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 y H\u01af\u1edaNG D\u1eaaN GI\u1ea2I 8 C\u00e2u 1. (1,5 \u0111i\u1ec3m) Cho (P) : y = \u2212x2 v\u00e0 \u0111\u01b0\u1eddng th\u1eb3ng (d) : y = 3x \u2212 4 . 2 a) V\u1ebd \u0111\u1ed3 th\u1ecb (P) v\u00e0 (d) tr\u00ean c\u00f9ng h\u1ec7 tr\u1ee5c t\u1ecda \u0111\u1ed9. -2 O 1 2 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 8 2 0 2 8 2 x1 2 -4 4x y = 2x \u2212 2 0 2 b) T\u00ecm t\u1ecda \u0111\u1ed9 giao \u0111i\u1ec3m c\u1ee7a (P) v\u00e0 (d) b\u1eb1ng ph\u00e9p t\u00ednh. Ph\u01b0\u01a1ng tr\u00ecnh ho\u00e0nh \u0111\u1ed9 giao \u0111i\u1ec3m c\u1ee7a ( P) v\u00e0 (d ) : x2 = 2x \u2212 2 \uf0db x2 \u2212 4x + 4 = 0 \uf0db x = 2. 2 Thay x = 2 v\u00e0o y = x2 , ta \u0111\u01b0\u1ee3c: y = 22 = 2. 22 V\u1eady (2; 2) l\u00e0 giao \u0111i\u1ec3m c\u1ea7n t\u00ecm. C\u00e2u 2. (1 \u0111i\u1ec3m) Cho ph\u01b0\u01a1ng tr\u00ecnh x2 \u2212 5x + 4 = 0 . Kh\u00f4ng gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh, h\u00e3y t\u00ednh gi\u00e1 tr\u1ecb bi\u1ec3u th\u1ee9c A = 5x1 \u2212 x2 \u2212 x1 \u2212 5x2 . x1 x2 L\u1eddi gi\u1ea3i x2 \u22125x + 4 = 0 . V\u00ec \uf044 = b2 \u2212 4ac = (\u22125)2 \u2212 4.1.4 = 9 \uf03e 0 . N\u00ean ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m ph\u00e2n bi\u1ec7t x1, x2 . \uf0ec S = x1 + x2 = \u2212b = 5 \uf0ef\uf0ef P = x1.x2 = a Theo \u0111\u1ecbnh l\u00ed Vi-et, ta c\u00f3: \uf0ed c =4 . \uf0ef a \uf0ef\uf0ee 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 = 5x1 \u2212 x2 \u2212 x1 \u2212 5x2 = (5x1 \u2212 x2 ).x2 \u2212 ( x1 \u2212 5x2 ).x1 x1 x2 x1.x2 ( )A = 5x1.x2 \u2212 x22 \u2212 x12 + 5x1.x2 = 10x1.x2 \u2212 x12 + x22 x1.x2 x1.x2 ( ) ( )10.P \u2212 A= S 2 \u2212 2.P 10.4 \u2212 52 \u2212 2.4 = 23 . = P 44 C\u00e2u 3. (1 \u0111i\u1ec3m) L\u00fac 6 gi\u1edd s\u00e1ng, m\u1ed9t xe \u00f4 t\u00f4 \u1edf v\u1ecb tr\u00ed c\u00e1ch th\u00e0nh ph\u1ed1 H\u1ed3 Ch\u00ed Minh 50km v\u00e0 kh\u1edfi h\u00e0nh \u0111i H\u00e0 N\u1ed9i (\u1edf ng\u01b0\u1ee3c chi\u1ec1u v\u1edbi TPHCM). G\u1ecdi y = ax + b l\u00e0 h\u00e0m s\u1ed1 bi\u1ec3u di\u1ec5n \u0111\u1ed9 d\u00e0i qu\u00e3ng \u0111\u01b0\u1eddng t\u1eeb TPHCM \u0111\u1ebfn v\u1ecb tr\u00ed c\u1ee7a xe \u00f4 t\u00f4 sau x gi\u1edd theo \u0111\u1ed3 th\u1ecb \u1edf h\u00ecnh sau. y (km) 230 50 50 km 0 3 x (gi\u1edd) TPHCM H\u00e0 N\u1ed9i a) T\u00ecm a v\u00e0 b . b) V\u00e0o l\u00fac m\u1ea5y gi\u1edd th\u00ec xe \u00f4 t\u00f4 c\u00e1ch TPHCM 410km ? L\u1eddi gi\u1ea3i a) T\u00ecm a v\u00e0 b . 50 = a.0 + b\uf0deb = 50 . 230 = a.3 + 50\uf0de x = 60 . b) V\u00e0o l\u00fac m\u1ea5y gi\u1edd th\u00ec xe \u00f4 t\u00f4 c\u00e1ch TPHCM 410km ? Xe \u00f4 t\u00f4 c\u00e1ch TPHCM 410 km \uf0de y = 410 . Thay v\u00e0o ta c\u00f3 x = 6 (gi\u1edd). V\u1eady l\u00fac 12 gi\u1edd th\u00ec xe \u00f4 t\u00f4 c\u00e1ch TPHCM 410km . C\u00e2u 4. (1 \u0111i\u1ec3m). Hai x\u00ed nghi\u1ec7p theo k\u1ebf ho\u1ea1ch ph\u1ea3i l\u00e0m t\u1ed5ng c\u1ed9ng 360 d\u1ee5ng c\u1ee5. Tr\u00ean th\u1ef1c t\u1ebf, x\u00ed nghi\u1ec7p A v\u01b0\u1ee3t m\u1ee9c 12% , x\u00ed nghi\u1ec7p B v\u01b0\u1ee3t m\u1ee9c 10% do \u0111\u00f3 c\u1ea3 hai x\u00ed nghi\u1ec7p l\u00e0m t\u1ed5ng c\u1ed9ng 400 d\u1ee5ng c\u1ee5. T\u00ednh s\u1ed1 d\u1ee5ng c\u1ee5 m\u1ed7i x\u00ed nghi\u1ec7p ph\u1ea3i l\u00e0m. L\u1eddi gi\u1ea3i G\u1ecdi x (d\u1ee5ng c\u1ee5) l\u00e0 s\u1ed1 d\u1ee5ng c\u1ee5 x\u00ed nghi\u1ec7p A l\u00e0m; y (d\u1ee5ng c\u1ee5) l\u00e0 s\u1ed1 d\u1ee5ng c\u1ee5 x\u00ed nghi\u1ec7p B l\u00e0m. (\u0110K: x, y \uf0ce N * ).(sai k\u00ed hi\u1ec7u t\u1eadp h\u1ee3p s\u1ed1 t\u1ef1 nhi\u00ean x, y \uf0ce * ). V\u00ec theo k\u1ebf ho\u1ea1ch ph\u1ea3i l\u00e0m t\u1ed5ng c\u1ed9ng l\u00e0 360 d\u1ee5ng c\u1ee5 n\u00ean ta c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh: x + y = 360 (1) T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N 4","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH V\u00ec tr\u00ean th\u1ef1c t\u1ebf, x\u00ed nghi\u1ec7p A v\u01b0\u1ee3t m\u1ee9c 12% , x\u00ed nghi\u1ec7p B v\u01b0\u1ee3t m\u1ee9c 10% do \u0111\u00f3 c\u1ea3 hai x\u00ed nghi\u1ec7p l\u00e0m t\u1ed5ng c\u1ed9ng 400 d\u1ee5ng c\u1ee5 n\u00ean ta c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh: 112%.x +110%.y = 400 (2) T\u1eeb (1) v\u00e0 (2) , ta c\u00f3 h\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh: \uf0ecx + y = 360 \uf0db \uf0ecx = 200 (N) \uf0ed\uf0ee112%.x +110%.y = 400 \uf0ed = 160 (N) \uf0ee y V\u1eady x\u00ed nghi\u1ec7p A l\u00e0m \u0111\u01b0\u1ee3c 200 d\u1ee5ng c\u1ee5; x\u00ed nghi\u1ec7p B l\u00e0m \u0111\u01b0\u1ee3c 160 d\u1ee5ng c\u1ee5. C\u00e2u 5. (0,75 \u0111i\u1ec3m) M\u1ed9t gi\u1ea3i b\u00f3ng \u0111\u00e1 g\u1ed3m 6 \u0111\u1ed9i b\u00f3ng thi \u0111\u1ea5u theo th\u1ec3 th\u1ee9c vo\u0300ng tro\u0300n 1 l\u01b0\u1ee3t. \u0110\u1ed9i th\u1eafng \u0111\u01b0\u1ee3c 3 \u0111i\u1ec3m, ho\u00e0 \u0111\u01b0\u1ee3c 1 \u0111i\u1ec3m, thua 0 \u0111i\u1ec3m. K\u1ebft thu\u0301c gi\u1ea3i \u0111\u1ea5u, t\u1ed5ng s\u1ed1 \u0111i\u1ec3m c\u1ee7a c\u1ea3 6 \u0111\u1ed9i l\u00e0 41 \u0111i\u1ec3m. a) Ho\u0309i gi\u1ea3i \u0111\u1ea5u c\u00f3 bao nhi\u00eau tr\u1eadn? b) T\u00ednh s\u1ed1 tr\u1eadn h\u00f2a c\u1ee7a gi\u1ea3i \u0111\u1ea5u? L\u1eddi gi\u1ea3i a) Ho\u0309i gi\u1ea3i \u0111\u1ea5u c\u00f3 bao nhi\u00eau tr\u1eadn? S\u1ed1 tr\u1eadn \u0111\u1ea5u c\u1ee7a gi\u1ea3i l\u00e0: 5 + 4 + 3+ 2 +1 =15 (tr\u1eadn). *) b) T\u00ednh s\u1ed1 tr\u1eadn h\u00f2a c\u1ee7a gi\u1ea3i \u0111\u1ea5u? G\u1ecdi x (tr\u1eadn) l\u00e0 s\u1ed1 tr\u1eadn h\u00f2a c\u1ee7a gi\u1ea3i \u0111\u1ea5u (\u0110K: x \uf0ce V\u00ec m\u1ed9t tr\u1eadn th\u1eafng \u0111\u01b0\u1ee3c 3 ho\u0300a \u0111\u01b0\u1ee3c 1 \u0111i\u1ec3m v\u00e0 c\u00f3 t\u1ed5ng c\u1ed9ng l\u00e0 41 \u0111i\u1ec3m n\u00ean ta c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh: 2.x + 3.(15 \u2212 x) = 41 \uf0db x = 4 (nh\u1eadn). V\u1eady s\u1ed1 tr\u1eadn h\u00f2a c\u1ee7a gi\u1ea3i \u0111\u1ea5u l\u00e0 4 . C\u00e2u 6. (1 \u0111i\u1ec3m) Ba b\u1ea1n D\u0169ng, T\u00e0i v\u00e0 Tr\u00ed \u0111\u1ee9ng \u1edf ba v\u1ecb tr\u00ed A , B , C tr\u00ean m\u1ed9t \u0111\u01b0\u1eddng tr\u00f2n t\u00e2m O \u0111\u1ec3 ch\u01a1i tr\u00f2 truy\u1ec1n c\u1ea7u. Bi\u1ebft kho\u1ea3ng c\u00e1ch t\u1eeb D\u0169ng \u0111\u1ebfn T\u00e0i b\u1eb1ng kho\u1ea3ng c\u00e1ch t\u1eeb D\u0169ng \u0111\u1ebfn Tr\u00ed l\u00e0 16 m ( AB = AC =16m), kho\u1ea3ng c\u00e1ch t\u1eeb T\u00e0i \u0111\u1ebfn Tr\u00ed l\u00e0 19, 2 m ( BC = 19, 2m ) (H\u00ecnh b\u00ean). Em h\u00e3y t\u00ednh b\u00e1n k\u00ednh c\u1ee7a \u0111\u01b0\u1eddng tr\u00f2n (O) . L\u1eddi gi\u1ea3i 5 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 + \u0110\u01b0\u1eddng th\u1eb3ng AO c\u1eaft BC v\u00e0 (O) l\u1ea7n l\u01b0\u1ee3t t\u1ea1i H v\u00e0 D . \uf0de \uf044ABD vu\u00f4ng t\u1ea1i B v\u00e0 H l\u00e0 trung \u0111i\u1ec3m c\u1ee7a BC . + Ta c\u00f3: H l\u00e0 trung \u0111i\u1ec3m c\u1ee7a BC . HB = HC = BC = 19, 2 = 9, 6m . 22 + \uf044ABH vu\u00f4ng t\u1ea1i H c\u00f3: AH = 162 \u2212 9, 62 = 12,8m + \uf044ABD vu\u00f4ng t\u1ea1i B c\u00f3 BH l\u00e0 \u0111\u01b0\u1eddng cao. \uf0de AB2 = AH.AD (h\u1ec7 th\u1ee9c l\u01b0\u1ee3ng). \uf0de AD = AB2 = 162 = 20m . AH 12,8 V\u1eady b\u00e1n k\u00ednh c\u1ee7a (O) l\u00e0: R = AD : 2 = 20 : 2 =10m . C\u00e2u 7. (0,75 \u0111i\u1ec3m) B\u1ea1n An \u0111i mua giu\u0301p b\u1ed1 c\u00e2y l\u0103n s\u01a1n \u1edf c\u1eeda h\u00e0ng nh\u00e0 b\u00e1c To\u00e0n. M\u1ed9t c\u00e2y l\u0103n s\u01a1n t\u01b0\u1eddng c\u00f3 d\u1ea1ng m\u1ed9t kh\u1ed1i tr\u1ee5 v\u1edbi b\u00e1n k\u00ednh \u0111\u00e1y l\u00e0 5cm v\u00e0 chi\u1ec1u cao l\u00e0 23cm (h\u00ecnh v\u1ebd b\u00ean). Nh\u00e0 s\u1ea3n xu\u1ea5t cho bi\u1ebft sau khi l\u0103n 1000 vo\u0300ng th\u00ec c\u00e2y s\u01a1n t\u01b0\u1eddng c\u00f3 th\u1ec3 b\u1ecb ho\u0309ng. Ho\u0309i b\u1ea1n An c\u1ea7n mua \u00edt nh\u1ea5t m\u1ea5y c\u00e2y l\u0103n s\u01a1n t\u01b0\u1eddng bi\u1ebft di\u1ec7n t\u00edch t\u01b0\u1eddng m\u00e0 b\u1ed1 b\u1ea1n An c\u1ea7n s\u01a1n l\u00e0 100 m2 ? L\u1eddi gi\u1ea3i + Di\u1ec7n t\u00edch xung quanh c\u1ee7a c\u00e2y l\u0103n s\u01a1n: S = 2\uf070 R.h = 2.\uf070 .2,5.23 = 115\uf070 cm2 = 115\uf070 .10\u22124 m2 . + S\u1ed1 c\u00e2y l\u0103n s\u01a1n c\u1ea7n mua \u0111\u1ec3 s\u01a1n \u0111\u01b0\u1ee3c 100 m2 : 100 \uf0bb 2, 77 =3 c\u00e2y. 115.\uf070 .10\u22124.1000 C\u00e2u 8. (3 \u0111i\u1ec3m) Cho \uf044ABC nh\u1ecdn ( AB \uf03c AC ) n\u1ed9i ti\u1ebfp \u0111\u01b0\u1eddng tr\u00f2n (O) . C\u00e1c \u0111\u01b0\u1eddng cao AD , BE , CF c\u1eaft nhau t\u1ea1i H . Tia EF c\u1eaft tia CB t\u1ea1i K . T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N 6","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH a) Ch\u1ee9ng minh t\u1ee9 gi\u00e1c BFEC n\u1ed9i ti\u1ebfp v\u00e0 KE.KF = KB.KC . b) \u0110\u01b0\u1eddng th\u1eb3ng KA c\u1eaft (O) t\u1ea1i M . Ch\u1ee9ng minh t\u1ee9 gi\u00e1c AEFM n\u1ed9i ti\u1ebfp. c) G\u1ecdi N l\u00e0 trung \u0111i\u1ec3m c\u1ee7a BC . Ch\u1ee9ng minh M , H , N th\u1eb3ng h\u00e0ng. L\u1eddi gi\u1ea3i A ME FHO K B DN C I a) Ch\u1ee9ng minh t\u1ee9 gi\u00e1c BFEC n\u1ed9i ti\u1ebfp v\u00e0 KE.KF = KB.KC . + X\u00e9t t\u1ee9 gi\u00e1c BFEC , c\u00f3: \uf0ec\uf0efBFC = 90\uf0b0(CF \u22a5 AB) \uf0ed \uf0ef\uf0eeBEC = 90\uf0b0( BE \u22a5 AC ) \uf0de BFEC l\u00e0 t\u1ee9 gi\u00e1c n\u1ed9i ti\u1ebfp. \uf0de KFB = KCE . M\u00e0 EKC l\u00e0 g\u00f3c chung. \uf0de \uf044KBF \u223d\uf044KEC ( g.g ) \uf0de KB = KF ( 2 c\u1ea1nh t\u01b0\u01a1ng \u1ee9ng t\u1ec9 l\u1ec7) KE KC \uf0de KB.KC = KE.KF (1) b) \u0110\u01b0\u1eddng th\u1eb3ng KA c\u1eaft (O) t\u1ea1i M . Ch\u1ee9ng minh t\u1ee9 gi\u00e1c AEFM n\u1ed9i ti\u1ebfp. + Ta c\u00f3 AMBC n\u1ed9i ti\u1ebfp (O) . \uf0de KMB = KCA . M\u00e0 AKC l\u00e0 g\u00f3c chung. \uf0de \uf044KMB\u223d\uf044KCA ( g.g ) . 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 \uf0de KB = KM ( 2 c\u1ea1nh t\u01b0\u01a1ng \u1ee9ng t\u1ec9 l\u1ec7). KA KC \uf0de KB.KC = KM.KA (2) . T\u1eeb (1) v\u00e0 (2) \uf0de KM.KA = KE.KF . \uf0de KM = KF . KE KA M\u00e0 AKE l\u00e0 g\u00f3c chung. \uf0de \uf044KMF \u223d\uf044KEA (c.g.c) \uf0de KMF = KEA ( 2 g\u00f3c t\u01b0\u01a1ng \u1ee9ng b\u1eb1ng nhau). \uf0de AEFM l\u00e0 t\u1ee9 gi\u00e1c n\u1ed9i ti\u1ebfp. c) G\u1ecdi N l\u00e0 trung \u0111i\u1ec3m c\u1ee7a BC . Ch\u1ee9ng minh M , H , N th\u1eb3ng h\u00e0ng. + X\u00e9t t\u1ee9 gi\u00e1c AFHE , ta c\u00f3: \uf0ef\uf0ecHFA = 90o ( HF \u22a5 AB) . \uf0ed \uf0ef\uf0eeHEA = 90o ( HE \u22a5 AC ) Suy ra: AFHE n\u1ed9i ti\u1ebfp \u0111\u01b0\u1eddng tro\u0300n \u0111\u01b0\u1eddng k\u00ednh AH . + M\u00e0 AEFM n\u1ed9i ti\u1ebfp (ch\u1ee9ng minh tr\u00ean) \uf0de A, M , F, H , E c\u00f9ng thu\u1ed9c \u0111\u01b0\u1eddng tro\u0300n \u0111\u01b0\u1eddng k\u00ednh AH . \uf0de AHFM l\u00e0 t\u1ee9 gi\u00e1c n\u1ed9i ti\u1ebfp. \uf0de HMA = HFA = 90\uf0b0 \uf0de HM \u22a5 AM (3) + K\u1ebb AI l\u00e0 \u0111\u01b0\u1eddng k\u00ednh c\u1ee7a (O) . \uf0de AMI = 90\uf0b0 ; ACI = 90\uf0b0 . \uf0de \uf0ef\uf0ecAM \u22a5 MI (4) \uf0ed \uf0ee\uf0efCI \u22a5 AC T\u1eeb (3) v\u00e0 (4) \uf0de M , H , I th\u1eb3ng h\u00e0ng. (5) Ta c\u00f3 : + IC\u2225 BH (v\u00ec c\u00f9ng \u22a5 AC ). + IB\u2225 CH (v\u00ec c\u00f9ng \u22a5 AB ) \uf0de BHCI l\u00e0 h\u00ecnh b\u00ecnh h\u00e0nh. M\u00e0 N l\u00e0 trung \u0111i\u1ec3m c\u1ee7a BC (gt). \uf0de N l\u00e0 trung \u0111i\u1ec3m c\u1ee7a HI . 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 \uf0de H , N, I th\u1eb3ng h\u00e0ng (6) T\u1eeb (5) v\u00e0 (6) \uf0de M , H , N th\u1eb3ng h\u00e0ng (\u0111pcm). ----H\u1ebeT--- T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N 9","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH S\u00d4\u00db GD&\u00d1T TP HO\u00c0 CH\u00cd MINH \u00d1E\u00c0 THAM KHA\u00dbO TUYE\u00c5N SINH 10 PHO\u00d8NG G\u00d1&\u00d1T QUA\u00c4N 5 NA\u00caM HO\u00cfC: 2023 - 2024 \u0110\u1ec0 THAM KH\u1ea2O M\u00d4N: TO\u00c1N 9 M\u00c3 \u0110\u1ec0: Qu\u1eadn 5 - 3 \u0110\u00ea thi g\u1ed3m 8 c\u00e2u ho\u0309i t\u01b0\u0323 lu\u00e2\u0323n. Th\u01a1\u0300i gian: 120 phu\u0301t (kh\u00f4ng k\u00ea\u0309 th\u01a1\u0300i gian pha\u0301t \u0111\u00ea\u0300) C\u00e2u 1. (1,5 \u0111i\u1ec3m). Cho (P) : y = \u2212 x2 v\u00e0 \u0111\u01b0\u1eddng th\u1eb3ng (d) : y = 3x + 4 . 2 a) V\u1ebd \u0111\u1ed3 th\u1ecb (P) v\u00e0 (d) tr\u00ean c\u00f9ng h\u1ec7 tr\u1ee5c t\u1ecda \u0111\u1ed9. b) T\u00ecm t\u1ecda \u0111\u1ed9 giao \u0111i\u1ec3m c\u1ee7a (P) v\u00e0 (d) b\u1eb1ng ph\u00e9p t\u00ednh. C\u00e2u 2. (1 \u0111i\u1ec3m). Ch\u1ee9ng minh ph\u01b0\u01a1ng tr\u00ecnh : 2x2 \u2212 5x + 1 = 0 c\u00f3 hai nghi\u1ec7m l\u00e0 x1 , x2 . Kh\u00f4ng gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh, h\u00e3y t\u00ednh gi\u00e1 tr\u1ecb c\u1ee7a bi\u1ec3u th\u1ee9c: A= x1 + x2 \u2212 x1x2 x2 x1 C\u00e2u 3. (1 \u0111i\u1ec3m). C\u00f4ng th\u1ee9c YMCA d\u00f9ng \u0111\u1ec3 \u0111o l\u01b0\u1ee3ng \u201cm\u1ee1 th\u1eeba\u201d trong c\u01a1 th\u1ec3 d\u1ef1a v\u00e0o c\u00e2n n\u1eb7ng v\u00e0 s\u1ed1 \u0111o v\u00f2ng 2 nh\u01b0 sau: a + 4,15.m \u2212 0,082.n n Trong \u0111\u00f3: H\u1ec7 s\u1ed1 a = \u221298,42 n\u1ebfu l\u00e0 nam v\u00e0 a = \u221276,76 n\u1ebfu l\u00e0 n\u1eef; m l\u00e0 s\u1ed1 \u0111o v\u00f2ng 2 t\u00ednh b\u1eb1ng inch, n l\u00e0 c\u00e2n n\u1eb7ng t\u00ednh b\u1eb1ng pound. ( 1inch = 2,54cm ; 1 kg = 2,2 pound ) B\u1ea3ng ph\u00e2n lo\u1ea1i \u0111\u00e1nh gi\u00e1 l\u01b0\u1ee3ng \u201cm\u1ee1 th\u1eeba\u201d trong c\u01a1 th\u1ec3 X\u1ebfp lo\u1ea1i N\u1eef ( % ch\u1ea5t b\u00e9o) Nam ( % ch\u1ea5t b\u00e9o) T\u1ed1i thi\u1ec3u 10% \u2212 13% 2% \u2212 5% \u00cdt m\u1ee1 14% \u2212 20% 6% \u2212 13% B\u00ecnh th\u01b0\u1eddng 21% \u2212 24% 14% \u2212 17% Th\u1eeba c\u00e2n 25% \u2212 31% 18% \u2212 25% B\u00e9o ph\u00ec 32% + 26% + a) Anh Ho\u00e0ng c\u00f3 s\u1ed1 \u0111o v\u00f2ng 2 l\u00e0 78cm , n\u1eb7ng 74kg . D\u1ef1a v\u00e0o c\u00e1ch t\u00ednh tr\u00ean h\u00e3y \u0111\u00e1nh gi\u00e1 l\u01b0\u1ee3ng \u201cm\u1ee1 th\u1eeba\u201d trong c\u01a1 th\u1ec3 c\u1ee7a anh Ho\u00e0ng. b) Ch\u1ecb Hoa c\u00e2n n\u1eb7ng 60kg . Ch\u1ecb Hoa n\u00ean c\u00f3 s\u1ed1 \u0111o v\u00f2ng 2 bao nhi\u00eau \u0111\u1ec3 % ch\u1ea5t b\u00e9o ch\u1ec9 t\u1eeb 21% \u0111\u1ebfn 24% . 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 C th\u00ec l\u01b0\u1ee3ng calo c\u1ea7n t\u0103ng th\u00eam kho\u1ea3ng 30 calo. T\u1ea1i 21 C , 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). T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N 1","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH a) X\u00e1c \u0111\u1ecbnh h\u1ec7 s\u1ed1 a,b . b) N\u1ebfu m\u1ed9t ng\u01b0\u1eddi l\u00e0m vi\u1ec7c \u1edf sa m\u1ea1c Sahara trong nhi\u1ec7t \u0111\u1ed9 50 C th\u00ec c\u1ea7n bao nhi\u00eau calo? C\u00e2u 5. (1 \u0111i\u1ec3m). M\u1ed9t ng\u01b0\u1eddi c\u00f3 100 tri\u1ec7u \u0111\u1ed3ng mu\u1ed1n g\u1edfi v\u00e0o ng\u00e2n h\u00e0ng A c\u00f3 hai l\u1ef1a ch\u1ecdn: Ng\u01b0\u1eddi g\u1eedi c\u00f3 th\u1ec3 nh\u1eadn l\u1ea1i su\u1ea5t 5% \/n\u0103m v\u00e0 2000000 \u0111\u1ed3ng ti\u1ec1n th\u01b0\u1edfng n\u1ebfu g\u1edfi b\u1eb1ng ti\u1ec1n Vi\u1ec7t Nam ho\u1eb7c l\u00e3i su\u1ea5t 2% \/n\u0103m n\u1ebfu g\u1edfi b\u1eb1ng \u0111\u1ed3ng \u0111\u00f4 la M\u1ef9. N\u1ebfu gi\u00e1 \u0111\u00f4 la \u1edf th\u1eddi \u0111i\u1ec3m g\u1edfi v\u00e0o v\u00e0 l\u1ea5y ra sau c\u00f9ng m\u1ed9t n\u0103m l\u1ea7n l\u01b0\u1ee3t l\u00e0: 21500 \u0111\u1ed3ng\/\u0111\u00f4 la v\u00e0 21800 \u0111\u1ed3ng\/\u0111\u00f4 la. N\u1ebfu ng\u01b0\u1eddi \u0111\u00f3 g\u1edfi c\u00f3 k\u1ef3 h\u1ea1n m\u1ed9t n\u0103m th\u00ec n\u00ean ch\u1ecdn gi\u1ea3i ph\u00e1p n\u00e0o c\u00f3 l\u1ee3i h\u01a1n (t\u00ednh theo ti\u1ec1n \u0111\u1ed3ng Vi\u1ec7t Nam). (L\u00e0m tr\u00f2n \u0111\u1ebfn h\u00e0ng tr\u0103m ngh\u00ecn) C\u00e2u 6. (1 \u0111i\u1ec3m). \u0110\u1ec3 l\u00e0m c\u1ed1ng tho\u00e1t n\u01b0\u1edbc cho m\u1ed9t khu v\u1ef1c d\u00e2n c\u01b0 ng\u01b0\u1eddi ta c\u1ea7n \u0111\u00fac 500 \u1ed1ng c\u1ed1ng b\u00ea t\u00f4ng h\u00ecnh tr\u1ee5 c\u00f3 \u0111\u01b0\u1eddng k\u00ednh trong l\u00e0 2 m v\u00e0 chi\u1ec1u d\u00e0i m\u1ed7i \u1ed1ng l\u00e0 1,6 m , \u0111\u1ed9 d\u00e0y th\u00e0nh \u1ed1ng l\u00e0 10 cm . H\u1ecfi c\u00e1c c\u00f4ng nh\u00e2n ph\u1ea3i chu\u1ea9n b\u1ecb bao nhi\u00eau bao xi m\u0103ng \u0111\u1ec3 l\u00e0m \u0111\u1ee7 s\u1ed1 \u1ed1ng n\u00f3i tr\u00ean ?. Bi\u1ebft m\u1ed7i m\u00e9t kh\u1ed1i b\u00ea t\u00f4ng c\u1ea7n 7 bao xi m\u0103ng. C\u00e2u 7. (1 \u0111i\u1ec3m). V\u00e0o ng\u00e0y l\u1ec5 \u201cBlack Friday\u201d, c\u1eeda h\u00e0ng \u0111\u1ed3ng lo\u1ea1t gi\u1ea3m gi\u00e1 to\u00e0n b\u1ed9 s\u1ea3n ph\u1ea9m trong c\u1eeda h\u00e0ng. M\u1ed9t \u00e1o th\u1ec3 thao gi\u1ea3m 10% , m\u1ed9t qu\u1ea7n th\u1ec3 thao gi\u1ea3m 20% , m\u1ed9t \u0111\u00f4i gi\u00e0y th\u1ec3 thao gi\u1ea3m 30% . \u0110\u1eb7c bi\u1ec7t n\u1ebfu mua \u0111\u1ee7 b\u1ed9 bao g\u1ed3m 1 qu\u1ea7n, 1\u00e1o, 1 \u0111\u00f4i gi\u00e0y th\u00ec s\u1ebd \u0111\u01b0\u1ee3c gi\u1ea3m ti\u1ebfp (t\u00ednh theo gi\u00e1 tr\u1ecb c\u1ee7a 3 m\u1eb7t h\u00e0ng tr\u00ean sau khi gi\u1ea3m gi\u00e1). B\u1ea1n An v\u00e0o c\u1eeda h\u00e0ng mua 3 \u00e1o gi\u00e1 300000 VN\u0110\/c\u00e1i, 2 qu\u1ea7n gi\u00e1 250000 \/c\u00e1i, 1 \u0111\u00f4i gi\u00e0y gi\u00e1 500000 VN\u0110\/ \u0111\u00f4i (gi\u00e1 tr\u00ean l\u00e0 gi\u00e1 ch\u01b0a gi\u1ea3m). V\u1eady s\u1ed1 ti\u1ec1n b\u1ea1n An ph\u1ea3i tr\u1ea3 l\u00e0 bao nhi\u00eau? C\u00e2u 8. (3 \u0111i\u1ec3m) Cho \uf044ABC ( AB \uf03c AC) nh\u1ecdn n\u1ed9i ti\u1ebfp (O) c\u00f3 AH l\u00e0 \u0111\u01b0\u1eddng cao v\u00e0 I l\u00e0 t\u00e2m \u0111\u01b0\u1eddng tr\u00f2n n\u1ed9i ti\u1ebfp \uf044ABC . G\u1ecdi T,D l\u1ea7n l\u01b0\u1ee3t l\u00e0 giao \u0111i\u1ec3m c\u1ee7a AI v\u1edbi BC v\u00e0 (O) . a) Ch\u1ee9ng minh: OD vu\u00f4ng g\u00f3c v\u1edbi BC v\u00e0 tam gi\u00e1c IBD c\u00e2n. b) Qua D v\u1ebd \u0111\u01b0\u1eddng th\u1eb3ng vu\u00f4ng g\u00f3c v\u1edbi AD , c\u1eaft AH v\u00e0 BC l\u1ea7n l\u01b0\u1ee3t t\u1ea1i P,R . Ch\u1ee9ng minh: IP \u22a5 IR . c) V\u1ebd IK \u22a5 BC t\u1ea1i K,DK c\u1eaft AH t\u1ea1i S . Ch\u1ee9ng minh: t\u1ee9 gi\u00e1c SIDP 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 (P) : y = \u2212 x2 v\u00e0 \u0111\u01b0\u1eddng th\u1eb3ng (d) : y = 3x + 4 . 2 a) V\u1ebd \u0111\u1ed3 th\u1ecb (P) v\u00e0 (d) tr\u00ean c\u00f9ng h\u1ec7 tr\u1ee5c t\u1ecda \u0111\u1ed9. b) T\u00ecm t\u1ecda \u0111\u1ed9 giao \u0111i\u1ec3m c\u1ee7a (P) v\u00e0 (d) b\u1eb1ng ph\u00e9p t\u00ednh. L\u1eddi gi\u1ea3i a) V\u1ebd \u0111\u1ed3 th\u1ecb (P) v\u00e0 (d) tr\u00ean c\u00f9ng h\u1ec7 tr\u1ee5c t\u1ecda \u0111\u1ed9. a) BGT: x \u22122 \u22121 0 1 2 y = \u2212 x2 \u22122 \u2212 1 0 \u2212 1 \u22122 2 22 x0 1 y = 3x + 4 4 7 b) T\u00ecm t\u1ecda \u0111\u1ed9 giao \u0111i\u1ec3m c\u1ee7a (P) v\u00e0 (d) b\u1eb1ng ph\u00e9p t\u00ednh. Ph\u01b0\u01a1ng tr\u00ecnh ho\u00e0nh \u0111\u1ed9 giao \u0111i\u1ec3m c\u1ee7a (P) v\u00e0 (d) : \u2212 x2 = 3x + 4 2 \uf0db \u2212 x2 \u2212 3x \u2212 4 = 0 2 \uf0db \uf0e9x = \u22122 \uf0ea\uf0ebx = \u22124 Thay x = \u22122 v\u00e0o y = \u2212 x2 , ta \u0111\u01b0\u1ee3c: y = \u22122 . 2 Thay x = \u22124 v\u00e0o y = \u2212 x2 , ta \u0111\u01b0\u1ee3c: y = \u22128 . 2 V\u1eady (\u22122; \u2212 2) , (\u22124; \u2212 8) l\u00e0 hai giao \u0111i\u1ec3m c\u1ea7n t\u00ecm. C\u00e2u 2. (1 \u0111i\u1ec3m) a) Ch\u1ee9ng minh ph\u01b0\u01a1ng tr\u00ecnh : 2x2 \u2212 5x + 1 = 0 c\u00f3 hai nghi\u1ec7m l\u00e0 x1 , x2 . T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N 3","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH b) Kh\u00f4ng gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh, h\u00e3y t\u00ednh gi\u00e1 tr\u1ecb c\u1ee7a bi\u1ec3u th\u1ee9c: A= x1 + x2 \u2212 x1x2 x2 x1 L\u1eddi gi\u1ea3i V\u00ec \uf044 = b2 \u2212 4ac = (\u22125)2 \u2212 4.2.1 = 17 \uf03e 0 N\u00ean ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m ph\u00e2n bi\u1ec7t x1 ,x2 . \uf0ec\uf0ef\uf0efS = x1 + x2 = \u2212b = 5 \uf0ed = x1 a 2 Theo \u0111\u1ecbnh l\u00ed Vi-et, ta c\u00f3: \uf0ef\uf0ef\uf0eeP 1 c = 2 .x2 = a Ta c\u00f3: A= x1 + x2 \u2212 x1x2 x2 x1 A = x12 + x22 \u2212 x1x2 x1x2 ( )A = x1 + x2 2 \u2212 2x1x2 \u2212 x1x2 x1x2 \uf0e6 5 \uf0f62 \u2212 2. 1 1 \uf0e7\uf0e8 2 \uf0f7\uf0f8 2 A = \u2212 12 2 A = 10 C\u00e2u 3. (1 \u0111i\u1ec3m). C\u00f4ng th\u1ee9c YMCA d\u00f9ng \u0111\u1ec3 \u0111o l\u01b0\u1ee3ng \u201cm\u1ee1 th\u1eeba\u201d trong c\u01a1 th\u1ec3 d\u1ef1a v\u00e0o c\u00e2n n\u1eb7ng v\u00e0 s\u1ed1 \u0111o v\u00f2ng 2 nh\u01b0 sau: a + 4,15.m \u2212 0,082.n n Trong \u0111\u00f3: H\u1ec7 s\u1ed1 a = \u221298,42 n\u1ebfu l\u00e0 nam v\u00e0 a = \u221276,76 n\u1ebfu l\u00e0 n\u1eef; m l\u00e0 s\u1ed1 \u0111o v\u00f2ng 2 t\u00ednh b\u1eb1ng inch, n l\u00e0 c\u00e2n n\u1eb7ng t\u00ednh b\u1eb1ng pound. (1inch = 2,54cm; 1 kg = 2,2 pound) B\u1ea3ng ph\u00e2n lo\u1ea1i \u0111\u00e1nh gi\u00e1 l\u01b0\u1ee3ng \u201cm\u1ee1 th\u1eeba\u201d trong c\u01a1 th\u1ec3 X\u1ebfp lo\u1ea1i N\u1eef (% ch\u1ea5t b\u00e9o) Nam (% ch\u1ea5t b\u00e9o) T\u1ed1i thi\u1ec3u 10% - 13% 2% - 5% \u00cdt m\u1ee1 14% - 20% 6% - 13% B\u00ecnh th\u01b0\u1eddng 21% - 24% 14% - 17% Th\u1eeba c\u00e2n 25% - 31% 18% - 25% B\u00e9o ph\u00ec 32%+ 26%+ L\u1eddi gi\u1ea3i a) Anh Ho\u00e0ng c\u00f3 s\u1ed1 \u0111o v\u00f2ng 2 l\u00e0 78cm, n\u1eb7ng 74kg. D\u1ef1a v\u00e0o c\u00e1ch t\u00ednh tr\u00ean h\u00e3y \u0111\u00e1nh gi\u00e1 l\u01b0\u1ee3ng \u201cm\u1ee1 th\u1eeba\u201d trong c\u01a1 th\u1ec3 c\u1ee7a anh Ho\u00e0ng. T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N 4","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH \u0110\u1ed5i \u0111\u01a1n v\u1ecb: 78 cm = 3900 inch; 74 kg = 162,8 pound . 127 L\u01b0\u1ee3ng m\u1ee1 th\u1eeba trong c\u01a1 th\u1ec3 anh Ho\u00e0ng l\u00e0: a + 4,15.m \u2212 0, 082.n \u221298, 42 + 4,15. 3900 \u2212 0, 082.162, 8 127 = \uf0bb 0,1 \uf0bb 10% n 162,8 V\u1eady anh Ho\u00e0ng thu\u1ed9c lo\u1ea1i \u00edt m\u1ee1. b) Ch\u1ecb Hoa c\u00e2n n\u1eb7ng 60kg . Ch\u1ecb Hoa n\u00ean c\u00f3 s\u1ed1 \u0111o v\u00f2ng 2 bao nhi\u00eau \u0111\u1ec3 % ch\u1ea5t b\u00e9o ch\u1ec9 t\u1eeb 21% \u0111\u1ebfn 24% . \u0110\u1ed5i \u0111\u01a1n v\u1ecb: 60 kg = 132 pound ; L\u01b0\u1ee3ng m\u1ee1 th\u1eeba trong c\u01a1 th\u1ec3 ch\u1ecb Hoa l\u00e0: 21% \uf0a3 a + 4,15.m \u2212 0, 082.n \uf0a3 24% n \uf0db 0, 21 \uf0a3 \u221276, 76 + 4,15.m \u2212 0, 082.132 \uf0a3 0, 24 132 \uf0db \uf0ec4,15.m \uf0a3 119, 264 \uf0ed\uf0ee115,304 \uf0a3 4,15.m \uf0db 27,784 \uf0a3 m \uf0a3 28,738(inch) \uf0db 70,572 \uf0a3 m \uf0a3 72,995(cm) 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\u02daC th\u00ec l\u01b0\u1ee3ng calo c\u1ea7n t\u0103ng th\u00eam kho\u1ea3ng 30 calo. T\u1ea1i 21\u02daC, 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\u02daC th\u00ec c\u1ea7n bao nhi\u00eau calo? L\u1eddi gi\u1ea3i a) Theo \u0111\u1ec1 b\u00e0i: T\u1ea1i 21\u02daC, m\u1ed9t ng\u01b0\u1eddi l\u00e0m vi\u1ec7c c\u1ea7n s\u1eed d\u1ee5ng kho\u1ea3ng 3000 calo m\u1ed7i ng\u00e0y, thay v\u00e0o h\u00e0m s\u1ed1 y ax b ta c\u00f3: 21a + b = 3000 1 Nhi\u1ec7t \u0111\u1ed9 m\u00f4i tr\u01b0\u1eddng gi\u1ea3m \u0111i 1\u02daC th\u00ec l\u01b0\u1ee3ng calo c\u1ea7n t\u0103ng th\u00eam kho\u1ea3ng 30 calo, ta c\u00f3: 20a + b = 3030 2 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 T\u1eeb 1 , 2 ta c\u00f3 hpt: \uf0ec21a + b = 3000 \uf0ed\uf0ee20a + b = 3030 Gi\u1ea3i hpt ta c\u00f3: a 30; b 3630 b). N\u1ebfu ng\u01b0\u1eddi \u0111\u00f3 \u1edf sa m\u1ea1c Sahara trong nhi\u1ec7t \u0111\u1ed9 50 C th\u00ec ng\u01b0\u1eddi \u0111\u00f3 c\u1ea7n n\u0103ng l\u01b0\u1ee3ng l\u00e0: y 30.50 3630 2130 calo C\u00e2u 5. (1 \u0111i\u1ec3m) M\u1ed9t ng\u01b0\u1eddi c\u00f3 100 tri\u1ec7u \u0111\u1ed3ng mu\u1ed1n g\u1edfi v\u00e0o ng\u00e2n h\u00e0ng A c\u00f3 hai l\u1ef1a ch\u1ecdn: Ng\u01b0\u1eddi g\u1eedi c\u00f3 th\u1ec3 nh\u1eadn l\u1ea1i su\u1ea5t 5%\/n\u0103m v\u00e0 2 000 000 \u0111\u1ed3ng ti\u1ec1n th\u01b0\u1edfng n\u1ebfu g\u1edfi b\u1eb1ng ti\u1ec1n Vi\u1ec7t Nam ho\u1eb7c l\u00e3i su\u1ea5t 2%\/n\u0103m n\u1ebfu g\u1edfi b\u1eb1ng \u0111\u1ed3ng \u0111\u00f4 la M\u1ef9. N\u1ebfu gi\u00e1 \u0111\u00f4 la \u1edf th\u1eddi \u0111i\u1ec3m g\u1edfi v\u00e0o v\u00e0 l\u1ea5y ra sau c\u00f9ng m\u1ed9t n\u0103m l\u1ea7n l\u01b0\u1ee3t l\u00e0: 21 500 \u0111\u1ed3ng\/\u0111\u00f4 la v\u00e0 21 800 \u0111\u1ed3ng\/\u0111\u00f4 la. N\u1ebfu ng\u01b0\u1eddi \u0111\u00f3 g\u1edfi c\u00f3 k\u1ef3 h\u1ea1n m\u1ed9t n\u0103m th\u00ec n\u00ean ch\u1ecdn gi\u1ea3i ph\u00e1p n\u00e0o c\u00f3 l\u1ee3i h\u01a1n (t\u00ednh theo ti\u1ec1n \u0111\u1ed3ng Vi\u1ec7t Nam). (L\u00e0m tr\u00f2n \u0111\u1ebfn h\u00e0ng tr\u0103m ngh\u00ecn) L\u1eddi gi\u1ea3i \u0110\u1ed5i 100 (tri\u1ec7u \u0111\u1ed3ng) = 4651 (\u0111\u00f4 la) V\u1edbi l\u00e3i su\u1ea5t 2% \/n\u0103m ta \u0111\u01b0\u1ee3c: 4651.(1 + 2%) = 4744 (\u0111\u00f4 la) \u0110\u1ed5i ra ti\u1ec1n Vi\u1ec7t \u1edf th\u1eddi \u0111i\u1ec3m l\u1ea5y ra, ta \u0111\u01b0\u1ee3c: 4744 . 21 800 = 103 400 000 (\u0111\u1ed3ng) N\u1ebfu g\u1edfi b\u1eb1ng ti\u1ec1n Vi\u1ec7t Nam, ta \u0111\u01b0\u1ee3c: 100 000 000.105% + 2 000 000 = 107 000 000 (\u0111\u1ed3ng). Nh\u01b0 v\u1eady, g\u1edfi b\u1eb1ng ti\u1ec1n Vi\u1ec7t Nam s\u1ebd c\u00f3 l\u1ee3i h\u01a1n (gi\u1ea3i th\u00edch) C\u00e2u 6. (1 \u0111i\u1ec3m) \u0110\u1ec3 l\u00e0m c\u1ed1ng tho\u00e1t n\u01b0\u1edbc cho m\u1ed9t khu v\u1ef1c d\u00e2n c\u01b0 ng\u01b0\u1eddi ta c\u1ea7n \u0111\u00fac 500 \u1ed1ng c\u1ed1ng b\u00ea t\u00f4ng h\u00ecnh tr\u1ee5 c\u00f3 \u0111\u01b0\u1eddng k\u00ednh trong l\u00e0 2 m v\u00e0 chi\u1ec1u d\u00e0i m\u1ed7i \u1ed1ng l\u00e0 1,6 m, \u0111\u1ed9 d\u00e0y th\u00e0nh \u1ed1ng l\u00e0 10 cm. H\u1ecfi c\u00e1c c\u00f4ng nh\u00e2n ph\u1ea3i chu\u1ea9n b\u1ecb bao nhi\u00eau bao xi m\u0103ng \u0111\u1ec3 l\u00e0m \u0111\u1ee7 s\u1ed1 \u1ed1ng n\u00f3i tr\u00ean ?. Bi\u1ebft m\u1ed7i m\u00e9t kh\u1ed1i b\u00ea t\u00f4ng c\u1ea7n 7 bao xi m\u0103ng. L\u1eddi gi\u1ea3i 6 \u0110\u1ed5i 10cm = 0,1 m . B\u00e1n k\u00ednh trong R\u2019 = 1m 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\u00e1n k\u00ednh ngo\u00e0i : R =0,1 + 1 = 1,1 m L\u01b0\u1ee3ng b\u00ea t\u00f4ng c\u1ea7n d\u00f9ng cho m\u1ed9t \u1ed1ng c\u1ed1ng h\u00ecnh tr\u1ee5 l\u00e0 : ( ) ( ) ( )V = \uf070 R2h \u2013 \uf070 R\u20192 h = \uf070 h R2 \u2013 R\u20192 = \uf070 .1,6. 1,12 \u2212 12 = 0,336\uf070 m3 ( )L\u01b0\u01a1ng b\u00ea t\u00f4ng c\u1ea7n d\u00f9ng cho 500 \u1ed1ng c\u1ed1ng l\u00e0: 500 . 0,366\uf070 = 168\uf070 m3 S\u1ed1 bao xi m\u0103ng \u0111\u1ec3 l\u00e0m \u0111\u1ee7 s\u1ed1 \u00f4ng c\u1ed1ng tr\u00ean l\u00e0 : 168\uf070 .7 \uf0bb 3695 (bao) C\u00e2u 7. (1 \u0111i\u1ec3m) ). V\u00e0o ng\u00e0y l\u1ec5 \u201cBlack Friday\u201d, c\u1eeda h\u00e0ng \u0111\u1ed3ng lo\u1ea1t gi\u1ea3m gi\u00e1 to\u00e0n b\u1ed9 s\u1ea3n ph\u1ea9m trong c\u1eeda h\u00e0ng. M\u1ed9t \u00e1o th\u1ec3 thao gi\u1ea3m 10%, m\u1ed9t qu\u1ea7n th\u1ec3 thao gi\u1ea3m 20%, m\u1ed9t \u0111\u00f4i gi\u00e0y th\u1ec3 thao gi\u1ea3m 30% . \u0110\u1eb7c bi\u1ec7t n\u1ebfu mua \u0111\u1ee7 b\u1ed9 bao g\u1ed3m 1 qu\u1ea7n, 1\u00e1o, 1\u0111\u00f4i gi\u00e0y th\u00ec s\u1ebd \u0111\u01b0\u1ee3c gi\u1ea3m ti\u1ebfp (t\u00ednh theo gi\u00e1 tr\u1ecb c\u1ee7a 3 m\u1eb7t h\u00e0ng tr\u00ean sau khi gi\u1ea3m gi\u00e1). B\u1ea1n An v\u00e0o c\u1eeda h\u00e0ng mua 3 \u00e1o gi\u00e1 300 000 VN\u0110\/c\u00e1i, 2 qu\u1ea7n gi\u00e1 250 000 \/c\u00e1i, 1 \u0111\u00f4i gi\u00e0y gi\u00e1 500 000 VN\u0110\/ \u0111\u00f4i (gi\u00e1 tr\u00ean l\u00e0 gi\u00e1 ch\u01b0a gi\u1ea3m). V\u1eady s\u1ed1 ti\u1ec1n b\u1ea1n An ph\u1ea3i tr\u1ea3 l\u00e0 bao nhi\u00eau? L\u1eddi gi\u1ea3i S\u1ed1 ti\u1ec1n b\u1ea1n An ph\u1ea3i tr\u1ea3 khi mua 3 \u00e1o l\u00e0 : 3.300000.90% = 810000 ( \u0111\u1ed3ng) S\u1ed1 ti\u1ec1n b\u1ea1n An ph\u1ea3i tr\u1ea3 khi mua 2 qu\u1ea7n l\u00e0 : 2.250000.80% = 400000 ( \u0111\u1ed3ng) S\u1ed1 ti\u1ec1n b\u1ea1n An ph\u1ea3i tr\u1ea3 khi mua 1 \u0111\u00f4i gi\u1ea7y l\u00e0 : 500000.70% = 350000 ( \u0111\u1ed3ng) S\u1ed1 ti\u1ec1n b\u1ea1n An \u0111\u01b0\u1ee3c gi\u1ea3m khi mua \u0111\u1ee7 b\u1ed9 l\u00e0 : (810000 : 3 + 400000 : 2 + 350000).5% = 41000 ( \u0111\u1ed3ng) V\u1eady, s\u1ed1 ti\u1ec1n b\u1ea1n An ph\u1ea3i tr\u1ea3 l\u00e0 : 810000 + 400000 + 350000 \u2212 41000 =1519000 ( \u0111\u1ed3ng) T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N 7","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH C\u00e2u 8. (3 \u0111i\u1ec3m) Cho \u0394ABC (AB < AC) nh\u1ecdn n\u1ed9i ti\u1ebfp (O) c\u00f3 AH l\u00e0 \u0111\u01b0\u1eddng cao v\u00e0 I l\u00e0 t\u00e2m \u0111\u01b0\u1eddng tr\u00f2n n\u1ed9i ti\u1ebfp \u0394ABC. G\u1ecdi T, D l\u1ea7n l\u01b0\u1ee3t l\u00e0 giao \u0111i\u1ec3m c\u1ee7a AI v\u1edbi BC v\u00e0 (O). a) Ch\u1ee9ng minh: OD vu\u00f4ng g\u00f3c v\u1edbi BC v\u00e0 tam gi\u00e1c IBD c\u00e2n. b) Qua D v\u1ebd \u0111\u01b0\u1eddng th\u1eb3ng vu\u00f4ng g\u00f3c v\u1edbi AD, c\u1eaft AH v\u00e0 BC l\u1ea7n l\u01b0\u1ee3t t\u1ea1i P, R. Ch\u1ee9ng minh r\u1eb1ng: IP vu\u00f4ng g\u00f3c IR. c) V\u1ebd IK \u23ca BC t\u1ea1i K, DK c\u1eaft AH t\u1ea1i S. Ch\u1ee9ng minh: t\u1ee9 gi\u00e1c SIDP n\u1ed9i ti\u1ebfp. L\u1eddi gi\u1ea3i a) * C\/m: OD \u22a5 BC I l\u00e0 t\u00e2m \u0111\u01b0\u1eddng tr\u00f2n n\u1ed9i ti\u1ebfp \uf044ABC n\u00ean AD l\u00e0 tia ph\u00e2n gi\u00e1c BAC \uf0de D l\u00e0 \u0111i\u1ec3m ch\u00ednh gi\u1eefa BD \uf0de OD \u22a5 BC * C\/m: \uf044DBI c\u00e2n t\u1ea1i D . \uf0ec \uf0efBID = ABI + BAI \uf0ef\uf0edABI = IBC ( BI l\u00e0 ph\u00e2n gi\u00e1c ABC ) \uf0ef = 1 sdBD = 1 sdBD = DBC \uf0efBAI \uf0ee2 2 \uf0de BID = IBC + DBC = IBD \uf0de \uf044BID c\u00e2n t\u1ea1i D b) C\/m: IP \u22a5 IR . 8 \u2022 D\u1ec5 d\u00e0ng ch\u1ee9ng minh \u0111\u01b0\u1ee3c: \uf044DBT \u223d \uf044DAB(g \u2212 g) T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH \uf0de DB2 = DT.DA \uf0de DI2 = DT.DA ( \uf044BID c\u00e2n t\u1ea1i D ) 1 \u2022 Ch\u1ee9ng minh d\u1ec5 d\u00e0ng: T\u1ee9 gi\u00e1c AHDR n\u1ed9i ti\u1ebfp \uf0de PAD = PRT \uf0de \uf044DRT \u223d \uf044DAP(g \u2212 g) \uf0de DT.DA = DP.DR 2 T\u1eeb 1 , 2 ta c\u00f3: DI2 = DP.DR \uf0de \uf044DIP \u223d \uf044DRI(g \u2212 g) \uf0ec\uf0ef\uf0de IPD = DIR \uf0ed \uf0ef\uf0eeIPD + DIP = 90o (\uf044IDP vuong tai D) \uf0de DIR + DIP = 90o hay IP \u22a5 IR (\u0111pcm) c) Ta c\u00f3: PSD = IKS (hai g\u00f3c so le trong) IKS = DRI (t\u1ee9 gi\u00e1c IKDR n\u1ed9i ti\u1ebfp) DRI = DIP (c\u00f9ng ph\u1ee5 IPR ) Suy ra: PSD = DIP Suy ra: t\u1ee9 gi\u00e1c SIDP l\u00e0 t\u1ee9 gi\u00e1c n\u1ed9i ti\u1ebfp. T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N 9","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH S\u1ede GD&\u0110T TP. H\u1ed2 CH\u00cd MINH \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH L\u1edaP 10 TR\u01af\u1edcNG THPT TH S\u00c0I G\u00d2N N\u0102M H\u1eccC 2023 - 2024 M\u00d4N: TO\u00c1N 9 \u0110\u1ec0 THAM KH\u1ea2O \u0110\u00ea thi g\u1ed3m 8 c\u00e2u ho\u0309i t\u01b0\u0323 lu\u00e2\u0323n. M\u00c3 \u0110\u1ec0: THSG - 1 Th\u01a1\u0300i gian: 120 phu\u0301t (kh\u00f4ng k\u00ea\u0309 th\u01a1\u0300i gian pha\u0301t \u0111\u00ea\u0300) C\u00e2u 1. (1,5 \u0111i\u1ec3m). Cho parabol (P) : y = 1 x2 v\u00e0 \u0111\u01b0\u1eddng th\u1eb3ng (d) : y = \u2212x + 4 2 a) V\u1ebd (P) v\u00e0 (d) tr\u00ean c\u00f9ng m\u1ed9t m\u1eb7t ph\u1eb3ng t\u1ecda \u0111\u1ed9 Oxy ; b) 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 7x2 \u2212 9x \u2212 5 = 0 c\u00f3 hai nghi\u1ec7m x1 , x2. Kh\u00f4ng gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh, ( ) ( )h\u00e3y t\u00ednh gi\u00e1 tr\u1ecb c\u1ee7a bi\u1ec3u th\u1ee9c A = x1+ x2 x1 x2 1 \u2212 x2 1 \u2212 x1 . C\u00e2u 3. (0,75 \u0111i\u1ec3m). M\u1ed9t b\u00e0i ki\u1ec3m tra tr\u1eafc nghi\u1ec7m kh\u00e1ch quan g\u1ed3m 20 c\u00e2u h\u1ecfi v\u1edbi c\u00e1ch th\u1ee9c t\u00ednh \u0111i\u1ec3m nh\u01b0 sau: m\u1ed7i c\u00e2u tr\u1ea3 l\u1eddi \u0111\u00fang \u0111\u01b0\u1ee3c c\u1ed9ng 5 \u0111i\u1ec3m, m\u1ed7i c\u00e2u tr\u1ea3 l\u1eddi sai b\u1ecb tr\u1eeb 2 \u0111i\u1ec3m v\u00e0 m\u1ed7i c\u00e2u h\u1ecfi kh\u00f4ng c\u00f3 c\u00e2u tr\u1ea3 l\u1eddi s\u1ebd kh\u00f4ng \u0111\u01b0\u1ee3c c\u1ed9ng c\u0169ng kh\u00f4ng b\u1ecb tr\u1eeb \u0111i\u1ec3m n\u00e0o. a) B\u00ecnh tr\u1ea3 l\u1eddi to\u00e0n b\u1ed9 c\u00e1c c\u00e2u trong b\u00e0i ki\u1ec3m tra v\u00e0 \u0111\u00fang 13 c\u00e2u. An kh\u00f4ng tr\u1ea3 l\u1eddi c\u00e2u v\u00e0 ch\u1ec9 \u0111\u00fang 12 c\u00e2u. Trong hai b\u1ea1n, ai \u0111\u1ea1t \u0111i\u1ec3m cao h\u01a1n, v\u00ec sao? b) K\u1ebft th\u00fac b\u00e0i ki\u1ec3m tra, B\u00e1ch \u0111\u1ea1t \u0111\u01b0\u1ee3c 69 \u0111i\u1ec3m. H\u1ecfi B\u00e1ch \u0111\u00e3 tr\u1ea3 l\u1eddi \u0111\u00fang bao nhi\u00eau c\u00e2u? C\u00e2u 4. (0,75 \u0111i\u1ec3m). S\u1ed1 ti\u1ec1n l\u01b0\u01a1ng m\u1ed7i th\u00e1ng y (tri\u1ec7u \u0111\u1ed3ng) c\u1ee7a anh H\u00e0o, nh\u00e2n vi\u00ean b\u00e1n h\u00e0ng c\u00f4ng ty A , ph\u1ee5 thu\u1ed9c v\u00e0o s\u1ed1 l\u01b0\u1ee3ng x s\u1ea3n ph\u1ea9m b\u00e1n ra v\u01b0\u1ee3t ch\u1ec9 ti\u00eau c\u1ee7a anh trong th\u00e1ng \u0111\u00f3. M\u1ed1i li\u00ean h\u1ec7 gi\u1eefa hai \u0111\u1ea1i l\u01b0\u1ee3ng n\u00e0y l\u00e0 m\u1ed9t h\u00e0m s\u1ed1 b\u1eadc nh\u1ea5t y = ax + b . a) T\u00ecm bi\u1ebft r\u1eb1ng trong th\u00e1ng 4 , anh H\u00e0o b\u00e1n v\u01b0\u1ee3t ch\u1ec9 ti\u00eau 9 s\u1ea3n ph\u1ea9m, nh\u1eadn ti\u1ec1n l\u01b0\u01a1ng l\u00e0 9,8 tri\u1ec7u \u0111\u1ed3ng v\u00e0 trong th\u00e1ng 5 , anh b\u00e1n v\u01b0\u1ee3t ch\u1ec9 ti\u00eau 7 s\u1ea3n ph\u1ea9m, nh\u1eadn ti\u1ec1n l\u01b0\u01a1ng l\u00e0 9,4 tri\u1ec7u \u0111\u1ed3ng; b) Anh H\u00e0o mu\u1ed1n d\u00f9ng ti\u1ec1n l\u01b0\u01a1ng th\u00e1ng 6 \u0111\u1ec3 l\u1eafp m\u1ed9t m\u00e1y \u0111i\u1ec1u h\u00f2a nhi\u1ec7t \u0111\u1ed9 v\u1edbi t\u1ed5ng chi ph\u00ed 10 480000 \u0111\u1ed3ng th\u00ec trong th\u00e1ng 6 n\u00e0y anh ph\u1ea3i b\u00e1n v\u01b0\u1ee3t ch\u1ec9 ti\u00eau \u00edt nh\u1ea5t bao nhi\u00eau s\u1ea3n ph\u1ea9m? C\u00e2u 5. (1 \u0111i\u1ec3m). Nh\u00e2n d\u1ecbp l\u1ec5 30 \/ 4 , c\u1eeda h\u00e0ng \u0111i\u1ec7n m\u00e1y B th\u1ef1c hi\u1ec7n khuy\u1ebfn m\u00e3i gi\u1ea3m 20% so v\u1edbi gi\u00e1 ni\u00eam y\u1ebft cho t\u1ea5t c\u1ea3 c\u00e1c s\u1ea3n ph\u1ea9m. a) M\u1eb9 Lan \u0111\u00e3 \u0111\u1ebfn c\u1eeda h\u00e0ng n\u00e0y \u0111\u1ec3 mua m\u1ed9t chi\u1ebfc tivi c\u00f3 gi\u00e1 ni\u00eam y\u1ebft l\u00e0 11 tri\u1ec7u \u0111\u1ed3ng. Sau khi \u0111\u01b0\u1ee3c h\u01b0\u1edfng ch\u01b0\u01a1ng tr\u00ecnh khuy\u1ebfn m\u00e3i tr\u00ean, do c\u00f3 th\u1ebb VIP, m\u1eb9 Lan c\u00f2n \u0111\u01b0\u1ee3c gi\u1ea3m 5% tr\u00ean t\u1ed5ng h\u00f3a \u0111\u01a1n. H\u1ecfi m\u1eb9 Lan ph\u1ea3i tr\u1ea3 bao nhi\u00eau ti\u1ec1n? T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N 1","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH b) Khi b\u00e1n tivi cho m\u1eb9 Lan, c\u1eeda h\u00e0ng v\u1eabn c\u00f2n l\u00e3i 10% so v\u1edbi gi\u00e1 v\u1ed1n. H\u1ecfi c\u1eeda h\u00e0ng \u0111\u00e3 nh\u1eadp l\u00f4 100 chi\u1ebfc tivi tr\u00ean v\u1edbi s\u1ed1 ti\u1ec1n v\u1ed1n l\u00e0 bao nhi\u00eau? C\u00e2u 6. (1 \u0111i\u1ec3m). M\u1ed9t lon n\u01b0\u1edbc ng\u1ecdt c\u1ee7a h\u00e3ng C c\u00f3 d\u1ea1ng h\u00ecnh tr\u1ee5 v\u1edbi \u0111\u01b0\u1eddng k\u00ednh \u0111\u00e1y l\u00e0 6,5 cm , ch\u1ee9a. 355 ml ml n\u01b0\u1edbc ng\u1ecdt. a) T\u00ednh chi\u1ec1u cao c\u1ee7a lon (l\u00e0m tr\u00f2n \u0111\u1ebfn cm ), bi\u1ebft r\u1eb1ng \u0111\u1ec3 tr\u00e1nh lon v\u1ee1 do s\u1ef1 gi\u00e3n n\u1edf v\u00ec nhi\u1ec7t c\u1ee7a ch\u1ea5t l\u1ecfng, l\u01b0\u1ee3ng n\u01b0\u1edbc ng\u1ecdt trong lon ch\u1ec9 chi\u1ebfm 90% th\u1ec3 t\u00edch lon. b) Ng\u01b0\u1eddi ta th\u01b0\u1eddng \u0111\u00f3ng m\u1ed9t l\u1ed1c 6 lon v\u00e0o m\u1ed9t th\u00f9ng c\u00e1c-t\u00f4ng d\u1ea1ng h\u00ecnh h\u1ed9p ch\u1eef nh\u1eadt v\u1eeba kh\u00edt nh\u01b0 h\u00ecnh b\u00ean, chi\u1ec1u cao c\u1ee7a th\u00f9ng b\u1eb1ng chi\u1ec1u cao c\u1ee7a lon (\u0111\u00e3 l\u00e0m tr\u00f2n \u1edf c\u00e2u a). T\u00ednh th\u1ec3 t\u00edch c\u1ee7a th\u00f9ng c\u00e1c-t\u00f4ng n\u00e0y. C\u00e2u 7. (1 \u0111i\u1ec3m). Ch\u00e0o \u0111\u00f3n SEA Games 31, m\u1ed9t ph\u00e2n x\u01b0\u1edfng nh\u1eadn may \u0111\u1ed3ng ph\u1ee5c cho c\u1ed5 \u0111\u1ed9ng vi\u00ean v\u1edbi d\u1ef1 ki\u1ebfn m\u1ed7i ng\u00e0y s\u1ebd may xong 90 b\u1ed9 qu\u1ea7n \u00e1o. Khi th\u1ef1c hi\u1ec7n, nh\u1edd c\u1ea3i ti\u1ebfn k\u1ef9 thu\u1eadt, m\u1ed7i ng\u00e0y x\u01b0\u1edfng \u0111\u00e3 may \u0111\u01b0\u1ee3c 120 b\u1ed9 qu\u1ea7n \u00e1o. Do \u0111\u00f3, x\u01b0\u1edfng \u0111\u00e3 ho\u00e0n th\u00e0nh tr\u01b0\u1edbc k\u1ebf ho\u1ea1ch 6 ng\u00e0y v\u00e0 may th\u00eam \u0111\u01b0\u1ee3c 60 b\u1ed9 qu\u1ea7n \u00e1o. H\u1ecfi theo k\u1ebf ho\u1ea1ch ban \u0111\u1ea7u, ph\u00e2n x\u01b0\u1edfng n\u00e0y c\u1ea7n may bao nhi\u00eau b\u1ed9 qu\u1ea7n \u00e1o? C\u00e2u 8. (3 \u0111i\u1ec3m) Cho tam gi\u00e1c ABC vu\u00f4ng t\u1ea1i B ( AB \uf03c BC) , n\u1ed9i ti\u1ebfp \u0111\u01b0\u1eddng tr\u00f2n (O; R) . D\u1ef1ng \u0111\u01b0\u1eddng k\u00ednh BD , ti\u1ebfp tuy\u1ebfn t\u1ea1i C c\u1ee7a (O) c\u1eaft c\u00e1c tia AB, AD l\u1ea7n l\u01b0\u1ee3t t\u1ea1i E, F. a) Ch\u1ee9ng minh AB.AE = AD.AF v\u00e0 t\u1ee9 gi\u00e1c BDFE n\u1ed9i ti\u1ebfp. b) D\u1ef1ng \u0111\u01b0\u1eddng th\u1eb3ng d qua A v\u00e0 vu\u00f4ng g\u00f3c v\u1edbi BD , d c\u1eaft (O) v\u00e0 EF theo th\u1ee9 t\u1ef1 t\u1ea1i M v\u00e0 N ( M \uf0b9 A) . Ch\u1ee9ng minh t\u1ee9 gi\u00e1c BMNE n\u1ed9i ti\u1ebfp v\u00e0 N l\u00e0 trung \u0111i\u1ec3m c\u1ee7a EF. c) G\u1ecdi I l\u00e0 t\u00e2m \u0111\u01b0\u1eddng tr\u00f2n ngo\u1ea1i ti\u1ebfp tam gi\u00e1c BDE . T\u00ednh kho\u1ea3ng c\u00e1ch t\u1eeb I \u0111\u1ebfn \u0111\u01b0\u1eddng th\u1eb3ng EF theo R . ----H\u1ebeT--- T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N 2","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH H\u01af\u1edaNG D\u1eaaN GI\u1ea2I C\u00e2u 1. (1,5 \u0111i\u1ec3m) Cho parabol (P) : y = 1 x2 v\u00e0 \u0111\u01b0\u1eddng th\u1eb3ng (d) : y = \u2212x + 4 2 a) V\u1ebd (P) v\u00e0 (d) tr\u00ean c\u00f9ng m\u1ed9t m\u1eb7t ph\u1eb3ng t\u1ecda \u0111\u1ed9 Oxy ; b) T\u00ecm t\u1ecda \u0111\u1ed9 giao \u0111i\u1ec3m c\u1ee7a (P) v\u00e0 (d) b\u1eb1ng ph\u00e9p t\u00ednh. L\u1eddi gi\u1ea3i a) V\u1ebd \u0111\u1ed3 th\u1ecb (P) v\u00e0 (d) tr\u00ean c\u00f9ng h\u1ec7 tr\u1ee5c t\u1ecda \u0111\u1ed9. BGT: x \u22124 \u22122 0 2 4 y = 1 x2 8 2 0 2 8 2 x 02 y = \u2212x + 4 4 2 b) T\u00ecm t\u1ecda \u0111\u1ed9 giao \u0111i\u1ec3m c\u1ee7a (P) v\u00e0 (d) b\u1eb1ng ph\u00e9p t\u00ednh. Ph\u01b0\u01a1ng tr\u00ecnh ho\u00e0nh \u0111\u1ed9 giao \u0111i\u1ec3m c\u1ee7a (P) v\u00e0 (d) : 1 x2 = \u2212x + 4 2 \uf0db 1 x2 + x \u2212 4 = 0 2 \uf0db \uf0e9 x = \u22124 \uf0ea x = 2 \uf0eb Thay x = 2 v\u00e0o y = 1 x2 , ta \u0111\u01b0\u1ee3c: y = 2 . 2 Thay x = \u22124 v\u00e0o y = 1 x2 , ta \u0111\u01b0\u1ee3c: y = 8 . 2 V\u1eady (2; 2) , (\u22124; 8) l\u00e0 hai giao \u0111i\u1ec3m c\u1ea7n t\u00ecm. C\u00e2u 2. (1 \u0111i\u1ec3m) Cho ph\u01b0\u01a1ng tr\u00ecnh 7x2 \u2212 9x \u2212 5 = 0 c\u00f3 hai nghi\u1ec7m x1 , x2. Kh\u00f4ng gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh, h\u00e3y t\u00ednh gi\u00e1 tr\u1ecb c\u1ee7a bi\u1ec3u th\u1ee9c A = x1 ( 1 \u2212 x2 )+ x2 ( 1 \u2212 x1 ) . x2 x1 L\u1eddi gi\u1ea3i V\u00ec \uf044 = b2 \u2212 4ac = (\u22129)2 \u2212 4.(\u22125).7 = 221 \uf03e 0 N\u00ean ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m ph\u00e2n bi\u1ec7t x1 , x2 . 3 T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH \uf0ec\uf0ef\uf0efS = x1 + x2 = \u2212b = 9 \uf0ed = x1 a 7 Theo \u0111\u1ecbnh l\u00ed Vi-et, ta c\u00f3: \uf0ef\uf0ef\uf0eeP \u22125 c = 7 .x2 = a A = x12 + x22 \u2212 x1x2 ( )Ta c\u00f3: x1 + x2 A = ( x1 + x2 )2 \u2212 2x1x2 \u2212( x1 + x2 ) x1x2 \uf0e6 9 \uf0f62 \u2212 2. \u22125 9 28 \uf0e7\uf0e8 7 \uf0f7\uf0f8 7 A = \u2212 = \u2212 \u22125 7 5 7 C\u00e2u 3. (0,75 \u0111i\u1ec3m) M\u1ed9t b\u00e0i ki\u1ec3m tra tr\u1eafc nghi\u1ec7m kh\u00e1ch quan g\u1ed3m 20 c\u00e2u h\u1ecfi v\u1edbi c\u00e1ch th\u1ee9c t\u00ednh \u0111i\u1ec3m nh\u01b0 sau: m\u1ed7i c\u00e2u tr\u1ea3 l\u1eddi \u0111\u00fang \u0111\u01b0\u1ee3c c\u1ed9ng 5 \u0111i\u1ec3m, m\u1ed7i c\u00e2u tr\u1ea3 l\u1eddi sai b\u1ecb tr\u1eeb 2 \u0111i\u1ec3m v\u00e0 m\u1ed7i c\u00e2u h\u1ecfi kh\u00f4ng c\u00f3 c\u00e2u tr\u1ea3 l\u1eddi s\u1ebd kh\u00f4ng \u0111\u01b0\u1ee3c c\u1ed9ng c\u0169ng kh\u00f4ng b\u1ecb tr\u1eeb \u0111i\u1ec3m n\u00e0o. a) B\u00ecnh tr\u1ea3 l\u1eddi to\u00e0n b\u1ed9 c\u00e1c c\u00e2u trong b\u00e0i ki\u1ec3m tra v\u00e0 \u0111\u00fang 13 c\u00e2u. An kh\u00f4ng tr\u1ea3 l\u1eddi c\u00e2u v\u00e0 ch\u1ec9 \u0111\u00fang 12 c\u00e2u. Trong hai b\u1ea1n, ai \u0111\u1ea1t \u0111i\u1ec3m cao h\u01a1n, v\u00ec sao? b) K\u1ebft th\u00fac b\u00e0i ki\u1ec3m tra, B\u00e1ch \u0111\u1ea1t \u0111\u01b0\u1ee3c 69 \u0111i\u1ec3m. H\u1ecfi B\u00e1ch \u0111\u00e3 tr\u1ea3 l\u1eddi \u0111\u00fang bao nhi\u00eau c\u00e2u? L\u1eddi gi\u1ea3i a) B\u00ecnh \u0111\u1ea1t \u0111\u01b0\u1ee3c 13.5 \u2212 7.2 = 51 \u0111i\u1ec3m, An \u0111\u1ea1t \u0111\u01b0\u1ee3c 12.5 \u2212 2.4 = 52 \u0111i\u1ec3m. V\u1eady An cao \u0111i\u1ec3m h\u01a1n. b) G\u1ecdi x, y l\u1ea7n l\u01b0\u1ee3t l\u00e0 s\u1ed1 c\u00e2u tr\u1ea3 l\u1eddi \u0111\u00fang v\u00e0 sai c\u1ee7a B\u00e1ch, x, y nguy\u00ean d\u01b0\u01a1ng, x \uf03e 13. \uf0ec 5x \u2212 2y = 69 ( 1 ) \uf0ed \uf0ee x + y \uf0a3 20 ( 1 ) \uf0de y = 2x + x \u2212 69 2 \u0110\u1eb7t t = x \u2212 69 \uf0ce \uf0de \uf0ec x = 2t + 69 2 \uf0ed y = 5t + 138 \uf0ee \uf0ecx \uf03e 13 \uf0de t \uf03e \u221228 \uf0de t = \u221227 \uf0de x = 15, y = 3 \uf0ed + y \uf0a3 20 \uf0de 7t + \uf0ee x 207 \uf0a3 20 \uf0de t \uf03c \u221226 V\u1eady B\u00e1ch tr\u1ea3 l\u1eddi \u0111\u00fang 15 c\u00e2u. C\u00e2u 4. (0,75 \u0111i\u1ec3m). S\u1ed1 ti\u1ec1n l\u01b0\u01a1ng m\u1ed7i th\u00e1ng y (tri\u1ec7u \u0111\u1ed3ng) c\u1ee7a anh H\u00e0o, nh\u00e2n vi\u00ean b\u00e1n h\u00e0ng c\u00f4ng ty A , ph\u1ee5 thu\u1ed9c v\u00e0o s\u1ed1 l\u01b0\u1ee3ng x s\u1ea3n ph\u1ea9m b\u00e1n ra v\u01b0\u1ee3t ch\u1ec9 ti\u00eau c\u1ee7a anh trong th\u00e1ng \u0111\u00f3. M\u1ed1i li\u00ean h\u1ec7 gi\u1eefa hai \u0111\u1ea1i l\u01b0\u1ee3ng n\u00e0y l\u00e0 m\u1ed9t h\u00e0m s\u1ed1 b\u1eadc nh\u1ea5t y = ax + b. 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 a) T\u00ecm bi\u1ebft r\u1eb1ng trong th\u00e1ng 4 , anh H\u00e0o b\u00e1n v\u01b0\u1ee3t ch\u1ec9 ti\u00eau 9 s\u1ea3n ph\u1ea9m, nh\u1eadn ti\u1ec1n l\u01b0\u01a1ng l\u00e0 9,8 tri\u1ec7u \u0111\u1ed3ng v\u00e0 trong th\u00e1ng 5 , anh b\u00e1n v\u01b0\u1ee3t ch\u1ec9 ti\u00eau 7 s\u1ea3n ph\u1ea9m, nh\u1eadn ti\u1ec1n l\u01b0\u01a1ng l\u00e0 9,4 tri\u1ec7u \u0111\u1ed3ng; b) Anh H\u00e0o mu\u1ed1n d\u00f9ng ti\u1ec1n l\u01b0\u01a1ng th\u00e1ng 6 \u0111\u1ec3 l\u1eafp m\u1ed9t m\u00e1y \u0111i\u1ec1u h\u00f2a nhi\u1ec7t \u0111\u1ed9 v\u1edbi t\u1ed5ng chi ph\u00ed 10 480000 \u0111\u1ed3ng th\u00ec trong th\u00e1ng 6 n\u00e0y anh ph\u1ea3i b\u00e1n v\u01b0\u1ee3t ch\u1ec9 ti\u00eau \u00edt nh\u1ea5t bao nhi\u00eau s\u1ea3n ph\u1ea9m? L\u1eddi gi\u1ea3i a) Trong th\u00e1ng 4 , anh H\u00e0o b\u00e1n v\u01b0\u1ee3t ch\u1ec9 ti\u00eau 9 s\u1ea3n ph\u1ea9m, nh\u1eadn ti\u1ec1n l\u01b0\u01a1ng l\u00e0 9,8 tri\u1ec7u \u0111\u1ed3ng, suy ra: 9a + b = 9,8 (1) . Trong th\u00e1ng 5 , anh b\u00e1n v\u01b0\u1ee3t ch\u1ec9 ti\u00eau 7 s\u1ea3n ph\u1ea9m, nh\u1eadn ti\u1ec1n l\u01b0\u01a1ng l\u00e0 9,4 tri\u1ec7u \u0111\u1ed3ng, suy ra: 7a + b = 9,4 (2) . T\u1eeb (1) v\u00e0 (2) , suy ra: \uf0ec9a + b = 9,8 \uf0db \uf0ec\uf0ef a = 1 \uf0ed = 9,4 \uf0ed = 5. \uf0ee 7 a + b \uf0ef\uf0ee b 8 b) 1 x + 8 \uf0b3 10,48 \uf0de x \uf0b3 12,4 5 V\u1eady anh H\u00e0o c\u1ea7n b\u00e1n v\u01b0\u1ee3t ch\u1ec9 ti\u00eau \u00edt nh\u1ea5t 13 s\u1ea3n ph\u1ea9m. C\u00e2u 5. (1 \u0111i\u1ec3m) Nh\u00e2n d\u1ecbp l\u1ec5 30 \/ 4 , c\u1eeda h\u00e0ng \u0111i\u1ec7n m\u00e1y B th\u1ef1c hi\u1ec7n khuy\u1ebfn m\u00e3i gi\u1ea3m 20% so v\u1edbi gi\u00e1 ni\u00eam y\u1ebft cho t\u1ea5t c\u1ea3 c\u00e1c s\u1ea3n ph\u1ea9m. a) M\u1eb9 Lan \u0111\u00e3 \u0111\u1ebfn c\u1eeda h\u00e0ng n\u00e0y \u0111\u1ec3 mua m\u1ed9t chi\u1ebfc tivi c\u00f3 gi\u00e1 ni\u00eam y\u1ebft l\u00e0 11 tri\u1ec7u \u0111\u1ed3ng. Sau khi \u0111\u01b0\u1ee3c h\u01b0\u1edfng ch\u01b0\u01a1ng tr\u00ecnh khuy\u1ebfn m\u00e3i tr\u00ean, do c\u00f3 th\u1ebb VIP, m\u1eb9 Lan c\u00f2n \u0111\u01b0\u1ee3c gi\u1ea3m 5% tr\u00ean t\u1ed5ng h\u00f3a \u0111\u01a1n. H\u1ecfi m\u1eb9 Lan ph\u1ea3i tr\u1ea3 bao nhi\u00eau ti\u1ec1n? b) Khi b\u00e1n tivi cho m\u1eb9 Lan, c\u1eeda h\u00e0ng v\u1eabn c\u00f2n l\u00e3i 10% so v\u1edbi gi\u00e1 v\u1ed1n. H\u1ecfi c\u1eeda h\u00e0ng \u0111\u00e3 nh\u1eadp l\u00f4 100 chi\u1ebfc tivi tr\u00ean v\u1edbi s\u1ed1 ti\u1ec1n v\u1ed1n l\u00e0 bao nhi\u00eau? L\u1eddi gi\u1ea3i a) S\u1ed1 ti\u1ec1n m\u1eb9 Lan ph\u1ea3i tr\u1ea3 l\u00e0 95%.80%.11 = 8,36 tri\u1ec7u \u0111\u1ed3ng. b) Gi\u00e1 v\u1ed1n c\u1ee7a m\u1ed7i tivi: 8,36 : 110% = 7,6 tri\u1ec7u \u0111\u1ed3ng Ti\u1ec1n v\u1ed1n nh\u1eadp l\u00f4 100 chi\u1ebfc tivi tr\u00ean: 100.7,6 = 760 tri\u1ec7u \u0111\u1ed3ng. C\u00e2u 6. (1 \u0111i\u1ec3m) M\u1ed9t lon n\u01b0\u1edbc ng\u1ecdt c\u1ee7a h\u00e3ng C c\u00f3 d\u1ea1ng h\u00ecnh tr\u1ee5 v\u1edbi \u0111\u01b0\u1eddng k\u00ednh \u0111\u00e1y l\u00e0 6,5 cm, ch\u1ee9a. 355 . ml n\u01b0\u1edbc ng\u1ecdt. T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N 5","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH a) T\u00ednh chi\u1ec1u cao c\u1ee7a lon (l\u00e0m tr\u00f2n \u0111\u1ebfn cm), bi\u1ebft r\u1eb1ng \u0111\u1ec3 tr\u00e1nh lon v\u1ee1 do s\u1ef1 gi\u00e3n n\u1edf v\u00ec nhi\u1ec7t c\u1ee7a ch\u1ea5t l\u1ecfng, l\u01b0\u1ee3ng n\u01b0\u1edbc ng\u1ecdt trong lon ch\u1ec9 chi\u1ebfm 90% th\u1ec3 t\u00edch lon. b) Ng\u01b0\u1eddi ta th\u01b0\u1eddng \u0111\u00f3ng m\u1ed9t l\u1ed1c 6 lon v\u00e0o m\u1ed9t th\u00f9ng c\u00e1c-t\u00f4ng d\u1ea1ng h\u00ecnh h\u1ed9p ch\u1eef nh\u1eadt v\u1eeba kh\u00edt nh\u01b0 h\u00ecnh b\u00ean, chi\u1ec1u cao c\u1ee7a th\u00f9ng b\u1eb1ng chi\u1ec1u cao c\u1ee7a lon (\u0111\u00e3 l\u00e0m tr\u00f2n \u1edf c\u00e2u a). T\u00ednh th\u1ec3 t\u00edch c\u1ee7a th\u00f9ng c\u00e1c-t\u00f4ng n\u00e0y. L\u1eddi gi\u1ea3i a) Th\u1ec3 t\u00edch lon: 355 : 90% = 3550 ml 9 Chi\u1ec1u cao c\u1ee7a lon: 3550 : \uf0e9 \uf070 \uf0e6 6,5 \uf0f62 \uf0f9 \uf0bb 12 cm . 9 \uf0ea \uf0e7\uf0e8 \uf0f7\uf0f8 \uf0fa \uf0ea\uf0eb 2 \uf0fa\uf0fb b) C\u00e1c k\u00edch th\u01b0\u1edbc c\u1ee7a th\u00f9ng: 6,5.3 = 19,5 cm; 6,5.2 = 13 cm; 12 cm Th\u1ec3 t\u00edch c\u1ee7a th\u00f9ng: 19,5.13.12 = 3042 cm3 . C\u00e2u 7. (1 \u0111i\u1ec3m) Ch\u00e0o \u0111\u00f3n SEA Games 31, m\u1ed9t ph\u00e2n x\u01b0\u1edfng nh\u1eadn may \u0111\u1ed3ng ph\u1ee5c cho c\u1ed5 \u0111\u1ed9ng vi\u00ean v\u1edbi d\u1ef1 ki\u1ebfn m\u1ed7i ng\u00e0y s\u1ebd may xong 90 b\u1ed9 qu\u1ea7n \u00e1o. Khi th\u1ef1c hi\u1ec7n, nh\u1edd c\u1ea3i ti\u1ebfn k\u1ef9 thu\u1eadt, m\u1ed7i ng\u00e0y x\u01b0\u1edfng \u0111\u00e3 may \u0111\u01b0\u1ee3c 120 b\u1ed9 qu\u1ea7n \u00e1o. Do \u0111\u00f3, x\u01b0\u1edfng \u0111\u00e3 ho\u00e0n th\u00e0nh tr\u01b0\u1edbc k\u1ebf ho\u1ea1ch 6 ng\u00e0y v\u00e0 may th\u00eam \u0111\u01b0\u1ee3c 60 b\u1ed9 qu\u1ea7n \u00e1o. H\u1ecfi theo k\u1ebf ho\u1ea1ch ban \u0111\u1ea7u, ph\u00e2n x\u01b0\u1edfng n\u00e0y c\u1ea7n may bao nhi\u00eau b\u1ed9 qu\u1ea7n \u00e1o? L\u1eddi gi\u1ea3i G\u1ecdi x l\u00e0 s\u1ed1 b\u1ed9 qu\u1ea7n \u00e1o x\u01b0\u1edfng c\u1ea7n may theo k\u1ebf ho\u1ea1ch ( x \uf0ce * ) . Th\u1eddi gian d\u1ef1 ki\u1ebfn may xong: x ng\u00e0y, th\u1eddi gian may th\u1ef1c t\u1ebf x + 60 ng\u00e0y 90 120 Ta c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh: x \u2212 x + 60 = 6 \uf0db ... \uf0db x = 2340 (nh\u1eadn) 90 120 V\u1eady, theo k\u1ebf ho\u1ea1ch, x\u01b0\u1edfng c\u1ea7n may 2340 b\u1ed9 qu\u1ea7n \u00e1o. C\u00e2u 8. (3 \u0111i\u1ec3m) Cho tam gi\u00e1c ABC vu\u00f4ng t\u1ea1i B ( AB \uf03c BC) , n\u1ed9i ti\u1ebfp \u0111\u01b0\u1eddng tr\u00f2n (O; R) . D\u1ef1ng \u0111\u01b0\u1eddng k\u00ednh BD , ti\u1ebfp tuy\u1ebfn t\u1ea1i C c\u1ee7a (O) c\u1eaft c\u00e1c tia AB, AD l\u1ea7n l\u01b0\u1ee3t t\u1ea1i E, F. a) Ch\u1ee9ng minh AB.AE = AD.AF v\u00e0 t\u1ee9 gi\u00e1c BDFE n\u1ed9i ti\u1ebfp. b) D\u1ef1ng \u0111\u01b0\u1eddng th\u1eb3ng d qua A v\u00e0 vu\u00f4ng g\u00f3c v\u1edbi BD, d c\u1eaft (O) v\u00e0 EF theo th\u1ee9 t\u1ef1 t\u1ea1i M v\u00e0 N (M \uf0b9 A). Ch\u1ee9ng minh t\u1ee9 gi\u00e1c BMNE n\u1ed9i ti\u1ebfp v\u00e0 N l\u00e0 trung \u0111i\u1ec3m c\u1ee7a EF. T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N 6","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH c) G\u1ecdi I l\u00e0 t\u00e2m \u0111\u01b0\u1eddng tr\u00f2n ngo\u1ea1i ti\u1ebfp tam gi\u00e1c BDE . T\u00ednh kho\u1ea3ng c\u00e1ch t\u1eeb I \u0111\u1ebfn \u0111\u01b0\u1eddng th\u1eb3ng EF theo R . L\u1eddi gi\u1ea3i a) AB.AE = AC2 = AD.AF AB.AE = AD.AF \uf0de AB = AD \uf0de \uf044ABD \u0111\u1ed3ng d\u1ea1ng \uf044AFE (c \u2212 g \u2212 c) AF AE ABD = AFE \uf0de BDFE n\u1ed9i ti\u1ebfp. b) AMB = ADB = BEN \uf0de BMNE n\u1ed9i ti\u1ebfp AEN = AMB = ADB = NAE (cung phu MAD) \uf0de \uf044ANE c\u00e2n t\u1ea1i N \uf0de NE = NA Ch\u1ee9ng minh NAF = AFN \uf0de \uf044ANF c\u00e2n t\u1ea1i N \uf0de NF = NA Suy ra N l\u00e0 trung \u0111i\u1ec3m c\u1ee7a EF. c) \u0110\u01b0\u1eddng tr\u00f2n ngo\u1ea1i ti\u1ebfp tam gi\u00e1c BDE l\u00e0 \u0111\u01b0\u1eddng tr\u00f2n ngo\u1ea1i ti\u1ebfp t\u1ee9 gi\u00e1c BDFE . Suy ra I l\u00e0 giao \u0111i\u1ec3m c\u1ee7a \u0111\u01b0\u1eddng trung tr\u1ef1c c\u1ee7a BD ( qua O ) v\u00e0 \u0111\u01b0\u1eddng trung tr\u1ef1c c\u1ee7a FE ( qua N ). Kho\u1ea3ng c\u00e1ch t\u1eeb I \u0111\u1ebfn \u0111\u01b0\u1eddng th\u1eb3ng EF l\u00e0 \u0111o\u1ea1n IN. Ch\u1ee9ng minh AOIN l\u00e0 h\u00ecnh b\u00ecnh h\u00e0nh. Suy ra IN = OA = R. ----H\u1ebeT--- T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N 7","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH S\u00d4\u00db GD&\u00d1T TP HO\u00c0 CH\u00cd MINH \u00d1E\u00c0 THAM KHA\u00dbO TUYE\u00c5N SINH 10 PHO\u00d8NG G\u00d1&\u00d1T QUA\u00c4N 6 NA\u00caM HO\u00cfC: 2023 - 2024 M\u00d4N: TO\u00c1N 9 \u0110\u1ec0 THAM KH\u1ea2O M\u00c3 \u0110\u1ec0: Qu\u1eadn 6 - 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) Gi\u00e1o vi\u00ean s\u1eeda ph\u1ea7n m\u00e0u \u0111\u1ecf theo m\u00e3 \u0111\u1ec1 c\u1ee7a m\u00ecnh C\u00e2u 1. (1,5 \u0111i\u1ec3m). Cho h\u00e0m s\u1ed1 y = x2 c\u00f3 h\u00e0m s\u1ed1 (P) v\u00e0 h\u00e0m s\u1ed1 y = x + 2 c\u00f3 h\u00e0m s\u1ed1 l\u00e0 (D) . a) V\u1ebd \u0111\u1ed3 th\u1ecb (P) v\u00e0 (D) tr\u00ean c\u00f9ng m\u1ed9t m\u1eb7t ph\u1eb3ng t\u1ecda \u0111\u1ed9. b) T\u00ecm t\u1ecda \u0111\u1ed9 giao \u0111i\u1ec3m c\u1ee7a (P) v\u00e0 (D) b\u1eb1ng ph\u00e9p to\u00e1n. C\u00e2u 2. (1 \u0111i\u1ec3m). Cho ph\u01b0\u01a1ng tr\u00ecnh x2 \u2212 2x \u2212 3 = 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 x1 \u2212 x2 + 2023 C\u00e2u 3. (0,5 \u0111i\u1ec3m). S\u00f3ng th\u1ea7n (Tsunami) l\u00e0 m\u1ed9t lo\u1ea1t c\u00e1c \u0111\u1ee3t s\u00f3ng t\u1ea1o n\u00ean khi m\u1ed9t th\u1ec3 t\u00edch l\u1edbn c\u1ee7a n\u01b0\u1edbc \u0111\u1ea1i d\u01b0\u01a1ng b\u1ecb d\u1ecbch chuy\u1ec3n ch\u1edbp nho\u00e1ng tr\u00ean m\u1ed9t quy m\u00f4 l\u1edbn. \u0110\u1ed9ng \u0111\u1ea5t c\u00f9ng nh\u1eefng d\u1ecbch chuy\u1ec3n \u0111\u1ecba ch\u1ea5t l\u1edbn b\u00ean tr\u00ean ho\u1eb7c b\u00ean d\u01b0\u1edbi m\u1eb7t n\u01b0\u1edbc, n\u00fai l\u1eeda phun v\u00e0 va ch\u1ea1m thi\u00ean th\u1ea1ch \u0111\u1ec1u c\u00f3 kh\u1ea3 n\u0103ng g\u00e2y ra s\u00f3ng th\u1ea7n. C\u01a1n s\u00f3ng th\u1ea7n kh\u1edfi ph\u00e1t t\u1eeb d\u01b0\u1edbi \u0111\u00e1y bi\u1ec3n s\u00e2u, khi c\u00f2n ngo\u00e0i xa kh\u01a1i, s\u00f3ng c\u00f3 bi\u00ean \u0111\u1ed9 (chi\u1ec1u cao s\u00f3ng) kh\u00e1 nh\u1ecf nh\u01b0ng chi\u1ec1u d\u00e0i c\u1ee7a c\u01a1n s\u00f3ng l\u00ean \u0111\u1ebfn h\u00e0ng tr\u0103m km. Con s\u00f3ng \u0111i qua \u0111\u1ea1i d\u01b0\u01a1ng v\u1edbi t\u1ed1c \u0111\u1ed9 trung b\u00ecnh 500 d\u1eb7m m\u1ed9t gi\u1edd. Khi ti\u1ebfn t\u1edbi \u0111\u1ea5t li\u1ec1n, \u0111\u00e1y bi\u1ec3n tr\u1edf n\u00ean n\u00f4ng, con s\u00f3ng kh\u00f4ng c\u00f2n d\u1ecbch chuy\u1ec3n nhanh \u0111\u01b0\u1ee3c n\u1eefa, v\u00ec th\u1ebf n\u00f3 b\u1eaft \u0111\u1ea7u \u201cd\u1ef1ng \u0111\u1ee9ng l\u00ean\u201d c\u00f3 th\u1ec3 \u0111\u1ea1t chi\u1ec1u cao m\u1ed9t t\u00f2a nh\u00e0 s\u00e1u t\u1ea7ng hay h\u01a1n n\u1eefa v\u00e0 t\u00e0n ph\u00e1 kh\u1ee7ng khi\u1ebfp. T\u1ed1c \u0111\u1ed9 c\u1ee7a con s\u00f3ng th\u1ea7n v\u00e0 chi\u1ec1u s\u00e2u c\u1ee7a \u0111\u1ea1i d\u01b0\u01a1ng li\u00ean h\u1ec7 b\u1edfi c\u00f4ng th\u1ee9c s = dg . Trong \u0111\u00f3, g = 9,81m \/ s2 , d (deep) l\u00e0 chi\u1ec1u s\u00e2u \u0111\u1ea1i d\u01b0\u01a1ng t\u00ednh b\u1eb1ng m , s l\u00e0 v\u1eadn t\u1ed1c c\u1ee7a s\u00f3ng th\u1ea7n t\u00ednh b\u1eb1ng m \/ s . Susan Kieffer, m\u1ed9t chuy\u00ean gia v\u1ec1 c\u01a1 h\u1ecdc ch\u1ea5t l\u1ecfng \u0111\u1ecba ch\u1ea5t c\u1ee7a \u0111\u1ea1i h\u1ecdc Illinois t\u1ea1i M\u1ef9, \u0111\u00e3 nghi\u00ean c\u1ee9u n\u0103ng l\u01b0\u1ee3ng c\u1ee7a tr\u1eadn s\u00f3ng th\u1ea7n Tohoku 2011 t\u1ea1i Nh\u1eadt B\u1ea3n. Nh\u1eefng t\u00ednh to\u00e1n c\u1ee7a Kieffer cho th\u1ea5y t\u1ed1c \u0111\u1ed9 s\u00f3ng th\u1ea7n v\u00e0o x\u1ea5p x\u1ec9 220m \/gi\u00e2y. H\u00e3y t\u00ednh \u0111\u1ed9 s\u00e2u c\u1ee7a \u0111\u1ea1i d\u01b0\u01a1ng n\u01a1i xu\u1ea5t ph\u00e1t con s\u00f3ng th\u1ea7n n\u00e0y (k\u1ebft qu\u1ea3 l\u00e0m tr\u00f2n \u0111\u1ebfn m\u00e9t). C\u00e2u 4. (1 \u0111i\u1ec3m). M\u1ed9t si\u00eau th\u1ecb ch\u1ea1y ch\u01b0\u01a1ng tr\u00ecnh khuy\u1ebfn m\u00e3i cho n\u01b0\u1edbc t\u0103ng l\u1ef1c c\u00f3 gi\u00e1 ni\u00eam y\u1ebft l\u00e0 9000 (\u0111\/lon) nh\u01b0 sau: - N\u1ebfu mua 1 lon th\u00ec kh\u00f4ng gi\u1ea3m gi\u00e1. - N\u1ebfu mua 2 lon th\u00ec lon th\u1ee9 hai \u0111\u01b0\u1ee3c gi\u1ea3m 500 \u0111\u1ed3ng - N\u1ebfu mua 3 lon th\u00ec lon th\u1ee9 hai \u0111\u01b0\u1ee3c gi\u1ea3m 500 \u0111\u1ed3ng v\u00e0 lon th\u1ee9 ba \u0111\u01b0\u1ee3c gi\u1ea3m gi\u00e1 10% . - N\u1ebfu mua tr\u00ean 3 lon th\u00ec lon th\u1ee9 hai \u0111\u01b0\u1ee3c gi\u1ea3m 500 \u0111\u1ed3ng, lon th\u1ee9 ba \u0111\u01b0\u1ee3c gi\u1ea3m 10% v\u00e0 nh\u1eefng lon th\u1ee9 t\u01b0 tr\u1edf \u0111i \u0111\u1ec1u \u0111\u01b0\u1ee3c gi\u1ea3m th\u00eam 2% tr\u00ean gi\u00e1 \u0111\u00e3 gi\u1ea3m c\u1ee7a lon th\u1ee9 ba. a) H\u00f9ng mua 3 lon n\u01b0\u1edbc t\u0103ng l\u1ef1c tr\u00ean th\u00ec ph\u1ea3i thanh to\u00e1n s\u1ed1 ti\u1ec1n l\u00e0 bao nhi\u00eau? b) V\u01b0\u01a1ng ph\u1ea3i tr\u1ea3 422 500 \u0111\u1ed3ng \u0111\u1ec3 thanh to\u00e1n khi mua nh\u1eefng lon n\u01b0\u1edbc t\u0103ng l\u1ef1c tr\u00ean. V\u01b0\u01a1ng \u0111\u00e3 mua bao nhi\u00eau lon n\u01b0\u1edbc? C\u00e2u 5. (1 \u0111i\u1ec3m). M\u1ed9t l\u1edbp h\u1ecdc 40 h\u1ecdc sinh, trong \u0111\u00f3 nam nhi\u1ec1u h\u01a1n n\u1eef. Trong gi\u1edd ra ch\u01a1i, c\u00f4 gi\u00e1o \u0111\u01b0a c\u1ea3 l\u1edbp 260000 \u0111\u1ed3ng \u0111\u1ec3 m\u1ed7i b\u1ea1n nam mua m\u1ed9t ly Coca gi\u00e1 5000 \u0111\u1ed3ng\/ly, m\u1ed7i b\u1ea1n n\u1eef T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N 1","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH mua m\u1ed9t b\u00e1nh ph\u00f4 mai gi\u00e1 8000 \u0111\u1ed3ng\/c\u00e1i v\u00e0 \u0111\u01b0\u1ee3c c\u0103n tin th\u1ed1i l\u1ea1i 3000 \u0111\u1ed3ng. H\u1ecfi l\u1edbp c\u00f3 bao nhi\u00eau h\u1ecdc sinh nam v\u00e0 bao nhi\u00eau h\u1ecdc sinh n\u1eef? C\u00e2u 6. (1 \u0111i\u1ec3m). Ng\u01b0\u1eddi ta \u0111un s\u00f4i n\u01b0\u1edbc b\u1eb1ng \u1ea5m \u0111i\u1ec7n. C\u00f4ng su\u1ea5t hao ph\u00ed P s\u1ebd ph\u1ee5 thu\u1ed9c v\u00e0o th\u1eddi gian t . Bi\u1ebft r\u1eb1ng m\u1ed1i li\u00ean h\u1ec7 gi\u1eefa P v\u00e0 t l\u00e0 m\u1ed9t h\u00e0m b\u1eadc nh\u1ea5t c\u00f3 d\u1ea1ng P = a.t + b \u0111\u01b0\u1ee3c bi\u1ec3u di\u1ec5n b\u1eb1ng \u0111\u1ed3 th\u1ecb h\u00ecnh b\u00ean. a) X\u00e1c \u0111\u1ecbnh c\u00e1c h\u1ec7 s\u1ed1 a v\u00e0 b . b) T\u00ednh c\u00f4ng su\u1ea5t hao ph\u00ed khi \u0111un n\u01b0\u1edbc trong 30 gi\u00e2y. C\u00e2u 7. (1 \u0111i\u1ec3m). a) M\u1ed9t b\u1ed3n n\u01b0\u1edbc inox h\u00ecnh tr\u1ee5 n\u1eb1m ngang c\u00f3 k\u00edch th\u01b0\u1edbc \u0111\u01b0\u1eddng k\u00ednh l\u00e0 1900 mm , chi\u1ec1u d\u00e0i 6300 mm ch\u1ee9a \u0111\u01b0\u1ee3c 15 000 l\u00edt n\u01b0\u1edbc. H\u1ecfi th\u1ec3 t\u00edch n\u01b0\u1edbc b\u1eb1ng bao nhi\u00eau ph\u1ea7n tr\u0103m th\u1ec3 t\u00edch b\u1ed3n (l\u00e0m tr\u00f2n t\u1edbi h\u00e0ng \u0111\u01a1n v\u1ecb). b) L\u00fac 1g30' s\u00e1ng ng\u00e0y 23 \/ 3 \/ 2019 . M\u1ed9t v\u1ee5 h\u1ecfa ho\u1ea1n \u0111\u00e3 b\u00f9ng ph\u00e1t t\u1ea1i chung c\u01b0 Carina Plaza (g\u1ed3m 3 t\u00f2a nh\u00e0), t\u1ecda l\u1ea1c t\u1ea1i 1648 \u0111\u1ea1i l\u1ed9 Mai Ch\u00ed Th\u1ecd - V\u00f5 V\u0103n Ki\u1ec7t, Ph\u01b0\u1eddng 16 , Qu\u1eadn 8 , th\u00e0nh ph\u1ed1 H\u1ed3 Ch\u00ed Minh. \u0110\u00e2y l\u00e0 v\u1ee5 h\u1ecfa ho\u1ea1n nghi\u00eam tr\u1ecdng nh\u1ea5t h\u01a1n 10 n\u0103m qua \u1edf th\u00e0nh ph\u1ed1 H\u1ed3 Ch\u00ed Minh. H\u1eadu qu\u1ea3 l\u00e0m 13 ng\u01b0\u1eddi ch\u1ebft, 91 ng\u01b0\u1eddi b\u1ecb th\u01b0\u01a1ng, g\u1ea7n 500 xe m\u00e1y, h\u01a1n 80 \u00f4 t\u00f4 b\u1ecb ch\u00e1y. Nguy\u00ean nh\u00e2n l\u00e0 m\u1ed9t chi\u1ebfc xe m\u00e1y b\u1ecb ch\u1eadp \u0111i\u1ec7n v\u00e0 ch\u00e1y trong t\u1ea7ng h\u1ea7m, trong khi h\u1ec7 th\u1ed1ng b\u00e1o v\u00e0 ch\u1eefa ch\u00e1y kh\u00f4ng ho\u1ea1t \u0111\u1ed9ng. H\u1ec7 th\u1ed1ng ch\u1eefa ch\u00e1y t\u1ef1 \u0111\u1ed9ng Sprinkler (xem h\u00ecnh) khi nhi\u1ec7t \u0111\u1ed9 ch\u00e1y s\u1ebd l\u00e0m nh\u1eefng Sprinkler t\u1ef1 \u0111\u1ed9ng phun n\u01b0\u1edbc ch\u1eefa ch\u00e1y, m\u1ed9t Sprinkler b\u1ea3o v\u1ec7 cho ph\u1ea7n di\u1ec7n t\u00edch t\u1ed1i \u0111a l\u00e0 12m2 , l\u01b0u l\u01b0\u1ee3ng t\u1ed1ithi\u1ec3u cho m\u1ed9t Sprinkler l\u00e0 3456 l\u00edt\/gi\u1edd. Theo ti\u00eau chu\u1ea9n ph\u00f2ng ch\u00e1y ch\u1eefa ch\u00e1y c\u1ee7a Vi\u1ec7t Nam th\u00ec 1 Sprinkler ho\u1ea1t \u0111\u1ed9ng t\u1ed1i thi\u1ec3u trong 0,5 gi\u1edd. Gi\u1ea3 s\u1eed t\u1ea7ng h\u1ea7m t\u00f2a nh\u00e0 chung c\u01b0 Carina Plaza r\u1ed9ng 1200m2 th\u00ec chung c\u01b0 c\u1ea7n bao nhi\u00eau b\u1ed3n inox \u1edf c\u00e2u a \u0111\u1ec3 tr\u1eef n\u01b0\u1edbc cho h\u1ec7 th\u1ed1ng ch\u1eefa ch\u00e1y? C\u00e2u 8. (3 \u0111i\u1ec3m) Cho tam gi\u00e1c ABC nh\u1ecdn ( AB \uf03c AC) n\u1ed9i ti\u1ebfp \u0111\u01b0\u1eddng tr\u00f2n (O) c\u00f3 hai \u0111\u01b0\u1eddng cao BE, CF c\u1eaft nhau t\u1ea1i H v\u00e0 c\u1eaft (O) l\u1ea7n l\u01b0\u1ee3t t\u1ea1i X v\u00e0 Y . K\u1ebb \u0111\u01b0\u1eddng k\u00ednh AK c\u1ee7a (O) , HK c\u1eaft (O) t\u1ea1i P . a) Ch\u1ee9ng minh t\u1ee9 gi\u00e1c APFE n\u1ed9i ti\u1ebfp \u0111\u01b0\u1eddng tr\u00f2n. b) Ch\u1ee9ng minh r\u1eb1ng PB.PE = PC.PF . c) G\u1ecdi M l\u00e0 \u0111i\u1ec3m ch\u00ednh gi\u1eefa c\u1ee7a cung nh\u1ecf BC , MX v\u00e0 MY c\u1eaft AB, AC l\u1ea7n l\u01b0\u1ee3t t\u1ea1i I v\u00e0 J . Ch\u1ee9ng minh: H,I, J 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 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 h\u00e0m s\u1ed1 y = x2 c\u00f3 h\u00e0m s\u1ed1 (P) v\u00e0 h\u00e0m s\u1ed1 y = x + 2 c\u00f3 h\u00e0m s\u1ed1 l\u00e0 (D) . a) V\u1ebd \u0111\u1ed3 th\u1ecb (P) v\u00e0 (D) tr\u00ean c\u00f9ng m\u1ed9t m\u1eb7t ph\u1eb3ng t\u1ecda \u0111\u1ed9. b) T\u00ecm t\u1ecda \u0111\u1ed9 giao \u0111i\u1ec3m c\u1ee7a (P) v\u00e0 (D) b\u1eb1ng ph\u00e9p to\u00e1n. 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 = x2 4 1 0 1 4 x 01 y=x+2 2 3 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 + 2 \uf0db x2 \u2212 x \u2212 2 = 0 \uf0db \uf0e9x = \u22121 \uf0ea\uf0ebx = 2 ( )Thay x = \u22121 v\u00e0o y = x2 , ta \u0111\u01b0\u1ee3c: y = \u22121 2 = 1 . Thay x = 2 v\u00e0o y = x2 , ta \u0111\u01b0\u1ee3c: y = 22 = 4 . V\u1eady (\u22121; 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 2x \u2212 3 = 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 x1 \u2212 x2 + 2023 . L\u1eddi gi\u1ea3i V\u00ec \uf044 = b2 \u2212 4ac = (\u22122)2 \u2212 4.1.(\u22123) = 16 \uf03e 0 N\u00ean ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m ph\u00e2n bi\u1ec7t x1 , x2 . T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N 4","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH \uf0ec\uf0ef\uf0efS = x1 + x2 = \u2212b = 2 \uf0ed = x1 a Theo \u0111\u1ecbnh l\u00ed Vi-et, ta c\u00f3: \uf0ef\uf0ef\uf0eeP c = \u22123 a .x2 = Ta c\u00f3: A = x12 + x22 \u2212 x1 \u2212 x2 + 2023 ( ) ( )2 A = x1 + x2 \u2212 x1 + x2 + 2023 A = 22 \u2212 (\u22123) + 2023 A = 4 + 3 + 2023 A = 2030. C\u00e2u 3. (1 \u0111i\u1ec3m) S\u00f3ng th\u1ea7n (Tsunami) l\u00e0 m\u1ed9t lo\u1ea1t c\u00e1c \u0111\u1ee3t s\u00f3ng t\u1ea1o n\u00ean khi m\u1ed9t th\u1ec3 t\u00edch l\u1edbn c\u1ee7a n\u01b0\u1edbc \u0111\u1ea1i d\u01b0\u01a1ng b\u1ecb d\u1ecbch chuy\u1ec3n ch\u1edbp nho\u00e1ng tr\u00ean m\u1ed9t quy m\u00f4 l\u1edbn. \u0110\u1ed9ng \u0111\u1ea5t c\u00f9ng nh\u1eefng d\u1ecbch chuy\u1ec3n \u0111\u1ecba ch\u1ea5t l\u1edbn b\u00ean tr\u00ean ho\u1eb7c b\u00ean d\u01b0\u1edbi m\u1eb7t n\u01b0\u1edbc, n\u00fai l\u1eeda phun v\u00e0 va ch\u1ea1m thi\u00ean th\u1ea1ch \u0111\u1ec1u c\u00f3 kh\u1ea3 n\u0103ng g\u00e2y ra s\u00f3ng th\u1ea7n. C\u01a1n s\u00f3ng th\u1ea7n kh\u1edfi ph\u00e1t t\u1eeb d\u01b0\u1edbi \u0111\u00e1y bi\u1ec3n s\u00e2u, khi c\u00f2n ngo\u00e0i xa kh\u01a1i, s\u00f3ng c\u00f3 bi\u00ean \u0111\u1ed9 (chi\u1ec1u cao s\u00f3ng) kh\u00e1 nh\u1ecf nh\u01b0ng chi\u1ec1u d\u00e0i c\u1ee7a c\u01a1n s\u00f3ng l\u00ean \u0111\u1ebfn h\u00e0ng tr\u0103m km. Con s\u00f3ng \u0111i qua \u0111\u1ea1i d\u01b0\u01a1ng v\u1edbi t\u1ed1c \u0111\u1ed9 trung b\u00ecnh 500 d\u1eb7m m\u1ed9t gi\u1edd. Khi ti\u1ebfn t\u1edbi \u0111\u1ea5t li\u1ec1n, \u0111\u00e1y bi\u1ec3n tr\u1edf n\u00ean n\u00f4ng, con s\u00f3ng kh\u00f4ng c\u00f2n d\u1ecbch chuy\u1ec3n nhanh \u0111\u01b0\u1ee3c n\u1eefa, v\u00ec th\u1ebf n\u00f3 b\u1eaft \u0111\u1ea7u \u201cd\u1ef1ng \u0111\u1ee9ng l\u00ean\u201d c\u00f3 th\u1ec3 \u0111\u1ea1t chi\u1ec1u cao m\u1ed9t t\u00f2a nh\u00e0 s\u00e1u t\u1ea7ng hay h\u01a1n n\u1eefa v\u00e0 t\u00e0n ph\u00e1 kh\u1ee7ng khi\u1ebfp. T\u1ed1c \u0111\u1ed9 c\u1ee7a con s\u00f3ng th\u1ea7n v\u00e0 chi\u1ec1u s\u00e2u c\u1ee7a \u0111\u1ea1i d\u01b0\u01a1ng li\u00ean h\u1ec7 b\u1edfi c\u00f4ng th\u1ee9c s = dg . Trong \u0111\u00f3, g = 9,81m \/ s2 , d (deep) l\u00e0 chi\u1ec1u s\u00e2u \u0111\u1ea1i d\u01b0\u01a1ng t\u00ednh b\u1eb1ng m, s l\u00e0 v\u1eadn t\u1ed1c c\u1ee7a s\u00f3ng th\u1ea7n t\u00ednh b\u1eb1ng m\/s. Susan Kieffer, m\u1ed9t chuy\u00ean gia v\u1ec1 c\u01a1 h\u1ecdc ch\u1ea5t l\u1ecfng \u0111\u1ecba ch\u1ea5t c\u1ee7a \u0111\u1ea1i h\u1ecdc Illinois t\u1ea1i M\u1ef9, \u0111\u00e3 nghi\u00ean c\u1ee9u n\u0103ng l\u01b0\u1ee3ng c\u1ee7a tr\u1eadn s\u00f3ng th\u1ea7n Tohoku 2011 t\u1ea1i Nh\u1eadt B\u1ea3n. Nh\u1eefng t\u00ednh to\u00e1n c\u1ee7a Kieffer cho th\u1ea5y t\u1ed1c \u0111\u1ed9 s\u00f3ng th\u1ea7n v\u00e0o x\u1ea5p x\u1ec9 220 m\/gi\u00e2y. H\u00e3y t\u00ednh \u0111\u1ed9 s\u00e2u c\u1ee7a \u0111\u1ea1i d\u01b0\u01a1ng n\u01a1i xu\u1ea5t ph\u00e1t con s\u00f3ng th\u1ea7n n\u00e0y (k\u1ebft qu\u1ea3 l\u00e0m tr\u00f2n \u0111\u1ebfn m\u00e9t). L\u1eddi gi\u1ea3i Thay s = 220 v\u00e0 g = 9,81 v\u00e0o c\u00f4ng th\u1ee9c s = dg , ta \u0111\u01b0\u1ee3c: 9,81d = 220 \uf0db 9,81d = 48400 \uf0db d = 48400 \uf0bb 4934 9 , 81 V\u1eady \u0111\u1ed9 s\u00e2u c\u1ee7a \u0111\u1ea1i d\u01b0\u01a1ng n\u01a1i xu\u1ea5t ph\u00e1t con s\u00f3ng th\u1ea7n n\u00e0y l\u00e0 4934(m) . T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N 5","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH C\u00e2u 4. (1 \u0111i\u1ec3m). M\u1ed9t si\u00eau th\u1ecb ch\u1ea1y ch\u01b0\u01a1ng tr\u00ecnh khuy\u1ebfn m\u00e3i cho n\u01b0\u1edbc t\u0103ng l\u1ef1c c\u00f3 gi\u00e1 ni\u00eam y\u1ebft l\u00e0 9000 (\u0111\/lon) nh\u01b0 sau: - N\u1ebfu mua 1 lon th\u00ec kh\u00f4ng gi\u1ea3m gi\u00e1. - N\u1ebfu mua 2 lon th\u00ec lon th\u1ee9 hai \u0111\u01b0\u1ee3c gi\u1ea3m 500 \u0111\u1ed3ng - N\u1ebfu mua 3 lon th\u00ec lon th\u1ee9 hai \u0111\u01b0\u1ee3c gi\u1ea3m 500 \u0111\u1ed3ng v\u00e0 lon th\u1ee9 ba \u0111\u01b0\u1ee3c gi\u1ea3m gi\u00e1 10%. - N\u1ebfu mua tr\u00ean 3 lon th\u00ec lon th\u1ee9 hai \u0111\u01b0\u1ee3c gi\u1ea3m 500 \u0111\u1ed3ng, lon th\u1ee9 ba \u0111\u01b0\u1ee3c gi\u1ea3m 10% v\u00e0 nh\u1eefng lon th\u1ee9 t\u01b0 tr\u1edf \u0111i \u0111\u1ec1u \u0111\u01b0\u1ee3c gi\u1ea3m th\u00eam 2% tr\u00ean gi\u00e1 \u0111\u00e3 gi\u1ea3m c\u1ee7a lon th\u1ee9 ba. a) H\u00f9ng mua 3 lon n\u01b0\u1edbc t\u0103ng l\u1ef1c tr\u00ean th\u00ec ph\u1ea3i thanh to\u00e1n s\u1ed1 ti\u1ec1n l\u00e0 bao nhi\u00eau? b) V\u01b0\u01a1ng ph\u1ea3i tr\u1ea3 422 500 \u0111\u1ed3ng \u0111\u1ec3 thanh to\u00e1n khi mua nh\u1eefng lon n\u01b0\u1edbc t\u0103ng l\u1ef1c tr\u00ean. V\u01b0\u01a1ng \u0111\u00e3 mua bao nhi\u00eau lon n\u01b0\u1edbc? L\u1eddi gi\u1ea3i a) H\u00f9ng mua 3 lon n\u01b0\u1edbc t\u0103ng l\u1ef1c tr\u00ean th\u00ec ph\u1ea3i thanh to\u00e1n s\u1ed1 ti\u1ec1n l\u00e0 bao nhi\u00eau? S\u1ed1 ti\u1ec1n ph\u1ea3i thanh to\u00e1n cho 3 lon n\u01b0\u1edbc t\u0103ng l\u1ef1c l\u00e0: 9000 + (9000 \u2212 500) + 90%.9000 = 25600 (\u0111\u1ed3ng) V\u01b0\u01a1ng ph\u1ea3i tr\u1ea3 422 500 \u0111\u1ed3ng \u0111\u1ec3 thanh to\u00e1n khi mua nh\u1eefng lon n\u01b0\u1edbc t\u0103ng l\u1ef1c tr\u00ean. V\u01b0\u01a1ng \u0111\u00e3 mua bao nhi\u00eau lon n\u01b0\u1edbc? b) G\u1ecdi s\u1ed1 lon n\u01b0\u1edbc V\u01b0\u01a1ng \u0111\u00e3 mua l\u00e0 x ( x \uf0ce * , lon) Theo \u0111\u1ec1 b\u00e0i ta c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh: 25600 + 98%.(90%.9000).(x \u2212 3) = 422500 \uf0db 25600 + 7938(x \u2212 3) = 422500 \uf0db 7938x = 420714 \uf0db x = 53 (nh\u1eadn) V\u1eady V\u01b0\u01a1ng \u0111\u00e3 mua 53 (lon). C\u00e2u 5. (1 \u0111i\u1ec3m) M\u1ed9t l\u1edbp h\u1ecdc 40 h\u1ecdc sinh, trong \u0111\u00f3 nam nhi\u1ec1u h\u01a1n n\u1eef. Trong gi\u1edd ra ch\u01a1i, c\u00f4 gi\u00e1o \u0111\u01b0a c\u1ea3 l\u1edbp 260000 \u0111\u1ed3ng \u0111\u1ec3 m\u1ed7i b\u1ea1n nam mua m\u1ed9t ly Coca gi\u00e1 5000 \u0111\u1ed3ng\/ly, m\u1ed7i b\u1ea1n n\u1eef mua m\u1ed9t b\u00e1nh ph\u00f4 mai gi\u00e1 8000 \u0111\u1ed3ng\/c\u00e1i v\u00e0 \u0111\u01b0\u1ee3c c\u0103n tin th\u1ed1i l\u1ea1i 3000 \u0111\u1ed3ng. H\u1ecfi l\u1edbp c\u00f3 bao nhi\u00eau h\u1ecdc sinh nam v\u00e0 bao nhi\u00eau h\u1ecdc sinh n\u1eef? L\u1eddi gi\u1ea3i G\u1ecdi s\u1ed1 h\u1ecdc sinh nam c\u1ee7a l\u1edbp l\u00e0 x ( x \uf0ce * , h\u1ecdc sinh) S\u1ed1 h\u1ecdc sinh n\u1eef c\u1ee7a l\u1edbp l\u00e0 y ( y \uf0ce * , h\u1ecdc sinh) V\u00ec l\u1edbp h\u1ecdc c\u00f3 40 h\u1ecdc sinh n\u00ean ta c\u00f3: x + y = 40 T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N 6","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH V\u00ec c\u00f4 gi\u00e1o \u0111\u01b0a c\u1ea3 l\u1edbp 260000 \u0111\u1ed3ng \u0111\u01b0\u1ee3c c\u0103n tin th\u1ed1i l\u1ea1i 3000 \u0111\u1ed3ng n\u00ean ta c\u00f3: 5000x + 8000y = 257 000 T\u1eeb \u0111\u00f3, ta c\u00f3 h\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh sau: \uf0ecx + y = 40 = 257 000 \uf0db \uf0ecx = 21 (nh\u1eadn) \uf0ed\uf0ee5000x + 8000y \uf0ed\uf0eey = 19 V\u1eady l\u1edbp c\u00f3 21 nam v\u00e0 19 n\u1eef. C\u00e2u 6. (1 \u0111i\u1ec3m) Ng\u01b0\u1eddi ta \u0111un s\u00f4i n\u01b0\u1edbc b\u1eb1ng \u1ea5m \u0111i\u1ec7n. C\u00f4ng su\u1ea5t hao ph\u00ed P s\u1ebd ph\u1ee5 thu\u1ed9c v\u00e0o th\u1eddi gian t. Bi\u1ebft r\u1eb1ng m\u1ed1i li\u00ean h\u1ec7 gi\u1eefa P v\u00e0 t l\u00e0 m\u1ed9t h\u00e0m b\u1eadc nh\u1ea5t c\u00f3 d\u1ea1ng P = a.t + b \u0111\u01b0\u1ee3c bi\u1ec3u di\u1ec5n b\u1eb1ng \u0111\u1ed3 th\u1ecb h\u00ecnh b\u00ean. a) X\u00e1c \u0111\u1ecbnh c\u00e1c h\u1ec7 s\u1ed1 a v\u00e0 b. b) T\u00ednh c\u00f4ng su\u1ea5t hao ph\u00ed khi \u0111un n\u01b0\u1edbc trong 30 gi\u00e2y. L\u1eddi gi\u1ea3i a) X\u00e1c \u0111\u1ecbnh c\u00e1c h\u1ec7 s\u1ed1 a v\u00e0 b . Theo \u0111\u1ec1 b\u00e0i, ta c\u00f3: V\u1edbi \uf0ect = 0 \uf0de 100 = 0.a + b . (1) \uf0ed\uf0eeP = 100 V\u1edbi \uf0ect = 200 \uf0de 200 = 200.a + b . (2) \uf0ed\uf0eeP = 200 T\u1eeb (1) v\u00e0 (2) ta c\u00f3 h\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh: \uf0ec0a + b = 100 \uf0db \uf0ec\uf0efa = 1 . \uf0ed\uf0ee200a + b = 200 \uf0ed = \uf0ef\uf0eeb 2 100 V\u1eady ta c\u00f3 c\u00f4ng th\u1ee9c: P = 1 t + 100 2 b) T\u00ednh c\u00f4ng su\u1ea5t hao ph\u00ed khi \u0111un n\u01b0\u1edbc trong 30 gi\u00e2y. Thay t = 30 v\u00e0o c\u00f4ng th\u1ee9c P = 1 t + 100 , ta \u0111\u01b0\u1ee3c: 2 P = 1 .30 + 100 = 115 2 V\u1eady c\u00f4ng su\u1ea5t hao ph\u00ed khi \u0111un n\u01b0\u1edbc trong 30 gi\u00e2y l\u00e0 115(W ) . T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N 7","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH C\u00e2u 7. (1 \u0111i\u1ec3m) a) M\u1ed9t b\u1ed3n n\u01b0\u1edbc inox h\u00ecnh tr\u1ee5 n\u1eb1m ngang c\u00f3 k\u00edch th\u01b0\u1edbc \u0111\u01b0\u1eddng k\u00ednh l\u00e0 1900 mm, chi\u1ec1u d\u00e0i 6300 mm ch\u1ee9a \u0111\u01b0\u1ee3c 15 000 l\u00edt n\u01b0\u1edbc. H\u1ecfi th\u1ec3 t\u00edch n\u01b0\u1edbc b\u1eb1ng bao nhi\u00eau ph\u1ea7n tr\u0103m th\u1ec3 t\u00edch b\u1ed3n (l\u00e0m tr\u00f2n t\u1edbi h\u00e0ng \u0111\u01a1n v\u1ecb). b) L\u00fac 1g30\u2019 s\u00e1ng ng\u00e0y 23\/3\/2018. M\u1ed9t v\u1ee5 h\u1ecfa ho\u1ea1n \u0111\u00e3 b\u00f9ng ph\u00e1t t\u1ea1i chung c\u01b0 Carina Plaza (g\u1ed3m 3 t\u00f2a nh\u00e0), t\u1ecda l\u1ea1c t\u1ea1i 1648 \u0111\u1ea1i l\u1ed9 Mai Ch\u00ed Th\u1ecd - V\u00f5 V\u0103n Ki\u1ec7t, Ph\u01b0\u1eddng 16, Qu\u1eadn 8, th\u00e0nh ph\u1ed1 H\u1ed3 Ch\u00ed Minh. \u0110\u00e2y l\u00e0 v\u1ee5 h\u1ecfa ho\u1ea1n nghi\u00eam tr\u1ecdng nh\u1ea5t h\u01a1n 10 n\u0103m qua \u1edf th\u00e0nh ph\u1ed1 H\u1ed3 Ch\u00ed Minh. H\u1eadu qu\u1ea3 l\u00e0m 13 ng\u01b0\u1eddi ch\u1ebft, 91 ng\u01b0\u1eddi b\u1ecb th\u01b0\u01a1ng, g\u1ea7n 500 xe m\u00e1y, h\u01a1n 80 \u00f4 t\u00f4 b\u1ecb ch\u00e1y. Nguy\u00ean nh\u00e2n l\u00e0 m\u1ed9t chi\u1ebfc xe m\u00e1y b\u1ecb ch\u1eadp \u0111i\u1ec7n v\u00e0 ch\u00e1y trong t\u1ea7ng h\u1ea7m, trong khi h\u1ec7 th\u1ed1ng b\u00e1o v\u00e0 ch\u1eefa ch\u00e1y kh\u00f4ng ho\u1ea1t \u0111\u1ed9ng. H\u1ec7 th\u1ed1ng ch\u1eefa ch\u00e1y t\u1ef1 \u0111\u1ed9ng Sprinkler (xem h\u00ecnh) khi nhi\u1ec7t \u0111\u1ed9 ch\u00e1y s\u1ebd l\u00e0m nh\u1eefng Sprinkler t\u1ef1 \u0111\u1ed9ng phun n\u01b0\u1edbc ch\u1eefa ch\u00e1y, m\u1ed9t Sprinkler b\u1ea3o v\u1ec7 cho ph\u1ea7n di\u1ec7n t\u00edch t\u1ed1i \u0111a l\u00e0 12 m2, l\u01b0u l\u01b0\u1ee3ng t\u1ed1i thi\u1ec3u cho m\u1ed9t Sprinkler l\u00e0 3456 l\u00edt\/gi\u1edd. Theo ti\u00eau chu\u1ea9n ph\u00f2ng ch\u00e1y ch\u1eefa ch\u00e1y c\u1ee7a Vi\u1ec7t Nam th\u00ec 1 Sprinkler ho\u1ea1t \u0111\u1ed9ng t\u1ed1i thi\u1ec3u trong 0,5 gi\u1edd. Gi\u1ea3 s\u1eed t\u1ea7ng h\u1ea7m t\u00f2a nh\u00e0 chung c\u01b0 Carina Plaza r\u1ed9ng 1200 m2 th\u00ec chung c\u01b0 c\u1ea7n bao nhi\u00eau b\u1ed3n inox \u1edf c\u00e2u a \u0111\u1ec3 tr\u1eef n\u01b0\u1edbc cho h\u1ec7 th\u1ed1ng ch\u1eefa ch\u00e1y? L\u1eddi gi\u1ea3i \uf0e6 1900 \uf0f62 ( )a) \uf0f7 Th\u1ec3 t\u00edch b\u1ed3n n\u01b0\u1edbc l\u00e0: \uf0e7 2 \uf0f8 .6300\uf070 = 1,8.1010 mm3 = 18000 (l\u00edt) \uf0e8 Ph\u1ea7n tr\u0103m th\u1ec3 t\u00edch n\u01b0\u1edbc l\u00e0: 15000 .100 \uf0bb 83(%) 18000 b) S\u1ed1 Sprinkler cho 1200m2 l\u00e0 1200 = 100 (Sprinkler) 12 S\u1ed1 l\u00edt n\u01b0\u1edbc t\u1ed1i thi\u1ec3u cho 1 Sprinkler ho\u1ea1t \u0111\u1ed9ng trong 1 gi\u1edd: 3456 . 0,5 = 1728 (l\u00edt) S\u1ed1 l\u00edt n\u01b0\u1edbc t\u1ed1i thi\u1ec3u cho 100 Sprinkler: 1728 . 100 = 172800 (l\u00edt_) S\u1ed1 b\u1ed3n inox l\u00e0: 172800 \uf0bb 12 (b\u1ed3n) 15000 V\u1eady chung c\u01b0 c\u1ea7n kho\u1ea3ng 12 b\u1ed3n inox. ( ) ( )C\u00e2u 9. (3 \u0111i\u1ec3m) Cho tam gi\u00e1c ABC nh\u1ecdn AB \uf03c AC n\u1ed9i ti\u1ebfp \u0111\u01b0\u1eddng tr\u00f2n O c\u00f3 hai \u0111\u01b0\u1eddng cao BE, ( ) ( ) ( )CF c\u1eaft nhau t\u1ea1i H v\u00e0 c\u1eaft O l\u1ea7n l\u01b0\u1ee3t t\u1ea1i X v\u00e0 Y . K\u1ebb \u0111\u01b0\u1eddng k\u00ednh AK c\u1ee7a O , HK c\u1eaft O t\u1ea1i P . a) Ch\u1ee9ng minh t\u1ee9 gi\u00e1c APFE n\u1ed9i ti\u1ebfp \u0111\u01b0\u1eddng tr\u00f2n. b) Ch\u1ee9ng minh r\u1eb1ng PB.PE = PC.PF . T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N 8","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH c) G\u1ecdi M l\u00e0 \u0111i\u1ec3m ch\u00ednh gi\u1eefa c\u1ee7a cung nh\u1ecf BC , MX v\u00e0 MY c\u1eaft AB, AC l\u1ea7n l\u01b0\u1ee3t t\u1ea1i I v\u00e0 J . Ch\u1ee9ng minh: H,I, J th\u1eb3ng h\u00e0ng. L\u1eddi gi\u1ea3i a) Ch\u1ee9ng minh t\u1ee9 gi\u00e1c APFE n\u1ed9i ti\u1ebfp \u0111\u01b0\u1eddng tr\u00f2n. X\u00e9t t\u1ee9 gi\u00e1c AFHE , c\u00f3: \uf0ec\uf0efAFH = 90\uf0b0(CF \u22a5 AB) \uf0ed \uf0ef\uf0ee AEH = 90\uf0b0( BE \u22a5 AC ) \uf0de AFH + AEH = 180\uf0b0 \uf0de T\u1ee9 gi\u00e1c AFHE n\u1ed9i ti\u1ebfp v\u00ec c\u00f3 hai g\u00f3c \u0111\u1ed1i b\u00f9 nhau. ( )\uf0de 4 \u0111i\u1ec3m A,F,H,E c\u00f9ng thu\u1ed9c m\u1ed9t \u0111\u01b0\u1eddng tr\u00f2n. 1 X\u00e9t t\u1ee9 gi\u00e1c APFH , c\u00f3: \uf0ef\uf0ecAPH = 90\uf0b0( goc noi tiep chan nua duong tron) \uf0ed \uf0ef\uf0ee AFH = 90\uf0b0(CF \u22a5 AB) \uf0de APH = AFH = 90\uf0b0 \uf0de T\u1ee9 gi\u00e1c APFH n\u1ed9i ti\u1ebfp v\u00ec c\u00f3 hai g\u00f3c \u0111\u1ed1i b\u00f9 nhau. ( )\uf0de 4 \u0111i\u1ec3m A,P,F ,H c\u00f9ng thu\u1ed9c m\u1ed9t \u0111\u01b0\u1eddng tr\u00f2n. 2 ( ) ( )T\u1eeb 1 v\u00e0 2 \uf0de 5 \u0111i\u1ec3m A,P,F ,H,E c\u00f9ng thu\u1ed9c m\u1ed9t \u0111\u01b0\u1eddng tr\u00f2n. N\u00ean: t\u1ee9 gi\u00e1c APFE n\u1ed9i ti\u1ebfp \u0111\u01b0\u1eddng tr\u00f2n. b) Ch\u1ee9ng minh r\u1eb1ng PB.PE = PC.PF . Ta c\u00f3: PEF = PAF (g\u00f3c n\u1ed9i ti\u1ebfp c\u00f9ng ch\u1eafn PF ) M\u00e0 PAF = PCB (g\u00f3c n\u1ed9i ti\u1ebfp c\u00f9ng ch\u1eafn PB ) N\u00ean: PEF = PCB 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 X\u00e9t \u0394 PEF v\u00e0 \u0394 PCB , c\u00f3: PEF = PCB (cmt) PFE = PBC (c\u00f9ng b\u00f9 v\u1edbi PAC ) \uf0de \u0394PEF \u223d\u0394PCB (g \u2013 g) \uf0de PE = PF PC PB \uf0de PB.PE = PC.PF (\u0111pcm) c) G\u1ecdi M l\u00e0 \u0111i\u1ec3m ch\u00ednh gi\u1eefa c\u1ee7a cung nh\u1ecf BC , MX v\u00e0 MY c\u1eaft AB, AC l\u1ea7n l\u01b0\u1ee3t t\u1ea1i I v\u00e0 J . Ch\u1ee9ng minh: H,I, J th\u1eb3ng h\u00e0ng. X\u00e9t t\u1ee9 gi\u00e1c BFEC c\u00f3 : BFE = BEC = 90\uf0b0 ( BE v\u00e0 CF l\u00e0 \u0111\u01b0\u1eddng cao c\u1ee7a \u0394 ABC\u00a0) \uf0de T\u1ee9 gi\u00e1c BFEC n\u1ed9i ti\u1ebfp \u0111\u01b0\u1eddng tr\u00f2n (T\u1ee9 gi\u00e1c c\u00f3 hai \u0111\u1ec9nh li\u00ean ti\u1ebfp c\u00f9ng nh\u00ecn m\u1ed9t c\u1ea1nh d\u01b0\u1edbi hai g\u00f3c b\u1eb1ng nhau) \uf0de FBE = FCE (c\u00f9ng ch\u1eafn cung FE ) ( )M\u00e0 FCE = FBX (2 g\u00f3c n\u1ed9i ti\u1ebfp c\u00f9ng ch\u1eafn cung AP c\u1ee7a O ) N\u00ean FBE = FBX \uf0de BF l\u00e0 tia ph\u00e2n gi\u00e1c c\u1ee7a XBH . V\u00ec MB = MC ( M l\u00e0 \u0111i\u1ec3m ch\u00ednh gi\u1eefa c\u1ee7a cung BC ) N\u00ean BXM = CXM (2 g\u00f3c n\u1ed9i ti\u1ebfp ch\u1eafn 2 cung b\u1eb1ng nhau) \uf0de XI l\u00e0 tia ph\u00e2n gi\u00e1c c\u1ee7a BXH . X\u00e9t tam gi\u00e1c BXH c\u00f3: BF l\u00e0 tia ph\u00e2n gi\u00e1c c\u1ee7a XBH . XI l\u00e0 tia ph\u00e2n gi\u00e1c c\u1ee7a BXH . BF c\u1eaft XI t\u1ea1i I . N\u00ean HI l\u00e0 tia ph\u00e2n gi\u00e1c XHB \uf0de IHB = 1 XHB . 2 * Ch\u1ee9ng minh t\u01b0\u01a1ng t\u1ef1 \u0111\u1ed1i v\u1edbi tam gi\u00e1c YHC ta c\u0169ng c\u00f3 HJ l\u00e0 tia ph\u00e2n gi\u00e1c c\u1ee7a YHC \uf0de YHJ = 1 YHC . 2 M\u00e0 XBH = YHC (\u0111\u1ed1i \u0111\u1ec9nh) n\u00ean BHI = YHJ Ta c\u00f3: IHB + IHE = 180\uf0b0 (k\u1ec1 b\u00f9) \uf0de IHE + YHJ = 180\uf0b0 \uf0de IHJ = 180\uf0b0 T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N 10","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH \uf0de H,I, J th\u1eb3ng h\u00e0ng. ----H\u1ebeT--- T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N 11","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH S\u00d4\u00db GD&\u00d1T TP HO\u00c0 CH\u00cd MINH \u00d1E\u00c0 THAM KHA\u00dbO TUYE\u00c5N SINH 10 PHO\u00d8NG G\u00d1&\u00d1T QUA\u00c4N 6 NA\u00caM HO\u00cfC: 2022 - 2023 M\u00d4N: TO\u00c1N 9 M\u0110\u00c3\u1ec0 \u0110T\u1ec0H: AQMu\u1eadnKTH\u00e2\u1ea2n POh\u00fa \u0110\u00ea thi g\u1ed3m 8 c\u00e2u ho\u0309i t\u01b0\u0323 lu\u00e2\u0323n. Th\u01a1\u0300i gian: 120 phu\u0301t (kh\u00f4ng k\u00ea\u0309 th\u01a1\u0300i gian pha\u0301t \u0111\u00ea\u0300) C\u00e2u 1. (1,5 \u0111i\u1ec3m). Cho (P) : y = \u2212x2 v\u00e0 \u0111\u01b0\u1eddng th\u1eb3ng (d) : y = x \u2212 2 . 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 \u2212 3x \u2212 4 = 0 c\u00f3 2 nghi\u1ec7m l\u00e0 x1 , x2 . Kh\u00f4ng gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh, h\u00e3y t\u00ednh gi\u00e1 tr\u1ecb c\u1ee7a bi\u1ec3u th\u1ee9c A = \uf0e6 1 \uf0f62 + \uf0e6 1 \uf0f62 \uf0e7 x1 \uf0f7 \uf0e7 x2 \uf0f7 \uf0e8 \uf0f8 \uf0e8 \uf0f8 C\u00e2u 3. (1 \u0111i\u1ec3m). B\u00e1c H\u00f9ng mua hai m\u00f3n h\u00e0ng t\u1ea1i m\u1ed9t c\u1eeda h\u00e0ng m\u00f3n h\u00e0ng th\u1ee9 nh\u1ea5t c\u00f3 gi\u00e1 ghi l\u00e0 400000 \u0111\u1ed3ng v\u00e0 B\u00e1c \u0111\u01b0\u1ee3c gi\u1ea3m 20% tr\u00ean gi\u00e1 tr\u1ecb m\u00f3n h\u00e0ng; m\u00f3n h\u00e0ng th\u1ee9 hai B\u00e1c \u0111\u01b0\u1ee3c gi\u1ea3m 30% tr\u00ean gi\u00e1 tr\u1ecb m\u00f3n h\u00e0ng. T\u1ed5ng s\u1ed1 ti\u1ec1n B\u00e1c ph\u1ea3i thanh to\u00e1n l\u00e0 740000 \u0111\u1ed3ng. H\u1ecfi n\u1ebfu B\u00e1c mua th\u00eam m\u1ed9t m\u00f3n h\u00e0ng th\u1ee9 hai th\u00ec B\u00e1c \u0111\u01b0\u1ee3c gi\u1ea3m t\u1ea5t c\u1ea3 bao nhi\u00eau ti\u1ec1n? C\u00e2u 4. (1 \u0111i\u1ec3m). M\u1ed9t th\u00f9ng ch\u1ee9a m\u1ed9t l\u01b0\u1ee3ng th\u1ec3 t\u00edch dung d\u1ecbch c\u1ed3n r\u1eeda tay 70\uf0b0 . L\u1ea7n \u0111\u1ea7u ng\u01b0\u1eddi ta s\u1eed d\u1ee5ng 1 th\u1ec3 t\u00edch dung d\u1ecbch; l\u1ea7n th\u1ee9 hai ng\u01b0\u1eddi ta s\u1eed d\u1ee5ng 1 c\u1ee7a th\u1ec3 t\u00edch dung d\u1ecbch l\u1ea7n \u0111\u1ea7u 62 s\u1eed d\u1ee5ng; l\u1ea7n th\u1ee9 ba ng\u01b0\u1eddi ta s\u1eed d\u1ee5ng 1 th\u1ec3 t\u00edch dung d\u1ecbch; l\u1ea7n th\u1ee9 t\u01b0 ng\u01b0\u1eddi ta s\u1eed d\u1ee5ng 7 nhi\u1ec1u h\u01a1n l\u1ea7n th\u1ee9 nh\u1ea5t v\u00e0 l\u1ea7n th\u1ee9 ba s\u1eed d\u1ee5ng 8 th\u1ec3 t\u00edch dung d\u1ecbch; l\u00fac n\u00e0y trong th\u00f9ng c\u00f2n 42 l\u1ea1i 0,9 l\u00edt dung d\u1ecbch. H\u1ecfi th\u1ec3 t\u00edch dung d\u1ecbch c\u00f3 trong th\u00f9ng l\u00e0 bao nhi\u00eau l\u00edt? C\u00e2u 5. (1 \u0111i\u1ec3m). \u0110\u1ec3 t\u00ecm hi\u1ec3u v\u1ec1 s\u1ef1 n\u1edf v\u00ec nhi\u1ec7t c\u1ee7a ch\u1ea5t r\u1eafn, B\u1ea1n An \u0111\u00e3 th\u1ef1c hi\u1ec7n m\u1ed9t th\u00ed nghi\u1ec7m \u0111\u01a1n gi\u1ea3n. Chu\u1ea9n b\u1ecb m\u1ed9t thanh kim lo\u1ea1i \u0111\u1ed3ng ch\u1ea5t, sau \u0111\u00f3 nung n\u00f3ng thanh kim lo\u1ea1i. Quan s\u00e1t s\u1ef1 thay \u0111\u1ed5i chi\u1ec1u d\u00e0i c\u1ee7a thanh kim lo\u1ea1i theo nhi\u1ec7t \u0111\u1ed9, b\u1ea1n th\u1ea5y r\u1eb1ng ban \u0111\u1ea7u khi \u1edf nhi\u1ec7t \u0111\u1ed9 400 thanh kim lo\u1ea1i c\u00f3 chi\u1ec1u d\u00e0i l\u00e0 5 m\u00e9t; khi nung n\u00f3ng thanh kim lo\u1ea1i \u1edf nhi\u1ec7t \u0111\u1ed9 1400 th\u00ec chi\u1ec1u d\u00e0i c\u1ee7a c\u1ee7a n\u00f3 t\u0103ng th\u00eam 6 mm. M\u1ed1i li\u00ean h\u1ec7 gi\u1eefa chi\u1ec1u d\u00e0i y (m\u00e9t) c\u1ee7a thanh kim v\u00e0 nhi\u1ec7t \u0111\u1ed9 ( )x 0C l\u00e0 m\u1ed9t h\u00e0m s\u1ed1 b\u1eadc nh\u1ea5t y = ax + b . a. X\u00e1c \u0111\u1ecbnh h\u1ec7 s\u1ed1 a v\u00e0 b . b. H\u00e3y t\u00ednh chi\u1ec1u d\u00e0i c\u1ee7a thanh kim lo\u1ea1i khi \u1edf nhi\u1ec7t \u0111\u1ed9 1000 . C\u00e2u 6. (1 \u0111i\u1ec3m). Bi\u1ec3u gi\u00e1 b\u00e1n l\u1ebb \u0111i\u1ec7n sinh ho\u1ea1t c\u1ee7a kh\u00e1ch h\u00e0ng n\u0103m 2022 \u0111\u01b0\u1ee3c \u00e1p d\u1ee5ng \u0111\u1ec3 t\u00ednh to\u00e1n ti\u1ec1n s\u1eed d\u1ee5ng \u0111i\u1ec7n nh\u01b0 sau: 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 Nh\u01b0 v\u1eady, s\u1ed1 \u0111i\u1ec7n ti\u00eau th\u1ee5 sinh ho\u1ea1t h\u00e0ng th\u00e1ng c\u00e0ng cao th\u00ec m\u1ee9c gi\u00e1 ti\u1ec1n \u0111i\u1ec7n c\u00e0ng cao. \u0110\u1ec3 t\u00ednh ti\u1ec1n \u0111i\u1ec7n h\u00e0ng th\u00e1ng \u0111\u01b0\u1ee3c th\u1ef1c hi\u1ec7n theo c\u00f4ng th\u1ee9c sau: Ti\u1ec1n \u0111i\u1ec7n b\u1eadc i = (s\u1ed1 kWh \u00e1p d\u1ee5ng gi\u00e1 b\u1eadc i) x (gi\u00e1 1 kWh b\u1eadc i) Ti\u1ec1n \u0111i\u1ec7n = t\u1ed5ng ti\u1ec1n \u0111i\u1ec7n c\u00e1c b\u1eadc V\u00ed d\u1ee5: n\u1ebfu s\u1eed d\u1ee5ng h\u1ebft 100kWh th\u00ec ti\u1ec1n \u0111i\u1ec7n l\u00e0 50.1678 + 50.1734 = 170600 \u0111\u1ed3ng Ngo\u00e0i ra, tr\u00ean h\u00f3a \u0111\u01a1n ti\u1ec1n \u0111i\u1ec7n ng\u01b0\u1eddi s\u1eed d\u1ee5ng \u0111i\u1ec7n c\u00f2n ph\u1ea3i tr\u1ea3 th\u00eam 8% thu\u1ebf gi\u00e1 tr\u1ecb gia t\u0103ng (VAT) tr\u00ean s\u1ed1 ti\u1ec1n \u0111i\u1ec7n. a. H\u00e3y t\u00ednh s\u1ed1 ti\u1ec1n \u0111i\u1ec7n kh\u00e1ch h\u00e0ng c\u1ea7n tr\u1ea3 n\u1ebfu s\u1eed d\u1ee5ng h\u1ebft 200kWh \u0111i\u1ec7n (bao g\u1ed3m c\u1ea3 VAT). b. Trong th\u00e1ng 4 \/ 2022 nh\u00e0 c\u00f4 B\u00ecnh \u0111\u00e3 tr\u1ea3 336506,4 \u0111\u1ed3ng cho h\u00f3a \u0111\u01a1n ti\u1ec1n \u0111i\u1ec7n. H\u1ecfi nh\u00e0 c\u00f4 B\u00ecnh ti\u00eau th\u1ee5 h\u1ebft bao nhi\u00eau kWh \u0111i\u1ec7n? C\u00e2u 7. (0,75 \u0111i\u1ec3m). M\u1ed9t h\u1ed9p kem h\u00ecnh tr\u1ee5 c\u00f3 \u0111\u01b0\u1eddng k\u00ednh 12cm v\u00e0 chi\u1ec1u cao 15cm \u0111\u1ef1ng \u0111\u1ea7y kem. Kem s\u1ebd \u0111\u01b0\u1ee3c chia v\u00e0o c\u00e1c b\u00e1nh \u1ed1c qu\u1ebf h\u00ecnh n\u00f3n c\u00f3 chi\u1ec1u cao 12cm v\u00e0 \u0111\u01b0\u1eddng k\u00ednh 6cm , c\u00f3 h\u00ecnh b\u00e1n c\u1ea7u tr\u00ean \u0111\u1ec9nh nh\u01b0 h\u00ecnh v\u1ebd. H\u00e3y t\u00ecm s\u1ed1 que kem c\u00f3 th\u1ec3 chia \u0111\u01b0\u1ee3c. C\u00e2u 8. (3 \u0111i\u1ec3m). T\u1eeb \u0111i\u1ec3m A \u1edf ngo\u00e0i (O) v\u1ebd 2 ti\u1ebfp tuy\u1ebfn AB v\u00e0 AC ( B , C l\u00e0 2 ti\u1ebfp \u0111i\u1ec3m) v\u00e0 c\u00e1t tuy\u1ebfn ADE( AD \uf03c AE) a) Ch\u1ee9ng minh: OA \u22a5 BC v\u00e0 t\u1ee9 gi\u00e1c ABOC n\u1ed9i ti\u1ebfp. b) \u0110\u01b0\u1eddng th\u1eb3ng \u0111i qua \u0111i\u1ec3m C , song song v\u1edbi DE v\u00e0 c\u1eaft \u0111\u01b0\u1eddng tr\u00f2n (O) t\u1ea1i F ( F kh\u00e1c C ). G\u1ecdi I l\u00e0 giao \u0111i\u1ec3m c\u1ee7a BF v\u00e0 DE . Ch\u1ee9ng minh: I l\u00e0 trung \u0111i\u1ec3m c\u1ee7a DE . c) Ch\u1ee9ng minh: BE.EF + BD.DF = BC.DE . ----H\u1ebeT--- T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N 2","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH H\u01af\u1edaNG D\u1eaaN GI\u1ea2I C\u00e2u 1. (1,5 \u0111i\u1ec3m) Cho (P) : y = \u2212x2 v\u00e0 \u0111\u01b0\u1eddng th\u1eb3ng (d) : y = x \u2212 2 . a) V\u1ebd \u0111\u1ed3 th\u1ecb (P) v\u00e0 (d) tr\u00ean c\u00f9ng h\u1ec7 tr\u1ee5c t\u1ecda \u0111\u1ed9. b) T\u00ecm t\u1ecda \u0111\u1ed9 giao \u0111i\u1ec3m c\u1ee7a (P) v\u00e0 (d) b\u1eb1ng ph\u00e9p t\u00ednh. L\u1eddi gi\u1ea3i a. V\u1ebd \u0111\u1ed3 th\u1ecb (P) v\u00e0 (d) tr\u00ean c\u00f9ng h\u1ec7 tr\u1ee5c t\u1ecda \u0111\u1ed9. BGT: x \u22122 \u22121 0 1 2 y = \u2212x2 \u22124 \u22121 0 \u22121 \u22124 x 01 y = x\u22122 \u22122 \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 = x \u2212 2 \uf0db x2 + x \u2212 2 = 0 \uf0db \uf0e9x = 1 \uf0eb\uf0eax = \u22122 Thay x = 1 v\u00e0o y = \u2212x2 , ta \u0111\u01b0\u1ee3c: y = \u221212 = \u22121 . Thay x = \u22122 v\u00e0o y = \u2212x2 , ta \u0111\u01b0\u1ee3c: y = \u2212(\u22122)2 = \u22124 . V\u1eady (1; \u2212 1) , (\u22122; \u2212 4) l\u00e0 hai giao \u0111i\u1ec3m c\u1ea7n t\u00ecm. C\u00e2u 2. (1 \u0111i\u1ec3m). Cho ph\u01b0\u01a1ng tr\u00ecnh 2x2 \u2212 3x \u2212 4 = 0 c\u00f3 2 nghi\u1ec7m l\u00e0 x1 , x2 . Kh\u00f4ng gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh, h\u00e3y t\u00ednh gi\u00e1 tr\u1ecb c\u1ee7a bi\u1ec3u th\u1ee9c A = \uf0e6 1 \uf0f62 + \uf0e6 1 \uf0f62 \uf0e7 x1 \uf0f7 \uf0e7 x2 \uf0f7 \uf0e8 \uf0f8 \uf0e8 \uf0f8 L\u1eddi gi\u1ea3i V\u00ec \uf044 = b2 \u2212 4ac = (\u22123)2 \u2212 4.(\u22124).2 = 41 \uf03e 0 N\u00ean ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m ph\u00e2n bi\u1ec7t x1 , x2 . T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N 3","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH \uf0ec = x1 + x2 = \u2212b = \u2212 ( \u22123) = 3 \uf0ef\uf0efS a 2 Theo \u0111\u1ecbnh l\u00ed Vi-et, ta c\u00f3: \uf0ed 2 \uf0ef\uf0ef\uf0eeP = x1.x2 = c = \u22124 = \u22122 a 2 \uf0e6 1 \uf0f62 \uf0e6 1 \uf0f62 x12 + x22 S2 \u2212 2P \uf0e6 3 \uf0f62 \u2212 2.( \u22122 ) 25 \uf0e7 x1 \uf0f7 \uf0e7 x2 \uf0f7 x12 .x22 P2 \uf0e7\uf0e8 2 \uf0f7\uf0f8 16 A = \uf0e8 \uf0f8 + \uf0f8 = = = = ( \u22122 )2 \uf0e8 C\u00e2u 3. (1 \u0111i\u1ec3m). B\u00e1c H\u00f9ng mua hai m\u00f3n h\u00e0ng t\u1ea1i m\u1ed9t c\u1eeda h\u00e0ng m\u00f3n h\u00e0ng th\u1ee9 nh\u1ea5t c\u00f3 gi\u00e1 ghi l\u00e0 400000 \u0111\u1ed3ng v\u00e0 B\u00e1c \u0111\u01b0\u1ee3c gi\u1ea3m 20% tr\u00ean gi\u00e1 tr\u1ecb m\u00f3n h\u00e0ng; m\u00f3n h\u00e0ng th\u1ee9 hai B\u00e1c \u0111\u01b0\u1ee3c gi\u1ea3m 30% tr\u00ean gi\u00e1 tr\u1ecb m\u00f3n h\u00e0ng. T\u1ed5ng s\u1ed1 ti\u1ec1n B\u00e1c ph\u1ea3i thanh to\u00e1n l\u00e0 740000 \u0111\u1ed3ng. H\u1ecfi n\u1ebfu B\u00e1c mua th\u00eam m\u1ed9t m\u00f3n h\u00e0ng th\u1ee9 hai th\u00ec B\u00e1c \u0111\u01b0\u1ee3c gi\u1ea3m t\u1ea5t c\u1ea3 bao nhi\u00eau ti\u1ec1n? L\u1eddi gi\u1ea3i G\u1ecdi x (\u0111\u1ed3ng) l\u00e0 gi\u00e1 ban \u0111\u1ea7u c\u1ee7a m\u00f3n h\u00e0ng th\u1ee9 hai ( x \uf03e 0 ) V\u00ec t\u1ed5ng s\u1ed1 ti\u1ec1n thanh to\u00e1n 740000 n\u00ean 400000.(1\u2212 20%) + x (1\u2212 30%) = 740000 \uf0db x = 600000 V\u1eady s\u1ed1 ti\u1ec1n B\u00e1c \u0111\u01b0\u1ee3c gi\u1ea3m l\u00e0 : 20%.400000 + 30%.600000 = 260000 \u0111\u1ed3ng. C\u00e2u 4. (1 \u0111i\u1ec3m). M\u1ed9t th\u00f9ng ch\u1ee9a m\u1ed9t l\u01b0\u1ee3ng th\u1ec3 t\u00edch dung d\u1ecbch c\u1ed3n r\u1eeda tay 70\uf0b0 . L\u1ea7n \u0111\u1ea7u ng\u01b0\u1eddi ta 11 s\u1eed d\u1ee5ng 6 th\u1ec3 t\u00edch dung d\u1ecbch; l\u1ea7n th\u1ee9 hai ng\u01b0\u1eddi ta s\u1eed d\u1ee5ng 2 c\u1ee7a th\u1ec3 t\u00edch dung d\u1ecbch l\u1ea7n 1 \u0111\u1ea7u s\u1eed d\u1ee5ng; l\u1ea7n th\u1ee9 ba ng\u01b0\u1eddi ta s\u1eed d\u1ee5ng th\u1ec3 t\u00edch dung d\u1ecbch; l\u1ea7n th\u1ee9 t\u01b0 ng\u01b0\u1eddi ta s\u1eed d\u1ee5ng 7 8 nhi\u1ec1u h\u01a1n l\u1ea7n th\u1ee9 nh\u1ea5t v\u00e0 l\u1ea7n th\u1ee9 ba s\u1eed d\u1ee5ng th\u1ec3 t\u00edch dung d\u1ecbch; l\u00fac n\u00e0y trong th\u00f9ng 42 c\u00f2n l\u1ea1i 0,9 l\u00edt dung d\u1ecbch. H\u1ecfi th\u1ec3 t\u00edch dung d\u1ecbch c\u00f3 trong th\u00f9ng l\u00e0 bao nhi\u00eau l\u00edt? L\u1eddi gi\u1ea3i G\u1ecdi x (l) l\u00e0 th\u1ec3 t\u00edch dung d\u1ecbch c\u00f3 trong th\u00f9ng l\u00fac \u0111\u1ea7u. ( x \uf03e 0) S\u1ed1 l\u00edt dung d\u1ecbch l\u1ea7n \u0111\u1ea7u s\u1eed d\u1ee5ng: x \u2212 1 x = 5 x (l\u00edt) 66 S\u1ed1 l\u00edt dung d\u1ecbch l\u1ea7n th\u1ee9 hai s\u1eed d\u1ee5ng: 1 . 1 x = 1 x (l\u00edt) 2 6 12 S\u1ed1 l\u00edt dung d\u1ecbch l\u1ea7n th\u1ee9 ba s\u1eed d\u1ee5ng: 1 x (l\u00edt) 7 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 S\u1ed1 l\u00edt dung d\u1ecbch l\u1ea7n th\u1ee9 t\u01b0 s\u1eed d\u1ee5ng: 1 x + 1 x + 8 x = 1 x (l\u00edt) 6 7 42 2 S\u1ed1 l\u00edt dung d\u1ecbch c\u00f2n l\u1ea1i: x \u2212 1 x \u2212 1 x \u2212 1 x \u2212 1 x = 3 x (l\u00edt) 6 12 7 2 28 Ta c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh: 3 x = 0,9 \uf0db x = 8, 4 (nh\u1eadn) 28 V\u1eady th\u1ec3 t\u00edch dung d\u1ecbch c\u00f3 trong th\u00f9ng l\u00fac \u0111\u1ea7u l\u00e0 8, 4 l\u00edt. C\u00e2u 5. (1 \u0111i\u1ec3m). \u0110\u1ec3 t\u00ecm hi\u1ec3u v\u1ec1 s\u1ef1 n\u1edf v\u00ec nhi\u1ec7t c\u1ee7a ch\u1ea5t r\u1eafn, B\u1ea1n An \u0111\u00e3 th\u1ef1c hi\u1ec7n m\u1ed9t th\u00ed nghi\u1ec7m \u0111\u01a1n gi\u1ea3n. Chu\u1ea9n b\u1ecb m\u1ed9t thanh kim lo\u1ea1i \u0111\u1ed3ng ch\u1ea5t, sau \u0111\u00f3 nung n\u00f3ng thanh kim lo\u1ea1i. Quan s\u00e1t s\u1ef1 thay \u0111\u1ed5i chi\u1ec1u d\u00e0i c\u1ee7a thanh kim lo\u1ea1i theo nhi\u1ec7t \u0111\u1ed9, b\u1ea1n th\u1ea5y r\u1eb1ng ban \u0111\u1ea7u khi \u1edf nhi\u1ec7t \u0111\u1ed9 400 thanh kim lo\u1ea1i c\u00f3 chi\u1ec1u d\u00e0i l\u00e0 5 m\u00e9t; khi nung n\u00f3ng thanh kim lo\u1ea1i \u1edf nhi\u1ec7t \u0111\u1ed9 1400 th\u00ec chi\u1ec1u d\u00e0i c\u1ee7a c\u1ee7a n\u00f3 t\u0103ng th\u00eam 6 ( )mm. M\u1ed1i li\u00ean h\u1ec7 gi\u1eefa chi\u1ec1u d\u00e0i y (m\u00e9t) c\u1ee7a thanh kim v\u00e0 nhi\u1ec7t \u0111\u1ed9 x 0C l\u00e0 m\u1ed9t h\u00e0m s\u1ed1 b\u1eadc nh\u1ea5t y = ax + b . a. X\u00e1c \u0111\u1ecbnh h\u1ec7 s\u1ed1 a v\u00e0 b . b. H\u00e3y t\u00ednh chi\u1ec1u d\u00e0i c\u1ee7a thanh kim lo\u1ea1i khi \u1edf nhi\u1ec7t \u0111\u1ed9 1000 . L\u1eddi gi\u1ea3i a. X\u00e1c \u0111\u1ecbnh c\u00e1c h\u1ec7 s\u1ed1 a v\u00e0 b . (1) Theo \u0111\u1ec1 b\u00e0i, ta c\u00f3: V\u1edbi \uf0ecx = 40 \uf0de 5 = 40.a +b . (1) \uf0ed\uf0eey = 5 V\u1edbi \uf0ecx = 140 \uf0de 5.006 = 140.a + b . (2) \uf0ed\uf0eey = 5 + 0,006 T\u1eeb (1) v\u00e0 (2) ta c\u00f3 h\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh: \uf0ec40a + b = 5 \uf0db \uf0ec\uf0efa = 3 . \uf0ed\uf0ee140a + b = 5,006 \uf0ed = \uf0ef\uf0eeb 50000 4 , 9976 V\u1eady: a = 3 , b = 4,9976 v\u00e0 y = 3 x + 4,9976 . 50000 50000 b. Chi\u1ec1u d\u00e0i c\u1ee7a thanh kim lo\u1ea1i khi \u1edf nhi\u1ec7t \u0111\u1ed9 100\uf0b0C Th\u1ebf x =100 v\u00e0o y = 3 x + 4,9976 = 3 .100 + 4,9976 = 5, 0036 50000 50000 V\u1eady chi\u1ec1u d\u00e0i c\u1ee7a thanh kim lo\u1ea1i c\u1ee7a thanh kim lo\u1ea1i khi \u1edf nhi\u1ec7t \u0111\u1ed9 1000C l\u00e0 5, 0036 m . 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. (1 \u0111i\u1ec3m). Bi\u1ec3u gi\u00e1 b\u00e1n l\u1ebb \u0111i\u1ec7n sinh ho\u1ea1t c\u1ee7a kh\u00e1ch h\u00e0ng n\u0103m 2022 \u0111\u01b0\u1ee3c \u00e1p d\u1ee5ng \u0111\u1ec3 t\u00ednh to\u00e1n ti\u1ec1n s\u1eed d\u1ee5ng \u0111i\u1ec7n nh\u01b0 sau: Nh\u01b0 v\u1eady, s\u1ed1 \u0111i\u1ec7n ti\u00eau th\u1ee5 sinh ho\u1ea1t h\u00e0ng th\u00e1ng c\u00e0ng cao th\u00ec m\u1ee9c gi\u00e1 ti\u1ec1n \u0111i\u1ec7n c\u00e0ng cao. \u0110\u1ec3 t\u00ednh ti\u1ec1n \u0111i\u1ec7n h\u00e0ng th\u00e1ng \u0111\u01b0\u1ee3c th\u1ef1c hi\u1ec7n theo c\u00f4ng th\u1ee9c sau: Ti\u1ec1n \u0111i\u1ec7n b\u1eadc i = (s\u1ed1 kWh \u00e1p d\u1ee5ng gi\u00e1 b\u1eadc i) x (gi\u00e1 1 kWh b\u1eadc i) Ti\u1ec1n \u0111i\u1ec7n = t\u1ed5ng ti\u1ec1n \u0111i\u1ec7n c\u00e1c b\u1eadc V\u00ed d\u1ee5: n\u1ebfu s\u1eed d\u1ee5ng h\u1ebft 100kWh th\u00ec ti\u1ec1n \u0111i\u1ec7n l\u00e0 50.1678 + 50.1734 = 170600 \u0111\u1ed3ng Ngo\u00e0i ra, tr\u00ean h\u00f3a \u0111\u01a1n ti\u1ec1n \u0111i\u1ec7n ng\u01b0\u1eddi s\u1eed d\u1ee5ng \u0111i\u1ec7n c\u00f2n ph\u1ea3i tr\u1ea3 th\u00eam 8% thu\u1ebf gi\u00e1 tr\u1ecb gia t\u0103ng (VAT) tr\u00ean s\u1ed1 ti\u1ec1n \u0111i\u1ec7n. a. H\u00e3y t\u00ednh s\u1ed1 ti\u1ec1n \u0111i\u1ec7n kh\u00e1ch h\u00e0ng c\u1ea7n tr\u1ea3 n\u1ebfu s\u1eed d\u1ee5ng h\u1ebft 200kWh \u0111i\u1ec7n (bao g\u1ed3m c\u1ea3 VAT). b. Trong th\u00e1ng 4 \/ 2022 nh\u00e0 c\u00f4 B\u00ecnh \u0111\u00e3 tr\u1ea3 336506,4 \u0111\u1ed3ng cho h\u00f3a \u0111\u01a1n ti\u1ec1n \u0111i\u1ec7n. H\u1ecfi nh\u00e0 c\u00f4 B\u00ecnh ti\u00eau th\u1ee5 h\u1ebft bao nhi\u00eau kWh \u0111i\u1ec7n? L\u1eddi gi\u1ea3i a. S\u1ed1 ti\u1ec1n \u0111i\u1ec7n kh\u00e1ch h\u00e0ng c\u1ea7n tr\u1ea3 n\u1ebfu s\u1eed d\u1ee5ng h\u1ebft 200kWh \u0111i\u1ec7n l\u00e0: (50.1678 + 50.1734 +100.2014).(1+ 8%) = 401760 \u0111\u1ed3ng b. V\u00ec s\u1ed1 ti\u1ec1n nh\u00e0 c\u00f4 B\u00ecnh ph\u1ea3i tr\u1ea3 l\u1edbn h\u01a1n s\u1ed1 ti\u1ec1n s\u1eed d\u1ee5ng 100kWh N\u00ean s\u1ed1 kWh \u0111i\u1ec7n nh\u00e0 c\u00f4 B\u00ecnh s\u1eed d\u1ee5ng \u1edf b\u1eadc 3 . G\u1ecdi x (kWh) l\u00e0 s\u1ed1 kWh \u0111i\u1ec7n nh\u00e0 c\u00f4 B\u00ecnh s\u1eed d\u1ee5ng. Theo \u0111\u1ec1 b\u00e0i, ta c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh: \uf0e9\uf0eb50.1678 + 50.1734 + ( x \u2212100) 2014\uf0f9\uf0fb.(1+ 8%) = 336506, 4 \uf0db 170600 + ( x \u2212100) 2014 = 336506, 4 \uf0db x = 170 V\u1eady nh\u00e0 c\u00f4 B\u00ecnh ti\u00eau th\u1ee5 h\u1ebft 170 kWh \u0111i\u1ec7n. T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N 6","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH C\u00e2u 7. (0,75 \u0111i\u1ec3m). M\u1ed9t h\u1ed9p kem h\u00ecnh tr\u1ee5 c\u00f3 \u0111\u01b0\u1eddng k\u00ednh 12cm v\u00e0 chi\u1ec1u cao 15cm \u0111\u1ef1ng \u0111\u1ea7y kem. Kem s\u1ebd \u0111\u01b0\u1ee3c chia v\u00e0o c\u00e1c b\u00e1nh \u1ed1c qu\u1ebf h\u00ecnh n\u00f3n c\u00f3 chi\u1ec1u cao 12cm v\u00e0 \u0111\u01b0\u1eddng k\u00ednh 6cm , c\u00f3 h\u00ecnh b\u00e1n c\u1ea7u tr\u00ean \u0111\u1ec9nh nh\u01b0 h\u00ecnh v\u1ebd. H\u00e3y t\u00ecm s\u1ed1 que kem c\u00f3 th\u1ec3 chia \u0111\u01b0\u1ee3c. L\u1eddi gi\u1ea3i ( )Th\u1ec3 \uf0e6 12 \uf0f62 t\u00edch kem trong h\u1ed9p h\u00ecnh tr\u1ee5 l\u00e0: VT = \uf070 RT2.hT =\uf070 \uf0e7\uf0e8 2 \uf0f7\uf0f8 .15 = 540\uf070 cm3 Th\u1ec3 t\u00edch kem trong h\u1ed9p h\u00ecnh tr\u1ee5 l\u00e0: ( )Vkem1 RN2 1 4 RC3 1\uf070 \uf0e6 6 \uf0f62 1 4 .\uf0e6\uf0e7\uf0e8 6 \uf0f63 = 3 \uf070 .hN + 2 . 3 \uf070 = 3 \uf0e7\uf0e8 2 \uf0f7\uf0f8 .12 + 2 . 3 .\uf070 2 \uf0f7\uf0f8 = 54\uf070 cm3 V\u1eady s\u1ed1 que kem c\u00f3 th\u1ec3 chia \u0111\u01b0\u1ee3c l\u00e0: 540\uf070 = 10 que. 54\uf070 C\u00e2u 8. (3 \u0111i\u1ec3m). T\u1eeb \u0111i\u1ec3m A \u1edf ngo\u00e0i (O) v\u1ebd 2 ti\u1ebfp tuy\u1ebfn AB v\u00e0 AC ( B , C l\u00e0 2 ti\u1ebfp \u0111i\u1ec3m) v\u00e0 c\u00e1t tuy\u1ebfn ADE( AD \uf03c AE) a. Ch\u1ee9ng minh: OA \u22a5 BC v\u00e0 t\u1ee9 gi\u00e1c ABOC n\u1ed9i ti\u1ebfp. b. \u0110\u01b0\u1eddng th\u1eb3ng \u0111i qua \u0111i\u1ec3m C , song song v\u1edbi DE v\u00e0 c\u1eaft \u0111\u01b0\u1eddng tr\u00f2n (O) t\u1ea1i F ( F kh\u00e1c C ). G\u1ecdi I l\u00e0 giao \u0111i\u1ec3m c\u1ee7a BF v\u00e0 DE . Ch\u1ee9ng minh: I l\u00e0 trung \u0111i\u1ec3m c\u1ee7a DE . c. Ch\u1ee9ng minh: BE.EF + BD.DF = BC.DE . L\u1eddi gi\u1ea3i B E I D A HO C F 7 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 a. Ch\u1ee9ng minh: OA \u22a5 BC v\u00e0 t\u1ee9 gi\u00e1c ABOC n\u1ed9i ti\u1ebfp. * \uf0ec\uf0efAB = AC (2 tiep tuyen cat nhau) \uf0ed\uf0ef\uf0eeOA = OB (= R) Suy ra OA l\u00e0 \u0111\u01b0\u1eddng trung tr\u1ef1c c\u1ee7a BC \uf0de OA \u22a5 BC *X\u00e9t t\u1ee9 gi\u00e1c ABOC c\u00f3: ABO = 900 (V\u00ec AB l\u00e0 ti\u1ebfp tuy\u1ebfn c\u1ee7a (O) ) ACO = 900 (V\u00ec AC l\u00e0 ti\u1ebfp tuy\u1ebfn c\u1ee7a (O) ) \uf0de ABO + ACO = 180\uf0b0 V\u1eady t\u1ee9 gi\u00e1c ABOC l\u00e0 t\u1ee9 gi\u00e1c n\u1ed9i ti\u1ebfp. b. Ch\u1ee9ng minh: I l\u00e0 trung \u0111i\u1ec3m c\u1ee7a DE . AIB = CFB ( AE \/ \/CF ) V\u00e0 CFB = ACB (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 1 cung) \uf0de AIB = BOA \uf0e6 = 1 BOC \uf0f6 \uf0e8\uf0e7 2 \uf0f8\uf0f7 Suy ra t\u1ee9 gi\u00e1c BIOA n\u1ed9i ti\u1ebfp Suy ra AIO = ABO = 90\uf0b0 ( 2 g\u00f3c n\u1ed9i ti\u1ebfp c\u00f9ng ch\u1eafn 1 cung) \uf0de OI \u22a5 DE t\u1ea1i I Suy ra I l\u00e0 trung \u0111i\u1ec3m c\u1ee7a DE (Quan h\u1ec7 vu\u00f4ng g\u00f3c gi\u1eefa \u0111\u01b0\u1eddng k\u00ednh v\u00e0 d\u00e2y cung) c. Ch\u1ee9ng minh: BE.EF + BD.DF = BC.DE . * BEC = ACB (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 1 cung BC ) V\u00e0 EIF = BIA (\u0111\u1ed1i \u0111\u1ec9nh) M\u1eb7t kh\u00e1c BIA = ACB ( 2 g\u00f3c n\u1ed9i ti\u1ebfp c\u00f9ng ch\u1eafn 1 cung c\u1ee7a \u0111\u01b0\u1eddng tr\u00f2n \u0111i qua 5 \u0111i\u1ec3m A, B,O,C, I ) Suy ra BEC = EIF * X\u00e9t \uf044BEC v\u00e0 \uf044EIF c\u00f3 : + BCE = BFE ( 2 g\u00f3c n\u1ed9i ti\u1ebfp c\u00f9ng ch\u1eafn 1 cung) + BEC = EIF (cmt) \uf0de \uf044CEB \uf044FIE ( g \u2212 g ) 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 \uf0de BE = BC \uf0de BE.FE = BC.IE (1) EI FE * DE \/ \/CF (GT) Suy ra t\u1ee9 gi\u00e1c DEFC l\u00e0 h\u00ecnh thang M\u00e0 DEFC n\u1ed9i ti\u1ebfp (O) Suy ra DEFC l\u00e0 h\u00ecnh thang c\u00e2n N\u00ean DC = EF \uf0de DC = EF \uf0de DBC = EDF * X\u00e9t \uf044CBD v\u00e0 \uf044FDI c\u00f3: + DCB = DFI ( 2 g\u00f3c n\u1ed9i ti\u1ebfp c\u00f9ng ch\u1eafn 1 cung BD ) + DBC = IDF (cmt) \uf0de \uf044CBD \uf044FDI ( g \u2212 g ) \uf0de CB = BD \uf0de BD.DF = BC.ID (2) DF DI * T\u1eeb (1) v\u00e0 (2) ta c\u00f3: BE.EF + BD.DF = BC.IE + BC.ID \uf0de BE.EF + BD.DF = BC.(IE + ID) \uf0de BE.EF + BD.DF = BC.DE ----H\u1ebeT--- T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N 9","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH S\u00d4\u00db GD&\u00d1T TP HO\u00c0 CH\u00cd MINH \u00d1E\u00c0 THAM KHA\u00dbO TUYE\u00c5N SINH 10 PHO\u00d8NG G\u00d1&\u00d1T QUA\u00c4N 6 NA\u00caM HO\u00cfC: 2023 - 2024 M\u00d4N: TO\u00c1N 9 \u0110\u1ec0 THAM KH\u1ea2O M\u00c3 \u0110\u1ec0: Qu\u1eadn 6 - 3 \u0110\u00ea thi g\u1ed3m 8 c\u00e2u ho\u0309i t\u01b0\u0323 lu\u00e2\u0323n. Th\u01a1\u0300i gian: 120 phu\u0301t (kh\u00f4ng k\u00ea\u0309 th\u01a1\u0300i gian pha\u0301t \u0111\u00ea\u0300) C\u00e2u 1 (1,5 \u0111i\u1ec3m). Cho parabol (P) : y = 1 x2 v\u00e0 \u0111\u01b0\u1eddng th\u1eb3ng (d) : y = 1 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 (1,0 \u0111i\u1ec3m). Cho ph\u01b0\u01a1ng tr\u00ecnh 2x2 \u2212 5x \u2212 1 = 0 c\u00f3 hai 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 (2x1 \u2212 x2 )(2x2 \u2212 x1 ) \u2212 2023. C\u00e2u 3 (1,0 \u0111i\u1ec3m). C\u00f4ng ty vi\u1ec5n th\u00f4ng c\u00f3 g\u00f3i c\u01b0\u1edbc \u0111\u01b0\u1ee3c t\u00ednh nh\u01b0 sau: - G\u00f3i I : 2000 \u0111\u1ed3ng\/ph\u00fat cho 30 ph\u00fat \u0111\u1ea7u ti\u00ean; 1800 \u0111\u1ed3ng\/ph\u00fat cho 30 ph\u00fat ti\u1ebfp theo; 1200 \u0111\u1ed3ng\/ph\u00fat cho 30 ph\u00fat ti\u1ebfp theo n\u1eefa v\u00e0 800 \u0111\u1ed3ng\/ph\u00fat cho th\u1eddi gian c\u00f2n l\u1ea1i. - G\u00f3i II : 1800 \u0111\u1ed3ng\/ph\u00fat cho 60 ph\u00fat \u0111\u1ea7u ti\u00ean; 1500 \u0111\u1ed3ng\/ph\u00fat cho 60 ph\u00fat ti\u1ebfp theo v\u00e0 1000 \u0111\u1ed3ng\/ ph\u00fat cho th\u1eddi gian c\u00f2n l\u1ea1i. a) T\u00ednh s\u1ed1 ti\u1ec1n ph\u1ea3i tr\u1ea3 c\u1ee7a g\u00f3i I khi g\u1ecdi 130 ph\u00fat v\u00e0 g\u00f3i II khi g\u1ecdi 130 ph\u00fat. b) B\u00e1c An nh\u1eadn th\u1ea5y r\u1eb1ng m\u1ed7i th\u00e1ng trung b\u00ecnh g\u1ecdi ch\u01b0a \u0111\u1ebfn 904 ph\u00fat. Sau khi c\u00e2n nh\u1eafc th\u00ec b\u00e1c An ch\u1ecdn g\u00f3i I v\u00ec s\u1ebd ti\u1ebft ki\u1ec7m \u0111\u01b0\u1ee3c 150000 \u0111\u1ed3ng so v\u1edbi g\u00f3i II . H\u1ecfi trung b\u00ecnh b\u00e1c An g\u1ecdi bao nhi\u00eau ph\u00fat m\u1ed7i th\u00e1ng? C\u00e2u 4 (0,75 \u0111i\u1ec3m). Do \u1ea3nh h\u01b0\u1edfng c\u1ee7a t\u00ecnh h\u00ecnh d\u1ecbch b\u1ec7nh, thu nh\u1eadp c\u1ee7a m\u1ed9t c\u00f4ng ty b\u1ecb gi\u1ea3m d\u1ea7n trong n\u0103m 2021. C\u00e1c s\u1ed1 li\u1ec7u th\u1ed1ng k\u00ea \u0111\u01b0\u1ee3c th\u1ec3 hi\u1ec7n b\u1eb1ng \u0111\u1ed3 th\u1ecb nh\u01b0 h\u00ecnh v\u1ebd. a) T\u00ecm h\u00e0m s\u1ed1 th\u1ec3 hi\u1ec7n s\u1ef1 li\u00ean quan c\u1ee7a \u0111\u1ea1i l\u01b0\u1ee3ng y (tr\u0103m tri\u1ec7u\/ th\u00e1ng) theo \u0111\u1ea1i l\u01b0\u1ee3ng x (th\u00e1ng). b) Bi\u1ebft m\u1ed9t s\u1ea3n ph\u1ea9m b\u00e1n \u0111\u01b0\u1ee3c th\u00ec c\u00f4ng ty c\u00f3 l\u1ee3i nhu\u1eadn l\u00e0 100 ng\u00e0n \u0111\u1ed3ng, em h\u00e3y t\u00ednh s\u1ed1 s\u1ea3n ph\u1ea9m m\u00e0 c\u00f4ng ty b\u00e1n \u0111\u01b0\u1ee3c trong th\u00e1ng 9 n\u0103m 2021 (l\u00e0m tr\u00f2n \u0111\u1ebfn h\u00e0ng \u0111\u01a1n v\u1ecb). C\u00e2u 5 (1,0 \u0111i\u1ec3m). C\u1eeda h\u00e0ng l\u1ea5y m\u1ed9t th\u00f9ng n\u01b0\u1edbc ng\u1ecdt ( 24 lon) c\u1ee7a \u0111\u1ea1i l\u00fd ph\u00e2n ph\u1ed1i v\u1edbi gi\u00e1 192000 \u0111\u1ed3ng v\u00e0 b\u00e1n l\u1ebb v\u1edbi gi\u00e1 10000 \u0111\u1ed3ng m\u1ed9t lon. T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N 1","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH a) H\u1ecfi khi b\u00e1n h\u1ebft m\u1ed9t th\u00f9ng n\u01b0\u1edbc ng\u1ecdt \u0111\u00f3 th\u00ec c\u1eeda h\u00e0ng thu \u0111\u01b0\u1ee3c l\u00e3i bao nhi\u00eau ph\u1ea7n tr\u0103m so v\u1edbi gi\u00e1 g\u1ed1c? b) Trong \u0111\u1ee3t khuy\u1ebfn m\u00e3i, do \u0111\u1ea1i l\u00fd ph\u00e2n ph\u1ed1i gi\u1ea3m gi\u00e1 n\u00ean c\u1eeda h\u00e0ng c\u0169ng gi\u1ea3m gi\u00e1 c\u00f2n 9500 \u0111\u1ed3ng m\u1ed9t lon v\u00e0 thu \u0111\u01b0\u1ee3c l\u00e3i su\u1ea5t nh\u01b0 c\u0169. H\u1ecfi trong \u0111\u1ee3t n\u00e0y c\u1eeda h\u00e0ng \u0111\u00e3 mua m\u1ed9t th\u00f9ng n\u01b0\u1edbc ng\u1ecdt v\u1edbi gi\u00e1 bao nhi\u00eau? C\u00e2u 6 (1,0 \u0111i\u1ec3m). M\u1ed9t b\u1ec3 ch\u1ee9a n\u01b0\u1edbc c\u00f3 d\u1ea1ng nh\u01b0 h\u00ecnh v\u1ebd. a) T\u00ednh th\u1ec3 t\u00edch c\u1ee7a b\u1ec3 (k\u1ebft qu\u1ea3 kh\u00f4ng l\u00e0m tr\u00f2n). b) Ban \u0111\u1ea7u, b\u1ec3 kh\u00f4ng c\u00f3 n\u01b0\u1edbc. Sau \u0111\u00f3 ng\u01b0\u1eddi ta b\u01a1m n\u01b0\u1edbc v\u00e0o b\u1ec3 v\u1edbi t\u1ed1c \u0111\u1ed9 1 l\u00edt\/gi\u00e2y. H\u1ecfi sau 20 ph\u00fat k\u1ec3 t\u1eeb khi b\u1eaft \u0111\u1ea7u b\u01a1m th\u00ec m\u1ef1c n\u01b0\u1edbc trong h\u1ed3 c\u00e1ch mi\u1ec7ng h\u1ed3 bao nhi\u00eau m\u00e9t (l\u00e0m tr\u00f2n \u0111\u1ebfn h\u00e0ng ph\u1ea7n tr\u0103m)? Bi\u1ebft th\u1ec3 t\u00edch h\u00ecnh tr\u1ee5 l\u00e0 V = \uf070 r2h v\u00e0 th\u1ec3 t\u00edch h\u00ecnh n\u00f3n l\u00e0 V = 1 \uf070 r2h. 3 C\u00e2u 7 (1,0 \u0111i\u1ec3m). Tr\u01b0\u1edbc ng\u00e0y k\u1ebft th\u00fac n\u0103m h\u1ecdc t\u1eadp th\u1ec3 c\u00e1c h\u1ecdc sinh l\u1edbp 9A mu\u1ed1n mua qu\u00e0 t\u1eb7ng cho c\u00e1c gi\u00e1o vi\u00ean gi\u1ea3ng d\u1ea1y l\u1edbp m\u00ecnh trong su\u1ed1t n\u0103m h\u1ecdc \u0111\u1ec3 t\u1ecf l\u00f2ng tri \u00e2n, m\u1ed7i m\u00f3n qu\u00e0 t\u1eb7ng cho th\u1ea7y v\u1edbi gi\u00e1 l\u00e0 50 ng\u00e0n \u0111\u1ed3ng, m\u1ed7i m\u00f3n qu\u00e0 t\u1eb7ng cho c\u00f4 c\u00f3 gi\u00e1 l\u00e0 65 ng\u00e0n \u0111\u1ed3ng, bi\u1ebft l\u1edbp t\u1eb7ng qu\u00e0 cho 15 gi\u00e1o vi\u00ean v\u00e0 t\u1ed5ng s\u1ed1 ti\u1ec1n m\u00e0 l\u1edbp mua qu\u00e0 l\u00e0 870 ng\u00e0n \u0111\u1ed3ng. Em h\u00e3y t\u00ednh s\u1ed1 th\u1ea7y gi\u00e1o v\u00e0 s\u1ed1 c\u00f4 gi\u00e1o l\u1edbp 9A d\u1ef1 \u0111\u1ecbnh mua qu\u00e0 t\u1eb7ng. C\u00e2u 8 (3,0 \u0111i\u1ec3m). Cho \u0111\u01b0\u1eddng tr\u00f2n (O) \u0111\u01b0\u1eddng k\u00ednh AB . L\u1ea5y \u0111i\u1ec3m S t\u00f9y \u00fd tr\u00ean tia \u0111\u1ed1i c\u1ee7a tia BA . V\u1ebd c\u00e1t tuy\u1ebfn SMC c\u1ee7a (O) . V\u1ebd d\u00e2y cung CD c\u1ee7a \u0111\u01b0\u1eddng tr\u00f2n (O) vu\u00f4ng g\u00f3c v\u1edbi AB . a) Ch\u1ee9ng minh tam gi\u00e1c SMA \u0111\u1ed3ng d\u1ea1ng v\u1edbi tam gi\u00e1c SBC . b) C\u00e1c d\u00e2y cung AM, BC c\u1eaft nhau t\u1ea1i N ; c\u00e1c d\u00e2y cung AB, DM c\u1eaft nhau t\u1ea1i P . Ch\u1ee9ng minh r\u1eb1ng t\u1ee9 gi\u00e1c BMNP n\u1ed9i ti\u1ebfp v\u00e0 NP song song v\u1edbi CD. c) Ch\u1ee9ng minh OP.OS = OM2. ---- 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 x v\u00e0 \u0111\u01b0\u1eddng th\u1eb3ng d : y x . 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) B\u1ea3ng gi\u00e1 tr\u1ecb: x yx x yx b) Ph\u01b0\u01a1ng tr\u00ecnh ho\u00e0nh \u0111\u1ed9 giao \u0111i\u1ec3m c\u1ee7a P v\u00e0 d l\u00e0 x x xx x ho\u1eb7c x 3 . V\u1edbi x ta \u0111\u01b0\u1ee3c y .( ) . V\u1edbi x 3 ta \u0111\u01b0\u1ee3c y 1 .32 9. 2 2 V\u1eady ; v\u00e0 ; l\u00e0 t\u1ecda \u0111\u1ed9 c\u1ee7a hai giao \u0111i\u1ec3m c\u1ea7n t\u00ecm. C\u00e2u 2 (1,0 \u0111i\u1ec3m). Cho ph\u01b0\u01a1ng tr\u00ecnh x x c\u00f3 hai nghi\u1ec7m l\u00e0 x ,x . Kh\u00f4ng gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh, h\u00e3y t\u00ednh gi\u00e1 tr\u1ecb c\u1ee7a bi\u1ec3u th\u1ee9c x x x x . V\u00ec a.c 2.( 1) L\u1eddi gi\u1ea3i 2 0 n\u00ean ph\u01b0\u01a1ng tr\u00ecnh \u0111\u00e3 cho c\u00f3 hai nghi\u1ec7m ph\u00e2n bi\u1ec7t x1, x2. Theo \u0111\u1ecbnh l\u00fd Vi\u00e8te, ta c\u00f3 S x1 x2 b 5 v\u00e0 P x 1x 2 c 1 . Khi \u0111\u00f3 a 2 a 2 2x1 x2 2x2 x1 2023 4x 1x 2 2x 2 2x 2 x1x2 2023 1 2 2023 5x1x2 2 x 2 x 2 4x1x2 2023 1 2 5x1x2 2 x1 x2 2 9P 2S 2 2023 9. 1 2. 5 2 2023 2040. 2 2 C\u00e2u 3 (1,0 \u0111i\u1ec3m). C\u00f4ng ty vi\u1ec5n th\u00f4ng c\u00f3 g\u00f3i c\u01b0\u1edbc \u0111\u01b0\u1ee3c t\u00ednh nh\u01b0 sau: 3 T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH - G\u00f3i I : 2000 \u0111\u1ed3ng\/ph\u00fat cho 30 ph\u00fat \u0111\u1ea7u ti\u00ean; 1800 \u0111\u1ed3ng\/ph\u00fat cho 30 ph\u00fat ti\u1ebfp theo; 1200 \u0111\u1ed3ng\/ph\u00fat cho 30 ph\u00fat ti\u1ebfp theo n\u1eefa v\u00e0 800 \u0111\u1ed3ng\/ph\u00fat cho th\u1eddi gian c\u00f2n l\u1ea1i. - G\u00f3i II : 1800 \u0111\u1ed3ng\/ph\u00fat cho 60 ph\u00fat \u0111\u1ea7u ti\u00ean; 1500 \u0111\u1ed3ng\/ph\u00fat cho 60 ph\u00fat ti\u1ebfp theo v\u00e0 1000 \u0111\u1ed3ng\/ ph\u00fat cho th\u1eddi gian c\u00f2n l\u1ea1i. a) T\u00ednh s\u1ed1 ti\u1ec1n ph\u1ea3i tr\u1ea3 c\u1ee7a g\u00f3i I khi g\u1ecdi 130 ph\u00fat v\u00e0 g\u00f3i II khi g\u1ecdi 130 ph\u00fat. b) B\u00e1c An nh\u1eadn th\u1ea5y r\u1eb1ng m\u1ed7i th\u00e1ng trung b\u00ecnh g\u1ecdi ch\u01b0a \u0111\u1ebfn 904 ph\u00fat. Sau khi c\u00e2n nh\u1eafc th\u00ec b\u00e1c An ch\u1ecdn g\u00f3i I v\u00ec s\u1ebd ti\u1ebft ki\u1ec7m \u0111\u01b0\u1ee3c 150000 \u0111\u1ed3ng so v\u1edbi g\u00f3i II . H\u1ecfi trung b\u00ecnh b\u00e1c An g\u1ecdi bao nhi\u00eau ph\u00fat m\u1ed7i th\u00e1ng? L\u1eddi gi\u1ea3i a) S\u1ed1 ti\u1ec1n ph\u1ea3i tr\u1ea3 c\u1ee7a g\u00f3i I khi g\u1ecdi 130 ph\u00fat l\u00e0 2000 30 1800 30 1200 30 800 (130 90) 182000 (\u0111\u1ed3ng). S\u1ed1 ti\u1ec1n ph\u1ea3i tr\u1ea3 c\u1ee7a g\u00f3i II khi g\u1ecdi 130 ph\u00fat l\u00e0 1800 60 1500 60 1000 (130 120) 208000 (\u0111\u1ed3ng). b) G\u1ecdi x (ph\u00fat) l\u00e0 th\u1eddi gian trung b\u00ecnh m\u00e0 b\u00e1c An g\u1ecdi m\u1ed7i th\u00e1ng v\u00e0 y (ngh\u00ecn \u0111\u1ed3ng) l\u00e0 s\u1ed1 ti\u1ec1n ch\u00eanh l\u1ec7ch gi\u1eefa g\u00f3i I v\u00e0 g\u00f3i II . Nh\u1eadn x\u00e9t r\u1eb1ng 60 x 904 . \u2022 N\u1ebfu 60 x 90 th\u00ec y 1, 8 60 1, 5(x 60) 2 30 1, 8 30 1,2(x 60) 0, 3(x 60) 6 , do \u0111\u00f3 y 150 x 580 (kh\u00f4ng th\u1ecfa m\u00e3n). 0, 7x 60, \u2022 N\u1ebfu 90 x 120 th\u00ec y 1, 8 60 1, 5(x 60) 2 30 1, 8 30 1,2 30 0, 8 (x 90) do \u0111\u00f3 y 150 x 300 (kh\u00f4ng th\u1ecfa m\u00e3n). 0, 2x, \u2022 N\u1ebfu x 120 th\u00ec y 1, 8 60 1, 5 60 (x 120) 2 30 1, 8 30 1,2 30 0, 8(x 90) do \u0111\u00f3 y 150 x 750 (th\u1ecfa m\u00e3n). V\u1eady trung b\u00ecnh m\u1ed7i th\u00e1ng b\u00e1c An g\u1ecdi 750 ph\u00fat. C\u00e2u 4 (0,75 \u0111i\u1ec3m). Do \u1ea3nh h\u01b0\u1edfng c\u1ee7a t\u00ecnh h\u00ecnh d\u1ecbch b\u1ec7nh, thu nh\u1eadp c\u1ee7a m\u1ed9t c\u00f4ng ty b\u1ecb gi\u1ea3m d\u1ea7n trong n\u0103m 2021 . C\u00e1c s\u1ed1 li\u1ec7u th\u1ed1ng k\u00ea \u0111\u01b0\u1ee3c th\u1ec3 hi\u1ec7n b\u1eb1ng \u0111\u1ed3 th\u1ecb nh\u01b0 h\u00ecnh v\u1ebd. 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 a) T\u00ecm h\u00e0m s\u1ed1 th\u1ec3 hi\u1ec7n s\u1ef1 li\u00ean quan c\u1ee7a \u0111\u1ea1i l\u01b0\u1ee3ng y (tr\u0103m tri\u1ec7u\/ th\u00e1ng) theo \u0111\u1ea1i l\u01b0\u1ee3ng x (th\u00e1ng). b) Bi\u1ebft m\u1ed9t s\u1ea3n ph\u1ea9m b\u00e1n \u0111\u01b0\u1ee3c th\u00ec c\u00f4ng ty c\u00f3 l\u1ee3i nhu\u1eadn l\u00e0 100 ng\u00e0n \u0111\u1ed3ng, em h\u00e3y t\u00ednh s\u1ed1 s\u1ea3n ph\u1ea9m m\u00e0 c\u00f4ng ty b\u00e1n \u0111\u01b0\u1ee3c trong th\u00e1ng 9 n\u0103m 2021 (l\u00e0m tr\u00f2n \u0111\u1ebfn h\u00e0ng \u0111\u01a1n v\u1ecb). L\u1eddi gi\u1ea3i a) G\u1ecdi y ax b a,b l\u00e0 h\u00e0m s\u1ed1 b\u1eadc nh\u1ea5t th\u1ec3 hi\u1ec7n s\u1ef1 li\u00ean quan gi\u1eefa \u0111\u1ea1i l\u01b0\u1ee3ng y theo x. V\u00ec c\u00e1c \u0111i\u1ec3m (1;5) v\u00e0 (12;2) thu\u1ed9c \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1 tr\u00ean n\u00ean ta c\u00f3 h\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh ab 5 a3 12a b 2 11 . 58 b 11 V\u1eady h\u00e0m s\u1ed1 c\u1ea7n t\u00ecm c\u00f3 d\u1ea1ng y 3 x 58 . 11 11 b) Thu nh\u1eadp c\u1ee7a c\u00f4ng ty trong th\u00e1ng 9 l\u00e0 3 .9 58 31 (tr\u0103m tri\u1ec7u). 11 11 11 S\u1ed1 s\u1ea3n ph\u1ea9m b\u00e1n \u0111\u01b0\u1ee3c trong th\u00e1ng 9 l\u00e0 31 : 0, 001 2818 (s\u1ea3n ph\u1ea9m). 11 C\u00e2u 5 (1,0 \u0111i\u1ec3m). C\u1eeda h\u00e0ng l\u1ea5y m\u1ed9t th\u00f9ng n\u01b0\u1edbc ng\u1ecdt (24 lon) c\u1ee7a \u0111\u1ea1i l\u00fd ph\u00e2n ph\u1ed1i v\u1edbi gi\u00e1 192000 \u0111\u1ed3ng v\u00e0 b\u00e1n l\u1ebb v\u1edbi gi\u00e1 10000 \u0111\u1ed3ng m\u1ed9t lon. a) H\u1ecfi khi b\u00e1n h\u1ebft m\u1ed9t th\u00f9ng n\u01b0\u1edbc ng\u1ecdt \u0111\u00f3 th\u00ec c\u1eeda h\u00e0ng thu \u0111\u01b0\u1ee3c l\u00e3i bao nhi\u00eau ph\u1ea7n tr\u0103m so v\u1edbi gi\u00e1 g\u1ed1c? b) Trong \u0111\u1ee3t khuy\u1ebfn m\u00e3i, do \u0111\u1ea1i l\u00fd ph\u00e2n ph\u1ed1i gi\u1ea3m gi\u00e1 n\u00ean c\u1eeda h\u00e0ng c\u0169ng gi\u1ea3m gi\u00e1 c\u00f2n 9500 \u0111\u1ed3ng m\u1ed9t lon v\u00e0 thu \u0111\u01b0\u1ee3c l\u00e3i su\u1ea5t nh\u01b0 c\u0169. H\u1ecfi trong \u0111\u1ee3t n\u00e0y c\u1eeda h\u00e0ng \u0111\u00e3 mua m\u1ed9t th\u00f9ng n\u01b0\u1edbc ng\u1ecdt v\u1edbi gi\u00e1 bao nhi\u00eau? L\u1eddi gi\u1ea3i a) S\u1ed1 ti\u1ec1n thu \u0111\u01b0\u1ee3c khi b\u00e1n h\u1ebft th\u00f9ng n\u01b0\u1edbc ng\u1ecdt l\u00e0 10000 24 240000 (\u0111\u1ed3ng). Ph\u1ea7n tr\u0103m l\u00e3i so v\u1edbi gi\u00e1 g\u1ed1c c\u1eeda h\u00e0ng thu \u0111\u01b0\u1ee3c l\u00e0 . % %. b) S\u1ed1 ti\u1ec1n thu \u0111\u01b0\u1ee3c khi b\u00e1n h\u1ebft th\u00f9ng n\u01b0\u1edbc ng\u1ecdt v\u1edbi gi\u00e1 gi\u1ea3m l\u00e0 (\u0111\u1ed3ng). G\u1ecdi x (\u0111\u1ed3ng) l\u00e0 s\u1ed1 ti\u1ec1n c\u1eeda h\u00e0ng \u0111\u00e3 mua c\u1ee7a \u0111\u1ea1i l\u00fd trong \u0111\u1ee3t n\u00e0y (x 0 ). Ta c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh 228000 x 0,25 228000 1,25x x 182400. x V\u1eady trong \u0111\u1ee3t n\u00e0y c\u1eeda h\u00e0ng \u0111\u00e3 mua c\u1ee7a \u0111\u1ea1i l\u00fd 182400 \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 (1,0 \u0111i\u1ec3m). M\u1ed9t b\u1ec3 ch\u1ee9a n\u01b0\u1edbc c\u00f3 d\u1ea1ng nh\u01b0 h\u00ecnh v\u1ebd. a) T\u00ednh th\u1ec3 t\u00edch c\u1ee7a b\u1ec3 (k\u1ebft qu\u1ea3 kh\u00f4ng l\u00e0m tr\u00f2n). b) Ban \u0111\u1ea7u, b\u1ec3 kh\u00f4ng c\u00f3 n\u01b0\u1edbc. Sau \u0111\u00f3 ng\u01b0\u1eddi ta b\u01a1m n\u01b0\u1edbc v\u00e0o b\u1ec3 v\u1edbi t\u1ed1c \u0111\u1ed9 1 l\u00edt\/gi\u00e2y. H\u1ecfi sau 20 ph\u00fat k\u1ec3 t\u1eeb khi b\u1eaft \u0111\u1ea7u b\u01a1m th\u00ec m\u1ef1c n\u01b0\u1edbc trong h\u1ed3 c\u00e1ch mi\u1ec7ng h\u1ed3 bao nhi\u00eau m\u00e9t (l\u00e0m tr\u00f2n \u0111\u1ebfn h\u00e0ng ph\u1ea7n tr\u0103m)? Bi\u1ebft th\u1ec3 t\u00edch h\u00ecnh tr\u1ee5 l\u00e0 V r 2h v\u00e0 th\u1ec3 t\u00edch h\u00ecnh n\u00f3n l\u00e0 V 1 r 2h. 3 L\u1eddi gi\u1ea3i a) Th\u1ec3 t\u00edch ph\u1ea7n kh\u1ed1i n\u00f3n l\u00e0 1 .0, 72.(1, 6 0, 7) 0,147 (m3) . 3 Th\u1ec3 t\u00edch ph\u1ea7n kh\u1ed1i tr\u1ee5 l\u00e0 .0, 72.0, 7 0, 343 (m3 ) . V\u1eady th\u1ec3 t\u00edch c\u1ee7a b\u1ec3 l\u00e0 0,147 0, 343 0, 49 (m3 ) . b) Sau 20 ph\u00fat th\u00ec th\u1ec3 t\u00edch n\u01b0\u1edbc trong b\u1ec3 l\u00e0 1 20 60 1200 (l\u00edt) 1,2 (m3 ) 0,147 (m3 ) . Do \u0111\u00f3 ph\u1ea7n n\u01b0\u1edbc trong b\u1ec3 \u0111\u1ea7y k\u00edn kh\u1ed1i n\u00f3n v\u00e0 th\u1ec3 t\u00edch n\u01b0\u1edbc trong kh\u1ed1i tr\u1ee5 l\u00e0 1,2 0,147 (m3 ) , suy ra chi\u1ec1u cao c\u1ee7a m\u1ef1c n\u01b0\u1edbc trong h\u1ed3 l\u00e0 1,2 0,147 (m) . .0, 72 V\u1eady m\u1ef1c n\u01b0\u1edbc trong h\u1ed3 c\u00e1ch mi\u1ec7ng h\u1ed3 m\u1ed9t kho\u1ea3ng l\u00e0 0, 7 1,2 0,147 0,22 (m) . .0, 72 C\u00e2u 7 (1,0 \u0111i\u1ec3m). Tr\u01b0\u1edbc ng\u00e0y k\u1ebft th\u00fac n\u0103m h\u1ecdc t\u1eadp th\u1ec3 c\u00e1c h\u1ecdc sinh l\u1edbp 9A mu\u1ed1n mua qu\u00e0 t\u1eb7ng cho c\u00e1c gi\u00e1o vi\u00ean gi\u1ea3ng d\u1ea1y l\u1edbp m\u00ecnh trong su\u1ed1t n\u0103m h\u1ecdc \u0111\u1ec3 t\u1ecf l\u00f2ng tri \u00e2n, m\u1ed7i m\u00f3n qu\u00e0 t\u1eb7ng cho th\u1ea7y v\u1edbi gi\u00e1 l\u00e0 50 ng\u00e0n \u0111\u1ed3ng, m\u1ed7i m\u00f3n qu\u00e0 t\u1eb7ng cho c\u00f4 c\u00f3 gi\u00e1 l\u00e0 65 ng\u00e0n \u0111\u1ed3ng, bi\u1ebft l\u1edbp t\u1eb7ng qu\u00e0 cho 15 gi\u00e1o vi\u00ean v\u00e0 t\u1ed5ng s\u1ed1 ti\u1ec1n m\u00e0 l\u1edbp mua qu\u00e0 l\u00e0 870 ng\u00e0n \u0111\u1ed3ng. Em h\u00e3y t\u00ednh s\u1ed1 th\u1ea7y gi\u00e1o v\u00e0 s\u1ed1 c\u00f4 gi\u00e1o l\u1edbp 9A d\u1ef1 \u0111\u1ecbnh mua qu\u00e0 t\u1eb7ng. L\u1eddi gi\u1ea3i G\u1ecdi s\u1ed1 th\u1ea7y gi\u00e1o v\u00e0 s\u1ed1 c\u00f4 gi\u00e1o c\u1ee7a l\u1edbp 9A d\u1ef1 \u0111\u1ecbnh mua qu\u00e0 t\u1eb7ng l\u1ea7n l\u01b0\u1ee3t l\u00e0 x, y (x,y *). V\u00ec l\u1edbp t\u1eb7ng qu\u00e0 cho 15 gi\u00e1o vi\u00ean v\u00e0 t\u1ed5ng s\u1ed1 ti\u1ec1n m\u00e0 l\u1edbp mua qu\u00e0 l\u00e0 870 ng\u00e0n \u0111\u1ed3ng n\u00ean t\u1eeb \u0111\u1ec1 b\u00e0i ta c\u00f3 h\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh x y 15 x7 50x 65y 870 y 8. V\u1eady l\u1edbp 9A d\u1ef1 \u0111\u1ecbnh mua qu\u00e0 t\u1eb7ng cho 7 th\u1ea7y gi\u00e1o v\u00e0 8 c\u00f4 gi\u00e1o. C\u00e2u 8 (3,0 \u0111i\u1ec3m). Cho \u0111\u01b0\u1eddng tr\u00f2n (O) \u0111\u01b0\u1eddng k\u00ednh AB . L\u1ea5y \u0111i\u1ec3m S t\u00f9y \u00fd tr\u00ean tia \u0111\u1ed1i c\u1ee7a tia BA . V\u1ebd c\u00e1t tuy\u1ebfn SMC c\u1ee7a (O) . V\u1ebd d\u00e2y cung CD c\u1ee7a \u0111\u01b0\u1eddng tr\u00f2n (O) vu\u00f4ng g\u00f3c v\u1edbi AB . a) Ch\u1ee9ng minh tam gi\u00e1c SMA \u0111\u1ed3ng d\u1ea1ng v\u1edbi tam gi\u00e1c SBC . b) C\u00e1c d\u00e2y cung AM, BC c\u1eaft nhau t\u1ea1i N ; c\u00e1c d\u00e2y cung AB, DM c\u1eaft nhau t\u1ea1i P . Ch\u1ee9ng minh r\u1eb1ng t\u1ee9 gi\u00e1c BMNP n\u1ed9i ti\u1ebfp v\u00e0 NP song song v\u1edbi CD. c) Ch\u1ee9ng minh OP.OS OM 2. 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 a) X\u00e9t SMA v\u00e0 SBC ta c\u00f3 S l\u00e0 g\u00f3c chung v\u00e0 SAM SCB (hai g\u00f3c n\u1ed9i ti\u1ebfp (O) c\u00f9ng ch\u1eafn BM ). D\u1eabn \u0111\u1ebfn SMA \u0111\u1ed3ng d\u1ea1ng SBC (g\u00f3c-g\u00f3c). b) \uf09f Ch\u1ee9ng minh BMNP l\u00e0 t\u1ee9 gi\u00e1c n\u1ed9i ti\u1ebfp. X\u00e9t \u0111\u01b0\u1eddng tr\u00f2n (O) \u0111\u01b0\u1eddng k\u00ednh AB c\u00f3 d\u00e2y cung CD AB , n\u00ean A l\u00e0 \u0111i\u1ec3m ch\u00ednh gi\u1eefa CD . D\u1eabn \u0111\u1ebfn AC AD , do \u0111\u00f3 ABC AMD (hai g\u00f3c n\u1ed9i ti\u1ebfp (O) ch\u1eafn hai cung b\u1eb1ng nhau). V\u00ec th\u1ebf t\u1ee9 gi\u00e1c BMNP n\u1ed9i ti\u1ebfp (do hai \u0111\u1ec9nh k\u1ec1 B, M c\u00f9ng nh\u00ecn NP d\u01b0\u1edbi hai g\u00f3c b\u1eb1ng nhau). \uf09f Ch\u1ee9ng minh NP song song CD . V\u00ec t\u1ee9 gi\u00e1c BMNP n\u1ed9i ti\u1ebfp n\u00ean BMD BNP (hai g\u00f3c n\u1ed9i ti\u1ebfp c\u00f9ng ch\u1eafn BP ). M\u1eb7t kh\u00e1c, BMD BCD (hai g\u00f3c n\u1ed9i ti\u1ebfp (O) c\u00f9ng ch\u1eafn BD ). Suy ra BNP BCD , m\u00e0 hai g\u00f3c n\u00e0y \u1edf v\u1ecb tr\u00ed \u0111\u1ed3ng v\u1ecb, n\u00ean NP song song CD . c) X\u00e9t (O) , c\u00f3 S s\u00f1AC s\u00f1BM (g\u00f3c c\u00f3 \u0111\u1ec9nh \u1edf b\u00ean ngo\u00e0i \u0111\u01b0\u1eddng tr\u00f2n). 2 C\u0169ng x\u00e9t (O) , l\u1ea1i c\u00f3 OMP AMP AMO s\u00f1AD MAO s\u00f1AD s\u00f1BM s\u00f1AD s\u00f1BM . 2 2 2 2 (v\u1edbi l\u01b0u \u00fd r\u1eb1ng AMO c\u00e2n t\u1ea1i O ). M\u00e0 AC AD (ch\u1ee9ng minh tr\u00ean), n\u00ean S OMP . X\u00e9t OSM v\u00e0 OMP c\u00f3 MOS l\u00e0 g\u00f3c chung v\u00e0 S OMP . D\u1eabn \u0111\u1ebfn OSM \u0111\u1ed3ng d\u1ea1ng v\u1edbi OMP OS OM OP 2 OS.OM. (g\u00f3c-g\u00f3c), suy ra OM OP ---- H\u1ebeT ---- T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N 7"]