed7i n\u0103m t\u0103ng th\u00eam 1, 5 kg . G\u1ecdi P kg l\u00e0 c\u00e2n n\u1eb7ng c\u1ee7a tr\u1ebb em d\u01b0\u1edbi 14 tu\u1ed5i; N l\u00e0 s\u1ed1 tu\u1ed5i (d\u1ef1a v\u00e0o h\u1eb1ng s\u1ed1 sinh h\u1ecdc ng\u01b0\u1eddi Vi\u1ec7t Nam n\u0103m 1975 ). a) L\u1eadp c\u00f4ng th\u1ee9c P theo N. T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N 6","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH b) D\u1ef1a v\u00e0o h\u1eb1ng s\u1ed1 sinh h\u1ecdc ng\u01b0\u1eddi Vi\u1ec7t Nam n\u0103m 1975 , m\u1ed1i li\u00ean h\u1ec7 s\u1ed1 tu\u1ed5i v\u00e0 chi\u1ec1u cao c\u1ee7a tr\u1ebb em tr\u00ean 1 tu\u1ed5i \u0111\u01b0\u1ee3c cho b\u1edfi c\u00f4ng th\u1ee9c: h 75 5 N 1 v\u1edbi h l\u00e0 chi\u1ec1u cao cm ; N l\u00e0 s\u1ed1 tu\u1ed5i. H\u1ecfi theo h\u1eb1ng s\u1ed1 sinh h\u1ecdc ng\u01b0\u1eddi Vi\u1ec7t Nam m\u1ed9t tr\u1ebb em n\u1eb7ng 16, 5 kg th\u00ec chi\u1ec1u cao t\u01b0\u01a1ng \u1ee9ng l\u00e0 bao nhi\u00eau? L\u1eddi gi\u1ea3i a) \u0110\u1ed5i: 3 000 g 3 kg 500 g 0, 5 kg C\u00e2n n\u1eb7ng c\u1ee7a tr\u1ebb m\u1ed9t tu\u1ed5i l\u00e0: 3.2 0, 5.6 9 kg C\u00f4ng th\u1ee9c P theo N l\u00e0 P 9 1, 5 N 1 b) S\u1ed1 tu\u1ed5i c\u1ee7a tr\u1ebb em n\u1eb7ng 16, 5 kg l\u00e0: P 9 1, 5 N 1 16, 5 9 1, 5 N 1 1, 5 N 1 7, 5 N1 5 N6 Chi\u1ec1u cao c\u1ee7a tr\u1ebb 6 tu\u1ed5i l\u00e0 h 75 5 N 1 75 5. 6 1 100 cm V\u1eady tr\u1ebb em n\u1eb7ng 16, 5 kg th\u00ec chi\u1ec1u cao t\u01b0\u01a1ng \u1ee9ng l\u00e0 100 cm C\u00e2u 6. (1 \u0111i\u1ec3m) M\u1ed9t b\u1ec3 ch\u1ee9a n\u01b0\u1edbc tr\u00ean n\u00f3c m\u1ed9t t\u00f2a chung c\u01b0 c\u00f3 d\u1ea1ng h\u00ecnh c\u1ea7u, \u0111\u01b0\u1eddng k\u00ednh b\u00ean trong b\u1ec3 c\u00f3 \u0111\u1ed9 d\u00e0i l\u00e0 8 m . Th\u1ec3 t\u00edch h\u00ecnh c\u1ea7u: V 4 R3 3 a) B\u1ec3 ch\u1ee9a \u0111\u01b0\u1ee3c bao nhi\u00eau l\u00edt n\u01b0\u1edbc khi \u0111\u01b0\u1ee3c b\u01a1m \u0111\u1ea7y (l\u00e0m tr\u00f2n \u0111\u1ebfn ch\u1eef s\u1ed1 th\u00e2\u0323p ph\u00e2n th\u1ee9 nh\u1ea5t). b) T\u00f2a nh\u00e0 chung c\u01b0 c\u00f3 608 ng\u01b0\u1eddi, trung b\u00ecnh m\u1ed7i ng\u00e0y m\u1ed7i ng\u01b0\u1eddi d\u00f9ng 18, 5 l\u00edt n\u01b0\u1edbc. H\u1ecfi khi \u0111\u01b0\u1ee3c b\u01a1m \u0111\u1ea7y th\u00ec l\u01b0\u1ee3ng n\u01b0\u1edbc trong b\u1ec3 \u0111\u1ee7 d\u00f9ng cho c\u00e1c h\u1ed9 d\u00e2n to\u00e0 nh\u00e0 chung c\u01b0 trong bao nhi\u00eau ng\u00e0y bi\u1ebft r\u1eb1ng l\u01b0\u1ee3ng n\u01b0\u1edbc hao h\u1ee5t trong qu\u00e1 tr\u00ecnh d\u1eabn t\u1eeb b\u1ec3 ch\u1ee9a xu\u1ed1ng h\u1ed9 d\u00e2n l\u00e0 0, 5% (l\u00e0m tr\u00f2n \u0111\u1ebfn ng\u00e0y). 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) B\u00e1n k\u00ednh c\u1ee7a b\u1ec3 ch\u1ee9a n\u01b0\u1edbc l\u00e0: 8:2 4 m 40 dm S\u1ed1 l\u00edt n\u01b0\u1edbc b\u1ec3 ch\u1ee9a \u0111\u01b0\u1ee3c khi b\u01a1m \u0111\u1ea7y l\u00e0: V 4 R3 4 403 256000 268 082,6 (l\u00edt) 33 3 b) S\u1ed1 l\u00edt n\u01b0\u1edbc to\u00e0 nh\u00e0 chung c\u01b0 d\u00f9ng m\u1ed9t ng\u00e0y l\u00e0: 608 . 18, 5 11248 (l\u00edt) S\u1ed1 ng\u00e0y m\u00e0 l\u01b0\u1ee3ng n\u01b0\u1edbc trong b\u1ec3 \u0111\u1ee7 d\u00f9ng cho c\u00e1c h\u1ed9 d\u00e2n to\u00e0 nh\u00e0 chung c\u01b0 l\u00e0: 256000 1 0, 5% : 11248 23 (ng\u00e0y) 3 C\u00e2u 7. (0,75 \u0111i\u1ec3m) B\u1ea1n A c\u00f3 m\u1ed9t \u1ed5 kh\u00f3a xe \u0111\u1ea1p nh\u01b0 trong h\u00ecnh. \u1ed5 kh\u00f3a c\u00f3 c\u00e1c s\u1ed1 t\u1eeb 0 \u0111\u1ebfn 9 tr\u00ean m\u1ed7i v\u00f2ng quay. Kh\u00f3a s\u1ebd k\u00eau t\u00e1ch nh\u1eb9 khi b\u1ea1n A quay l\u00ean hay quay xu\u1ed1ng m\u1ed9t s\u1ed1 tr\u00ean m\u1ed7i v\u00f2ng k\u1ec3 c\u1ea3 khi quay t\u1eeb 0 \u0111\u1ebfn 9 hay ng\u01b0\u1ee3c l\u1ea1i. Khi nh\u00ecn v\u00e0o \u1ed5 kh\u00f3a th\u00ec A th\u1ea5y c\u00f3 c\u00e1c s\u1ed1 m\u1ed7i v\u00f2ng \u0111ang \u1edf v\u1ecb tr\u00ed 9 0 4 nh\u01b0 h\u00ecnh v\u1ebd. M\u00e3 kh\u00f3a A \u0111\u00e3 c\u00e0i l\u00e0 5 8 7. a) Em h\u00e3y t\u00ednh s\u1ed1 ti\u1ebfng t\u00e1ch \u00edt nh\u1ea5t khi A c\u1ea7n \u0111\u1ec3 m\u1edf \u1ed5 kh\u00f3a. b) B\u1ea1n c\u1ee7a A c\u0169ng \u0111\u00e3 m\u1edf \u0111\u01b0\u1ee3c kh\u00f3a t\u1eeb v\u1ecb tr\u00ed 9 0 4 v\u1edbi s\u1ed1 ti\u1ebfng t\u00e1ch l\u00e0 nhi\u1ec1u nh\u1ea5t. T\u00ednh s\u1ed1 ti\u1ebfng t\u00e1ch trung b\u00ecnh c\u1ea7n \u0111\u1ec3 m\u1edf \u0111\u01b0\u1ee3c \u1ed5 kh\u00f3a. Xem nh\u01b0 g\u1ea7n v\u1edbi trung b\u00ecnh c\u1ed9ng c\u1ee7a s\u1ed1 ti\u1ebfng \u00edt nh\u1ea5t v\u00e0 nhi\u1ec1u nh\u1ea5t. L\u1eddi gi\u1ea3i a) T\u1eeb s\u1ed1 9 \u0111\u1ec3 quay \u0111\u1ebfn s\u1ed1 5 c\u00e1ch \u0111i \u00edt nh\u1ea5t 9 8 7 6 5 c\u00f3 4 ti\u1ebfng t\u00e1ch T\u1eeb s\u1ed1 0 \u0111\u1ec3 quay \u0111\u1ebfn s\u1ed1 8 c\u00e1ch \u0111i \u00edt nh\u1ea5t 0 9 8 c\u00f3 2 ti\u1ebfng t\u00e1ch 5 c\u00f3 6 ti\u1ebfng T\u1eeb s\u1ed1 4 \u0111\u1ec3 quay \u0111\u1ebfn s\u1ed1 7 c\u00e1ch \u0111i \u00edt nh\u1ea5t 4 5 6 7 c\u00f3 3 ti\u1ebfng t\u00e1ch C\u1ea7n \u00edt nh\u1ea5t 4 2 3 9 ti\u1ebfng t\u00e1ch khi A c\u1ea7n \u0111\u1ec3 m\u1edf \u1ed5 kh\u00f3a. b) T\u1eeb s\u1ed1 9 \u0111\u1ec3 quay \u0111\u1ebfn s\u1ed1 5 c\u00e1ch \u0111i nhi\u1ec1u nh\u1ea5t 9 0 1 2 3 4 t\u00e1ch T\u1eeb s\u1ed1 0 \u0111\u1ec3 quay \u0111\u1ebfn s\u1ed1 8 c\u00e1ch \u0111i nhi\u1ec1u nh\u1ea5t 0 1 2 3 4 5 6 7 8 c\u00f3 8 ti\u1ebfng t\u00e1ch T\u1eeb s\u1ed1 4 \u0111\u1ec3 quay \u0111\u1ebfn s\u1ed1 7 c\u00e1ch \u0111i nhi\u1ec1u nh\u1ea5t 4 3 2 1 0 9 8 7 c\u00f3 7 ti\u1ebfng t\u00e1ch C\u1ea7n nhi\u1ec1u nh\u1ea5t 6 8 7 21ti\u1ebfng t\u00e1ch khi b\u1ea1n A c\u1ea7n \u0111\u1ec3 m\u1edf \u1ed5 kh\u00f3a. T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N 8","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH S\u1ed1 ti\u1ebfng t\u00e1ch trung b\u00ecnh c\u1ea7n \u0111\u1ec3 m\u1edf \u0111\u01b0\u1ee3c \u1ed5 kh\u00f3a l\u00e0: 9 21 : 2 15 (ti\u1ebfng t\u00e1ch) C\u00e2u 8. (3 \u0111i\u1ec3m) Cho \u0111i\u1ec3m A n\u1eb1m ngo\u00e0i \u0111\u01b0\u1eddng tr\u00f2n (O; R) , k\u1ebb c\u00e1c ti\u1ebfp tuy\u1ebfn AB , AC v\u1edbi (O) (l\u00e0 c\u00e1c ti\u1ebfp \u0111i\u1ec3m). V\u1ebd \u0111\u01b0\u1eddng k\u00ednh CE c\u1ee7a (O) . G\u1ecdi H l\u00e0 giao \u0111i\u1ec3m c\u1ee7a OA v\u00e0 BC . a) Ch\u1ee9ng minh t\u1ee9 gi\u00e1c ABOC n\u1ed9i ti\u1ebfp v\u00e0 BE\/\/OA . b) AE c\u1eaft (O) t\u1ea1i D (kh\u00e1c E ), BD c\u1eaft OA t\u1ea1i M . Ch\u1ee9ng minh r\u1eb1ng MAD = MBA v\u00e0 AHD = ACD. c) V\u1ebd EI vu\u00f4ng g\u1edbi v\u1edbi OA t\u1ea1i I ; v\u1ebd DK l\u00e0 \u0111\u01b0\u1eddng k\u00ednh c\u1ee7a (O) . Ch\u1ee9ng minh K ,I ,B th\u1eb3ng h\u00e0ng. L\u1eddi gi\u1ea3i a) Ch\u1ee9ng minh t\u1ee9 gi\u00e1c ABOC n\u1ed9i ti\u1ebfp v\u00e0 BE\/\/OA . X\u00e9t t\u1ee9 gi\u00e1c ABOC , c\u00f3: \uf0ef\uf0ecABO = 90\uf0b0(OB \u22a5 AB) \uf0ed \uf0ef\uf0ee ACO = 90\uf0b0(OC \u22a5 AC ) \uf0de ABO + ACO = 180\uf0b0 \uf0de T\u1ee9 gi\u00e1c ABOC n\u1ed9i ti\u1ebfp v\u00ec c\u00f3 hai g\u00f3c \u0111\u1ed1i b\u00f9 nhau. ( )Ta c\u00f3: CBE l\u00e0 g\u00f3c n\u1ed9i ti\u1ebfp ch\u1eafc n\u1eeda \u0111\u01b0\u1eddng tr\u00f2n O N\u00ean: CBE = 90\uf0b0 . \uf0de BE \u22a5 BC . Ta c\u00f3: + AB = AC (t\u00ednh ch\u1ea5t hai ti\u1ebfp tuy\u1ebfn c\u1eaft nhau t\u1ea1i A ) + OB = OC = R T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N 9","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH Suy ra: OA l\u00e0 \u0111\u01b0\u1eddng trung tr\u1ef1c c\u1ee7a BC \uf0de OA \u22a5 BC M\u00e0: BE \u22a5 BC(cmt) N\u00ean: BE\/\/OA . b) AE c\u1eaft (O) t\u1ea1i D (kh\u00e1c E ), BD c\u1eaft OA t\u1ea1i M . Ch\u1ee9ng minh r\u1eb1ng MAD = MBA v\u00e0 AHD = ACD. Ta c\u00f3: + MAD = DEB (hai g\u00f3c so le trong v\u00e0 BE\/\/OA ) + MBA = DEB (g\u00f3c t\u1ea1o b\u1edfi ti\u1ebfp tuy\u1ebfn v\u00e0 d\u00e2y cung v\u00e0 g\u00f3c n\u1ed9i ti\u1ebfp c\u00f9ng ch\u1eafn BD ) Suy ra: MAD = MBA . X\u00e9t \uf044ABD v\u00e0 \uf044AEB , c\u00f3: + BAD chung + MBA = DEB (g\u00f3c t\u1ea1o b\u1edfi ti\u1ebfp tuy\u1ebfn v\u00e0 d\u00e2y cung v\u00e0 g\u00f3c n\u1ed9i ti\u1ebfp c\u00f9ng ch\u1eafn BD ) Suy ra: \uf044ABD \uf044AEB (g \u2013 g) \uf0de AB = AD AE AB \uf0de AB2 = AD.AE (1) . X\u00e9t \uf044ABO vu\u00f4ng t\u1ea1i B c\u00f3 BH l\u00e0 \u0111\u01b0\u1eddng cao \uf0de AB2 = AH.AO (2) T\u1eeb (1) , (2) suy ra: AD.AE = AH.AO . X\u00e9t \uf044ADH v\u00e0 \uf044AOE , c\u00f3: + EAO chung + AD = AH ( AD.AE = AH.AO ). AO AE Suy ra: \uf044ADH \uf044AOE (c \u2013 g \u2013 c) \uf0de AHD = AEO. M\u00e0 ACD = AEO. ( g\u00f3c t\u1ea1o b\u1edfi ti\u1ebfp tuy\u1ebfn v\u00e0 d\u00e2y cung v\u00e0 g\u00f3c n\u1ed9i ti\u1ebfp c\u00f9ng ch\u1eafn CD ) Suy ra: AHD = ACD. c) V\u1ebd EI vu\u00f4ng g\u1edbi v\u1edbi OA t\u1ea1i I ; v\u1ebd DK l\u00e0 \u0111\u01b0\u1eddng k\u00ednh c\u1ee7a (O) . Ch\u1ee9ng minh K ,I ,B th\u1eb3ng h\u00e0ng. T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N 10","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH X\u00e9t t\u1ee9 gi\u00e1c OHDE , ta c\u00f3: AHD = AEO. Suy ra: t\u1ee9 gi\u00e1c OHDE n\u1ed9i ti\u1ebfp v\u00ec g\u00f3c ngo\u00e0i b\u1eb1ng g\u00f3c trong \u0111\u1ed1i di\u1ec7n. D\u1ec5 d\u00e0ng ch\u1ee9ng minh t\u1ee9 gi\u00e1c BEIH l\u00e0 h\u00ecnh ch\u1eef nh\u1eadt Suy ra: IBE = IHE . M\u00e0 ODE = IHE (t\u1ee9 gi\u00e1c OHDE n\u1ed9i ti\u1ebfp) \uf0de IBE = ODE M\u1eb7t kh\u00e1c: KBE = ODE (hai g\u00f3c n\u1ed9i ti\u1ebfp c\u00f9ng ch\u1eafn EK ). N\u00ean IBE = KBE \uf0de BI \uf0ba BK V\u1eady K ,I ,B th\u1eb3ng h\u00e0ng. ----H\u1ebeT--- T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N 11","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH S\u1ede GD&\u0110T TP H\u1ed2 CH\u00cd MINH \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 PH\u00d2NG GD&\u0110T QU\u1eacN B\u00ccNH T\u00c2N N\u0102M H\u1eccC: 2023 \u2013 2024 \u0110\u1ec0 THAM KH\u1ea2O M\u00d4N: TO\u00c1N 9 \u0110\u1ec1 thi g\u1ed3m 8 c\u00e2u h\u1ecfi t\u1ef1 lu\u1eadn. M\u00c3 \u0110\u1ec0: Qu\u1eadn B\u00ecnh T\u00e2n \u2013 2 Th\u1eddi gian: 120 ph\u00fat (kh\u00f4ng k\u1ec3 th\u1eddi gian ph\u00e1t \u0111\u1ec1) C\u00e2u 1. (1,5 \u0111i\u1ec3m). Cho parabol (P) : y = x2 v\u00e0 \u0111\u01b0\u1eddng th\u1eb3ng (d) : y = 3x \u2212 2 . a) V\u1ebd (P) v\u00e0 (d) tr\u00ean c\u00f9ng h\u1ec7 tr\u1ee5c t\u1ecda \u0111\u1ed9. b) T\u00ecm t\u1ecda \u0111\u1ed9 giao \u0111i\u1ec3m c\u1ee7a (P) v\u00e0 (d) b\u1eb1ng ph\u00e9p t\u00ednh. C\u00e2u 2. (1 \u0111i\u1ec3m). Cho ph\u01b0\u01a1ng tr\u00ecnh: x2 \u2212 5x + 5 = 0 . G\u1ecdi x1 , x2 l\u00e0 hai nghi\u1ec7m (n\u1ebfu c\u00f3). 3 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 x1 C\u00e2u 3. (0,75 \u0111i\u1ec3m). B\u00ean c\u1ea1nh vi\u1ec7c xem \u0111ua ch\u00f3, nhi\u1ec1u ng\u01b0\u1eddi th\u00edch tham gia ch\u01b0\u01a1ng tr\u00ecnh d\u1ef1 th\u01b0\u1edfng \u0111\u1ec3 cu\u1ed9c \u0111ua th\u00eam ph\u1ea5n kh\u00edch. Tr\u01b0\u1edbc khi ch\u1ecdn ch\u00f3 \u0111\u1ec3 \u0111\u1ea1t th\u01b0\u1edfng, b\u1ea1n s\u1ebd \u0111\u01b0\u1ee3c ph\u00e1t m\u1ed9t cu\u1ed1n t\u00e0i li\u1ec7u c\u00f3 gi\u1edbi thi\u1ec7u kh\u00e1 chi ti\u1ebft v\u1ec1 t\u1eebng ch\u00fa ch\u00f3, t\u1eeb nh\u1eefng th\u00e0nh t\u00edch t\u1ed1t, x\u1ea5u \u0111\u1ebfn t\u00ecnh tr\u1ea1ng s\u1ee9c kh\u1ecfe, cu\u1ed9c \u0111ua g\u1ea7n nh\u1ea5t c\u00f3 tham gia\u2026s\u1ebd gi\u00fap b\u1ea1n d\u1ec5 d\u00e0ng ch\u1ecdn l\u1ef1a. V\u00e0 c\u00f3 5 ki\u1ec3u th\u1eafng gi\u1ea3i \u00e1p d\u1ee5ng cho c\u00e1c \u0111\u1ee3t \u0111ua l\u00e0: - Th\u1eafng nh\u1ea5t l\u00e0 Win n\u1ebfu con ch\u00f3 m\u00e0 b\u1ea1n ch\u1ecdn v\u1ec1 nh\u1ea5t. - Th\u1eafng nh\u1ea5t \u2013 nh\u00ec (Exacta) l\u00e0 2 con ch\u00f3 b\u1ea1n ch\u1ecdn v\u1ec1 nh\u1ea5t \u2013 nh\u00ec theo \u0111\u00fang th\u1ee9 t\u1ef1. - Th\u1eafng nh\u1ea5t \u2013 nh\u00ec \u2013 ba (Trifecta) l\u00e0 c\u1ea3 3 con ch\u00f3 b\u1ea1n ch\u1ecdn \u0111\u1ec1u v\u1ec1 3 th\u1ee9 h\u1ea1ng \u0111\u1ea7u theo \u0111\u00fang th\u1ee9 t\u1ef1. C\u00e1c gi\u1ea3i ti\u1ebfp theo l\u00e0 Quartet (4) , Superfecta (6) . C\u00e0ng l\u00ean cao c\u00e0ng tr\u00fang \u0111\u1eadm. H\u1ecfi trong m\u1ed9t \u0111\u1ee3t \u0111ua c\u00f3 8 ch\u00fa khuy\u1ec3n, ta c\u00f3 bao nhi\u00eau c\u00e1ch ch\u1ecdn v\u00e9 d\u1ef1 th\u01b0\u1edfng theo gi\u1ea3i nh\u1ea5t \u2013 nh\u00ec \u2013 ba? C\u00e2u 4. C\u00e2u 4. (1 \u0111i\u1ec3m). Cu\u1ed1i n\u0103m 2009 , m\u1ed9t b\u1ea3n b\u00e1o c\u00e1o \u0111\u01b0\u1ee3c tr\u00ecnh l\u00ean ch\u00ednh ph\u1ee7 Anh. Theo \u0111\u00f3, n\u1ebfu nhi\u1ec7t \u0111\u1ed9 tr\u00e1i \u0111\u1ea5t t\u0103ng l\u00ean 2\uf0b0C th\u00ec t\u1ed5ng gi\u00e1 tr\u1ecb kinh t\u1ebf th\u1ebf gi\u1edbi s\u1ebd b\u1ecb gi\u1ea3m \u0111i 3% n\u1ebfu nhi\u1ec7t \u0111\u1ed9 t\u0103ng l\u00ean 5\uf0b0C kinh t\u1ebf s\u1ebd gi\u1ea3m \u0111i 10% . T\u1eeb \u0111\u00f3, th\u00f4ng qua nghi\u00ean c\u1ee9u m\u1ed9t nh\u00f3m nh\u00e0 kinh t\u1ebf h\u1ecdc \u0111\u00e3 \u0111\u01b0a ra d\u1ef1 \u0111o\u00e1n v\u1ec1 m\u1ed1i li\u00ean h\u1ec7 gi\u1eefa nhi\u1ec7t \u0111\u1ed9 th\u1ebf gi\u1edbi v\u00e0 t\u1ed5ng gi\u00e1 tr\u1ecb kinh t\u1ebf c\u1ee7a th\u1ebf gi\u1edbi. K\u1ebft qu\u1ea3 nghi\u00ean c\u1ee9u \u0111\u01b0a ra r\u1eb1ng t\u1ed5ng gi\u00e1 tr\u1ecb kinh t\u1ebf b\u1ecb gi\u1ea3m y% l\u00e0 h\u00e0m s\u1ed1 b\u1eadc nh\u1ea5t theo x l\u00e0 nhi\u1ec7t \u0111\u1ed9 t\u0103ng l\u00ean c\u1ee7a Tr\u00e1i \u0111\u1ea5t (t\u00ednh theo \uf0b0C ). a) X\u00e1c \u0111\u1ecbnh m\u1ed1i li\u00ean h\u1ec7 gi\u1eefa y v\u00e0 x . b) Theo nghi\u00ean c\u1ee9u \u0111\u00f3, t\u1ed5ng gi\u00e1 tr\u1ecb kinh t\u1ebf s\u1ebd gi\u1ea3m bao nhi\u00eau n\u1ebfu th\u1ebf gi\u1edbi t\u0103ng th\u00eam 10\uf0b0C (l\u00e0m tr\u00f2n \u0111\u1ebfn h\u00e0ng \u0111\u01a1n v\u1ecb). C\u00e2u 5. C\u00e2u 5. (0,75 \u0111i\u1ec3m). L\u1edbp 9A d\u1ef1 \u0111\u1ecbnh t\u1ed5 ch\u1ee9c li\u00ean hoan l\u1edbp cu\u1ed1i n\u0103m, trong ph\u1ea7n n\u01b0\u1edbc c\u1ea7n 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 chu\u1ea9n b\u1ecb 42 ly tr\u00e0 s\u1eefa. \u0110\u1ec3 ti\u1ebft ki\u1ec7m chi ph\u00ed l\u1edbp \u0111\u00e3 t\u00ecm hi\u1ec3u gi\u00e1 c\u1ee7a hai c\u1eeda h\u00e0ng A v\u00e0 B nh\u01b0 sau: - C\u1eeda h\u00e0ng A : mua 5 ly b\u1ea5t k\u00ec s\u1ebd \u0111\u01b0\u1ee3c t\u1eb7ng 1 ly (c\u00f9ng lo\u1ea1i) v\u00e0 h\u00f3a \u0111\u01a1n tr\u00ean 400000 \u0111\u1ed3ng th\u00ec s\u1ebd gi\u1ea3m th\u00eam 10% tr\u00ean h\u00f3a \u0111\u01a1n. - C\u1eeda h\u00e0ng B : ch\u1ec9 khuy\u1ebfn m\u00e3i khi \u0111\u1eb7t qua App Grab Food mua t\u1eeb 10 ly tr\u1edf l\u00ean th\u00ec gi\u1ea3m 30% m\u1ed7i ly so v\u1edbi gi\u00e1 ni\u00eam y\u1ebft v\u00e0 ph\u00ed giao h\u00e0ng th\u00ec kh\u00e1ch t\u1ef1 tr\u1ea3 theo kho\u1ea3ng c\u00e1ch t\u1eeb c\u1eeda h\u00e0ng \u0111\u1ebfn n\u01a1i nh\u1eadn h\u00e0ng. H\u1ecfi l\u1edbp 9A n\u00ean mua \u1edf c\u1eeda h\u00e0ng n\u00e0o s\u1ebd ti\u1ebft ki\u1ec7m h\u01a1n v\u00e0 ti\u1ebft ki\u1ec7m h\u01a1n bao nhi\u00eau ti\u1ec1n? Bi\u1ebft gi\u00e1 ni\u00eam y\u1ebft 1 ly tr\u00e0 s\u1eefa \u1edf hai c\u1eeda h\u00e0ng \u0111\u1ec1u l\u00e0 30000 \u0111\u1ed3ng, kho\u1ea3ng c\u00e1ch t\u1eeb \u0111\u1ecba \u0111i\u1ec3m li\u00ean hoan \u0111\u1ebfn c\u1eeda h\u00e0ng B l\u00e0 9A . Ph\u00ed giao h\u00e0ng \u0111\u01b0\u1ee3c t\u00ednh theo b\u1ea3ng sau: Kho\u1ea3ng c\u00e1ch Gi\u00e1 ti\u1ec1n (\u0111\u1ed3ng) D\u01b0\u1edbi 10km 25000 T\u1eeb 10km \u0111\u1ebfn 20km 26000 T\u1eeb 20km \u0111\u1ebfn 40km 30000 Tr\u00ean 40km 5% gi\u00e1 tr\u1ecb \u0111\u01a1n h\u00e0ng C\u00e2u 6. C\u00e2u 6. (1 \u0111i\u1ec3m). \u0110\u1ec3 l\u00e0m m\u1ed9t c\u00e1i g\u00e0u t\u00e1t n\u01b0\u1edbc c\u00f3 d\u1ea1ng h\u00ecnh n\u00f3n (h\u00ecnh 1), b\u00e1c An d\u00f9ng m\u1ed9t t\u1ea5m t\u00f4n h\u00ecnh tam gi\u00e1c OMN c\u00e2n t\u1ea1i O c\u00f3 c\u1ea1nh b\u00ean OM = 6dm , g\u00f3c MON = 1200 (h\u00ecnh 2 ). B\u00e1c x\u00e1c \u0111\u1ecbnh trung \u0111i\u1ec3m H c\u1ee7a MN , v\u1ebd cung tr\u00f2n t\u00e2m O b\u00e1n k\u00ednh OH c\u1eaft c\u00e1c c\u1ea1nh OM , ON l\u1ea7n l\u01b0\u1ee3t t\u1ea1i A , B . Sau \u0111\u00f3 b\u00e1c c\u1eaft b\u1ecf ph\u1ea7n g\u1ea1ch s\u1ecdc, cu\u1ed9n ph\u1ea7n c\u00f2n l\u1ea1i c\u1ee7a t\u1ea5m t\u00f4n sao cho m\u00e9p OA tr\u00f9ng kh\u00edt v\u1edbi m\u00e9p OB t\u1ea1o th\u00e0nh chi\u1ebfc g\u00e0u (gi\u1ea3 s\u1eed ph\u1ea7n di\u1ec7n t\u00edch c\u1ee7a m\u00e9p n\u1ed1i kh\u00f4ng \u0111\u00e1ng k\u1ec3). H\u1ecfi khi m\u00fac \u0111\u1ea7y th\u00ec chi\u1ebfc g\u00e0u ch\u1ee9a \u0111\u01b0\u1ee3c bao nhi\u00eau l\u00edt n\u01b0\u1edbc? (k\u1ebft qu\u1ea3 l\u00e0m tr\u00f2n \u0111\u1ebfn m\u1ed9t ch\u1eef s\u1ed1 th\u1eadp ph\u00e2n, l\u1ea5y \uf070 \uf0bb 3,14 ) C\u00e2u 7. C\u00e2u 7. (1 \u0111i\u1ec3m). V\u00e0o d\u1ecbp khai tr\u01b0\u01a1ng, nh\u00e0 s\u00e1ch khuy\u1ebfn m\u00e3i m\u1ed7i c\u00e2y vi\u1ebft bi \u0111\u01b0\u1ee3c gi\u1ea3m 20% so v\u1edbi gi\u00e1 ni\u00eam y\u1ebft, c\u00f2n m\u1ed7i quy\u1ec3n t\u1eadp gi\u1ea3m 10% so v\u1edbi gi\u00e1 ni\u00eam y\u1ebft. B\u1ea1n An v\u00e0o nh\u00e0 s\u00e1ch mua 20 quy\u1ec3n t\u1eadp v\u00e0 10 c\u00e2y vi\u1ebft bi. Khi t\u00ednh ti\u1ec1n, b\u1ea1n An \u0111\u01b0a 175000 \u0111\u1ed3ng v\u00e0 \u0111\u01b0\u1ee3c th\u1ed1i l\u1ea1i 3000 \u0111\u1ed3ng. T\u00ednh gi\u00e1 ni\u00eam y\u1ebft c\u1ee7a m\u1ed7i quy\u1ec3n t\u1eadp v\u00e0 m\u1ed7i c\u00e2y vi\u1ebft bi m\u00e0 b\u1ea1n An \u0111\u00e3 mua. Bi\u1ebft r\u1eb1ng khi An nh\u00ecn v\u00e0o h\u00f3a \u0111\u01a1n, t\u1ed5ng s\u1ed1 ti\u1ec1n ph\u1ea3i tr\u1ea3 khi ch\u01b0a gi\u1ea3m gi\u00e1 l\u00e0 195000 \u0111\u1ed3ng. C\u00e2u 8. C\u00e2u 8. (3 \u0111i\u1ec3m). Cho tam gi\u00e1c ABC c\u00f3 3 g\u00f3c nh\u1ecdn v\u00e0 AB \uf03c AC . V\u1ebd c\u00e1c \u0111\u01b0\u1eddng cao AD , BE , CF c\u1ee7a tam gi\u00e1c \u0111\u00f3. G\u1ecdi H l\u00e0 giao \u0111i\u1ec3m c\u1ee7a c\u00e1c \u0111\u01b0\u1eddng cao v\u1eeba v\u1ebd. a) Ch\u1ee9ng minh: c\u00e1c t\u1ee9 gi\u00e1c AEHF v\u00e0 BFEC n\u1ed9i ti\u1ebfp. 2 T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH b) G\u1ecdi M , N l\u1ea7n l\u01b0\u1ee3t l\u00e0 trung \u0111i\u1ec3m c\u1ee7a c\u00e1c \u0111o\u1ea1n AH , BC . Ch\u1ee9ng minh: FM.FC = FN.FA. c) G\u1ecdi P , Q l\u1ea7n l\u01b0\u1ee3t l\u00e0 ch\u00e2n c\u00e1c \u0111\u01b0\u1eddng vu\u00f4ng g\u00f3c k\u1ebb t\u1eeb M , N \u0111\u1ebfn \u0111\u01b0\u1eddng th\u1eb3ng DF . Ch\u1ee9ng minh r\u1eb1ng \u0111\u01b0\u1eddng tr\u00f2n \u0111\u01b0\u1eddng k\u00ednh PQ \u0111i qua giao \u0111i\u1ec3m c\u1ee7a FE v\u00e0 MN . ----H\u1ebeT---- H\u01af\u1edaNG D\u1eaaN GI\u1ea2I C\u00e2u 1. (1,5 \u0111i\u1ec3m). Cho parabol ( P) : y = x2 v\u00e0 \u0111\u01b0\u1eddng th\u1eb3ng (d ) : y = 3x \u2212 2 . a) V\u1ebd ( P) v\u00e0 (d ) tr\u00ean c\u00f9ng h\u1ec7 tr\u1ee5c t\u1ecda \u0111\u1ed9. b) T\u00ecm t\u1ecda \u0111\u1ed9 giao \u0111i\u1ec3m c\u1ee7a ( P) v\u00e0 (d ) b\u1eb1ng ph\u00e9p t\u00ednh. L\u1eddi gi\u1ea3i T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N 3","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH a) V\u1ebd ( P) v\u00e0 (d ) tr\u00ean c\u00f9ng h\u1ec7 tr\u1ee5c t\u1ecda \u0111\u1ed9. BGT: x \u22122 \u22121 0 1 2 y = x2 4 1 0 1 4 x0 1 y = 3x \u2212 2 \u22122 1 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 ) l\u00e0: x2 = 3x \u2212 2 \uf0db x2 \u2212 3x + 2 = 0 \uf0db \uf0e9 x = 1 \uf0ea\uf0eb x = 2 Thay x = 1 v\u00e0o y = x2 ta \u0111\u01b0\u1ee3c: y = 12 = 1 Thay x = 2 v\u00e0o y = x2 ta \u0111\u01b0\u1ee3c: y = 22 = 4 V\u1eady (1;1) , (2;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 5x + 5 = 0. G\u1ecdi x1, x2 l\u00e0 hai nghi\u1ec7m (n\u1ebfu c\u00f3). 3 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 x1 L\u1eddi gi\u1ea3i Ph\u01b0\u01a1ng tr\u00ecnh: x2 \u2212 5x + 5 = 0 \uf0e6 a = 1;b = \u2212 5;c = 5\uf0f6 3 \uf0e7\uf0e8 3 \uf0f8\uf0f7 V\u00ec \uf044 = b2 \u2212 4ac = (\u22125)2 \u2212 4.1. 5 = 55 \uf03e 0 33 N\u00ean ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m x1, x2 T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N 4","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH \uf0ec = x1 + x2 = \u2212b = \u2212 ( \u22125) = 5 \uf0efS a 5 Theo \u0111\u1ecbnh \u2013 \uf0ef 1 \uf0ed l\u00fd Vi et, ta c\u00f3: \uf0ef = x1.x2 = C = 3 = 5 \uf0efP a 1 3 \uf0ee A = x13 + x23 x1.x2 Ta c\u00f3: A = ( x1 + x2 )3 \u2212 3x1.x2 ( x1 + x2 ) x1.x2 53 \u2212 3. 5 .5 A = 3 = 60 5 3 V\u1eady A = 60 C\u00e2u 3. (0,75 \u0111i\u1ec3m). B\u00ean c\u1ea1nh vi\u1ec7c xem \u0111ua ch\u00f3, nhi\u1ec1u ng\u01b0\u1eddi th\u00edch tham gia ch\u01b0\u01a1ng tr\u00ecnh d\u1ef1 th\u01b0\u1edfng \u0111\u1ec3 cu\u1ed9c \u0111ua th\u00eam ph\u1ea5n kh\u00edch. Tr\u01b0\u1edbc khi ch\u1ecdn ch\u00f3 \u0111\u1ec3 \u0111\u1ea1t th\u01b0\u1edfng, b\u1ea1n s\u1ebd \u0111\u01b0\u1ee3c ph\u00e1t m\u1ed9t cu\u1ed1n t\u00e0i li\u1ec7u c\u00f3 gi\u1edbi thi\u1ec7u kh\u00e1 chi ti\u1ebft v\u1ec1 t\u1eebng ch\u00fa ch\u00f3, t\u1eeb nh\u1eefng th\u00e0nh t\u00edch t\u1ed1t, x\u1ea5u \u0111\u1ebfn t\u00ecnh tr\u1ea1ng s\u1ee9c kh\u1ecfe, cu\u1ed9c \u0111ua g\u1ea7n nh\u1ea5t c\u00f3 tham gia\u2026s\u1ebd gi\u00fap b\u1ea1n d\u1ec5 d\u00e0ng ch\u1ecdn l\u1ef1a. V\u00e0 c\u00f3 5 ki\u1ec3u th\u1eafng gi\u1ea3i \u00e1p d\u1ee5ng cho c\u00e1c \u0111\u1ee3t \u0111ua l\u00e0: - Th\u1eafng nh\u1ea5t l\u00e0 Win n\u1ebfu con ch\u00f3 m\u00e0 b\u1ea1n ch\u1ecdn v\u1ec1 nh\u1ea5t. - Th\u1eafng nh\u1ea5t \u2013 nh\u00ec (Exacta) l\u00e0 2 con ch\u00f3 b\u1ea1n ch\u1ecdn v\u1ec1 nh\u1ea5t \u2013 nh\u00ec theo \u0111\u00fang th\u1ee9 t\u1ef1. - Th\u1eafng nh\u1ea5t \u2013 nh\u00ec \u2013 ba (Trifecta) l\u00e0 c\u1ea3 3 con ch\u00f3 b\u1ea1n ch\u1ecdn \u0111\u1ec1u v\u1ec1 3 th\u1ee9 h\u1ea1ng \u0111\u1ea7u theo \u0111\u00fang th\u1ee9 t\u1ef1. C\u00e1c gi\u1ea3i ti\u1ebfp theo l\u00e0 Quartet (4) , Superfecta (6) . C\u00e0ng l\u00ean cao c\u00e0ng tr\u00fang \u0111\u1eadm. H\u1ecfi trong m\u1ed9t \u0111\u1ee3t \u0111ua c\u00f3 8 ch\u00fa khuy\u1ec3n, ta c\u00f3 bao nhi\u00eau c\u00e1ch ch\u1ecdn v\u00e9 d\u1ef1 th\u01b0\u1edfng theo gi\u1ea3i nh\u1ea5t \u2013 nh\u00ec \u2013 ba? L\u1eddi gi\u1ea3i Con v\u1ec1 nh\u1ea5t c\u00f3: 8 c\u00e1ch ch\u1ecdn Con v\u1ec1 nh\u00ec c\u00f3: 7 c\u00e1ch ch\u1ecdn Con v\u1ec1 ba c\u00f3: 6 c\u00e1ch ch\u1ecdn V\u1eady c\u00f3: 8.7.6 = 336 c\u00e1ch ch\u1ecdn v\u00e9 d\u1ef1 th\u01b0\u1edfng theo gi\u1ea3i nh\u1ea5t \u2013 nh\u00ec \u2013 ba. C\u00e2u 4. (1 \u0111i\u1ec3m). Cu\u1ed1i n\u0103m 2009 , m\u1ed9t b\u1ea3n b\u00e1o c\u00e1o \u0111\u01b0\u1ee3c tr\u00ecnh l\u00ean ch\u00ednh ph\u1ee7 Anh. Theo \u0111\u00f3, n\u1ebfu nhi\u1ec7t \u0111\u1ed9 tr\u00e1i \u0111\u1ea5t t\u0103ng l\u00ean 2\uf0b0C th\u00ec t\u1ed5ng gi\u00e1 tr\u1ecb kinh t\u1ebf th\u1ebf gi\u1edbi s\u1ebd b\u1ecb gi\u1ea3m \u0111i 3% n\u1ebfu nhi\u1ec7t \u0111\u1ed9 t\u0103ng l\u00ean 5\uf0b0C kinh t\u1ebf s\u1ebd gi\u1ea3m \u0111i 10% . T\u1eeb \u0111\u00f3, th\u00f4ng qua nghi\u00ean c\u1ee9u m\u1ed9t nh\u00f3m nh\u00e0 kinh t\u1ebf h\u1ecdc \u0111\u00e3 \u0111\u01b0a ra d\u1ef1 \u0111o\u00e1n v\u1ec1 m\u1ed1i li\u00ean h\u1ec7 gi\u1eefa nhi\u1ec7t \u0111\u1ed9 th\u1ebf gi\u1edbi v\u00e0 t\u1ed5ng gi\u00e1 tr\u1ecb kinh t\u1ebf c\u1ee7a th\u1ebf gi\u1edbi. K\u1ebft qu\u1ea3 nghi\u00ean c\u1ee9u \u0111\u01b0a ra r\u1eb1ng t\u1ed5ng gi\u00e1 tr\u1ecb kinh t\u1ebf b\u1ecb gi\u1ea3m y% l\u00e0 h\u00e0m s\u1ed1 b\u1eadc nh\u1ea5t theo x l\u00e0 nhi\u1ec7t \u0111\u1ed9 t\u0103ng 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\u00ean c\u1ee7a Tr\u00e1i \u0111\u1ea5t (t\u00ednh theo \uf0b0C ). a) X\u00e1c \u0111\u1ecbnh m\u1ed1i li\u00ean h\u1ec7 gi\u1eefa y v\u00e0 x . b) Theo nghi\u00ean c\u1ee9u \u0111\u00f3, t\u1ed5ng gi\u00e1 tr\u1ecb kinh t\u1ebf s\u1ebd gi\u1ea3m bao nhi\u00eau n\u1ebfu th\u1ebf gi\u1edbi t\u0103ng th\u00eam 10\uf0b0C (l\u00e0m tr\u00f2n \u0111\u1ebfn h\u00e0ng \u0111\u01a1n v\u1ecb). L\u1eddi gi\u1ea3i a) X\u00e1c \u0111\u1ecbnh m\u1ed1i li\u00ean h\u1ec7 gi\u1eefa y v\u00e0 x . G\u1ecdi h\u00e0m s\u1ed1 bi\u1ec3u th\u1ecb m\u1ed1i li\u00ean h\u1ec7 gi\u1eefa y v\u00e0 x l\u00e0: y = ax + b(a \uf0b9 0) Khi x = 2 th\u00ec y = 3n\u00ean ta c\u00f3: 3 = a.2 + b \uf0db 2.a + b = 3(1) Khi x = 5th\u00ec y = 10 n\u00ean ta c\u00f3: 10 = a.5 + b \uf0db 5.a + b = 10(2) T\u1eeb (1) v\u00e0 (2) ta c\u00f3 h\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh: \uf0ec2a + b = 3 \uf0ed\uf0ee5a + b = 10 Gi\u1ea3i h\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh ta \u0111\u01b0\u1ee3c: a = 7 ;b = \u22125 33 V\u1eady h\u00e0m s\u1ed1 bi\u1ec3u th\u1ecb m\u1ed1i li\u00ean h\u1ec7 gi\u1eefa y v\u00e0 x l\u00e0: y = 7 x \u2212 5 33 y= 7x\u22125 y = 7 .10 \u2212 5 \uf0bb 22 b) Thay x = 10 v\u00e0o 3 3 ta \u0111\u01b0\u1ee3c: 3 3 V\u1eady t\u1ed5ng gi\u00e1 tr\u1ecb kinh t\u1ebf s\u1ebd gi\u1ea3m kho\u1ea3ng 22% n\u1ebfu th\u1ebf gi\u1edbi t\u0103ng th\u00eam 10\uf0b0C C\u00e2u 5. (0,75 \u0111i\u1ec3m). L\u1edbp 9A d\u1ef1 \u0111\u1ecbnh t\u1ed5 ch\u1ee9c li\u00ean hoan l\u1edbp cu\u1ed1i n\u0103m, trong ph\u1ea7n n\u01b0\u1edbc c\u1ea7n chu\u1ea9n b\u1ecb 42 ly tr\u00e0 s\u1eefa. \u0110\u1ec3 ti\u1ebft ki\u1ec7m chi ph\u00ed l\u1edbp \u0111\u00e3 t\u00ecm hi\u1ec3u gi\u00e1 c\u1ee7a hai c\u1eeda h\u00e0ng A v\u00e0 B nh\u01b0 sau: - C\u1eeda h\u00e0ng A : mua 5 ly b\u1ea5t k\u00ec s\u1ebd \u0111\u01b0\u1ee3c t\u1eb7ng 1 ly (c\u00f9ng lo\u1ea1i) v\u00e0 h\u00f3a \u0111\u01a1n tr\u00ean 400000 \u0111\u1ed3ng th\u00ec s\u1ebd gi\u1ea3m th\u00eam 10% tr\u00ean h\u00f3a \u0111\u01a1n. - C\u1eeda h\u00e0ng B : ch\u1ec9 khuy\u1ebfn m\u00e3i khi \u0111\u1eb7t qua App Grab Food mua t\u1eeb 10 ly tr\u1edf l\u00ean th\u00ec gi\u1ea3m 30% m\u1ed7i ly so v\u1edbi gi\u00e1 ni\u00eam y\u1ebft v\u00e0 ph\u00ed giao h\u00e0ng th\u00ec kh\u00e1ch t\u1ef1 tr\u1ea3 theo kho\u1ea3ng c\u00e1ch t\u1eeb c\u1eeda h\u00e0ng \u0111\u1ebfn n\u01a1i nh\u1eadn h\u00e0ng. H\u1ecfi l\u1edbp 9A n\u00ean mua \u1edf c\u1eeda h\u00e0ng n\u00e0o s\u1ebd ti\u1ebft ki\u1ec7m h\u01a1n v\u00e0 ti\u1ebft ki\u1ec7m h\u01a1n bao nhi\u00eau ti\u1ec1n? Bi\u1ebft gi\u00e1 ni\u00eam y\u1ebft 1 ly tr\u00e0 s\u1eefa \u1edf hai c\u1eeda h\u00e0ng \u0111\u1ec1u l\u00e0 30000 \u0111\u1ed3ng, kho\u1ea3ng c\u00e1ch t\u1eeb \u0111\u1ecba \u0111i\u1ec3m li\u00ean hoan \u0111\u1ebfn c\u1eeda h\u00e0ng B l\u00e0 9A . Ph\u00ed giao h\u00e0ng \u0111\u01b0\u1ee3c t\u00ednh theo b\u1ea3ng sau: Kho\u1ea3ng c\u00e1ch Gi\u00e1 ti\u1ec1n (\u0111\u1ed3ng) D\u01b0\u1edbi 10km 25000 T\u1eeb 10km \u0111\u1ebfn 20km 26000 T\u1eeb 20km \u0111\u1ebfn 40km 30000 5% gi\u00e1 tr\u1ecb \u0111\u01a1n h\u00e0ng Tr\u00ean 40km T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N 6","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH L\u1eddi gi\u1ea3i S\u1ed1 ti\u1ec1n ph\u1ea3i tr\u1ea3 n\u1ebfu mua 42 ly tr\u00e0 s\u1eefa \u1edf c\u1eeda h\u00e0ng A l\u00e0: (42 \u2013 7).30000(1\u201310%) = 945000 (\u0111\u1ed3ng) S\u1ed1 ti\u1ec1n ph\u1ea3i tr\u1ea3 n\u1ebfu mua 42 ly tr\u00e0 s\u1eefa \u1edf c\u1eeda h\u00e0ng B l\u00e0: 42.30000(1\u2013 30%) + 26000 = 908000 (\u0111\u1ed3ng) L\u1edbp 9A n\u00ean mua \u1edf c\u1eeda h\u00e0ng B s\u1ebd ti\u1ebft ki\u1ec7m h\u01a1n v\u00ec 908000 \uf03c 945000 V\u00e0 ti\u1ebft ki\u1ec7m h\u01a1n: 945000 \u2013 908000 = 37000 (\u0111\u1ed3ng) C\u00e2u 6. (1 \u0111i\u1ec3m). \u0110\u1ec3 l\u00e0m m\u1ed9t c\u00e1i g\u00e0u t\u00e1t n\u01b0\u1edbc c\u00f3 d\u1ea1ng h\u00ecnh n\u00f3n (h\u00ecnh 1 ), b\u00e1c An d\u00f9ng m\u1ed9t t\u1ea5m t\u00f4n h\u00ecnh tam gi\u00e1c OMN c\u00e2n t\u1ea1i O c\u00f3 c\u1ea1nh b\u00ean OM = 6dm, g\u00f3c MON = 1200 (h\u00ecnh 2 ). B\u00e1c x\u00e1c \u0111\u1ecbnh trung \u0111i\u1ec3m H c\u1ee7a MN , v\u1ebd cung tr\u00f2n t\u00e2m O b\u00e1n k\u00ednh OH c\u1eaft c\u00e1c c\u1ea1nh OM , ON l\u1ea7n l\u01b0\u1ee3t t\u1ea1i A , B . Sau \u0111\u00f3 b\u00e1c c\u1eaft b\u1ecf ph\u1ea7n g\u1ea1ch s\u1ecdc, cu\u1ed9n ph\u1ea7n c\u00f2n l\u1ea1i c\u1ee7a t\u1ea5m t\u00f4n sao cho m\u00e9p OA tr\u00f9ng kh\u00edt v\u1edbi m\u00e9p OB t\u1ea1o th\u00e0nh chi\u1ebfc g\u00e0u (gi\u1ea3 s\u1eed ph\u1ea7n di\u1ec7n t\u00edch c\u1ee7a m\u00e9p n\u1ed1i kh\u00f4ng \u0111\u00e1ng k\u1ec3). H\u1ecfi khi m\u00fac \u0111\u1ea7y th\u00ec chi\u1ebfc g\u00e0u ch\u1ee9a \u0111\u01b0\u1ee3c bao nhi\u00eau l\u00edt n\u01b0\u1edbc? (k\u1ebft qu\u1ea3 l\u00e0m tr\u00f2n \u0111\u1ebfn m\u1ed9t ch\u1eef s\u1ed1 th\u1eadp ph\u00e2n, l\u1ea5y \uf070 \uf0bb 3,14 ) L\u1eddi gi\u1ea3i - Tam gi\u00e1c OMN c\u00e2n, H l\u00e0 trung \u0111i\u1ec3m \uf0de OH l\u00e0 \u0111\u01b0\u1eddng cao, l\u00e0 ph\u00e2n gi\u00e1c X\u00e9t tam gi\u00e1c OHM vu\u00f4ng t\u1ea1i H , ta c\u00f3: cos 600 = OH \uf0de OH = 6.cos 600 = 3(dm) OM \u0110\u1ed9 d\u00e0i cung tr\u00f2n AB l\u00e0: l = \uf070 R.n = \uf070 .3.120 = 2\uf070 (dm) 180 180 B\u00e1n k\u00ednh \u0111\u01b0\u1eddng tr\u00f2n mi\u1ec7ng c\u1ee7a chi\u1ebfc g\u00e0u l\u00e0: R = C : 2\uf070 = 2\uf070 : 2\uf070 = 1(dm) Chi\u1ec1u cao c\u1ee7a chi\u1ebfc g\u00e0u l\u00e0: h = 32 \u221212 = 2 2 (dm) Khi m\u00fac \u0111\u1ea7y th\u00ec chi\u1ebfc g\u00e0u ch\u1ee9a \u0111\u01b0\u1ee3c s\u1ed1 l\u00edt n\u01b0\u1edbc l\u00e0: V = 1 \uf070 R2h = 1 .3,14.12.2 2 = 3,0(dm3) = 3(l) 33 C\u00e2u 7. (1 \u0111i\u1ec3m). V\u00e0o d\u1ecbp khai tr\u01b0\u01a1ng, nh\u00e0 s\u00e1ch khuy\u1ebfn m\u00e3i m\u1ed7i c\u00e2y vi\u1ebft bi \u0111\u01b0\u1ee3c gi\u1ea3m 20% so v\u1edbi gi\u00e1 ni\u00eam y\u1ebft, c\u00f2n m\u1ed7i quy\u1ec3n t\u1eadp gi\u1ea3m 10% so v\u1edbi gi\u00e1 ni\u00eam y\u1ebft. B\u1ea1n An v\u00e0o nh\u00e0 s\u00e1ch mua 20 quy\u1ec3n t\u1eadp v\u00e0 10 c\u00e2y vi\u1ebft bi. Khi t\u00ednh ti\u1ec1n, b\u1ea1n An \u0111\u01b0a 175000\u0111\u1ed3ng v\u00e0 \u0111\u01b0\u1ee3c th\u1ed1i l\u1ea1i 3000 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 \u0111\u1ed3ng. T\u00ednh gi\u00e1 ni\u00eam y\u1ebft c\u1ee7a m\u1ed7i quy\u1ec3n t\u1eadp v\u00e0 m\u1ed7i c\u00e2y vi\u1ebft bi m\u00e0 b\u1ea1n An \u0111\u00e3 mua. Bi\u1ebft r\u1eb1ng khi An nh\u00ecn v\u00e0o h\u00f3a \u0111\u01a1n, t\u1ed5ng s\u1ed1 ti\u1ec1n ph\u1ea3i tr\u1ea3 khi ch\u01b0a gi\u1ea3m gi\u00e1 l\u00e0 195000\u0111\u1ed3ng. L\u1eddi gi\u1ea3i G\u1ecdi gi\u00e1 ni\u00eam y\u1ebft c\u1ee7a m\u1ed7i quy\u1ec3n t\u1eadp l\u00e0 x ( x \uf03e 0 ; \u0111\u1ed3ng) Gi\u00e1 ni\u00eam y\u1ebft c\u1ee7a m\u1ed7i c\u00e2y vi\u1ebft bi l\u00e0 y ( y \uf03e 0 ; \u0111\u1ed3ng) V\u00ec t\u1ed5ng s\u1ed1 ti\u1ec1n mua 20 quy\u1ec3n t\u1eadp v\u00e0 10 c\u00e2y vi\u1ebft bi khi ch\u01b0a gi\u1ea3m gi\u00e1 l\u00e0 195000 \u0111\u1ed3ng n\u00ean ta c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh: 20x +10y = 195000(1) V\u00ec t\u1ed5ng s\u1ed1 ti\u1ec1n mua 20 quy\u1ec3n t\u1eadp v\u00e0 10 c\u00e2y vi\u1ebft bi khi \u0111\u00e3 gi\u1ea3m gi\u00e1 l\u00e0 175000 \u2013 3000 =172000 \u0111\u1ed3ng n\u00ean ta c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh: 20(1 \u201310%) x +10(1 \u2013 20%) y = 172000 \uf0db 18x + 8y = 172000(2) T\u1eeb (1) v\u00e0 (2) ta c\u00f3 h\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh: \uf0ec20x +10y = 195000 \uf0ed\uf0ee18x + 8y = 172000 Gi\u1ea3i h\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh, ta \u0111\u01b0\u1ee3c: \uf0ecx = 8000(n) \uf0ed = 3500(n) \uf0ee y V\u1eady gi\u00e1 ni\u00eam y\u1ebft c\u1ee7a m\u1ed7i quy\u1ec3n t\u1eadp l\u00e0 8000 \u0111\u1ed3ng v\u00e0 m\u1ed7i c\u00e2y vi\u1ebft bi 3500\u0111\u1ed3ng. C\u00e2u 8. (3 \u0111i\u1ec3m). Cho tam gi\u00e1c ABC c\u00f3 3 g\u00f3c nh\u1ecdn v\u00e0 AB \uf03c AC . V\u1ebd c\u00e1c \u0111\u01b0\u1eddng cao AD , BE , CF c\u1ee7a tam gi\u00e1c \u0111\u00f3. G\u1ecdi H l\u00e0 giao \u0111i\u1ec3m c\u1ee7a c\u00e1c \u0111\u01b0\u1eddng cao v\u1eeba v\u1ebd. a) Ch\u1ee9ng minh: c\u00e1c t\u1ee9 gi\u00e1c AEHF v\u00e0 BFEC n\u1ed9i ti\u1ebfp. b) G\u1ecdi M , N l\u1ea7n l\u01b0\u1ee3t l\u00e0 trung \u0111i\u1ec3m c\u1ee7a c\u00e1c \u0111o\u1ea1n AH , BC . Ch\u1ee9ng minh: FM.FC = FN.FA. c) G\u1ecdi P , Q l\u1ea7n l\u01b0\u1ee3t l\u00e0 ch\u00e2n c\u00e1c \u0111\u01b0\u1eddng vu\u00f4ng g\u00f3c k\u1ebb t\u1eeb M , N \u0111\u1ebfn \u0111\u01b0\u1eddng th\u1eb3ng DF . Ch\u1ee9ng minh r\u1eb1ng \u0111\u01b0\u1eddng tr\u00f2n \u0111\u01b0\u1eddng k\u00ednh PQ \u0111i qua giao \u0111i\u1ec3m c\u1ee7a FE v\u00e0 MN . L\u1eddi gi\u1ea3i a) Ch\u1ee9ng minh: c\u00e1c t\u1ee9 gi\u00e1c AEHF v\u00e0 BFEC n\u1ed9i ti\u1ebfp. 8 T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH \u2022 Ch\u1ee9ng minh: T\u1ee9 gi\u00e1c AEHF n\u1ed9i ti\u1ebfp X\u00e9t t\u1ee9 gi\u00e1c AEHF c\u00f3: \uf0ec\uf0ef AEH = 90\uf0b0 (V\u00ec BE , CF l\u00e0 \u0111\u01b0\u1eddng cao) \uf0ed \uf0ef\uf0eeAFH = 90\uf0b0 \uf0de AEH + AFH = 90\uf0b0 + 90\uf0b0 = 180\uf0b0 \uf0de T\u1ee9 gi\u00e1c AEHF n\u1ed9i ti\u1ebfp \u2022 Ch\u1ee9ng minh: T\u1ee9 gi\u00e1c BFEC n\u1ed9i ti\u1ebfp X\u00e9t t\u1ee9 gi\u00e1c BFEC c\u00f3: \uf0ec\uf0efBEC = 900 BE , CF l\u00e0 \u0111\u01b0\u1eddng cao) \uf0ed (V\u00ec \uf0ef\uf0eeBFC = 900 \uf0de BEC = BFC \uf0de T\u1ee9 gi\u00e1c BFEC n\u1ed9i ti\u1ebfp b) G\u1ecdi M , N l\u1ea7n l\u01b0\u1ee3t l\u00e0 trung \u0111i\u1ec3m c\u1ee7a c\u00e1c \u0111o\u1ea1n AH , BC . Ch\u1ee9ng minh: FM.FC = FN.FA. - Tam gi\u00e1c AFH vu\u00f4ng t\u1ea1i F , FM l\u00e0 trung tuy\u1ebfn \uf0de MF = MA (t\u00ednh ch\u1ea5t \u0111\u01b0\u1eddng trung tuy\u1ebfn trong tam gi\u00e1c vu\u00f4ng) \uf0de \uf044AMF c\u00e2n t\u1ea1i M \uf0de MAF = MFA - Tam gi\u00e1c BFC vu\u00f4ng t\u1ea1i F , FN l\u00e0 trung tuy\u1ebfn \uf0de NF = NC (t\u00ednh ch\u1ea5t \u0111\u01b0\u1eddng trung tuy\u1ebfn trong tam gi\u00e1c vu\u00f4ng) \uf0de \uf044CNF c\u00e2n t\u1ea1i N \uf0de NCF = NFC \u2022 X\u00e9t t\u1ee9 gi\u00e1c AFDC c\u00f3: \uf0ec\uf0ef AFC = 900 (V\u00ec BE , CF l\u00e0 \u0111\u01b0\u1eddng cao) \uf0ed \uf0ef\uf0ee ADC = 900 \uf0de AFC = ADC \uf0de T\u1ee9 gi\u00e1c AFDC n\u1ed9i ti\u1ebfp \uf0de FAD = FCD (c\u00f9ng ch\u1eafn cung FD ) \uf0de MAF = MFA = NCF = NFC \u2022 X\u00e9t \uf044AFM v\u00e0 \uf044CFN c\u00f3: \uf0ec\uf0efFAM = FCN \uf0ed \uf0ef\uf0eeAFM = CFN \uf0f0 \uf044AFM \u0111\u1ed3ng d\u1ea1ng \uf044CFN (g \u2013 g) 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 \uf0f0 FM = FA =\uf03e FM .FC = FN.FA FN FC c) G\u1ecdi P , Q l\u1ea7n l\u01b0\u1ee3t l\u00e0 ch\u00e2n c\u00e1c \u0111\u01b0\u1eddng vu\u00f4ng g\u00f3c k\u1ebb t\u1eeb M , N \u0111\u1ebfn \u0111\u01b0\u1eddng th\u1eb3ng DF . Ch\u1ee9ng minh r\u1eb1ng \u0111\u01b0\u1eddng tr\u00f2n \u0111\u01b0\u1eddng k\u00ednh PQ \u0111i qua giao \u0111i\u1ec3m c\u1ee7a FE v\u00e0 MN . G\u1ecdi K l\u00e0 giao \u0111i\u1ec3m c\u1ee7a EF v\u00e0 MN Ta c\u00f3 \uf044MAF \u0111\u1ed3ng d\u1ea1ng \uf044NCF Suy ra: MFA = NFC M\u00e0 AFC = 900 \uf0de MFN = 900 \uf0de MFP + NFQ = 900 Ta l\u1ea1i c\u00f3 MFP + FMQ = 900 Suy ra NFQ = FMP(1) V\u00ec M l\u00e0 t\u00e2m \u0111\u01b0\u1eddng tr\u00f2n ngo\u1ea1i ti\u1ebfp t\u1ee9 gi\u00e1c AEHF , N l\u00e0 t\u00e2m \u0111\u01b0\u1eddng tr\u00f2n ngo\u1ea1i ti\u1ebfp t\u1ee9 gi\u00e1c BFEC do \u0111\u00f3 MN \u22a5 EF t\u1ea1i K Suy ra t\u1ee9 gi\u00e1c MKFP n\u1ed9i ti\u1ebfp \uf0de FMP = PKF (2) T\u1eeb (1) v\u00e0 (2) suy ra NFQ = PKF V\u00e0 t\u1ee9 gi\u00e1c NKFQ n\u1ed9i ti\u1ebfp \uf0de QNF = QKF M\u00e0 QNF + NFQ = 900 suy ra QKF + PKF = 900 Suy ra QKP = 900 Do \u0111\u00f3 K thu\u1ed9c \u0111\u01b0\u1eddng tr\u00f2n \u0111\u01b0\u1eddng k\u00ednh PQ . T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N 10","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH S\u00d4\u00db GD&\u00d1T TP HO\u00c0 CH\u00cd MINH \u00d1E\u00c0 THAM KHA\u00dbO TUYE\u00c5N SINH 10 PHO\u00d8NG G\u00d1&\u00d1T QUA\u00c4N B\u00ccNH NA\u00caM HO\u00cfC: 2021 - 2022 M\u00d4N: TO\u00c1N 9 \u0110\u1ec0 THTHAM\u1ea0NKHH\u1ea2O \u0110\u00ea thi g\u1ed3m 8 c\u00e2u ho\u0309i t\u01b0\u0323 lu\u00e2\u0323n. Th\u01a1\u0300i gian: 120 phu\u0301t (kh\u00f4ng k\u00ea\u0309 th\u01a1\u0300i gian pha\u0301t \u0111\u00ea\u0300) M\u00c3 \u0110\u1ec0: Qu\u1eadn B\u00ecnh Th\u1ea1nh - 3 C\u00e2u 1. (1,5 \u0111i\u1ec3m). Cho (P) : y = \u2212x2 v\u00e0 \u0111\u01b0\u1eddng th\u1eb3ng (d) : y = \u22122x + 3 . 4 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 \u2212x2 + 4x + 3 = 0 c\u00f3 2 nghi\u1ec7m l\u00e0 x1 , x2 . Kh\u00f4ng gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh, h\u00e3y t\u00ednh gi\u00e1 tr\u1ecb c\u1ee7a bi\u1ec3u th\u1ee9c A = x1 \u2212 2x2 \u2212 1 + x2 + 2x2 \u2212 1 x2 x1 C\u00e2u 3. (0,75 \u0111i\u1ec3m). \u0110\u1ec3 \u01b0\u1edbc t\u00ednh t\u1ed1c \u0111\u1ed9 S (d\u1eb7m\/gi\u1edd) c\u1ee7a m\u1ed9t chi\u1ebfc xe, c\u1ea3nh s\u00e1t s\u1eed d\u1ee5ng c\u00f4ng th\u1ee9c S = 30 fd , \u1edf \u0111\u00f3 d l\u00e0 \u0111\u1ed9 d\u00e0i v\u1ebft tr\u01b0\u1ee3t c\u1ee7a b\u00e1nh xe t\u00ednh b\u1eb1ng feet v\u00e0 f l\u00e0 h\u1ec7 s\u1ed1 ma s\u00e1t. a) Tr\u00ean m\u1ed9t \u0111o\u1ea1n \u0111\u01b0\u1eddng (c\u00f3 g\u1eafn b\u1ea3ng b\u00e1o t\u1ed1c \u0111\u1ed9 b\u00ean tr\u00ean) c\u00f3 h\u1ec7 s\u1ed1 ma s\u00e1t l\u00e0 0,73 v\u00e0 v\u1ebft tr\u01b0\u1ee3t c\u1ee7a m\u1ed9t xe 4 b\u00e1nh sau khi th\u1eafng l\u1ea1i l\u00e0 49,7 feet . H\u1ecfi xe c\u00f3 v\u01b0\u1ee3t qu\u00e1 t\u1ed1c \u0111\u1ed9 theo bi\u1ec3n b\u00e1o tr\u00ean \u0111o\u1ea1n \u0111\u01b0\u1eddng \u0111\u00f3 kh\u00f4ng? (Cho bi\u1ebft 1 feet b\u1eb1ng 1,61km ). b) N\u1ebfu xe ch\u1ea1y v\u1edbi t\u1ed1c \u0111\u1ed9 48 km \/ h tr\u00ean \u0111o\u1ea1n \u0111\u01b0\u1eddng c\u00f3 h\u1ec7 s\u1ed1 ma s\u00e1t l\u00e0 0,45 th\u00ec khi th\u1eafng l\u1ea1i v\u1ebft tr\u01b0\u1ee3t tr\u00ean n\u1ec1n \u0111\u01b0\u1eddng d\u00e0i bao nhi\u00eau feet ? C\u00e2u 4. (0,75 \u0111i\u1ec3m). Qua nghi\u00ean c\u1ee9u, ng\u01b0\u1eddi ta nh\u1eadn th\u1ea5y r\u1eb1ng v\u1edbi m\u1ed7i ng\u01b0\u1eddi trung b\u00ecnh nhi\u1ec7t \u0111\u1ed9 m\u00f4i tr\u01b0\u1eddng gi\u1ea3m \u0111i 1\uf0b0C th\u00ec l\u01b0\u1ee3ng calo c\u1ea7n t\u0103ng th\u00eam kho\u1ea3ng 30 calo. T\u1ea1i 21\uf0b0C , m\u1ed9t ng\u01b0\u1eddi l\u00e0m vi\u1ec7c c\u1ea7n s\u1eed d\u1ee5ng kho\u1ea3ng 3000 calo m\u1ed7i ng\u00e0y. Ng\u01b0\u1eddi ta th\u1ea5y m\u1ed1i quan h\u1ec7 gi\u1eefa hai \u0111\u1ea1i l\u01b0\u1ee3ng n\u00e0y l\u00e0 m\u1ed9t h\u00e0m s\u1ed1 b\u1eadc nh\u1ea5t y = ax + b ( x : \u0111\u1ea1i l\u01b0\u1ee3ng bi\u1ec3u th\u1ecb cho nhi\u1ec7t \u0111\u1ed9 m\u00f4i tr\u01b0\u1eddng v\u00e0 y : \u0111\u1ea1i l\u01b0\u1ee3ng bi\u1ec3u th\u1ecb cho l\u01b0\u1ee3ng calo). a) X\u00e1c \u0111\u1ecbnh h\u1ec7 s\u1ed1 a,b . b) N\u1ebfu m\u1ed9t ng\u01b0\u1eddi l\u00e0m vi\u1ec7c \u1edf sa m\u1ea1c Sahara trong nhi\u1ec7t \u0111\u1ed9 50\uf0b0C th\u00ec c\u1ea7n bao nhi\u00eau calo? C\u00e2u 5. (1 \u0111i\u1ec3m). Nh\u00e0 anh B\u00ecnh l\u00e0m n\u00f4ng nghi\u1ec7p tr\u1ed3ng l\u00faa \u0111\u1ec3 b\u00e1n. Nh\u01b0ng n\u0103m nay ch\u1ecbu \u0111\u1ee3t s\u00e2u h\u1ea1i n\u00ean s\u1ed1 l\u01b0\u1ee3ng l\u00faa thu v\u1ec1 gi\u1ea3m 20% so v\u1edbi d\u1ef1 t\u00ednh v\u00e0 ch\u1ea5t l\u01b0\u1ee3ng l\u00faa c\u0169ng th\u1ea5p n\u00ean ch\u1ec9 b\u00e1n 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 \u0111\u01b0\u1ee3c v\u1edbi gi\u00e1 b\u00e1n b\u1eb1ng 3 gi\u00e1 b\u00e1n d\u1ef1 \u0111\u1ecbnh l\u00fac \u0111\u1ea7u. N\u1ebfu b\u00e1n h\u1ebft ph\u1ea7n c\u00f2n l\u1ea1i n\u00e0y v\u1edbi gi\u00e1 nh\u01b0 4 tr\u00ean th\u00ec s\u1ed1 ti\u1ec1n s\u1ebd \u00edt h\u01a1n 80 tri\u1ec7u \u0111\u1ed3ng so v\u1edbi d\u1ef1 t\u00ednh l\u00fac \u0111\u1ea7u. H\u1ecfi n\u1ebfu kh\u00f4ng b\u1ecb h\u01b0 h\u1ea1i v\u00e0 kh\u00f4ng gi\u1ea3m gi\u00e1 th\u00ec theo d\u1ef1 t\u00ednh, nh\u00e0 anh B\u00ecnh s\u1ebd thu v\u1ec1 bao nhi\u00eau ti\u1ec1n t\u1eeb vi\u1ec7c tr\u1ed3ng l\u00faa tr\u00ean? C\u00e2u 6. (1 \u0111i\u1ec3m). M\u1ed9t v\u00e9 xem phim c\u00f3 gi\u00e1 60000 \u0111\u1ed3ng. Khi c\u00f3 \u0111\u1ee3t gi\u1ea3m gi\u00e1, m\u1ed7i ng\u00e0y s\u1ed1 l\u01b0\u1ee3ng ng\u01b0\u1eddi xem t\u0103ng l\u00ean 50% , do \u0111\u00f3 doanh thu c\u0169ng t\u0103ng 25% . H\u1ecfi gi\u00e1 v\u00e9 khi \u0111\u01b0\u1ee3c gi\u1ea3m l\u00e0 bao nhi\u00eau? C\u00e2u 7. (1 \u0111i\u1ec3m). Qu\u1ea3 c\u1ea7u tuy\u1ebft l\u00e0 m\u1ed9t trong nh\u1eefng m\u00f3n qu\u00e0 l\u01b0u ni\u1ec7m \u0111\u01b0\u1ee3c r\u1ea5t nhi\u1ec1u ng\u01b0\u1eddi \u01b0a th\u00edch. Qu\u1ea3 c\u1ea7u \u0111\u01b0\u1ee3c l\u00e0m b\u1eb1ng th\u1ee7y tinh m\u1ecfng v\u00e0 \u0111\u01b0\u1ee3c \u0111\u1eb7t c\u1ed1 \u0111\u1ecbnh tr\u00ean m\u1ed9t gi\u00e1 \u0111\u1ee1, b\u00ean trong qu\u1ea3 c\u1ea7u l\u00e0 dung d\u1ecbch trong su\u1ed1t v\u00e0 m\u1ed9t s\u1ed1 ph\u1ee5 ki\u1ec7n trang tr\u00ed kh\u00e1c. Gi\u1ea3 s\u1eed m\u1ed9t qu\u1ea3 c\u1ea7u tuy\u1ebft \u0111\u01b0\u1ee3c l\u00e0m b\u1eb1ng th\u1ee7y tinh kh\u00f3 v\u1ee1 c\u00f3 \u0111\u1ed9 d\u00e0y l\u1edbp v\u1ecf 1,2 mm v\u00e0 \u0111\u01b0\u1eddng k\u00ednh c\u1ee7a c\u1ea3 qu\u1ea3 c\u1ea7u l\u00e0 6 cm . B\u00ean trong qu\u1ea3 c\u1ea7u \u0111\u01b0\u1ee3c b\u1ecf m\u1ed9t s\u1ed1 ph\u1ee5 ki\u1ec7n trang tr\u00ed chi\u1ebfm kho\u1ea3ng 10% th\u1ec3 t\u00edch. a) H\u00e3y t\u00ednh th\u1ec3 t\u00edch th\u1ee7y t\u00ednh \u0111\u01b0\u1ee3c d\u00f9ng \u0111\u1ec3 l\u00e0m qu\u1ea3 c\u1ea7u tuy\u1ebft (K\u1ebft qu\u1ea3 l\u00e0m tr\u00f2n \u0111\u1ebfn h\u00e0ng \u0111\u01a1n v\u1ecb). b) C\u1ea7n m\u1ed9t th\u1ec3 t\u00edch n\u01b0\u1edbc l\u00e0 bao nhi\u00eau \u0111\u1ec3 l\u00e0m \u0111\u1ea7y ph\u1ea7n ru\u1ed9t b\u00ean trong qu\u1ea3 c\u1ea7u (K\u1ebft qu\u1ea3 l\u00e0m tr\u00f2n \u0111\u1ebfn ch\u1eef s\u1ed1 th\u1eadp ph\u00e2n th\u1ee9 nh\u1ea5t). C\u00e2u 8. (3 \u0111i\u1ec3m) T\u1eeb m\u1ed9t \u0111i\u1ec3m A ngo\u00e0i \u0111\u01b0\u1eddng tr\u00f2n (O) k\u1ebb hai ti\u1ebfp tuy\u1ebfn AB , AC v\u00e0 m\u1ed9t c\u00e1t tuy\u1ebfn AEF kh\u00f4ng \u0111i qua (O) ( E n\u1eb1m gi\u1eefa A v\u00e0 F , tia AE v\u00e0 tia AC n\u1eb1m kh\u00e1c ph\u00eda so v\u1edbi tia AO ). G\u1ecdi H l\u00e0 giao \u0111i\u1ec3m c\u1ee7a AO v\u00e0 BC . a) Ch\u1ee9ng minh t\u1ee9 gi\u00e1c OBAC n\u1ed9i ti\u1ebfp v\u00e0 HB.HC = HA.HO b) Ch\u1ee9ng minh r\u1eb1ng t\u1ee9 gi\u00e1c OHEF n\u1ed9i ti\u1ebfp. c) EH k\u00e9o d\u00e0i c\u1eaft (O) t\u1ea1i D . FH c\u1eaft (O) t\u1ea1i K . Ch\u1ee9ng minh r\u1eb1ng FD song song v\u1edbi BC v\u00e0 3 \u0111i\u1ec3m A , K , D th\u1eb3ng h\u00e0ng. ----H\u1ebeT--- T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N 2","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH H\u01af\u1edaNG D\u1eaaN GI\u1ea2I C\u00e2u 1. (1,5 \u0111i\u1ec3m) Cho (P) : y = \u2212x2 v\u00e0 \u0111\u01b0\u1eddng th\u1eb3ng (d) : y = \u22122x + 3 . 4 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 \u22124 \u22121 0 \u22121 \u22124 4 x 12 y = \u22122x + 3 1 \u22121 b) T\u00ecm t\u1ecda \u0111\u1ed9 giao \u0111i\u1ec3m c\u1ee7a (P) v\u00e0 (d) b\u1eb1ng ph\u00e9p t\u00ednh. Ph\u01b0\u01a1ng tr\u00ecnh ho\u00e0nh \u0111\u1ed9 giao \u0111i\u1ec3m c\u1ee7a (P) v\u00e0 (d) : \u2212x2 = \u22122x + 3 4 \uf0db \u2212x2 + 2x \u2212 3 = 0 4 \uf0db \uf0e9x = 6 \uf0ea\uf0ebx = 2 Thay x = 6 v\u00e0o y = \u2212 1 x2 , ta \u0111\u01b0\u1ee3c: y = \u2212 1 .62 = \u22129 . 44 Thay x = 2 v\u00e0o y = \u2212 1 x2 , ta \u0111\u01b0\u1ee3c: y = \u2212 1 .22 = \u22121 . 3 44 T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH V\u1eady (6; \u2212 9) , (2; \u2212 1) l\u00e0 hai giao \u0111i\u1ec3m c\u1ea7n t\u00ecm. C\u00e2u 2. (1 \u0111i\u1ec3m) Cho ph\u01b0\u01a1ng tr\u00ecnh \u2212x2 + 4x + 3 = 0 c\u00f3 2 nghi\u1ec7m l\u00e0 x1 , x2 . Kh\u00f4ng gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh, h\u00e3y t\u00ednh gi\u00e1 tr\u1ecb c\u1ee7a bi\u1ec3u th\u1ee9c A = x1 \u2212 2x2 \u2212 1 + x2 + 2x2 \u2212 1 x2 x1 L\u1eddi gi\u1ea3i V\u00ec \uf044 = b2 \u2212 4ac = 42 \u2212 4.(\u22121).3 = 28 \uf03e 0 N\u00ean ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m ph\u00e2n bi\u1ec7t x1 ,x2 . \uf0ef\uf0ec\uf0efS = x1 + x2 = \u2212b = 4 \uf0ed = x1 a Theo \u0111\u1ecbnh l\u00ed Vi-et, ta c\u00f3: \uf0ef\uf0ef\uf0eeP c = \u22123 a .x2 = Ta c\u00f3: A = x1 \u2212 2x2 \u2212 1 + x2 + 2x2 \u2212 1 x2 x1 A = x1 \u2212 2x2 \u2212 1 + x2 + 2x2 \u2212 1 x2 x1 ( ) ( )= x1 x1 \u2212 2x2 \u2212 1 + x2 x2 + 2x2 \u2212 1 x1x2 = x12 + x22 \u2212 x1 \u2212 x2 = S2 \u2212 2P \u2212 S x1x2 P = 42 \u2212 2.(\u22123) \u2212 4 = \u22126 \u22123 C\u00e2u 3. (1 \u0111i\u1ec3m) \u0110\u1ec3 \u01b0\u1edbc t\u00ednh t\u1ed1c \u0111\u1ed9 S (d\u1eb7m\/gi\u1edd) c\u1ee7a m\u1ed9t chi\u1ebfc xe, c\u1ea3nh s\u00e1t s\u1eed d\u1ee5ng c\u00f4ng th\u1ee9c S = 30 fd , \u1edf \u0111\u00f3 d l\u00e0 \u0111\u1ed9 d\u00e0i v\u1ebft tr\u01b0\u1ee3t c\u1ee7a b\u00e1nh xe t\u00ednh b\u1eb1ng feet v\u00e0 f l\u00e0 h\u1ec7 s\u1ed1 ma s\u00e1t. a) Tr\u00ean m\u1ed9t \u0111o\u1ea1n \u0111\u01b0\u1eddng (c\u00f3 g\u1eafn b\u1ea3ng b\u00e1o t\u1ed1c \u0111\u1ed9 b\u00ean tr\u00ean) c\u00f3 h\u1ec7 s\u1ed1 ma s\u00e1t l\u00e0 0,73 v\u00e0 v\u1ebft tr\u01b0\u1ee3t c\u1ee7a m\u1ed9t xe 4 b\u00e1nh sau khi th\u1eafng l\u1ea1i l\u00e0 49,7 feet. H\u1ecfi xe c\u00f3 v\u01b0\u1ee3t qu\u00e1 t\u1ed1c \u0111\u1ed9 theo bi\u1ec3n b\u00e1o tr\u00ean \u0111o\u1ea1n \u0111\u01b0\u1eddng \u0111\u00f3 kh\u00f4ng? (Cho bi\u1ebft 1 feet b\u1eb1ng 1,61 km). b) N\u1ebfu xe ch\u1ea1y v\u1edbi t\u1ed1c \u0111\u1ed9 48 km\/h tr\u00ean \u0111o\u1ea1n \u0111\u01b0\u1eddng c\u00f3 h\u1ec7 s\u1ed1 ma s\u00e1t l\u00e0 0,45 th\u00ec khi th\u1eafng l\u1ea1i v\u1ebft tr\u01b0\u1ee3t tr\u00ean n\u1ec1n \u0111\u01b0\u1eddng d\u00e0i bao nhi\u00eau feet? T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N 4","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH L\u1eddi gi\u1ea3i a) Thay f = 0,73 v\u00e0 d = 49,7 v\u00e0o S = 30 fd S = 30 fd \uf0db S = 30.0,73.49,7 \uf0bb 33( feet \/ h) \uf0bb 53,13(km \/ h) V\u1eady xe v\u01b0\u1ee3t qu\u00e1 t\u1ed1c \u0111\u1ed9 cho ph\u00e9p tr\u00ean bi\u1ec3n b\u00e1o b) 48km \/ h = 4800 ( feet \/ h) 161 Thay f = 0,45 v\u00e0 S = 4800 v\u00e0o S = 30 fd 161 S = 30 fd \uf0db 4800 = 30.0,45.d 161 \uf0db d = 48002 \uf0bb 65, 8 ( feet ) 1612.30.0 , 45 V\u1eady n\u1ebfu xe ch\u1ea1y v\u1edbi t\u1ed1c \u0111\u1ed9 48 km \/ h tr\u00ean \u0111o\u1ea1n \u0111\u01b0\u1eddng c\u00f3 h\u1ec7 s\u1ed1 ma s\u00e1t l\u00e0 0,45 th\u00ec khi th\u1eafng l\u1ea1i v\u1ebft tr\u01b0\u1ee3t tr\u00ean n\u1ec1n \u0111\u01b0\u1eddng d\u00e0i kho\u1ea3ng 65,8 feet . C\u00e2u 4. Qua nghi\u00ean c\u1ee9u, ng\u01b0\u1eddi ta nh\u1eadn th\u1ea5y r\u1eb1ng v\u1edbi m\u1ed7i ng\u01b0\u1eddi trung b\u00ecnh nhi\u1ec7t \u0111\u1ed9 m\u00f4i tr\u01b0\u1eddng gi\u1ea3m \u0111i 10C th\u00ec l\u01b0\u1ee3ng calo c\u1ea7n t\u0103ng th\u00eam kho\u1ea3ng 30 calo. T\u1ea1i 210C , m\u1ed9t ng\u01b0\u1eddi l\u00e0m vi\u1ec7c c\u1ea7n s\u1eed d\u1ee5ng kho\u1ea3ng 3000 calo m\u1ed7i ng\u00e0y. Ng\u01b0\u1eddi ta th\u1ea5y m\u1ed1i quan h\u1ec7 gi\u1eefa hai \u0111\u1ea1i l\u01b0\u1ee3ng n\u00e0y l\u00e0 m\u1ed9t h\u00e0m s\u1ed1 b\u1eadc nh\u1ea5t y = ax + b ( x : \u0111\u1ea1i l\u01b0\u1ee3ng bi\u1ec3u th\u1ecb cho nhi\u1ec7t \u0111\u1ed9 m\u00f4i tr\u01b0\u1eddng v\u00e0 y : \u0111\u1ea1i l\u01b0\u1ee3ng bi\u1ec3u th\u1ecb cho l\u01b0\u1ee3ng calo). a) X\u00e1c \u0111\u1ecbnh h\u1ec7 s\u1ed1 a,b . b) N\u1ebfu m\u1ed9t ng\u01b0\u1eddi l\u00e0m vi\u1ec7c \u1edf sa m\u1ea1c Sahara trong nhi\u1ec7t \u0111\u1ed9 500C th\u00ec c\u1ea7n bao nhi\u00eau calo? L\u1eddi gi\u1ea3i a) X\u00e1c \u0111\u1ecbnh h\u1ec7 s\u1ed1 a,b . T\u1ea1i 21\uf0b0C , m\u1ed9t ng\u01b0\u1eddi l\u00e0m vi\u1ec7c c\u1ea7n s\u1eed d\u1ee5ng kho\u1ea3ng 3000 calo m\u1ed7i ng\u00e0y Thay x = 21 v\u00e0 y = 3000 v\u00e0o y = ax + b \uf0db 3000 = a.21 + b \uf0db 21a + b = 3000 (1) Trung b\u00ecnh nhi\u1ec7t \u0111\u1ed9 m\u00f4i tr\u01b0\u1eddng gi\u1ea3m \u0111i 1\uf0b0C th\u00ec l\u01b0\u1ee3ng calo c\u1ea7n t\u0103ng th\u00eam kho\u1ea3ng 30 calo Thay x = 20 v\u00e0 y = 2970 v\u00e0o y = ax + b 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 \uf0db 2970 = a.20 + b \uf0db 20a + b = 2970 (2) Ta c\u00f3 h\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh: \uf0ec21a + b = 3000 \uf0db \uf0eca = 20 \uf0ed\uf0ee20a + b = 2970 \uf0ed\uf0eeb = 2370 V\u1eady y = 20x + 2370 b) Thay x = 50 v\u00e0o y = 20x + 2370 \uf0db y = 20.50 + 2730 = 3730 calo. V\u1eady n\u1ebfu m\u1ed9t ng\u01b0\u1eddi l\u00e0m vi\u1ec7c \u1edf sa m\u1ea1c Sahara trong nhi\u1ec7t \u0111\u1ed9 50\uf0b0C th\u00ec c\u1ea7n 3730 calo C\u00e2u 5. Nh\u00e0 anh B\u00ecnh l\u00e0m n\u00f4ng nghi\u1ec7p tr\u1ed3ng l\u00faa \u0111\u1ec3 b\u00e1n. Nh\u01b0ng n\u0103m nay ch\u1ecbu \u0111\u1ee3t s\u00e2u h\u1ea1i n\u00ean s\u1ed1 l\u01b0\u1ee3ng l\u00faa thu v\u1ec1 gi\u1ea3m 20% so v\u1edbi d\u1ef1 t\u00ednh v\u00e0 ch\u1ea5t l\u01b0\u1ee3ng l\u00faa c\u0169ng th\u1ea5p n\u00ean ch\u1ec9 b\u00e1n \u0111\u01b0\u1ee3c v\u1edbi gi\u00e1 b\u00e1n b\u1eb1ng 3 gi\u00e1 b\u00e1n d\u1ef1 \u0111\u1ecbnh l\u00fac \u0111\u1ea7u. N\u1ebfu b\u00e1n h\u1ebft ph\u1ea7n c\u00f2n l\u1ea1i n\u00e0y v\u1edbi gi\u00e1 nh\u01b0 tr\u00ean th\u00ec s\u1ed1 4 ti\u1ec1n s\u1ebd \u00edt h\u01a1n 80 tri\u1ec7u \u0111\u1ed3ng so v\u1edbi d\u1ef1 t\u00ednh l\u00fac \u0111\u1ea7u. H\u1ecfi n\u1ebfu kh\u00f4ng b\u1ecb h\u01b0 h\u1ea1i v\u00e0 kh\u00f4ng gi\u1ea3m gi\u00e1 th\u00ec theo d\u1ef1 t\u00ednh, nh\u00e0 anh B\u00ecnh s\u1ebd thu v\u1ec1 bao nhi\u00eau ti\u1ec1n t\u1eeb vi\u1ec7c tr\u1ed3ng l\u00faa tr\u00ean? L\u1eddi gi\u1ea3i ( )Goi x kg l\u00e0 s\u1ed1 l\u01b0\u1ee3ng l\u00faa thu v\u1ec1 theo d\u1ef1 t\u00ednh y (tri\u1ec7u \u0111\u1ed3ng) l\u00e0 gi\u00e1 b\u00e1n d\u1ef1 t\u00ednh l\u00fac \u0111\u1ea7u , \u0110k: x, y \uf03e 0 S\u1ed1 ti\u1ec1n thu v\u1ec1 t\u1eeb vi\u1ec7c tr\u1ed3ng l\u00faa theo d\u1ef1 t\u00ednh l\u00fac \u0111\u1ea7u l\u00e0: xy (tri\u1ec7u \u0111\u1ed3ng) S\u1ed1 l\u01b0\u1ee3ng l\u00faa th\u1ef1c t\u1ebf thu v\u1ec1: 80%x = 0,8x(kg) Gi\u00e1 b\u00e1n th\u1ef1c t\u1ebf: 3 y (tri\u1ec7u \u0111\u1ed3ng) 4 N\u1ebfu b\u00e1n h\u1ebft ph\u1ea7n c\u00f2n l\u1ea1i n\u00e0y v\u1edbi gi\u00e1 nh\u01b0 tr\u00ean th\u00ec s\u1ed1 ti\u1ec1n s\u1ebd \u00edt h\u01a1n 80 tri\u1ec7u \u0111\u1ed3ng so v\u1edbi d\u1ef1 t\u00ednh l\u00fac \u0111\u1ea7u n\u00ean: 0,8x. 3 y + 80 = xy 4 \uf0db 0,6xy + 80 = xy \uf0db xy = 200 (n) S\u1ed1 ti\u1ec1n thu v\u1ec1 t\u1eeb vi\u1ec7c tr\u1ed3ng l\u00faa theo d\u1ef1 t\u00ednh l\u00fac \u0111\u1ea7u l\u00e0: 200 (tri\u1ec7u \u0111\u1ed3ng) C\u00e2u 6. M\u1ed9t v\u00e9 xem phim c\u00f3 gi\u00e1 60000 \u0111\u1ed3ng. Khi c\u00f3 \u0111\u1ee3t gi\u1ea3m gi\u00e1, m\u1ed7i ng\u00e0y s\u1ed1 l\u01b0\u1ee3ng ng\u01b0\u1eddi xem t\u0103ng l\u00ean 50% , do \u0111\u00f3 doanh thu c\u0169ng t\u0103ng 25% . H\u1ecfi gi\u00e1 v\u00e9 khi \u0111\u01b0\u1ee3c gi\u1ea3m l\u00e0 bao nhi\u00eau? L\u1eddi gi\u1ea3i Goi x (ng\u01b0\u1eddi) l\u00e0 s\u1ed1 l\u01b0\u1ee3ng ng\u01b0\u1eddi xem tr\u01b0\u1edbc khi gi\u1ea3m gi\u00e1 v\u00e9, \u0110k: x \uf0ce * S\u1ed1 l\u01b0\u1ee3ng ng\u01b0\u1eddi xem sau khi gi\u1ea3m gi\u00e1 l\u00e0 150%x = 1,5x (ng\u01b0\u1eddi) Doanh thu tr\u01b0\u1edbc gi\u1ea3m gi\u00e1: 60000x (\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 Doanh thu sau gi\u1ea3m gi\u00e1: 125%.60000x = 75000x (\u0111\u1ed3ng) V\u1eady gi\u00e1 v\u00e9 khi \u0111\u01b0\u1ee3c gi\u1ea3m l\u00e0: 75000x : 1,5x = 50000 (\u0111\u1ed3ng). C\u00e2u 7. Qu\u1ea3 c\u1ea7u tuy\u1ebft l\u00e0 m\u1ed9t trong nh\u1eefng m\u00f3n qu\u00e0 l\u01b0u ni\u1ec7m \u0111\u01b0\u1ee3c r\u1ea5t nhi\u1ec1u ng\u01b0\u1eddi \u01b0a th\u00edch. Qu\u1ea3 c\u1ea7u \u0111\u01b0\u1ee3c l\u00e0m b\u1eb1ng th\u1ee7y tinh m\u1ecfng v\u00e0 \u0111\u01b0\u1ee3c \u0111\u1eb7t c\u1ed1 \u0111\u1ecbnh tr\u00ean m\u1ed9t gi\u00e1 \u0111\u1ee1, b\u00ean trong qu\u1ea3 c\u1ea7u l\u00e0 dung d\u1ecbch trong su\u1ed1t v\u00e0 m\u1ed9t s\u1ed1 ph\u1ee5 ki\u1ec7n trang tr\u00ed kh\u00e1c. Gi\u1ea3 s\u1eed m\u1ed9t qu\u1ea3 c\u1ea7u tuy\u1ebft \u0111\u01b0\u1ee3c l\u00e0m b\u1eb1ng th\u1ee7y tinh kh\u00f3 v\u1ee1 c\u00f3 \u0111\u1ed9 d\u00e0y l\u1edbp v\u1ecf 1,2 mm v\u00e0 \u0111\u01b0\u1eddng k\u00ednh c\u1ee7a c\u1ea3 qu\u1ea3 c\u1ea7u l\u00e0 6 cm. B\u00ean trong qu\u1ea3 c\u1ea7u \u0111\u01b0\u1ee3c b\u1ecf m\u1ed9t s\u1ed1 ph\u1ee5 ki\u1ec7n trang tr\u00ed chi\u1ebfm kho\u1ea3ng 10% th\u1ec3 t\u00edch. a) H\u00e3y t\u00ednh th\u1ec3 t\u00edch th\u1ee7y tinh \u0111\u01b0\u1ee3c d\u00f9ng \u0111\u1ec3 l\u00e0m qu\u1ea3 c\u1ea7u tuy\u1ebft (K\u1ebft qu\u1ea3 l\u00e0m tr\u00f2n \u0111\u1ebfn h\u00e0ng \u0111\u01a1n v\u1ecb). b) C\u1ea7n m\u1ed9t th\u1ec3 t\u00edch n\u01b0\u1edbc l\u00e0 bao nhi\u00eau \u0111\u1ec3 l\u00e0m \u0111\u1ea7y ph\u1ea7n ru\u1ed9t b\u00ean trong qu\u1ea3 c\u1ea7u (K\u1ebft qu\u1ea3 l\u00e0m tr\u00f2n \u0111\u1ebfn ch\u1eef s\u1ed1 th\u1eadp ph\u00e2n th\u1ee9 nh\u1ea5t). L\u1eddi gi\u1ea3i a) Th\u1ec3 t\u00edch th\u1ee7y tinh \u0111\u01b0\u1ee3c d\u00f9ng \u0111\u1ec3 l\u00e0m qu\u1ea3 c\u1ea7u tuy\u1ebft: ( )4 \uf070 6,123 \u2212 63 \uf0bb 55cm3 3 b) Th\u1ec3 t\u00edch n\u01b0\u1edbc l\u00e0 bao nhi\u00eau \u0111\u1ec3 l\u00e0m \u0111\u1ea7y ph\u1ea7n ru\u1ed9t b\u00ean trong qu\u1ea3 c\u1ea7u: 90%. 4 .\uf070 .63 \uf0bb 814,3cm3 3 C\u00e2u 8. T\u1eeb m\u1ed9t \u0111i\u1ec3m A ngo\u00e0i \u0111\u01b0\u1eddng tr\u00f2n (O) k\u1ebb hai ti\u1ebfp tuy\u1ebfn AB , AC v\u00e0 m\u1ed9t c\u00e1t tuy\u1ebfn AEF kh\u00f4ng \u0111i qua (O) ( E n\u1eb1m gi\u1eefa A v\u00e0 F , tia AE v\u00e0 tia AC n\u1eb1m kh\u00e1c ph\u00eda so v\u1edbi tia AO ). G\u1ecdi H l\u00e0 giao \u0111i\u1ec3m c\u1ee7a AO v\u00e0 BC . a) Ch\u1ee9ng minh t\u1ee9 gi\u00e1c OBAC n\u1ed9i ti\u1ebfp v\u00e0 HB.HC = HA.HO b) Ch\u1ee9ng minh r\u1eb1ng t\u1ee9 gi\u00e1c OHEF n\u1ed9i ti\u1ebfp. c) EH k\u00e9o d\u00e0i c\u1eaft (O) t\u1ea1i D . FH c\u1eaft (O) t\u1ea1i K . Ch\u1ee9ng minh r\u1eb1ng FD song song v\u1edbi BC v\u00e0 3 \u0111i\u1ec3m A , K , D 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 v\u00e0 HB.HC = HA.HO X\u00e9t t\u1ee9 gi\u00e1c OBAC , c\u00f3: \uf0ec \uf0efOBA = 90\uf0b0 \uf0ed ( AB , AC l\u00e0 ti\u1ebfp tuy\u1ebfn c\u1ee7a (O) ) \uf0ee\uf0efOCA = 90\uf0b0 \uf0de OBA + OCA = 180\uf0b0 \uf0de T\u1ee9 gi\u00e1c OBAC n\u1ed9i ti\u1ebfp v\u00ec c\u00f3 hai g\u00f3c \u0111\u1ed1i b\u00f9 nhau. \uf0de ABC = AOC (c\u00f9ng nh\u00ecn c\u1ea1nh AC ). 8 X\u00e9t \uf044ABH v\u00e0 \uf044COH c\u00f3: ABH = HOC (cmt) AHB = OHC (\u0111\u1ed1i \u0111\u1ec9nh) V\u1eady \uf044ABH \uf044COH (g-g) \uf0de BH = AH \uf0de HB.HC = HA.HO OH CH b) Ch\u1ee9ng minh r\u1eb1ng t\u1ee9 gi\u00e1c OHEF n\u1ed9i ti\u1ebfp. Ta c\u00f3: OB = OC = R AB = AC (t\u00ednh ch\u1ea5t hai ti\u1ebfp tuy\u1ebfn c\u1eaft nhau) V\u1eady OA l\u00e0 \u0111\u01b0\u1eddng trung tr\u1ef1c c\u1ee7a BC X\u00e9t \uf044ABE v\u00e0 \uf044AFB c\u00f3: A : g\u00f3c chung ABE = BFE (c\u00f9ng ch\u1eafn cung BE ) 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 9 V\u1eady \uf044ABE \uf044AFB (g-g) \uf0de AE = AB \uf0de AE.AF = AB2 AB AF M\u00e0 AH.AO = AB2 (H\u1ec7 th\u1ee9c l\u01b0\u1ee3ng trong \uf044ABO vu\u00f4ng t\u1ea1i B , \u0111\u01b0\u1eddng cao BH ) N\u00ean AH.AO = AE.AF X\u00e9t \uf044AEH v\u00e0 \uf044AOF c\u00f3: A : g\u00f3c chung AH = AE (v\u00ec AH.AO = AE.AF ) AF AO V\u1eady \uf044AEH \uf044AOF (c-g-c) \uf0de AHE = AFO (Hai g\u00f3c t\u01b0\u01a1ng \u1ee9ng) \uf0de T\u1ee9 gi\u00e1c OHEF n\u1ed9i ti\u1ebfp (g\u00f3c ngo\u00e0i b\u1eb1ng g\u00f3c \u0111\u1ed1i trong) c) Ta c\u00f3: AB = AC (t\u00ednh ch\u1ea5t 2 ti\u1ebfp tuy\u1ebfn c\u1eaft nhau) OB = OC = R V\u1eady OA l\u00e0 \u0111\u01b0\u1eddng trung tr\u1ef1c c\u1ee7a BC (1) Ta c\u00f3: OHF = OEF (t\u1ee9 gi\u00e1c OHEF n\u1ed9i ti\u1ebfp) OFE = AHE M\u00e0 AHE + EHB = BHE + DHF = 90\uf0b0 N\u00ean EHB = BHF Suy ra HB l\u00e0 tia ph\u00e2n gi\u00e1c c\u1ee7a EHF \uf0de EHB = BHF = 1 EHF 2 M\u00e0 EOF = EHF (cmt) n\u00ean \uf0de EHB = 1 EOF 2 M\u00e0 EDF = 1 EOF (c\u00f9ng ch\u1eafn cung EF ) 2 N\u00ean EDF = EHB M\u00e0 CHD = EHB ( 2 g\u00f3c \u0111\u1ed1i \u0111\u1ec9nh) N\u00ean EDF = CHD M\u00e0 hai g\u00f3c n\u00e0y n\u1eb1m \u1edf v\u1ecb tr\u00ed so le trong n\u00ean BC \/ \/DF V\u1eady t\u1ee9 gi\u00e1c BCDF l\u00e0 h\u00ecnh thang c\u00e2n Suy ra OA l\u00e0 \u0111\u01b0\u1eddng trung tr\u1ef1c c\u1ee7a FD (2) 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 Ta l\u1ea1i c\u00f3: KHC = BHF = 1 EHF = 1 EOF 22 N\u00ean EKH = KHC M\u00e0 hai g\u00f3c n\u00e0y n\u1eb1m \u1edf v\u1ecb tr\u00ed \u0111\u1ed3ng v\u1ecb n\u00ean BC \/ \/EK Suy ra OA l\u00e0 \u0111\u01b0\u1eddng trung tr\u1ef1c c\u1ee7a EK (3) M\u00e0 A,E,F th\u1eb3ng h\u00e0ng (4) T\u1eeb (1) , (2) , (3) \u00b8 (4) suy ra A , K , D th\u1eb3ng h\u00e0ng. ----H\u1ebeT--- T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N 10","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH S\u00d4\u00db GD&\u00d1T TP HO\u00c0 CH\u00cd MINH \u00d1E\u00c0 THAM KHA\u00dbO TUYE\u00c5N SINH 10 PHO\u00d8NG G\u00d1&\u00d1T QUA\u00c4N B\u00ccNH TH\u1ea0NH NA\u00caM HO\u00cfC: 2023 - 2024 M\u00d4N: TO\u00c1N 9 \u0110\u1ec0 THAM KH\u1ea2O M\u00c3 \u0110\u1ec0: Qu\u1eadn BTH - 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 Parabol P : y 1 x2 v\u00e0 \u0111\u01b0\u1eddng th\u1eb3ng d :y 1x 3 2 2 a) V\u1ebd \u0111\u1ed3 th\u1ecb (P) v\u00e0 (d) tr\u00ean c\u00f9ng h\u1ec7 tr\u1ee5c t\u1ecda \u0111\u1ed9. b) T\u00ecm c\u00e1c t\u1ecda \u0111\u1ed9 giao \u0111i\u1ec3m c\u1ee7a (P) v\u00e0 (d) b\u1eb1ng ph\u00e9p to\u00e1n. C\u00e2u 2. (1,0 \u0111i\u1ec3m). Cho ph\u01b0\u01a1ng tr\u00ecnh: 2x 2 7x 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 x12 1 x 22 1 x2 x1 . C\u00e2u 3. (1,0 \u0111i\u1ec3m). Trong m\u1ed9t th\u00e1ng kho\u1ea3ng l\u1ee3i nhu\u1eadn y (\u0111\u1ed3ng) c\u1ee7a c\u1eeda h\u00e0ng A thu \u0111\u01b0\u1ee3c khi b\u00e1n x h\u1ed9p s\u1eefa lo\u1ea1i 500 g \u0111\u01b0\u1ee3c cho b\u1edfi ph\u01b0\u01a1ng tr\u00ecnh y ax b . Bi\u1ebft r\u1eb1ng trong th\u00e1ng 11 c\u1eeda h\u00e0ng b\u00e1n \u0111\u01b0\u1ee3c 96 h\u1ed9p s\u1eefa thu l\u1ee3i nhu\u1eadn 3 660 000 \u0111\u1ed3ng, th\u00e1ng 12 b\u00e1n \u0111\u01b0\u1ee3c 150 h\u1ed9p s\u1eefa thu \u0111\u01b0\u1ee3c l\u1ee3i nhu\u1eadn 6 000 000 \u0111\u1ed3ng. T\u00ecm h\u1ec7 s\u1ed1 a v\u00e0 b . C\u00e2u 4. (0,75 \u0111i\u1ec3m). C\u01a1 s\u1edf bu\u00f4n tr\u00e1i c\u00e2y nh\u00e0 b\u1ea1n Linh \u0111\u1eb7t h\u00e0ng 5, 5 t\u1ea5n cam v\u1edbi gi\u00e1 17 500 \u0111\u1ed3ng m\u1ed9t ki-l\u00f4-gam. Ti\u1ec1n v\u1eadn chuy\u1ec3n l\u00e0 6 875 000 \u0111\u1ed3ng. Trong qu\u00e1 tr\u00ecnh v\u1eadn chuy\u1ec3n th\u00ec 10% s\u1ed1 cam b\u1ecb h\u1ecfng, t\u1ea5t c\u1ea3 s\u1ed1 cam c\u00f2n l\u1ea1i \u0111\u1ec1u \u0111\u01b0\u1ee3c b\u00e1n h\u1ebft. H\u00e3y t\u00ednh xem c\u01a1 s\u1edf tr\u00e1i c\u00e2y nh\u00e0 b\u1ea1n Linh s\u1ebd b\u00e1n m\u1ed7i ki-l\u00f4-gam cam v\u1edbi gi\u00e1 bao nhi\u00eau \u0111\u1ec3 thu l\u00e3i 20% . C\u00e2u 5. (1,0 \u0111i\u1ec3m). Ng\u01b0\u1eddi ta mu\u1ed1n x\u00e2y m\u1ed9t b\u1ec3 ch\u1ee9a n\u01b0\u1edbc d\u1ea1ng hinh h\u1ed9p ch\u1eef nh\u1eadt kh\u00f4ng c\u00f3 n\u1eafp cao 1, 5m c\u00f3 th\u1ec3 t\u00edch l\u00e0 12 m3 v\u00e0 chi\u1ec1u d\u00e0i g\u1ea5p \u0111\u00f4i chi\u1ec1u r\u1ed9ng. a) T\u00ednh chi\u1ec1u d\u00e0i v\u00e0 chi\u1ec1u r\u1ed9ng c\u1ee7a b\u1ec3 n\u01b0\u1edbc. b) Gi\u00e1 thu\u00ea c\u00f4ng nh\u00e2n x\u00e2y b\u1ec3 l\u00e0 500 000 \u0111\u1ed3ng\/m2 . Chi ph\u00ed thu\u00ea nh\u00e2n c\u00f4ng l\u00e0 bao nhi\u00eau? C\u00e2u 6. (0,75 \u0111i\u1ec3m). Anh Minh sinh v\u00e0o th\u1ebf k\u1ef7 20 . \u0110\u1ebfn ng\u00e0y sinh nh\u1eadt c\u1ee7a m\u00ecnh trong n\u0103m 2021 , anh nh\u1eadn th\u1ea5y tu\u1ed5i c\u1ee7a m\u00ecnh b\u1eb1ng \u0111\u00fang t\u1ed5ng c\u00e1c ch\u1eef s\u1ed1 trong n\u0103m sinh c\u1ee7a m\u00ecnh. T\u00ecm n\u0103m sinh c\u1ee7a anh Minh. C\u00e2u 7. (1 \u0111i\u1ec3m). Hai xe m\u00e1y kh\u1edfi h\u00e0nh c\u00f9ng l\u00fac t\u1eeb A \u0111\u1ebfn B . Xe th\u1ee9 nh\u1ea5t ch\u1ea1y v\u1edbi v\u1eadn t\u1ed1c 32km \/ h , xe th\u1ee9 hai ch\u1ea1y v\u1edbi v\u1eadn t\u1ed1c l\u1edbn h\u01a1n xe th\u1ee9 nh\u1ea5t 8 km\/h. Tr\u00ean \u0111\u01b0\u1eddng \u0111i xe th\u1ee9 hai d\u1eebng l\u1ea1i ngh\u1ec9 45 ph\u00fat r\u1ed3i l\u1ea1i ti\u1ebfp t\u1ee5c ch\u1ea1y v\u1edbi v\u1eadn t\u1ed1c c\u0169. T\u00ednh qu\u00e3ng \u0111\u01b0\u1eddng AB bi\u1ebft hai xe \u0111\u1ebfn B c\u00f9ng m\u1ed9t l\u00fac. C\u00e2u 8. (3,0 \u0111i\u1ec3m) Cho ABC nh\u1ecdn AB AC n\u1ed9i ti\u1ebfp \u0111\u01b0\u1eddng tr\u00f2n O , \u0111\u01b0\u1eddng cao CE c\u1ee7a ABC c\u1eaft O t\u1ea1i M . V\u1ebd d\u00e2y MN vu\u00f4ng g\u00f3c v\u1edbi OA t\u1ea1i T , AO c\u1eaft O t\u1ea1i D . a) Ch\u1ee9ng minh AM AN v\u00e0 AM 2 AT.AD . b) G\u1ecdi F v\u00e0 H l\u1ea7n l\u01b0\u1ee3t l\u00e0 giao \u0111i\u1ec3m c\u1ee7a BN v\u1edbi AC v\u00e0 CE . Ch\u1ee9ng minh EF \/\/ MN v\u00e0 H l\u00e0 tr\u1ef1c t\u00e2m c\u1ee7a ABC . 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) G\u1ecdi K l\u00e0 trung \u0111i\u1ec3m c\u1ee7a AH . Ch\u1ee9ng minh AH AM v\u00e0 t\u1ee9 gi\u00e1c OHKT n\u1ed9i ti\u1ebfp. ----H\u1ebeT--- T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N 2","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH H\u01af\u1edaNG D\u1eaaN GI\u1ea2I C\u00e2u 1. (1,5 \u0111i\u1ec3m). Cho Parabol P : y 1 x2 v\u00e0 \u0111\u01b0\u1eddng th\u1eb3ng d :y 1x 3 2 2 a) V\u1ebd \u0111\u1ed3 th\u1ecb (P) v\u00e0 (d) tr\u00ean c\u00f9ng h\u1ec7 tr\u1ee5c t\u1ecda \u0111\u1ed9. b) T\u00ecm c\u00e1c t\u1ecda \u0111\u1ed9 giao \u0111i\u1ec3m c\u1ee7a (P) v\u00e0 (d) b\u1eb1ng ph\u00e9p to\u00e1n. L\u1eddi gi\u1ea3i a) V\u1ebd \u0111\u1ed3 th\u1ecb (P) v\u00e0 (d) tr\u00ean c\u00f9ng h\u1ec7 tr\u1ee5c t\u1ecda \u0111\u1ed9. BGT: 0 12 x \u22122 \u22121 0 12 2 y 1x2 2 1 22 x 02 y 1 x 33 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) : 1x2 1x 3 22 \uf0db \u2212 1 x2 \u2212 1 x + 3 = 0 22 \uf0db \uf0e9x = 2 \uf0ea\uf0ebx = \u22123 Thay x 2 v\u00e0o y 1 x2 , ta \u0111\u01b0\u1ee3c: y 1 .22 2. 2 2 Thay x 3 v\u00e0o y 1 x2, ta \u0111\u01b0\u1ee3c: y 1 32 9 2 2 2. V\u1eady 2; 2 , 3; 9 l\u00e0 hai giao \u0111i\u1ec3m c\u1ea7n t\u00ecm. 2 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. (1,0 \u0111i\u1ec3m). Cho ph\u01b0\u01a1ng tr\u00ecnh: 2x 2 7x 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 x12 1 x 2 1 x2 2 x1 . L\u1eddi gi\u1ea3i V\u00ec b2 4ac 72 4.2.4 17 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 = \u22127 \uf0ed = x1 a 2 Theo \u0111\u1ecbnh l\u00ed Vi-et, ta c\u00f3: \uf0ef\uf0ef\uf0eeP =2 c .x2 = a Ta c\u00f3: A x12 1 x 22 1 x2 x1 A x1x2 2 x1 x2 1 x1x2 1 A x1x2 2 x1 x2 x 1x 2 A 22 7 1 8. 2 2 C\u00e2u 3. (1,0 \u0111i\u1ec3m). Trong m\u1ed9t th\u00e1ng kho\u1ea3ng l\u1ee3i nhu\u1eadn y (\u0111\u1ed3ng) c\u1ee7a c\u1eeda h\u00e0ng A thu \u0111\u01b0\u1ee3c khi b\u00e1n x h\u1ed9p s\u1eefa lo\u1ea1i 500 g \u0111\u01b0\u1ee3c cho b\u1edfi ph\u01b0\u01a1ng tr\u00ecnh y ax b . Bi\u1ebft r\u1eb1ng trong th\u00e1ng 11 c\u1eeda h\u00e0ng b\u00e1n \u0111\u01b0\u1ee3c 96 h\u1ed9p s\u1eefa thu l\u1ee3i nhu\u1eadn 3 660 000 \u0111\u1ed3ng, th\u00e1ng 12 b\u00e1n \u0111\u01b0\u1ee3c 150 h\u1ed9p s\u1eefa thu \u0111\u01b0\u1ee3c l\u1ee3i nhu\u1eadn 6 000 000 \u0111\u1ed3ng. T\u00ecm h\u1ec7 s\u1ed1 a v\u00e0 b . L\u1eddi gi\u1ea3i Ta c\u00f3 kho\u1ea3ng l\u1ee3i nhu\u1eadn y (\u0111\u1ed3ng) c\u1ee7a c\u1eeda h\u00e0ng A thu \u0111\u01b0\u1ee3c khi b\u00e1n x h\u1ed9p s\u1eefa lo\u1ea1i 500 g \u0111\u01b0\u1ee3c cho b\u1edfi ph\u01b0\u01a1ng tr\u00ecnh y ax b . Trong th\u00e1ng 11 : c\u1eeda h\u00e0ng b\u00e1n \u0111\u01b0\u1ee3c 96 h\u1ed9p s\u1eefa thu l\u1ee3i nhu\u1eadn 3 660 000 \u0111\u1ed3ng. Suy ra: a.96 b 3 660 000 1 . Trong th\u00e1ng 12 : b\u00e1n \u0111\u01b0\u1ee3c 150 h\u1ed9p s\u1eefa thu \u0111\u01b0\u1ee3c l\u1ee3i nhu\u1eadn 6 000 000 \u0111\u1ed3ng. Suy ra: a.150 b 6 000 000 2 . 96a b 3 660 000 a 130000 T\u1eeb 1 v\u00e0 2 suy ra 150a b 6 000 000 3. b 500000 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). C\u01a1 s\u1edf bu\u00f4n tr\u00e1i c\u00e2y nh\u00e0 b\u1ea1n Linh \u0111\u1eb7t h\u00e0ng 5, 5 t\u1ea5n cam v\u1edbi gi\u00e1 17 500 \u0111\u1ed3ng m\u1ed9t ki-l\u00f4-gam. Ti\u1ec1n v\u1eadn chuy\u1ec3n l\u00e0 6 875 000 \u0111\u1ed3ng. Trong qu\u00e1 tr\u00ecnh v\u1eadn chuy\u1ec3n th\u00ec 10% s\u1ed1 cam b\u1ecb h\u1ecfng, t\u1ea5t c\u1ea3 s\u1ed1 cam c\u00f2n l\u1ea1i \u0111\u1ec1u \u0111\u01b0\u1ee3c b\u00e1n h\u1ebft. H\u00e3y t\u00ednh xem c\u01a1 s\u1edf tr\u00e1i c\u00e2y nh\u00e0 b\u1ea1n Linh s\u1ebd b\u00e1n m\u1ed7i ki-l\u00f4-gam cam v\u1edbi gi\u00e1 bao nhi\u00eau \u0111\u1ec3 thu l\u00e3i 20% . L\u1eddi gi\u1ea3i 5, 5 t\u1ea5n 5500kg . Trong qu\u00e1 tr\u00ecnh v\u1eadn chuy\u1ec3n 10% s\u1ed1 c\u1ea3m h\u1ecfng t\u01b0\u01a1ng \u1ee9ng v\u1edbi 550 kg; s\u1ed1 cam c\u00f2n l\u1ea1i \u0111\u01b0\u1ee3c b\u00e1n h\u1ebft l\u00e0 4950 kg. G\u1ecdi x (\u0111\u1ed3ng) l\u00e0 s\u1ed1 ti\u1ec1n m\u00e0 nh\u00e0 b\u1ea1n Linh b\u00e1n m\u1ed7i ki-l\u00f4-gam cam \u0111\u1ec3 thu l\u00e3i 20% x 0 . T\u1ed5ng ti\u1ec1n v\u1ed1n l\u00e0: 5500.17 500 6 875 000 103 125 000 (\u0111\u1ed3ng). T\u1eeb y\u00eau c\u1ea7u b\u00e0i to\u00e1n ta c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh sau: 4950x 103125000 0.2x x 20832, 5 (nh\u1eadn). V\u1eady c\u01a1 s\u1edf tr\u00e1i c\u00e2y nh\u00e0 b\u1ea1n Linh s\u1ebd b\u00e1n m\u1ed7i ki-l\u00f4-gam cam v\u1edbi gi\u00e1 20 832,5 \u0111\u1ed3ng \u0111\u1ec3 thu l\u00e3i 20% . C\u00e2u 5. (1,0 \u0111i\u1ec3m). Ng\u01b0\u1eddi ta mu\u1ed1n x\u00e2y m\u1ed9t b\u1ec3 ch\u1ee9a n\u01b0\u1edbc d\u1ea1ng hinh h\u1ed9p ch\u1eef nh\u1eadt kh\u00f4ng c\u00f3 n\u1eafp cao 1, 5 m c\u00f3 th\u1ec3 t\u00edch l\u00e0 12 m3 v\u00e0 chi\u1ec1u d\u00e0i g\u1ea5p \u0111\u00f4i chi\u1ec1u r\u1ed9ng. a) T\u00ednh chi\u1ec1u d\u00e0i v\u00e0 chi\u1ec1u r\u1ed9ng c\u1ee7a b\u1ec3 n\u01b0\u1edbc. b) Gi\u00e1 thu\u00ea c\u00f4ng nh\u00e2n x\u00e2y b\u1ec3 l\u00e0 500 000 \u0111\u1ed3ng\/ m2 . Chi ph\u00ed thu\u00ea nh\u00e2n c\u00f4ng l\u00e0 bao nhi\u00eau? L\u1eddi gi\u1ea3i a) G\u1ecdi a , b l\u1ea7n l\u01b0\u1ee3t l\u00e0 chi\u1ec1u d\u00e0i v\u00e0 chi\u1ec1u r\u1ed9ng c\u1ee7a b\u1ec3 n\u01b0\u1edbc a,b 0 . Chi\u1ec1u d\u00e0i g\u1ea5p \u0111\u00f4i chi\u1ec1u r\u1ed9ng n\u00ean: a 2b 1 . Th\u1ec3 t\u00edch c\u1ee7a b\u1ec3 l\u00e0 12 m3 n\u00ean: 1, 5.ab 12 2 . a 2b a 2b a 2b a 2b 2.2 4 T\u1eeb 1 v\u00e0 2 ta c\u00f3: 1, 5.ab 12 b2 4 b2 . b 2n b 2l T\u1eeb \u0111\u00e2y ta \u0111\u01b0\u1ee3c. chi\u1ec1u d\u00e0i v\u00e0 chi\u1ec1u r\u1ed9ng c\u1ee7a b\u1ec3 n\u01b0\u1edbc l\u1ea7n l\u01b0\u1ee3t l\u00e0 4 m v\u00e0 2 m. b) Ta c\u00f3 di\u1ec7n t\u00edch xung quanh c\u1ee7a b\u1ec3 n\u01b0\u1edbc l\u00e0: Sxq a b .2.1, 5 2 4 .2.1, 5 18 m2 , l\u1ea1i c\u00f3 chi ph\u00ed thu\u00ea nh\u00e2n c\u00f4ng l\u00e0 500 000 \u0111\u1ed3ng\/ m2 n\u00ean s\u1ed1 ti\u1ec1n ph\u1ea3i tr\u1ea3 l\u00e0: 500000.18 9 000 000 (\u0111\u1ed3ng). T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N 5","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH C\u00e2u 6. (0,75 \u0111i\u1ec3m). Anh Minh sinh v\u00e0o th\u1ebf k\u1ef7 20 . \u0110\u1ebfn ng\u00e0y sinh nh\u1eadt c\u1ee7a m\u00ecnh trong n\u0103m 2021 , anh nh\u1eadn th\u1ea5y tu\u1ed5i c\u1ee7a m\u00ecnh b\u1eb1ng \u0111\u00fang t\u1ed5ng c\u00e1c ch\u1eef s\u1ed1 trong n\u0103m sinh c\u1ee7a m\u00ecnh. T\u00ecm n\u0103m sinh c\u1ee7a anh Minh. L\u1eddi gi\u1ea3i Anh Minh sinh th\u1ebf k\u1ef7 20 suy ra n\u0103m sinh c\u1ee7a anh Minh t\u1eeb 1901 \u0111\u1ebfn 2000 . G\u1ecdi 19ab l\u00e0 n\u0103m sinh c\u1ee7a anh Minh a ; b ; a,b 10 . Tu\u1ed5i c\u1ee7a anh Minh v\u00e0o n\u0103m 2021 : 121 10a b . 2021 19ab 2.1000 2.10 1 1.1000 9.100 10.a b Theo \u0111\u1ec1 b\u00e0i ta c\u00f3: 121 11a 2. 5 b 11 11 a 2. 5 b . 121 10a b 1 9 a b V\u00ec 2. 5 b l\u00e0 s\u1ed1 ch\u1eb5n n\u00ean 11. 11 a l\u00e0 s\u1ed1 ch\u1eb5n. Ta c\u00f3: a 1a 3 a 5 a 7 a 9 b 50 39 28 17 6 Suy ra a 9 v\u00e0 b 6 V\u1eady anh Minh sinh n\u0103m 1996 . C\u00e2u 7. (1 \u0111i\u1ec3m). Hai xe m\u00e1y kh\u1edfi h\u00e0nh c\u00f9ng l\u00fac t\u1eeb A \u0111\u1ebfn B. Xe th\u1ee9 nh\u1ea5t ch\u1ea1y v\u1edbi v\u1eadn t\u1ed1c 32 km\/h, xe th\u1ee9 hai ch\u1ea1y v\u1edbi v\u1eadn t\u1ed1c l\u1edbn h\u01a1n xe th\u1ee9 nh\u1ea5t 8 km\/h. Tr\u00ean \u0111\u01b0\u1eddng \u0111i xe th\u1ee9 hai d\u1eebng l\u1ea1i ngh\u1ec9 45 ph\u00fat r\u1ed3i l\u1ea1i ti\u1ebfp t\u1ee5c ch\u1ea1y v\u1edbi v\u1eadn t\u1ed1c c\u0169. T\u00ednh qu\u00e3ng \u0111\u01b0\u1eddng AB bi\u1ebft hai xe \u0111\u1ebfn B c\u00f9ng m\u1ed9t l\u00fac. L\u1eddi gi\u1ea3i Xe th\u1ee9 nh\u1ea5t c\u00f3 v\u1eadn t\u1ed1c 32 km \/ h ; xe th\u1ee9 hai c\u00f3 v\u1eadn t\u1ed1c 32 8 40 km \/ h . 45 ph\u00fat 3 4 gi\u1edd. G\u1ecdi x km l\u00e0 qu\u00e3ng \u0111\u01b0\u1eddng AB x 0 . x Th\u1eddi gian xe th\u1ee9 nh\u1ea5t di chuy\u1ec3n \u0111\u1ebfn B l\u00e0: 32 (gi\u1edd). Th\u1eddi gian xe th\u1ee9 hai di chuy\u1ec3n \u0111\u1ebfn B l\u00e0: x 3 (gi\u1edd) 40 4 Do xe th\u1ee9 nh\u1ea5t v\u00e0 xe th\u1ee9 hai \u0111\u1ebfn B c\u00f9ng m\u1ed9t l\u00fac n\u00ean ta c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh: T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N 6","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH x x3 x 120 (nh\u1eadn). 32 40 4 V\u1eady qu\u00e3ng \u0111\u01b0\u1eddng AB d\u00e0i 120km . C\u00e2u 8. (3,0 \u0111i\u1ec3m) Cho ABC nh\u1ecdn AB AC n\u1ed9i ti\u1ebfp \u0111\u01b0\u1eddng tr\u00f2n O , \u0111\u01b0\u1eddng cao CE c\u1ee7a ABC c\u1eaft O t\u1ea1i M . V\u1ebd d\u00e2y MN vu\u00f4ng g\u00f3c v\u1edbi OA t\u1ea1i T , AO c\u1eaft O t\u1ea1i D . a) Ch\u1ee9ng minh AM AN v\u00e0 AM 2 AT.AD . b) G\u1ecdi F v\u00e0 H l\u1ea7n l\u01b0\u1ee3t l\u00e0 giao \u0111i\u1ec3m c\u1ee7a BN v\u1edbi AC v\u00e0 CE . Ch\u1ee9ng minh EF \u2016 MN v\u00e0 H l\u00e0 tr\u1ef1c t\u00e2m c\u1ee7a ABC . AM v\u00e0 t\u1ee9 gi\u00e1c OHKT n\u1ed9i ti\u1ebfp. c) G\u1ecdi K l\u00e0 trung \u0111i\u1ec3m c\u1ee7a AH . Ch\u1ee9ng minh AH L\u1eddi gi\u1ea3i a) *Ch\u1ee9ng minh AM AN AN . X\u00e9t tam gi\u00e1c vu\u00f4ng ATM v\u00e0 tam gi\u00e1c vu\u00f4ng ATN c\u00f3: AT : c\u1ea1nh chung; TM TN (do OA MN ). Suy ta vATM vATN (hai c\u1ea1nh g\u00f3c vu\u00f4ng) suy ra AM *Ch\u1ee9ng minh AM 2 AT.AD T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N 7","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH X\u00e9t AMD c\u00f3 AMD 90 (g\u00f3c n\u1ed9i ti\u1ebfp ch\u1eafn n\u1eeda \u0111\u01b0\u1eddng tr\u00f2n); l\u1ea1i c\u00f3 MT l\u00e0 \u0111\u01b0\u1eddng cao (gi\u1ea3 thi\u1ebft) n\u00ean ta suy ra AM 2 AT.AD . b) *Ch\u1ee9ng minh: EF \/\/MN X\u00e9t t\u1ee9 gi\u00e1c BEFC c\u00f3: EBF ECF (do ch\u1eafn hai cung b\u1eb1ng nhau); EBF , ECF c\u00f9ng nh\u00ecn c\u1ea1nh EF . Suy ra t\u1ee9 gi\u00e1c BEFC n\u1ed9i ti\u1ebfp. Do \u0111\u00f3 FEC FBC (c\u00f9ng ch\u1eafn cung FC ) 1 . M\u1eb7t kh\u00e1c NMC FBC (c\u00f9ng ch\u1eafn NC ) 2 . T\u1eeb 1 v\u00e0 2 suy ra NMC FEC EF \/\/MN ( 2 g\u00f3c so le trong). *Ch\u1ee9ng minh H l\u00e0 tr\u1ef1c t\u00e2m c\u1ee7a tam gi\u00e1c ABC . Do t\u1ee9 gi\u00e1c BEFC n\u1ed9i ti\u1ebfp n\u00ean BEC BFC 90 (c\u00f9ng ch\u1eafn cung BC ). Suy ra BF AC t\u1ea1i F ; L\u1ea1i c\u00f3 H l\u00e0 giao \u0111i\u1ec3m BF v\u00e0 CE n\u00ean H l\u00e0 tr\u1ef1c t\u00e2m c\u1ee7a tam gi\u00e1c ABC . c) *Ch\u1ee9ng minh AH AM G\u1ecdi P l\u00e0 giao \u0111i\u1ec3m c\u1ee7a AH v\u00e0 BC suy ra AH BC ( H l\u00e0 tr\u1ef1c t\u00e2m c\u1ee7a ABC ). Ta c\u00f3: AHE CHP (\u0111\u1ed1i \u0111\u1ec9nh) AHE HCP 90 1 . M\u1eb7t kh\u00e1c AEM vu\u00f4ng t\u1ea1i E n\u00ean AME MAE 90 2 , H\u01a1n n\u1eefa ta c\u00f3: MAE HCP (g\u00f3c n\u1ed9i ti\u1ebfp c\u00f9ng ch\u1eafn cung MB ) 3 . T\u1eeb 1 , 2 v\u00e0 3 ta suy ra MAE AHE AMH c\u00e2n t\u1ea1i A AM AH . * Ch\u1ee9ng minh t\u1ee9 gi\u00e1c OHKT n\u1ed9i ti\u1ebfp AM 2 AT.AD cmt Ta c\u00f3 AH AM cmt Suy ra AH 2 AT.AD 2AO ( AD l\u00e0 \u0111\u01b0\u1eddng k\u00ednh c\u1ee7a O ) M\u00e0 AH 2AK ( K l\u00e0 trung \u0111i\u1ec3m c\u1ee7a AH ) v\u00e0 AD N\u00ean AH .2AK AT.2AO hay AH .AK AT.AO 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 Suy ra AH AT AO AK AH AT X\u00e9t ATH v\u00e0 AKO c\u00f3 AA\u02c6 OchunAg K Suy ra ATH AKO (g-g) Suy ra AHT AOK T\u1ee9 gi\u00e1c OHKT c\u00f3: AHT AOK Suy ra t\u1ee9 gi\u00e1c OHKT n\u1ed9i ti\u1ebfp (hai g\u00f3c c\u00f9ng nh\u00ecn c\u1ea1nh KT b\u1eb1ng nhau) ----H\u1ebeT--- T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N 9"]