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 V\u1edbi \uf0ect = 0 \uf0de 39,3 = 0.a + b . (1) \uf0ed\uf0ees = 39,3 V\u1edbi \uf0ect = 2 \uf0de 159,3 = 2.a + b . (2) \uf0ed\uf0ees = 159,3 T\u1eeb (1) v\u00e0 (2) ta c\u00f3 h\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh: \uf0ec0a + b = 39, 3 \uf0db \uf0eca = 60 . \uf0ed\uf0ee2a + b = 159 , 3 \uf0ed\uf0eeb = 39, 3 V\u1eady: a = 60 , b = 39,3 v\u00e0 s = 60t + 39,3 . b) Sau 4 gi\u1edd, xe \u0111\u00e3 c\u00e1ch trung t\u00e2m Th\u00e0nh ph\u1ed1 H\u1ed3 Ch\u00ed Minh bao nhi\u00eau km; bi\u1ebft xe c\u00f3 gh\u00e9 ngh\u1ec9 ng\u01a1i t\u1ea1i tr\u1ea1m d\u1eebng ch\u00e2n 45 ph\u00fat Th\u1eddi gian xe ch\u1ea1y l\u00e0: 4 \u2212 45 = 4 \u2212 3 = 13 (h) 60 4 4 Th\u1ebf t = 13 (h) v\u00e0o h\u00e0m s\u1ed1 s = 60t + 39,3 , ta \u0111\u01b0\u1ee3c: 4 s = 60.13 + 39,3 = 234,3(km) 4 V\u1eady xe \u0111\u00e3 c\u00e1ch trung t\u00e2m Th\u00e0nh ph\u1ed1 H\u1ed3 Ch\u00ed Minh 234,3(km) C\u00e2u 4. (0,75 \u0111i\u1ec3m). Cu\u1ed1i tu\u1ea7n m\u1ed9t nh\u00f3m b\u1ea1n mu\u1ed1n \u0111i th\u01b0 gi\u00e3n b\u1eb1ng c\u00e1ch c\u1eafm tr\u1ea1i ngo\u00e0i tr\u1eddi. \u0110\u1ec3 che n\u1eafng che m\u01b0a trong l\u00fac c\u1eafm tr\u1ea1i, c\u00e1c b\u1ea1n quy\u1ebft \u0111\u1ecbnh d\u1ef1ng l\u1ec1u ch\u1eef A . Theo t\u00ednh to\u00e1n c\u1ee7a nh\u00f3m, c\u00e1c b\u1ea1n c\u00f3 s\u1eb5n hai c\u00e2y c\u1ecdc c\u00f3 chi\u1ec1u cao 2m . Nh\u00f3m c\u00f3 t\u1ea5m b\u1ea1t d\u00e0i 6m th\u00ec c\u00f3 th\u1ec3 d\u1ef1ng l\u1ec1u ch\u1eef A v\u1edbi g\u00f3c t\u1ea1o b\u1edfi t\u1ea5m b\u1ea1t v\u00e0 m\u1eb7t \u0111\u1ea5t l\u00e0 bao nhi\u00eau \u0111\u1ed9? A 2m C BH L\u1eddi gi\u1ea3i Ta c\u00f3: \uf044ABC c\u00e2n tai A \uf0de \uf0ec\uf0ef AB = AC = 6 : 2 = 3(m) \uf0ed \uf0ef\uf0eeHB = HC X\u00e9t \uf044ABH vu\u00f4ng \u1edf H , ta c\u00f3: sin ABH = AH = 2 \uf0de ABH \uf0bb 42\uf0b0 AB 3 V\u1eady g\u00f3c t\u1ea1o b\u1edfi t\u1ea5m b\u1ea1t v\u00e0 m\u1eb7t \u0111\u1ea5t l\u00e0 42\uf0b0 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 5. (1 \u0111i\u1ec3m) M\u1ed9t ng\u01b0\u1eddi \u0111i si\u00eau th\u1ecb mua hai m\u00f3n h\u00e0ng. M\u00f3n th\u1ee9 nh\u1ea5t c\u00f3 gi\u00e1 ni\u00eam y\u1ebft l\u00e0 4 tri\u1ec7u. Nh\u01b0ng \u0111\u00fang v\u00e0o \u0111\u1ee3t khuy\u1ebfn m\u00e3i n\u00ean m\u00f3n h\u00e0ng th\u1ee9 nh\u1ea5t \u0111\u01b0\u1ee3c gi\u1ea3m 10% , c\u00f2n m\u00f3n h\u00e0ng th\u1ee9 hai \u0111\u01b0\u1ee3c gi\u1ea3m 8% n\u00ean ng\u01b0\u1eddi \u0111\u00f3 ph\u1ea3i tr\u1ea3 6,36 tri\u1ec7u \u0111\u1ed3ng. H\u1ecfi gi\u00e1 ni\u00eam y\u1ebft c\u1ee7a m\u00f3n h\u00e0ng th\u1ee9 hai l\u00e0 bao nhi\u00eau? L\u1eddi gi\u1ea3i Gi\u00e1 ti\u1ec1n c\u1ee7a m\u00f3n h\u00e0ng th\u1ee9 nh\u1ea5t sau khi \u0111\u01b0\u1ee3c gi\u1ea3m gi\u00e1 l\u00e0: 4.90% = 3, 6 (tri\u1ec7u \u0111\u1ed3ng) Gi\u00e1 ti\u1ec1n c\u1ee7a m\u00f3n hang th\u1ee9 hai sau khi \u0111\u01b0\u1ee3c gi\u1ea3m gi\u00e1 l\u00e0: 6,36 \u2212 3, 6 = 2, 76 (tri\u1ec7u \u0111\u1ed3ng) Gi\u00e1 ni\u00eam y\u1ebft c\u1ee7a m\u00f3n h\u00e0ng th\u1ee9 hai l\u00e0: 2, 76 : 92% = 3 (tri\u1ec7u \u0111\u1ed3ng) C\u00e2u 6. (1 \u0111i\u1ec3m) M\u1ed9t qu\u00e1n tr\u00e0 s\u1eefa c\u00f3 ch\u01b0\u01a1ng tr\u00ecnh khuy\u1ebfn m\u00e3i mua 4 t\u1eb7ng 1 v\u1edbi m\u1eb7t h\u00e0ng tr\u00e0 s\u1eefa gi\u00e1 30000 \u0111\u1ed3ng \/ly Trong m\u1ed9t ng\u00e0y, s\u1ed1 l\u01b0\u1ee3ng ng\u01b0\u1eddi mua tr\u00e0 s\u1eefa \u0111\u01b0\u1ee3c th\u1ed1ng k\u00ea l\u1ea1i qua b\u1ea3ng sau: S\u1ed1 ly S\u1ed1 ng\u01b0\u1eddi 1 ly 15 2 ly 20 3 ly 10 4 ly 20 5 ly 10 6 ly 5 a) T\u00ednh s\u1ed1 ly tr\u00e0 s\u1eefa m\u00e0 qu\u00e1n \u0111\u00e3 l\u00e0m cho kh\u00e1ch h\u00e0ng? b) N\u1ebfu gi\u00e1 v\u1ed1n \u0111\u1ec3 l\u00e0m m\u1ed9t ly tr\u00e0 s\u1eefa l\u00e0 25000 \u0111\u1ed3ng, thu\u1ebf ph\u1ea3i n\u1ed9p l\u00e0 10% tr\u00ean t\u1ed5ng s\u1ed1 ti\u1ec1n b\u00e1n th\u00ec ng\u00e0y h\u00f4m \u0111\u00f3 qu\u00e1n c\u00f3 l\u1eddi hay kh\u00f4ng? V\u00e0 l\u1eddi bao nhi\u00eau ti\u1ec1n? L\u1eddi gi\u1ea3i a) T\u00ednh s\u1ed1 ly tr\u00e0 s\u1eefa m\u00e0 qu\u00e1n \u0111\u00e3 l\u00e0m cho kh\u00e1ch h\u00e0ng? T\u1ed5ng s\u1ed1 ly tr\u00e0 s\u1eefa qu\u00e1n \u0111\u00e3 l\u00e0m cho kh\u00e1ch l\u00e0: 1.15 + 2.20 + 3.10 + 4.20 + 5.10 + 6.5 = 245(ly) b) N\u1ebfu gi\u00e1 v\u1ed1n \u0111\u1ec3 l\u00e0m m\u1ed9t ly tr\u00e0 s\u1eefa l\u00e0 25000 \u0111\u1ed3ng, thu\u1ebf ph\u1ea3i n\u1ed9p l\u00e0 10% tr\u00ean t\u1ed5ng s\u1ed1 ti\u1ec1n b\u00e1n th\u00ec ng\u00e0y h\u00f4m \u0111\u00f3 qu\u00e1n c\u00f3 l\u1eddi hay kh\u00f4ng? V\u00e0 l\u1eddi bao nhi\u00eau ti\u1ec1n? Kh\u00e1ch h\u00e0ng \u0111\u01b0\u1ee3c khuy\u1ebfn m\u00e3i mua 4 t\u1eb7ng 1 th\u00ec s\u1ed1 ti\u1ec1n ph\u1ea3i tr\u1ea3 cho 5 ly tr\u00e0 s\u1eefa l\u00e0: 30000.4 =120000 (\u0111\u1ed3ng) S\u1ed1 ti\u1ec1n c\u1eeda h\u00e0ng thu v\u1ec1 ng\u00e0y h\u00f4m \u0111\u00f3 l\u00e0: 120000. 245 = 5880000 (\u0111\u1ed3ng) 6 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 S\u1ed1 ti\u1ec1n c\u1eeda h\u00e0ng \u0111\u00f3ng thu\u1ebf v\u00e0 ti\u1ec1n v\u1ed1n c\u1ee7a ng\u00e0y h\u00f4m \u0111\u00f3 l\u00e0: 5880000.10% + 25000.245 = 6713000 (\u0111\u1ed3ng) V\u1eady ng\u00e0y h\u00f4m \u0111\u00f3 qu\u00e1n b\u1ecb l\u1ed7 C\u00e2u 7. (1 \u0111i\u1ec3m) Tr\u00f2 Ch\u01a1i X\u00c2Y TH\u00c1P WOODY TOWER WD012 bao g\u1ed3m 48 thanh 6 m\u00e0u \u0111\u01b0\u1ee3c l\u00e0m t\u1eeb ch\u1ea5t li\u1ec7u g\u1ed7, c\u00f3 th\u1ec3 ch\u01a1i m\u1ed9t ho\u1eb7c nhi\u1ec1u ng\u01b0\u1eddi, m\u1ed7i l\u01b0\u1ee3t ng\u01b0\u1eddi ch\u01a1i r\u00fat ra m\u1ed9t thanh g\u1ed7 t\u1eeb t\u00f2a th\u00e1p v\u00e0 \u0111\u1eb7t thanh g\u1ed7 r\u00fat ra \u0111\u00f3 l\u00ean tr\u00ean \u0111\u1ec9nh m\u00e0 kh\u00f4ng l\u00e0m \u0111\u1ed5 th\u00e1p. Tr\u00f2 ch\u01a1i ti\u1ebfp t\u1ee5c nh\u01b0 v\u1eady, v\u1edbi m\u1ed7i l\u01b0\u1ee3t c\u1ee7a t\u1eebng ng\u01b0\u1eddi ch\u01a1i cho \u0111\u1ebfn khi th\u00e1p \u0111\u1ed5. M\u1ed7i thanh l\u00e0 m\u1ed9t h\u00ecnh h\u1ed9p ch\u1eef nh\u1eadt c\u00f3 k\u00edch th\u01b0\u1edbc 1,5cm x 2cm x 7,5cm ; \u0111\u01b0\u1ee3c \u0111\u1ef1ng v\u00e0o m\u1ed9t h\u1ed9p c\u00f3 k\u00edch th\u01b0\u1edbc 8cm x 8cm x 27cm . c) T\u00ednh th\u1ec3 t\u00edch c\u1ee7a h\u1ed9p \u0111\u1ef1ng 48 thanh g\u1ed7? d) H\u1ecfi th\u1ec3 t\u00edch c\u00e1c thanh g\u1ed7 m\u00e0u \u0111\u1ecf v\u00e0 t\u00edm trong h\u1ed9p chi\u1ebfm bao nhi\u00eau ph\u1ea7n tr\u0103m th\u1ec3 t\u00edch h\u1ed9p (C\u00e1c k\u1ebft qu\u1ea3 l\u00e0m tr\u00f2n \u0111\u1ebfn ch\u1eef s\u1ed1 th\u1eadp ph\u00e2n th\u1ee9 hai sau d\u1ea5u ph\u1ea9y) L\u1eddi gi\u1ea3i a) T\u00ednh th\u1ec3 t\u00edch c\u1ee7a h\u1ed9p \u0111\u1ef1ng 48 thanh g\u1ed7? ( )Th\u1ec3 t\u00edch c\u1ee7a h\u1ed9p \u0111\u1ef1ng 48 thanh g\u1ed7 l\u00e0: 8.8.27 = 1728 cm3 b) H\u1ecfi th\u1ec3 t\u00edch c\u00e1c thanh g\u1ed7 m\u00e0u \u0111\u1ecf v\u00e0 t\u00edm trong h\u1ed9p chi\u1ebfm bao nhi\u00eau ph\u1ea7n tr\u0103m th\u1ec3 t\u00edch h\u1ed9p (C\u00e1c k\u1ebft qu\u1ea3 l\u00e0m tr\u00f2n \u0111\u1ebfn ch\u1eef s\u1ed1 th\u1eadp ph\u00e2n th\u1ee9 hai sau d\u1ea5u ph\u1ea9y) ( )Th\u1ec3 t\u00edch m\u1ed9t thanh g\u1ed7 l\u00e0: 1,5.2.7,5 = 22,5 cm3 ( )T\u1ed5ng th\u1ec3 t\u00edch c\u00e1c kh\u1ed1i g\u1ed7 m\u00e0u \u0111\u1ecf v\u00e0 t\u00edm c\u00f3 trong h\u1ed9p l\u00e0: (48 : 6).2.22,5 = 360 cm3 V\u1eady th\u1ec3 t\u00edch c\u00e1c thanh g\u1ed7 m\u00e0u \u0111\u1ecf v\u00e0 t\u00edm trong h\u1ed9p chi\u1ebfm 360 .100 \uf0bb 20,83% th\u1ec3 t\u00edch 1728 h\u1ed9p C\u00e2u 8. (3 \u0111i\u1ec3m) Cho \u0111\u01b0\u1eddng tr\u00f2n (O) v\u00e0 \u0111i\u1ec3m A n\u1eb1m ngo\u00e0i (O) . T\u1eeb A v\u1ebd hai ti\u1ebfp tuy\u1ebfn AB, AC c\u1ee7a (O) ( B,C l\u00e0 hai ti\u1ebfp \u0111i\u1ec3m). G\u1ecdi H l\u00e0 giao \u0111i\u1ec3m c\u1ee7a AO v\u00e0 BC . Qua A v\u1ebd c\u00e1t tuy\u1ebfn ADE c\u1ee7a \u0111\u01b0\u1eddng tr\u00f2n (O) (D, E thu\u1ed9c (O)) sao cho \u0111\u01b0\u1eddng th\u1eb3ng AE c\u1eaft \u0111o\u1ea1n th\u1eb3ng HB t\u1ea1i I . G\u1ecdi M l\u00e0 trung \u0111i\u1ec3m d\u00e2y cung DE T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N 7","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH a) Ch\u1ee9ng minh: 5 \u0111i\u1ec3m A, B,O, M ,C c\u00f9ng thu\u1ed9c m\u1ed9t \u0111\u01b0\u1eddng tr\u00f2n; suy ra MA l\u00e0 tia ph\u00e2n gi\u00e1c g\u00f3c BMC (1\u0111) b) Ch\u1ee9ng minh: t\u1ee9 gi\u00e1c OHDE n\u1ed9i ti\u1ebfp (1\u0111) c) Tr\u00ean tia \u0111\u1ed1i c\u1ee7a tia HD l\u1ea5y \u0111i\u1ec3m F sao cho H l\u00e0 trung \u0111i\u1ec3m DF . Tia AO c\u1eaft \u0111\u01b0\u1eddng th\u1eb3ng EF t\u1ea1i K . Ch\u1ee9ng minh: IK \/ \/DF (1\u0111) L\u1eddi gi\u1ea3i B E M I D KO H A F C a) Ch\u1ee9ng minh: 5 \u0111i\u1ec3m A, B,O, M ,C c\u00f9ng thu\u1ed9c m\u1ed9t \u0111\u01b0\u1eddng tr\u00f2n; suy ra MA l\u00e0 tia ph\u00e2n gi\u00e1c g\u00f3c BMC (1\u0111) Ta c\u00f3: M l\u00e0 trung \u0111i\u1ec3m d\u00e2y cung DE(gt) \uf0de OM \u22a5 DE (quan h\u1ec7 \u0111\u01b0\u1eddng k\u00ednh d\u00e2y cung) \uf0de OMA = OBA = OCA = 90\uf0b0 \uf0de A, B,O, M ,C c\u00f9ng thu\u1ed9c m\u1ed9t \u0111\u01b0\u1eddng tr\u00f2n \u0111\u01b0\u1eddng k\u00ednh OA X\u00e9t \u0111\u01b0\u1eddng tr\u00f2n \u0111\u01b0\u1eddng k\u00ednh OA , ta c\u00f3: D\u00e2y cung AB = d\u00e2y cung AC (t\u00ednh ch\u1ea5t 2 ti\u1ebfp tuy\u1ebfn c\u1eaft nhau t\u1ea1i A) \uf0de AB = AC \uf0de BMA = CMA ( 2 g\u00f3c n\u1ed9i ti\u1ebfp ch\u1eafn AB = AC) \uf0de MA l\u00e0 tia ph\u00e2n gi\u00e1c g\u00f3c BMC b) Ch\u1ee9ng minh: t\u1ee9 gi\u00e1c OHDE n\u1ed9i ti\u1ebfp (1\u0111) Ta c\u00f3: \uf0ec AB = AC(cmt) \uf0ed\uf0eeOA = OC = R \uf0de OA l\u00e0 trung tr\u1ef1c BC 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 OA \u22a5 BC \u1edf H X\u00e9t \uf044OBA vu\u00f4ng t\u1ea1i B c\u00f3 AH l\u00e0 \u0111\u01b0\u1eddng cao: \uf0de AB2 = AH.AO (htl )(1) X\u00e9t \uf044ABD & \uf044AEB c\u00f3: BAD = EAB (g\u00f3c chung) ABD = AEB (g\u00f3c t\u1ea1o b\u1edfi ti\u1ebfp tuy\u1ebfn v\u00e0 d\u00e2y cung v\u1edbi g\u00f3c n\u1ed9i ti\u1ebfp c\u00f9ng ch\u1eafn BD ) \uf0de \uf044ABD \u223d \uf044AEB ( gg ) \uf0de AB = AD \uf0de AB2 = AE.AD (2) AE AB T\u1eeb (1) & (2) \uf0de AH.AO = AE.AD \uf0de AH = AD AE AO X\u00e9t \uf044AHD & \uf044AEO c\u00f3: \uf0ec\uf0efHAD = EAO (chung ) \uf0ed AH \uf0ef\uf0ee AE = AD (cmt ) AO \uf0de \uf044AHD \u223d \uf044AEO (cgc) \uf0de AHD = AEO (3) X\u00e9t t\u1ee9 gi\u00e1c OHDE ta c\u00f3: AHD = AEO (cmt ) \uf0de t\u1ee9 gi\u00e1c OHDE n\u1ed9i ti\u1ebfp (g\u00f3c ngo\u00e0i = g\u00f3c trong \u0111\u1ed1i di\u1ec7n) c) Ch\u1ee9ng minh: IK \/ \/DF (1\u0111) X\u00e9t \uf044ODE ta c\u00f3: OD = OE = R \uf0de \uf044ODE c\u00e2n t\u1ea1i O \uf0de ODE = OED (4) Do t\u1ee9 gi\u00e1c OHDE n\u1ed9i ti\u1ebfp n\u00ean ODE = OHE (5) T\u1eeb (3),(4),(5) \uf0de AHD = OHE M\u00e0 AHD = FHK (\u0111\u1ed1i \u0111\u1ec9nh) N\u00ean OHE = FHK \uf0de HK l\u00e0 ph\u00e2n gi\u00e1c c\u1ee7a EHF \uf0de KF = HF (t\u00ednh ch\u1ea5t \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c trong \uf044EHF ) KE HE M\u00e0 HF = HD (do H l\u00e0 trung \u0111i\u1ec3m DF ) N\u00ean KF = HD (6) KE HE Ta c\u00f3: 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 \uf0ec AHD = OHE (cmt ) \uf0ef\uf0ef \uf0ed AHD + IHD = BHD = 90\uf0b0 \uf0ef\uf0ef\uf0eeOHE + IHE = BHO = 90\uf0b0 \uf0de IHD = IHE \uf0de HI l\u00e0 ph\u00e2n gi\u00e1c c\u1ee7a EHD \uf0de ID = HD (7) (t\u00ednh ch\u1ea5t \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c trong \uf044EHD ) IE HE T\u1eeb (6),(7) \uf0de KF = ID KE IE X\u00e9t \uf044EFD ta c\u00f3: KF = ID (cmt ) \uf0de IK \/ \/DF (\u0111\u1ecbnh l\u00fd Talet \u0111\u1ea3o) KE IE ----H\u1ebeT--- T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N 10","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH S\u00d4\u00db GD&\u00d1T TP HO\u00c0 CH\u00cd MINH \u00d1E\u00c0 THAM KHA\u00dbO TUYE\u00c5N SINH 10 PHO\u00d8NG G\u00d1&\u00d1T QUA\u00c4N G\u00f2 v\u1ea5p NA\u00caM HO\u00cfC: 2021 - 2022 \u0110\u1ec0 THAM KH\u1ea2O M\u00d4N: TO\u00c1N 9 \u0110\u00ea thi g\u1ed3m 8 c\u00e2u ho\u0309i t\u01b0\u0323 lu\u00e2\u0323n. M\u00c3 \u0110\u1ec0: Qu\u1eadn G\u00f2 V\u1ea5p - 2 Th\u01a1\u0300i gian: 120 phu\u0301t (kh\u00f4ng k\u00ea\u0309 th\u01a1\u0300i gian pha\u0301t \u0111\u00ea\u0300) C\u00e2u 1. (1,5 \u0111i\u00ea\u0309m): Cho parabol (P) : y = x2 v\u00e0 \u0111\u01b0\u1eddng th\u1eb3ng (d) : y = \u2212x + 2 a) V\u1ebd (P) v\u00e0 (d) tr\u00ean c\u00f9ng h\u1ec7 tr\u1ee5c to\u1ea1 \u0111\u1ed9. b) T\u00ecm to\u1ea1 \u0111\u1ed9 giao \u0111i\u1ec3m c\u1ee7a (P) v\u00e0 (d) b\u1eb1ng ph\u00e9p t\u00ednh. C\u00e2u 2. (1 \u0111i\u00ea\u0309m) Cho ph\u01b0\u01a1ng tr\u00ecnh 3x2 + 5x \u2212 1 = 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 = (3x1 \u2212 2x2 )(3x2 \u2212 2x1 ) \u2212 2023 . C\u00e2u 3. (0,75 \u0111i\u00ea\u0309m) \u0110\u1ec3 \u01b0\u1edbc t\u00ednh chi\u1ec1u cao t\u1ed1i \u0111a c\u1ee7a tr\u1ebb em khi \u0111\u1ea1t \u0111\u1ebfn \u0111\u1ed9 tr\u01b0\u1edfng th\u00e0nh, ho\u00e0n to\u00e0n c\u00f3 th\u1ec3 d\u1ef1a v\u00e0o chi\u1ec1u cao c\u1ee7a b\u1ed1 m\u1eb9. C\u00e1ch t\u00ednh chi\u1ec1u cao c\u1ee7a con theo b\u1ed1 m\u1eb9 \u0111\u01b0\u1ee3c c\u00e1c chuy\u00ean gia \u0111\u00e1nh gi\u00e1 cao b\u1edfi th\u1ef1c t\u1ebf, s\u1ef1 di truy\u1ec1n c\u00e1c th\u1ebf h\u1ec7 c\u00f3 \u1ea3nh h\u01b0\u1edfng nh\u1ea5t \u0111\u1ecbnh \u0111\u1ebfn chi\u1ec1u cao c\u1ee7a tr\u1ebb. Ta c\u00f3 c\u00f4ng th\u1ee9c t\u00ednh nh\u01b0 sau: C = (B + M + 13A) : 2 . Trong \u0111\u00f3: C l\u00e0 chi\u1ec1u cao c\u1ee7a ng\u01b0\u1eddi con (cm) B l\u00e0 chi\u1ec1u cao c\u1ee7a ng\u01b0\u1eddi b\u1ed1 (cm) M l\u00e0 chi\u1ec1u cao c\u1ee7a ng\u01b0\u1eddi m\u1eb9 (cm) A = 1 khi ng\u01b0\u1eddi con c\u00f3 gi\u1edbi t\u00ednh l\u00e0 Nam A = \u22121 khi ng\u01b0\u1eddi con c\u00f3 gi\u1edbi t\u00ednh l\u00e0 N\u1eef a) Em h\u00e3y d\u00f9ng c\u00f4ng th\u1ee9c tr\u00ean \u0111\u1ec3 t\u00ecm chi\u1ec1u cao t\u1ed1i \u0111a c\u1ee7a b\u1ea1n Nam (gi\u1edbi t\u00ednh l\u00e0 nam) bi\u1ebft Ba c\u1ee7a b\u1ea1n Nam c\u00f3 chi\u1ec1u cao l\u00e0 175 cm v\u00e0 M\u1eb9 c\u1ee7a b\u1ea1n Nam c\u00f3 chi\u1ec1u cao l\u00e0 168 cm . b) B\u1ea1n H\u01b0\u01a1ng (gi\u1edbi t\u00ednh l\u00e0 n\u1eef) c\u00f3 chi\u1ec1u cao l\u00e0 164 cm . Em h\u00e3y t\u00ednh xem chi\u1ec1u cao t\u1ed1i \u0111a c\u1ee7a M\u1eb9 b\u1ea1n H\u01b0\u01a1ng khi bi\u1ebft chi\u1ec1u cao c\u1ee7a Ba b\u1ea1n H\u01b0\u01a1ng l\u00e0 180 cm . C\u00e2u 4. (1,0 \u0111i\u00ea\u0309m) M\u1ed9t h\u00e3ng h\u00e0ng kh\u00f4ng quy \u0111\u1ecbnh x\u1eed ph\u1ea1t h\u00e0nh l\u00fd k\u00ed g\u1eedi v\u01b0\u1ee3t qu\u00e1 quy \u0111\u1ecbnh mi\u1ec5n ph\u00ed (h\u00e0nh l\u00fd qu\u00e1 c\u01b0\u1edbc). C\u1ee9 v\u01b0\u1ee3t qu\u00e1 x(kg) h\u00e0nh l\u00fd th\u00ec kh\u00e1ch h\u00e0ng ph\u1ea3i tr\u1ea3 ti\u1ec1n ph\u1ea1t y(USD) . 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 c\u00f3 \u0111\u1ed3 th\u1ecb nh\u01b0 h\u00ecnh b\u00ean: a) X\u00e1c \u0111\u1ecbnh c\u00e1c h\u1ec7 s\u1ed1 a v\u00e0 b b) H\u00e3y t\u00ednh s\u1ed1 ti\u1ec1n ph\u1ea1t c\u1ee7a m\u1ed9t h\u00e0nh kh\u00e1ch c\u00f3 20 kg h\u00e0nh l\u00fd qu\u00e1 c\u01b0\u1edbc. C\u00e2u 5. (0,75 \u0111i\u00ea\u0309m) \u0110\u1ea7u m\u1ed7i th\u00e1ng \u00f4ng M\u1ea1nh g\u1eedi v\u00e0o ng\u00e2n h\u00e0ng 2000000 \u0111\u1ed3ng v\u1edbi l\u00e3i su\u1ea5t 0,65% \/th\u00e1ng v\u00e0 kh\u00f4ng r\u00fat g\u1ed1c, l\u00e3i th\u00e1ng tr\u01b0\u1edbc. Sau 3 th\u00e1ng th\u00ec s\u1ed1 ti\u1ec1n \u00f4ng M\u1ea1nh nh\u1eadn \u0111\u01b0\u1ee3c c\u1ea3 g\u1ed1c l\u1eabn l\u00e3i (sau khi ng\u00e2n h\u00e0ng \u0111\u00e3 t\u00ednh l\u00e3i th\u00e1ng cu\u1ed1i c\u00f9ng) l\u00e0 bao nhi\u00eau? T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N 1","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH C\u00e2u 6. (1 \u0111i\u00ea\u0309m) \u0110\u1ec3 \u0111o chi\u1ec1u cao m\u1ed9t ng\u1ecdn \u0111\u1ed3i, ng\u01b0\u1eddi ta \u0111\u1eb7t gi\u00e1c k\u1ebf t\u1ea1i hai v\u1ecb tr\u00ed l\u00e0 A (ch\u00e2n to\u00e0 nh\u00e0) v\u00e0 B (s\u00e2n th\u01b0\u1ee3ng to\u00e0 nh\u00e0). Th\u00f4ng qua gi\u00e1c k\u1ebf ng\u01b0\u1eddi ta \u0111o \u0111\u01b0\u1ee3c CAH = 45\uf0b0 v\u00e0 CBE = 30\uf0b0 . T\u00ednh \u0111\u1ed9 cao c\u1ee7a ng\u1ecdn \u0111\u1ed3i? Bi\u1ebft to\u00e0 nh\u00e0 cao 50 m . (K\u1ebft qu\u1ea3 l\u00e0m tr\u00f2n \u0111\u1ebfn h\u00e0ng \u0111\u01a1n v\u1ecb). C\u00e2u 7. (0,75 \u0111i\u00ea\u0309m) B\u00e1nh xe \u0111\u1ea1p b\u01a1m c\u0103ng c\u00f3 \u0111\u01b0\u1eddng k\u00ednh l\u00e0 73 cm . a) H\u1ecfi xe \u0111\u1ea1p \u0111i \u0111\u01b0\u1ee3c bao nhi\u00eau {km} n\u1ebfu b\u00e1nh xe quay \u0111\u01b0\u1ee3c 1000 v\u00f2ng? K\u1ebft qu\u1ea3 l\u00e0m tr\u00f2n \u0111\u1ebfn ch\u1eef s\u1ed1 th\u1eadp ph\u00e2n th\u1ee9 ba. b) H\u1ecfi b\u00e1nh xe quay \u0111\u01b0\u1ee3c bao nhi\u00eau v\u00f2ng khi xe \u0111i \u0111\u01b0\u1ee3c 4,64km ? K\u1ebft qu\u1ea3 l\u00e0m tr\u00f2n \u0111\u1ebfn h\u00e0ng \u0111\u01a1n v\u1ecb. C\u00e2u 8. Cho \uf044ABC nh\u1ecdn ( AB \uf03c AC) , n\u1ed9i ti\u1ebfp \u0111\u01b0\u1eddng tr\u00f2n (O; R) c\u00f3 hai \u0111\u01b0\u1eddng cao BE , CF c\u1eaft nhau t\u1ea1i H v\u00e0 c\u1eaft \u0111\u01b0\u1eddng tr\u00f2n (O) l\u1ea7n l\u01b0\u1ee3t t\u1ea1i Y v\u00e0 X . 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: 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 H\u01af\u1edaNG D\u1eaaN GI\u1ea2I C\u00e2u 1. (1,5 \u0111i\u00ea\u0309m): Cho parabol (P) : y = x2 v\u00e0 \u0111\u01b0\u1eddng th\u1eb3ng (d) : y = \u2212x + 2 a) V\u1ebd (P) v\u00e0 (d) tr\u00ean c\u00f9ng h\u1ec7 tr\u1ee5c to\u1ea1 \u0111\u1ed9. b) T\u00ecm to\u1ea1 \u0111\u1ed9 giao \u0111i\u1ec3m c\u1ee7a (P) v\u00e0 (d) b\u1eb1ng ph\u00e9p t\u00ednh. L\u1eddi gi\u1ea3i a) V\u1ebd \u0111\u1ed3 th\u1ecb (P) v\u00e0 (d) tr\u00ean c\u00f9ng h\u1ec7 tr\u1ee5c t\u1ecda \u0111\u1ed9. BGT: x \u22122 \u22121 0 1 2 y = x2 4 1 0 1 4 x \u22122 1 2 y = \u2212x + 2 4 1 0 a) 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 (n\u1ebfu c\u00f3) c\u1ee7a (P) v\u00e0 (d) : x2 = \u2212x + 2 \uf0db x2 + x \u2212 2 = 0 \uf0db \uf0e9x = \u22122 \uf0ea\uf0ebx = 1 Thay x = \u22122 v\u00e0o y = x2 , ta \u0111\u01b0\u1ee3c: y = (\u22122)2 \uf0db y = 4 Thay x = 1 v\u00e0o y = x2 , ta \u0111\u01b0\u1ee3c: y = 12 \uf0db y = 1 V\u1eady t\u1ecda \u0111\u1ed9 giao \u0111i\u1ec3m c\u1ee7a (P) v\u00e0 (d) l\u00e0 A(\u22122; 4) v\u00e0 B(1;1) . C\u00e2u 2. (1 \u0111i\u00ea\u0309m) Cho ph\u01b0\u01a1ng tr\u00ecnh 3x2 + 5x \u2212 1 = 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 = (3x1 \u2212 2x2 )(3x2 \u2212 2x1 ) \u2212 2023 . L\u1eddi gi\u1ea3i Ta c\u00f3: (1) 3x2 + 5x \u2212 1 = 0 (a = 3,b = 5,c = \u22121) V\u00ec ph\u01b0\u01a1ng tr\u00ecnh (1) c\u00f3 hai nghi\u1ec7m x` ,x2 1 \u00c1p d\u1ee5ng \u0111\u1ecbnh l\u00fd Vi \u2013 \u00e9t, ta c\u00f3: 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 S = x1 + x2 = \u2212b = \u22125 a 3 P = x1.x2 = c = \u22121 a 3 Ta c\u00f3: A = (3x1 \u2212 2x2 )(3x2 \u2212 2x1 ) \u2212 2023 = 9x1.x2 \u2212 6x12 \u2212 6x22 + 4x1.x2 \u2212 2023 ( )= 13x1.x2 \u2212 6 x12 + x22 \u2212 2023 ( )= \u22121 \u2212 \uf0e9 2 \uf0f9 13. 3 6 \uf0ea\uf0eb x1 + x2 \u2212 2x1.x2 \uf0fa\uf0fb = \u221213 \u2212 6 \uf0e9\uf0ea\uf0ea\uf0eb\uf0e6\uf0e7\uf0e8 \u22125 \uf0f62 \u2212 2. \u22121 \uf0f9\uf0fa \u2212 2023 = \u22122048 3 3 \uf0f7 3 \uf0fa\uf0fb \uf0f8 V\u1eady A = \u22122048 . C\u00e2u 3. (0,75 \u0111i\u00ea\u0309m) \u0110\u1ec3 \u01b0\u1edbc t\u00ednh chi\u1ec1u cao t\u1ed1i \u0111a c\u1ee7a tr\u1ebb em khi \u0111\u1ea1t \u0111\u1ebfn \u0111\u1ed9 tr\u01b0\u1edfng th\u00e0nh, ho\u00e0n to\u00e0n c\u00f3 th\u1ec3 d\u1ef1a v\u00e0o chi\u1ec1u cao c\u1ee7a b\u1ed1 m\u1eb9. C\u00e1ch t\u00ednh chi\u1ec1u cao c\u1ee7a con theo b\u1ed1 m\u1eb9 \u0111\u01b0\u1ee3c c\u00e1c chuy\u00ean gia \u0111\u00e1nh gi\u00e1 cao b\u1edfi th\u1ef1c t\u1ebf, s\u1ef1 di truy\u1ec1n c\u00e1c th\u1ebf h\u1ec7 c\u00f3 \u1ea3nh h\u01b0\u1edfng nh\u1ea5t \u0111\u1ecbnh \u0111\u1ebfn chi\u1ec1u cao c\u1ee7a tr\u1ebb. Ta c\u00f3 c\u00f4ng th\u1ee9c t\u00ednh nh\u01b0 sau: C = (B + M + 13A) : 2 . Trong \u0111\u00f3: C l\u00e0 chi\u1ec1u cao c\u1ee7a ng\u01b0\u1eddi con (cm) B l\u00e0 chi\u1ec1u cao c\u1ee7a ng\u01b0\u1eddi b\u1ed1 (cm) M l\u00e0 chi\u1ec1u cao c\u1ee7a ng\u01b0\u1eddi m\u1eb9 (cm) A = 1 khi ng\u01b0\u1eddi con c\u00f3 gi\u1edbi t\u00ednh l\u00e0 Nam A = \u22121 khi ng\u01b0\u1eddi con c\u00f3 gi\u1edbi t\u00ednh l\u00e0 N\u1eef a) Em h\u00e3y d\u00f9ng c\u00f4ng th\u1ee9c tr\u00ean \u0111\u1ec3 t\u00ecm chi\u1ec1u cao t\u1ed1i \u0111a c\u1ee7a b\u1ea1n Nam (gi\u1edbi t\u00ednh l\u00e0 nam) bi\u1ebft Ba c\u1ee7a b\u1ea1n Nam c\u00f3 chi\u1ec1u cao l\u00e0 175cm v\u00e0 M\u1eb9 c\u1ee7a b\u1ea1n Nam c\u00f3 chi\u1ec1u cao l\u00e0 168 cm . b) B\u1ea1n H\u01b0\u01a1ng (gi\u1edbi t\u00ednh l\u00e0 n\u1eef) c\u00f3 chi\u1ec1u cao l\u00e0 164 cm . Em h\u00e3y t\u00ednh xem chi\u1ec1u cao t\u1ed1i \u0111a c\u1ee7a M\u1eb9 b\u1ea1n H\u01b0\u01a1ng khi bi\u1ebft chi\u1ec1u cao c\u1ee7a Ba b\u1ea1n H\u01b0\u01a1ng l\u00e0 180 cm . L\u1eddi gi\u1ea3i a) Em h\u00e3y d\u00f9ng c\u00f4ng th\u1ee9c tr\u00ean \u0111\u1ec3 t\u00ecm chi\u1ec1u cao t\u1ed1i \u0111a c\u1ee7a b\u1ea1n Nam (gi\u1edbi t\u00ednh l\u00e0 nam) bi\u1ebft Ba c\u1ee7a b\u1ea1n Nam c\u00f3 chi\u1ec1u cao l\u00e0 175cm v\u00e0 M\u1eb9 c\u1ee7a b\u1ea1n Nam c\u00f3 chi\u1ec1u cao l\u00e0 168 cm . Chi\u1ec1u cao t\u1ed1i \u0111a c\u1ee7a Nam l\u00e0: 175 + 168 + 13.1 = 178(cm) 2 V\u1eady chi\u1ec1u cao t\u1ed1i \u0111a c\u1ee7a Nam l\u00e0: 178cm b) B\u1ea1n H\u01b0\u01a1ng (gi\u1edbi t\u00ednh l\u00e0 n\u1eef) c\u00f3 chi\u1ec1u cao l\u00e0 164 cm . Em h\u00e3y t\u00ednh xem chi\u1ec1u cao t\u1ed1i \u0111a c\u1ee7a M\u1eb9 b\u1ea1n H\u01b0\u01a1ng khi bi\u1ebft chi\u1ec1u cao c\u1ee7a Ba b\u1ea1n H\u01b0\u01a1ng l\u00e0 180 cm . Chi\u1ec1u cao t\u1ed1i \u0111a c\u1ee7a m\u1eb9 b\u1ea1n H\u01b0\u01a1ng l\u00e0: 1809 + M + 13.(\u22121) 164 = 2 \uf0db 180 + M \u2212 13 = 328 \uf0db M + 167 = 328 \uf0db M = 161(cm) V\u1eady chi\u1ec1u cao t\u1ed1i \u0111a c\u1ee7a m\u1eb9 b\u1ea1n H\u01b0\u01a1ng l\u00e0: 161cm . T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N 4","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH C\u00e2u 4. (1,0 \u0111i\u00ea\u0309m) M\u1ed9t h\u00e3ng h\u00e0ng kh\u00f4ng quy \u0111\u1ecbnh x\u1eed ph\u1ea1t h\u00e0nh l\u00fd k\u00ed g\u1eedi v\u01b0\u1ee3t qu\u00e1 quy \u0111\u1ecbnh mi\u1ec5n ph\u00ed (h\u00e0nh l\u00fd qu\u00e1 c\u01b0\u1edbc). C\u1ee9 v\u01b0\u1ee3t qu\u00e1 x(kg) h\u00e0nh l\u00fd th\u00ec kh\u00e1ch h\u00e0ng ph\u1ea3i tr\u1ea3 ti\u1ec1n ph\u1ea1t y(USD) . 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 c\u00f3 \u0111\u1ed3 th\u1ecb nh\u01b0 h\u00ecnh b\u00ean: a) X\u00e1c \u0111\u1ecbnh c\u00e1c h\u1ec7 s\u1ed1 a v\u00e0 b b) H\u00e3y t\u00ednh s\u1ed1 ti\u1ec1n ph\u1ea1t c\u1ee7a m\u1ed9t h\u00e0nh kh\u00e1ch c\u00f3 20 kg h\u00e0nh l\u00fd qu\u00e1 c\u01b0\u1edbc. L\u1eddi gi\u1ea3i a) X\u00e1c \u0111\u1ecbnh c\u00e1c h\u1ec7 s\u1ed1 a v\u00e0 b . Theo \u0111\u1ec1 b\u00e0i, ta c\u00f3: V\u1edbi \uf0ecx = 0 \uf0de 20 = 0.a + b . (1) \uf0ed\uf0eey = 20 V\u1edbi \uf0ecx = 12,5 \uf0de 30 = 12,5.a + b . (2) \uf0ed\uf0eey = 30 T\u1eeb (1) v\u00e0 (2) ta c\u00f3 h\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh: \uf0ec0a + b = 20 30 \uf0db \uf0ec\uf0efa = 4 \uf0ed\uf0ee12,5a + b = \uf0ed = 5. \uf0ef\uf0eeb 20 V\u1eady: a = 4 , b = 20 v\u00e0 y = 4 x + 20 . 55 b) H\u00e3y t\u00ednh s\u1ed1 ti\u1ec1n ph\u1ea1t c\u1ee7a m\u1ed9t h\u00e0nh kh\u00e1ch c\u00f3 20kg h\u00e0nh l\u00fd qu\u00e1 c\u01b0\u1edbc. V\u00ec m\u1ed9t h\u00e0nh kh\u00e1ch c\u00f3 20kg h\u00e0nh l\u00fd qu\u00e1 c\u01b0\u1edbc n\u00ean ta thay x = 20 v\u00e0o y = 4 x + 20 , ta \u0111\u01b0\u1ee3c: 5 y = 4 .20 + 20 \uf0db y = 36 5 V\u1eady s\u1ed1 ti\u1ec1n ph\u1ea1t c\u1ee7a m\u1ed9t h\u00e0nh kh\u00e1ch c\u00f3 20kg h\u00e0nh l\u00fd qu\u00e1 c\u01b0\u1edbc l\u00e0 36USD . C\u00e2u 5. (0,75 \u0111i\u00ea\u0309m) \u0110\u1ea7u m\u1ed7i th\u00e1ng \u00f4ng M\u1ea1nh g\u1eedi v\u00e0o ng\u00e2n h\u00e0ng 2000000 \u0111\u1ed3ng v\u1edbi l\u00e3i su\u1ea5t 0,65% \/th\u00e1ng v\u00e0 kh\u00f4ng r\u00fat g\u1ed1c, l\u00e3i th\u00e1ng tr\u01b0\u1edbc. Sau 3 th\u00e1ng th\u00ec s\u1ed1 ti\u1ec1n \u00f4ng M\u1ea1nh nh\u1eadn \u0111\u01b0\u1ee3c c\u1ea3 g\u1ed1c l\u1eabn l\u00e3i (sau khi ng\u00e2n h\u00e0ng \u0111\u00e3 t\u00ednh l\u00e3i th\u00e1ng cu\u1ed1i c\u00f9ng) l\u00e0 bao nhi\u00eau? L\u1eddi gi\u1ea3i S\u1ed1 ti\u1ec1n l\u00e3i \u00f4ng M\u1ea1nh nh\u1eadn \u0111\u01b0\u1ee3c sau ba th\u00e1ng l\u00e0: 2000000.(1+ 0, 65%)3 = 2 039 254 049 (\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 \u0111i\u00ea\u0309m) \u0110\u1ec3 \u0111o chi\u1ec1u cao m\u1ed9t ng\u1ecdn \u0111\u1ed3i, ng\u01b0\u1eddi ta \u0111\u1eb7t gi\u00e1c k\u1ebf t\u1ea1i hai v\u1ecb tr\u00ed l\u00e0 A (ch\u00e2n to\u00e0 nh\u00e0) v\u00e0 B (s\u00e2n th\u01b0\u1ee3ng to\u00e0 nh\u00e0). Th\u00f4ng qua gi\u00e1c k\u1ebf ng\u01b0\u1eddi ta \u0111o \u0111\u01b0\u1ee3c CAH = 45\uf0b0 v\u00e0 CBE = 30\uf0b0 . T\u00ednh \u0111\u1ed9 cao c\u1ee7a ng\u1ecdn \u0111\u1ed3i? Bi\u1ebft to\u00e0 nh\u00e0 cao 50 m . (K\u1ebft qu\u1ea3 l\u00e0m tr\u00f2n \u0111\u1ebfn h\u00e0ng \u0111\u01a1n v\u1ecb) . L\u1eddi gi\u1ea3i X\u00e9t t\u1ee9 gi\u00e1c ABEH , ta c\u00f3: \uf0ec\uf0ef\uf0ef\uf0edBBAEHH = 90\uf0b0 ( gt ) = 90\uf0b0 ( gt ) \uf0ef = 90\uf0b0( gt ) \uf0ef\uf0ee AHE \uf0de T\u1ee9 gi\u00e1c ABEH l\u00e0 h\u00ecnh ch\u1eef nh\u1eadt (t\u1ee9 gi\u00e1c c\u00f3 3 g\u00f3c vu\u00f4ng) \uf0de \uf0ecBE = AH \uf0ed\uf0ee AB = EH M\u00e0 AB = 50m ( AB l\u00e0 chi\u1ec1u cao t\u00f2a nh\u00e0 b\u1eb1ng 50m ) N\u00ean EH = 50m X\u00e9t \uf044ACH vu\u00f4ng t\u1ea1i H , ta c\u00f3: CAH = 45\uf0b0(gt) \uf0de \uf044ACH vu\u00f4ng c\u00e2n t\u1ea1i H \uf0de AH = CH M\u00e0 AH = BE(cmt) N\u00ean CH = BE X\u00e9t \uf044BEC vu\u00f4ng t\u1ea1i E , ta c\u00f3: tanCBE = FC (t\u1ec9 s\u1ed1 l\u01b0\u1ee3ng gi\u00e1c) BE ( )\uf0de tan 30\uf0b0 = FC CBE = 30\uf0b0,BE = CH CH \uf0de 1 = EC \uf0de CH = EC 3 CH 3 1 \u00c1p d\u1ee5ng t\u00ednh ch\u1ea5t d\u00e3y t\u1ec9 s\u1ed1 b\u1eb1ng nhau, ta c\u00f3: CH = EC = CH \u2212 EC = EH = 50 (CH = CE + EH,E \uf0ceCH,EH = 50m) 31 5 \u22121 3 \u22121 3 \u22121 \uf0de CH = 50 \uf0de CH = 50 3 \uf0de CH \uf0bb 118(m) 3 3 \u22121 3 \u22121 V\u1eady chi\u1ec1u cao c\u1ee7a ng\u1ecdn \u0111\u1ed3i kho\u1ea3ng 118m 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\u00ea\u0309m) B\u00e1nh xe \u0111\u1ea1p b\u01a1m c\u0103ng c\u00f3 \u0111\u01b0\u1eddng k\u00ednh l\u00e0 73cm . a) H\u1ecfi xe \u0111\u1ea1p \u0111i \u0111\u01b0\u1ee3c bao nhi\u00eau {km} n\u1ebfu b\u00e1nh xe quay \u0111\u01b0\u1ee3c 1000 v\u00f2ng? K\u1ebft qu\u1ea3 l\u00e0m tr\u00f2n \u0111\u1ebfn ch\u1eef s\u1ed1 th\u1eadp ph\u00e2n th\u1ee9 ba. b) H\u1ecfi b\u00e1nh xe quay \u0111\u01b0\u1ee3c bao nhi\u00eau v\u00f2ng khi xe \u0111i \u0111\u01b0\u1ee3c 4,64km ? K\u1ebft qu\u1ea3 l\u00e0m tr\u00f2n \u0111\u1ebfn h\u00e0ng \u0111\u01a1n v\u1ecb. L\u1eddi gi\u1ea3i a) Xe \u0111\u1ea1p \u0111i \u0111\u01b0\u1ee3c bao nhi\u00eau km n\u1ebfu b\u00e1nh xe quay 1000 v\u00f2ng? Chu vi c\u1ee7a b\u00e1nh xe \u0111\u1ea1p l\u00e0: C = 73.\uf070 (cm) = 0,00073\uf070 (km) . Qu\u00e3ng \u0111\u01b0\u1eddng xe m\u00e1y \u0111i \u0111\u01b0\u1ee3c n\u1ebfu b\u00e1nh xe quay 1000 v\u00f2ng: L = N.C = 1000.0,00073\uf070 \uf0bb 2,293(km) . V\u1eady xe \u0111\u1ea1p \u0111i \u0111\u01b0\u1ee3c kho\u1ea3ng 2,293km n\u1ebfu b\u00e1nh xe quay 1000 v\u00f2ng b) B\u00e1nh xe quay \u0111\u01b0\u1ee3c bao nhi\u00eau v\u00f2ng khi xe \u0111i 4,64km ? S\u1ed1 v\u00f2ng b\u00e1nh xe quay \u0111\u01b0\u1ee3c khi xe \u0111i 4,64m l\u00e0: N = L = 4,64 \uf0bb 2023 (v\u00f2ng). C 0,00073\uf070 V\u1eady b\u00e1nh xe quay \u0111\u01b0\u1ee3c kho\u1ea3ng 2023 v\u00f2ng khi xe \u0111i \u0111\u01b0\u1ee3c 4,64km . C\u00e2u 8. Cho \uf044ABC nh\u1ecdn ( AB \uf03c AC) , n\u1ed9i ti\u1ebfp \u0111\u01b0\u1eddng tr\u00f2n (O; R) c\u00f3 hai \u0111\u01b0\u1eddng cao BE , CF c\u1eaft nhau t\u1ea1i H v\u00e0 c\u1eaft \u0111\u01b0\u1eddng tr\u00f2n (O) l\u1ea7n l\u01b0\u1ee3t t\u1ea1i Y v\u00e0 X . 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: 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. L\u1eddi gi\u1ea3i T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N 7","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH A Y P XF E J O IH C B MK a) Ch\u1ee9ng minh: t\u1ee9 gi\u00e1c APFE n\u1ed9i ti\u1ebfp \u0111\u01b0\u1eddng tr\u00f2n. X\u00e9t (O) , ta c\u00f3: HPA l\u00e0 g\u00f3c n\u1ed9i ti\u1ebfp ch\u1eafn n\u1eeda \u0111\u01b0\u1eddng tr\u00f2n \u0111\u01b0\u1eddng k\u00ednh AK (gt) \uf0de HPA = 90\uf0b0 X\u00e9t t\u1ee9 gi\u00e1c APHE , ta c\u00f3: \uf0ec\uf0efAPH = 90\uf0b0(cmt) \uf0ed \uf0ef\uf0ee AEH = 90\uf0b0( BE \u22a5 AC ta\u00efi E; H \uf0ce BE) \uf0de APH + AEH = 180\uf0b0 \uf0de T\u1ee9 gi\u00e1c APHE n\u1ed9i ti\u1ebfp \u0111\u01b0\u1eddng tr\u00f2n \u0111\u01b0\u1eddng k\u00ednh AH (t\u1ee9 gi\u00e1c c\u00f3 2 g\u00f3c \u0111\u1ed1i b\u00f9 nhau) \uf0de A , P , H , E n\u1ed9i ti\u1ebfp \u0111\u01b0\u1eddng tr\u00f2n \u0111\u01b0\u1eddng k\u00ednh AH (1) X\u00e9t t\u1ee9 gi\u00e1c AFHE , ta c\u00f3: \uf0ec\uf0efAFH = 90\uf0b0(CF \u22a5 AB ta\u00efi F; H \uf0ceCF ) \uf0ed \uf0ef\uf0ee AEH = 90\uf0b0( BE \u22a5 AC ta\u00efi E; H \uf0ce BE) \uf0de AFH + AEH = 90\uf0b0 + 90\uf0b0 = 180\uf0b0 \uf0de T\u1ee9 gi\u00e1c AFHE n\u1ed9i ti\u1ebfp \u0111\u01b0\u1eddng tr\u00f2n \u0111\u01b0\u1eddng k\u00ednh AH (t\u1ee9 gi\u00e1c c\u00f3 hai g\u00f3c \u0111\u1ed1i b\u00f9 nhau) \uf0de A , F , H , E thu\u1ed9c \u0111\u01b0\u1eddng tr\u00f2n \u0111\u01b0\u1eddng k\u00ednh AH (2) T\u1eeb (1) v\u00e0 (2) suy ra: A, P, F, E thu\u1ed9c \u0111\u01b0\u1eddng tr\u00f2n \u0111\u01b0\u1eddng k\u00ednh AH \uf0de T\u1ee9 gi\u00e1c APFE n\u1ed9i ti\u1ebfp \u0111\u01b0\u1eddng tr\u00f2n \u0111\u01b0\u1eddng k\u00ednh AH b) Ch\u1ee9ng minh: PB.PE = PC.PF . Ta c\u00f3: BPC = BAC (2 g\u00f3c n\u1ed9i ti\u1ebfp c\u00f9ng ch\u1eafn BC c\u1ee7a (O) ) M\u00e0 BAC = FPE (2 g\u00f3c n\u1ed9i ti\u1ebfp c\u00f9ng ch\u1eafn EF c\u1ee7a \u0111\u01b0\u1eddng tr\u00f2n \u0111\u01b0\u1eddng k\u00ednh AH) ( )N\u00ean BPC = FPE = BAC Ta c\u00f3: BCP = BAP (2 g\u00f3c n\u1ed9i ti\u1ebfp c\u00f9ng ch\u1eafn BP c\u1ee7a (O) ) 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 M\u00e0 BAP = FEP (2 g\u00f3c n\u1ed9i ti\u1ebfp c\u00f9ng ch\u1eafn PF c\u1ee7a \u0111\u01b0\u1eddng tr\u00f2n \u0111\u01b0\u1eddng k\u00ednh AH) ( )N\u00ean BCP = FEP = BAP X\u00e9t \uf044BPC v\u00e0 \uf044FPE , ta c\u00f3: \uf0ec\uf0efBPC = FPE(cmt) \uf0ed \uf0ef\uf0eeBCP = FPE ( cmt ) \uf0de \uf044BPC \u223d \uf044FPE(g.g) \uf0de PB = PC (ts\u0111d) PF PE \uf0de PB.PE = PC.PF c) Ch\u1ee9ng minh: H , I , J th\u1eb3ng h\u00e0ng. X\u00e9t t\u1ee9 gi\u00e1c BFEC , ta c\u00f3: \uf0ec\uf0efBFC = 90\uf0b0(CF \u22a5 AB ta\u00efi F ) \uf0ed \uf0ef\uf0eeBEC = 90\uf0b0 ( BE \u22a5 AC ta\u00efi E ) \uf0de BFC = BEC (= 90\uf0b0) \uf0de T\u1ee9 gi\u00e1c BFEC n\u1ed9i ti\u1ebfp (t\u1ee9 gi\u00e1c c\u00f3 hai \u0111\u1ec9nh li\u00ean ti\u1ebfp c\u00f9ng nh\u00ecn c\u1ea1nh d\u01b0\u1edbi hai g\u00f3c b\u1eb1ng nhau) \uf0de FBE = FCE M\u00e0 XBF = FCE (hai g\u00f3c n\u1ed9i ti\u1ebfp c\u00f9ng ch\u1eafn PX c\u1ee7a O ) ( )N\u00ean FBE = XPF = FCE \uf0de BF l\u00e0 tia ph\u00e2n gi\u00e1c c\u1ee7a XBH X\u00e9t (O) , ta c\u00f3: ( )\uf0ec\uf0efBXI \uf0ef = 1 s\u00f1 BM go\u00f9c no\u00e4i tie\u00e1p cha\u00e9n BM 2 = 1 s\u00f1 CM 2 ( )\uf0ef\uf0edFXI \uf0ef go\u00f9c no\u00e4i tie\u00e1p cha\u00e9n CM ( )\uf0ef\uf0ef\uf0ees\u00f1 BM = s\u00f1 CM M la\u00f8\u00f1ie\u00e5m ch\u00ednh gi\u00f6\u00f5a BC \uf0de BXI = FXI \uf0de XI l\u00e0 tia ph\u00e2n gi\u00e1c c\u1ee7a BXF X\u00e9t BXH , ta c\u00f3 \uf0ec\uf0ef\uf0ef\uf0edXBFI lala\u00f8 \u00f8titaiapphhaa\u00e2n\u00e2nggiaia\u00f9c\u00f9cccuu\u00fba\u00fbaBXXBFH(c(mcmt)t) \uf0ef\uf0ef\uf0eeXI ca\u00e9t BF ta\u00efi I ( gt) \uf0de HI l\u00e0 tia ph\u00e2n gi\u00e1c c\u1ee7a BHX \uf0de BHI = 1 BHX 2 Ch\u1ee9ng minh t\u01b0\u01a1ng t\u1ef1: ta \u0111\u01b0\u1ee3c EHJ = 1 EHC 2 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 Ta c\u00f3: \uf0ec\uf0efBHI = 1 BHX (cmt) \uf0ef 2 \uf0ef\uf0edEHJ = 1 EHC ( cmt ) \uf0ef 2 \uf0efBHX = EHC (hai go\u00f9c \u00f1o\u00e1i \u00f1\u00e6nh) \uf0ef\uf0ee \uf0de BHI = EHJ M\u00e0 EHJ + BHJ = 180\uf0b0(hai go\u00f9c ke\u00e0 bu)\u00f8 N\u00ean BHI + BHJ = 180\uf0b0 \uf0de IHJ = 180\uf0b0 \uf0de I,H, 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 10","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH S\u00d4\u00db GD&\u00d1T TP HO\u00c0 CH\u00cd MINH \u00d1E\u00c0 THAM KHA\u00dbO TUYE\u00c5N SINH 10 PHO\u00d8NG G\u00d1&\u00d1T QUA\u00c4N G\u00d2 V\u1ea4P NA\u00caM HO\u00cfC: 2023 \u2013 2024 M\u00d4N: TO\u00c1N 9 \u0110\u1ec0 THAM KH\u1ea2O M\u00c3 \u0110\u1ec0: Qu\u1eadn G\u00f2 V\u1ea5p - 3 \u0110\u00ea thi g\u1ed3m 8 c\u00e2u ho\u0309i t\u01b0\u0323 lu\u00e2\u0323n. Th\u01a1\u0300i gian: 120 phu\u0301t (kh\u00f4ng k\u00ea\u0309 th\u01a1\u0300i gian pha\u0301t \u0111\u00ea\u0300) C\u00e2u 1. (1,5 \u0111i\u1ec3m). Cho (P) : y = 1 x2 v\u00e0 \u0111\u01b0\u1eddng th\u1eb3ng (d) : y = x + 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). Cho ph\u01b0\u01a1ng tr\u00ecnh 2x2 \u2212 7x \u2212 3 = 0 c\u00f3 2 nghi\u1ec7m l\u00e0 x1 , x2 . Kh\u00f4ng gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh, h\u00e3y t\u00ednh gi\u00e1 tr\u1ecb c\u1ee7a bi\u1ec3u th\u1ee9c A = x1x2 \u2212 x12 \u2212 x22 . L\u01b0u \u00fd: T\u1eeb b\u00e0i n\u00e0y, c\u00e1c s\u1ed1 li\u1ec7u t\u00ednh to\u00e1n v\u1ec1 \u0111\u1ed9 d\u00e0i khi l\u00e0m tr\u00f2n (n\u1ebfu c\u00f3) l\u1ea5y \u0111\u1ebfn m\u1ed9t ch\u1eef s\u1ed1 th\u1eadp ph\u00e2n, s\u1ed1 \u0111o g\u00f3c l\u00e0m tr\u00f2n \u0111\u1ebfn ph\u00fat. C\u00e2u 3. (0,75 \u0111i\u1ec3m). a) N\u01b0\u1edbc \u0111\u00f3ng vai tr\u00f2 c\u1ea5p thi\u1ebft cho m\u1ed9t s\u1ee9c kh\u1ecfe \u1ed5n \u0111\u1ecbnh. N\u00f3 gi\u00fap cho m\u00e1u \u0111\u01b0\u1ee3c l\u01b0u th\u00f4ng tu\u1ea7n ho\u00e0n, t\u0103ng c\u01b0\u1eddng \u0111\u00e0o th\u1ea3i \u0111\u1ed9c t\u1ed1 trong c\u01a1 th\u1ec3. S\u1ed1 l\u00edt n\u01b0\u1edbc c\u1ea7n n\u1ea1p m\u1ed7i ng\u00e0y s\u1ebd theo c\u00e2n n\u1eb7ng l\u00e0 l\u1ea5y s\u1ed1 c\u00e2n n\u1eb7ng (theo kg ) nh\u00e2n v\u1edbi 0,033 . B\u1ea1n Nga n\u1eb7ng 50 kg th\u00ec ph\u1ea3i n\u1ea1p bao nhi\u00eau l\u00edt n\u01b0\u1edbc m\u1ed7i ng\u00e0y? b) Ch\u1ec9 s\u1ed1 kh\u1ed1i c\u01a1 th\u1ec3 hay c\u00f2n g\u1ecdi l\u00e0 BMI \u0111\u01b0\u1ee3c s\u1eed d\u1ee5ng \u0111\u1ec3 x\u00e1c \u0111\u1ecbnh xem b\u1ea1n c\u00f3 \u0111ang \u1edf ph\u1ea1m vi c\u00e2n n\u1eb7ng kh\u1ecfe m\u1ea1nh so v\u1edbi chi\u1ec1u cao c\u1ee7a b\u1ea1n hay kh\u00f4ng. Ch\u1ec9 s\u1ed1 kh\u1ed1i c\u01a1 th\u1ec3 (Body Mass Index \u2013 BMI ), \u0111\u01b0\u1ee3c t\u00ednh b\u1eb1ng c\u00e2n n\u1eb7ng (kg) chia cho b\u00ecnh ph\u01b0\u01a1ng c\u1ee7a chi\u1ec1u cao (m\u00e9t), \u0111\u1ec3 \u0111\u00e1nh gi\u00e1t\u00ecnh tr\u1ea1ng dinh d\u01b0\u1ee1ng c\u1ee7a ng\u01b0\u1eddi tr\u01b0\u1edfng th\u00e0nh. Ph\u00e2n lo\u1ea1i ( )IDI & WPRO, 2000 BMI kg \/ m2 Nh\u1eb9 c\u00e2n (CED) T\u00ecnh tr\u1ea1ng dinh d\u01b0\u1ee1ng b\u00ecnhth\u01b0\u1eddng \uf03c 18,5 18,5 \u2212 22,9 Th\u1eeba c\u00e2n \uf0b3 23,0 Ti\u1ec1n b\u00e9o ph\u00ec 23,0 \u2212 24,9 B\u00e9o ph\u00ec \u0111\u1ed9 I 25,0 \u2212 29,9 B\u00e9o ph\u00ec \u0111\u1ed9 II \uf0b3 30,0 B\u1ea1n H\u1eb1ng n\u1eb7ng 50 kg ; cao 160cm th\u00ec thu\u1ed9c lo\u1ea1i th\u1ec3 tr\u1ea1ng n\u00e0o? c) BFP (Body Fat Percentage) \u2013 Ph\u1ea7n tr\u0103m M\u1ee1 c\u01a1 th\u1ec3 l\u00e0 t\u1ef7 l\u1ec7 m\u1ee1 so v\u1edbi kh\u1ed1i l\u01b0\u1ee3ng c\u01a1 th\u1ec3. C\u00f4ng th\u1ee9c ph\u1ea7n tr\u0103m m\u1ee1 c\u01a1 th\u1ec3 (BFP) \u0111\u1ed1i v\u1edbi nam gi\u1edbi tr\u01b0\u1edfng th\u00e0nh: BFP(%) = 1,20 \uf0b4 BMI + 0,23\uf0b4Tuoi \u2013 16,2 T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N 1","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH C\u00f4ng th\u1ee9c t\u1ef7 l\u1ec7 ph\u1ea7n tr\u0103m m\u1ee1 c\u01a1 th\u1ec3 (BFP) \u0111\u1ed1i v\u1edbi ph\u1ee5 n\u1eef tr\u01b0\u1edfng th\u00e0nh: BFP(%) = 1,20 \uf0b4 BMI + 0,23\uf0b4Tuoi \u2013 5,4 M\u1ed9t b\u1ea1n nam 20 tu\u1ed5i; cao 162cm ; c\u00f3 t\u1ec9 l\u1ec7 m\u1ee1 l\u00e0 25% . H\u1ecfi b\u1ea1n n\u00e0y c\u1ea7n n\u1ea1p v\u00e0o c\u01a1 th\u1ec3 bao nhi\u00eau l\u00edt n\u01b0\u1edbc m\u1ed7i ng\u00e0y (k\u1ebft qu\u1ea3 l\u00e0m tr\u00f2n 1 s\u1ed1 th\u1eadp ph\u00e2n). C\u00e2u 4. (0,75 \u0111i\u1ec3m). V\u00e0o \u0111\u1ea7u n\u0103m 2021 , anh Ho\u00e0ng chia kho\u1ea3n ti\u1ec1n 1 t\u1ec9 500 tri\u1ec7u \u0111\u1ed3ng m\u00e0 anh \u0111ang c\u00f3 th\u00e0nh hai th\u00e0nh ph\u1ea7n: m\u1ed9t ph\u1ea7n anh g\u1eedi ti\u1ebft ki\u1ec7m k\u00ec h\u1ea1n 1 n\u0103m v\u00e0o ng\u00e2n h\u00e0ng v\u1edbi m\u1ee9c l\u00e3i su\u1ea5t 7 % \/n\u0103m (theo h\u00ecnh th\u1ee9c l\u00e3i k\u00e9p); m\u1ed9t ph\u1ea7n anh \u0111\u1ea7u t\u01b0 v\u00e0o m\u1ed9t c\u00f4ng ty th\u01b0\u01a1ng m\u1ea1i. Bi\u1ebft r\u1eb1ng sau \u0111\u00fang 1 n\u0103m, d\u01b0\u1edbi s\u1ef1 \u1ea3nh h\u01b0\u1edfng c\u1ee7a d\u1ecbch COVID-19 th\u00ec t\u00ecnh h\u00ecnh kinh doanh kh\u00f3 kh\u0103n, kho\u1ea3n \u0111\u1ea7u t\u01b0 v\u00e0o c\u00f4ng ty \u0111\u00e3 b\u1ecb l\u1ed7 10,5% . Anh Ho\u00e0ng r\u00fat kho\u1ea3n ti\u1ec1n l\u00e3i t\u1eeb ng\u00e2n h\u00e0ng ra th\u00ec v\u1eeba \u0111\u1ee7 \u0111\u1ec3 b\u00f9 l\u1ed7. T\u00ednh s\u1ed1 ti\u1ec1n anh Ho\u00e0ng \u0111\u00e3 \u0111\u1ea7u t\u01b0 v\u00e0o c\u00f4ng ty th\u01b0\u01a1ng m\u1ea1i. C\u00e2u 5. (1 \u0111i\u1ec3m). Trong ti\u1ebft th\u1ef1c h\u00e0nh v\u1eadt l\u00fd; nh\u00f3m b\u1ea1n An \u0111\u01b0\u1ee3c c\u00f4 giao ghi l\u1ea1i th\u1eddi gian \u0111un s\u00f4i c\u1ee7a n\u01b0\u1edbc \u0111\u00e1 l\u00e0m t\u1eeb n\u01b0\u1edbc c\u1ea5t (b\u1ecf qua s\u1ef1 ph\u1ee5 thu\u1ed9c \u0111\u1ed9 cao). Nh\u00f3m b\u1ea1n ghi l\u1ea1i nh\u01b0 sau: T\u1ea1i ph\u00fat th\u1ee9 10 n\u01b0\u1edbc \u0111\u00e1 \u0111\u00e3 chuy\u1ec3n ho\u00e0n to\u00e0n t\u1eeb th\u1ec3 r\u1eafn sang th\u1ec3 l\u1ecfng v\u00e0 nhi\u1ec7t \u0111\u1ed9 \u0111o \u0111\u01b0\u1ee3c t\u1eeb nhi\u1ec7t k\u1ebf l\u00e0 0\uf0b0C . C\u1ee9 m\u1ed7i m\u1ed9t ph\u00fat \u0111un ti\u1ebfp theo v\u1edbi c\u00f9ng nhi\u1ec7t \u0111\u1ed9 l\u1eeda th\u00ec nh\u00f3m b\u1ea1n ghi nh\u1eadn nhi\u1ec7t \u0111\u1ed9 c\u1ee7a n\u01b0\u1edbc t\u0103ng th\u00eam 10\uf0b0C . G\u1ecdi h(\uf0b0C) l\u00e0 nhi\u1ec7t \u0111\u1ed9 n\u01b0\u1edbc \u0111o \u0111\u01b0\u1ee3c t\u1ea1i t (ph\u00fat) t\u1eeb l\u00fac n\u01b0\u1edbc \u1edf 0\uf0b0C \u0111\u1ebfn khi n\u01b0\u1edbc s\u00f4i c\u00f3 li\u00ean h\u1ec7 b\u1edfi h\u00e0m s\u1ed1 h = at + b(t \uf0b3 10) . a) X\u00e1c \u0111\u1ecbnh h\u1ec7 s\u1ed1 a, b c\u1ee7a h\u00e0m s\u1ed1 n\u00e0y. b) \u0110\u1ed9 F \u0111\u01b0\u1ee3c ra \u0111\u1eddi v\u00e0o n\u0103m 1724 b\u1edfi nh\u00e0 v\u1eadt l\u00fd h\u1ecdc ng\u01b0\u1eddi \u0110\u1ee9c Daniel Gabriel ( )Fahrenheit 1686 \u2212 1736 \u0111\u01b0\u1ee3c k\u00fd hi\u1ec7u l\u00e0 \uf0b0F . G\u1ecdi TC l\u00e0 nhi\u1ec7t \u0111\u1ed9 C ; TF l\u00e0 nhi\u1ec7t \u0111\u1ed9 F c\u00f3 c\u00f4ng th\u1ee9c chuy\u1ec3n \u0111\u1ed5i nh\u01b0 sau: TC = 5 (TF \u2212 32) . H\u1ecfi sau khi \u0111un 18 ph\u00fat th\u00ec n\u01b0\u1edbc \u0111\u01b0\u1ee3c 9 bao nhi\u00eau \u0111\u1ed9 F . C\u00e2u 6. (1 \u0111i\u1ec3m). C\u1eeda h\u00e0ng ABC nh\u1eadp v\u1ec1 m\u1ed9t s\u1ed1 \u00e1o v\u1edbi gi\u00e1 v\u1ed1n l\u00e0 300 000 \u0111\u1ed3ng\/c\u00e1i. Tu\u1ea7n th\u1ee9 nh\u1ea5t c\u1eeda h\u00e0ng b\u00e1n \u0111\u01b0\u1ee3c m\u1ed9t n\u1eeda s\u1ed1 l\u01b0\u1ee3ng \u00e1o th\u00ec l\u1eddi \u0111\u01b0\u1ee3c 40% gi\u00e1 v\u1ed1n. Tu\u1ea7n th\u1ee9 hai c\u1eeda h\u00e0ng 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 b\u00e1n 3 s\u1ed1 \u00e1o c\u00f2n l\u1ea1i v\u1edbi gi\u00e1 v\u1ed1n. Tu\u1ea7n th\u1ee9 ba c\u1eeda h\u00e0ng b\u00e1n h\u1ebft s\u1ed1 \u00e1o c\u00f2n l\u1ea1i th\u00ec l\u1ed7 20% gi\u00e1 5 v\u1ed1n. Sau khi th\u1ed1ng k\u1ebf th\u00ec c\u1ee7a h\u00e0ng thu l\u1ee3i nhu\u1eadn t\u1eeb vi\u1ec7c b\u00e1n h\u1ebft s\u1ed1 \u00e1o tr\u00ean l\u00e0 4 800 000 \u0111\u1ed3ng. H\u1ecfi c\u1ee7a h\u00e0ng \u0111\u00e3 nh\u1eadp v\u1ec1 bao nhi\u00eau c\u00e1i \u00e1o? C\u00e2u 7. (1 \u0111i\u1ec3m). \u0110\u1ec3 l\u00e0m th\u00ed nghi\u1ec7m v\u1ec1 s\u1ef1 n\u1ed5i c\u1ee7a c\u00e1c v\u1eadt th\u1ec3, Minh chu\u1ea9n b\u1ecb m\u1ed9t c\u00e1i c\u1ed1c th\u1ee7y tinh c\u00f3 d\u1ea1ng l\u00f2ng trong h\u00ecnh tr\u1ee5 c\u00f3 \u0111\u01b0\u1eddng k\u00ednh \u0111\u00e1y 6 cm v\u00e0 chi\u1ec1u cao l\u00e0 10cm ; m\u1ed9t qu\u1ea3 b\u00f3ng b\u00e0n ti\u00eau chu\u1ea9n c\u1ee7a c\u00e1c gi\u1ea3i \u0111\u1ea5u qu\u1ed1c t\u1ebf c\u00f3 d\u1ea1ng h\u00ecnh c\u1ea7u \u0111\u01b0\u1eddng k\u00ednh 40 mm . Minh b\u1ecf qu\u1ea3 b\u00f3ng b\u00e0n v\u00e0o trong c\u1ed1c, r\u00f3t t\u1eeb t\u1eeb 200cm3 n\u01b0\u1edbc v\u00e0 \u0111o \u0111\u01b0\u1ee3c m\u1ef1c n\u01b0\u1edbc d\u00e2ng l\u00ean cao 7,2cm . a) T\u00ednh th\u1ec3 t\u00edch c\u1ee7a qu\u1ea3 b\u00f3ng b\u00e0n. b) T\u00ednh t\u1ec9 l\u1ec7 ph\u1ea7n tr\u0103m th\u1ec3 t\u00edch ph\u1ea7n n\u1ed5i c\u1ee7a qu\u1ea3 b\u00f3ng b\u00e0n trong th\u00ed nghi\u1ec7m tr\u00ean. Bi\u1ebft c\u00f4ng th\u1ee9c x\u00e1c \u0111\u1ecbnh th\u1ec3 t\u00edch c\u1ee7a kh\u1ed1i c\u1ea7u b\u00e1n k\u00ednh R l\u00e0 V = 4 \uf070 R3 v\u00e0 c\u00f4ng th\u1ee9c t\u00ednh th\u1ec3 t\u00edch 3 h\u00ecnh tr\u1ee5 b\u00e1n k\u00ednh r ,chi\u1ec1u cao h l\u00e0 V = \uf070 r2h . L\u1ea5y \uf070 \uf0bb 3,14 v\u00e0 c\u00e1c k\u1ebft qu\u1ea3 l\u00e0m tr\u00f2n \u0111\u1ebfn ch\u1eef s\u1ed1 th\u1eadp ph\u00e2n th\u1ee9 hai. C\u00e2u 8. (3 \u0111i\u1ec3m) Cho tam gi\u00e1c ABC ( AB \uf03c AC) c\u00f3 ba g\u00f3c nh\u1ecdn n\u1ed9i ti\u1ebfp \u0111\u01b0\u1eddng tr\u00f2n (O) . \u0110\u01b0\u1eddng cao AD, BE, CF c\u1eaft nhau t\u1ea1i H . L\u1ea5y M t\u00f9y \u00fd thu\u1ed9c cung nh\u1ecf BC . a) Ch\u1ee9ng minh t\u1ee9 gi\u00e1c CDHE n\u1ed9i ti\u1ebfp v\u00e0 AMB = BHD . b) G\u1ecdi I l\u00e0 \u0111i\u1ec3m \u0111\u1ed1i x\u1ee9ng cu\u1ea3 M qua AB . Ch\u1ee9ng minh t\u1ee9 gi\u00e1c AHBI n\u1ed9i ti\u1ebfp v\u00e0 MAB = BHI . c) G\u1ecdi K l\u00e0 \u0111i\u1ec3m \u0111\u1ed1i x\u1ee9ng c\u1ee7a M qua \u0111\u01b0\u1eddng th\u1eb3ng AC . Ch\u1ee9ng minh ba \u0111i\u1ec3m I, H, K th\u1eb3ng h\u00e0ng. ----H\u1ebeT--- T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N 3","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH H\u01af\u1edaNG D\u1eaaN GI\u1ea2I C\u00e2u 1. (1,5 \u0111i\u1ec3m) Cho (P) : y = 1 x2 v\u00e0 \u0111\u01b0\u1eddng th\u1eb3ng (d) : y = x + 4 . 2 a) V\u1ebd \u0111\u1ed3 th\u1ecb (P) v\u00e0 (d) tr\u00ean c\u00f9ng h\u1ec7 tr\u1ee5c t\u1ecda \u0111\u1ed9. b) T\u00ecm t\u1ecda \u0111\u1ed9 giao \u0111i\u1ec3m c\u1ee7a (P) v\u00e0 (d) b\u1eb1ng ph\u00e9p t\u00ednh. L\u1eddi gi\u1ea3i a) V\u1ebd \u0111\u1ed3 th\u1ecb (P) v\u00e0 (d) tr\u00ean c\u00f9ng h\u1ec7 tr\u1ee5c t\u1ecda \u0111\u1ed9. BGT: x \u22124 \u22122 0 1 2 y = 1 x2 8 2 0 2 8 2 x 01 y=x+4 4 5 b) T\u00ecm t\u1ecda \u0111\u1ed9 giao \u0111i\u1ec3m c\u1ee7a (P) v\u00e0 (d) b\u1eb1ng ph\u00e9p t\u00ednh. Ph\u01b0\u01a1ng tr\u00ecnh ho\u00e0nh \u0111\u1ed9 giao \u0111i\u1ec3m c\u1ee7a (P) v\u00e0 (d) : 1 x2 = x + 4 2 \uf0db 1 x2 \u2212 x \u2212 4 = 0 2 \uf0db \uf0e9x = \u22122 \uf0ea\uf0ebx = 4 Thay x = \u22122 v\u00e0o y = x + 4 , ta \u0111\u01b0\u1ee3c: y = \u22122 + 4 = 2 . Thay x = 4 v\u00e0o y = x + 4 , ta \u0111\u01b0\u1ee3c: y = 4 + 4 = 8 . V\u1eady (\u22122; 2) , (4; 8) l\u00e0 hai giao \u0111i\u1ec3m c\u1ea7n t\u00ecm. C\u00e2u 2. (1 \u0111i\u1ec3m) Cho ph\u01b0\u01a1ng tr\u00ecnh 2x2 \u2212 7x \u2212 3 = 0 c\u00f3 2 nghi\u1ec7m l\u00e0 x1 , x2 . Kh\u00f4ng gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh, h\u00e3y t\u00ednh gi\u00e1 tr\u1ecb c\u1ee7a bi\u1ec3u th\u1ee9c A = x1x2 \u2212 x12 \u2212 x22 . L\u1eddi gi\u1ea3i 2x2 \u2212 7x \u2212 3 = 0 V\u00ec \uf044 = b2 \u2212 4ac = (\u22127)2 \u2212 4.2.(\u22123) = 73 \uf03e 0 T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N 4","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH N\u00ean ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m ph\u00e2n bi\u1ec7t x1 ,x2 . \uf0ec\uf0ef\uf0efS = x1 + x2 = \u2212b =\u2212 7 \uf0ed = x1 a 2 Theo \u0111\u1ecbnh l\u00ed Vi-et, ta c\u00f3: \uf0ef\uf0ef\uf0eeP \u22123 c = 2 .x2 = a Ta c\u00f3: A = x1x2 \u2212 x12 \u2212 x22 ( )A = x1x2 \u2212 x12 + x22 A = P \u2212 S2 + 2P A = \u22123 \u2212 \uf0e6 \u22127 \uf0f62 + 2 \uf0e6\uf0e7 \u22123 \uf0f6 2 \uf0e7 2 \uf0f7 \uf0e8 2 \uf0f7 \uf0e8 \uf0f8 \uf0f8 A = \u221267 4 L\u01b0u \u00fd: T\u1eeb b\u00e0i n\u00e0y, c\u00e1c s\u1ed1 li\u1ec7u t\u00ednh to\u00e1n v\u1ec1 \u0111\u1ed9 d\u00e0i khi l\u00e0m tr\u00f2n (n\u1ebfu c\u00f3) l\u1ea5y \u0111\u1ebfn m\u1ed9t ch\u1eef s\u1ed1 th\u1eadp ph\u00e2n, s\u1ed1 \u0111o g\u00f3c l\u00e0m tr\u00f2n \u0111\u1ebfn ph\u00fat. C\u00e2u 3. (1 \u0111i\u1ec3m) a) N\u01b0\u1edbc \u0111\u00f3ng vai tr\u00f2 c\u1ea5p thi\u1ebft cho m\u1ed9t s\u1ee9c kh\u1ecfe \u1ed5n \u0111\u1ecbnh. N\u00f3 gi\u00fap cho m\u00e1u \u0111\u01b0\u1ee3c l\u01b0u th\u00f4ng tu\u1ea7n ho\u00e0n, t\u0103ng c\u01b0\u1eddng \u0111\u00e0o th\u1ea3i \u0111\u1ed9c t\u1ed1 trong c\u01a1 th\u1ec3. S\u1ed1 l\u00edt n\u01b0\u1edbc c\u1ea7n n\u1ea1p m\u1ed7i ng\u00e0y s\u1ebd theo c\u00e2n n\u1eb7ng l\u00e0 l\u1ea5y s\u1ed1 c\u00e2n n\u1eb7ng (theo kg) nh\u00e2n v\u1edbi 0,033 . B\u1ea1n Nga n\u1eb7ng 50 kg th\u00ec ph\u1ea3i n\u1ea1p bao nhi\u00eau l\u00edt n\u01b0\u1edbc m\u1ed7i ng\u00e0y? b) Ch\u1ec9 s\u1ed1 kh\u1ed1i c\u01a1 th\u1ec3 hay c\u00f2n g\u1ecdi l\u00e0 BMI \u0111\u01b0\u1ee3c s\u1eed d\u1ee5ng \u0111\u1ec3 x\u00e1c \u0111\u1ecbnh xem b\u1ea1n c\u00f3 \u0111ang \u1edf ph\u1ea1m vi c\u00e2n n\u1eb7ng kh\u1ecfe m\u1ea1nh so v\u1edbi chi\u1ec1u cao c\u1ee7a b\u1ea1n hay kh\u00f4ng. Ch\u1ec9 s\u1ed1 kh\u1ed1i c\u01a1 th\u1ec3 (Body Mass Index \u2013 BMI), \u0111\u01b0\u1ee3c t\u00ednh b\u1eb1ng c\u00e2n n\u1eb7ng (kg) chia cho b\u00ecnh ph\u01b0\u01a1ng c\u1ee7a chi\u1ec1u cao (m\u00e9t), \u0111\u1ec3 \u0111\u00e1nh gi\u00e1 t\u00ecnh tr\u1ea1ng dinh d\u01b0\u1ee1ng c\u1ee7a ng\u01b0\u1eddi tr\u01b0\u1edfng th\u00e0nh. Ph\u00e2n lo\u1ea1i ( )IDI & WPRO, 2000 BMI kg \/ m2 Nh\u1eb9 c\u00e2n (CED) T\u00ecnh tr\u1ea1ng dinh d\u01b0\u1ee1ng b\u00ecnhth\u01b0\u1eddng \uf03c 18,5 18,5 \u2212 22,9 Th\u1eeba c\u00e2n \uf0b3 23,0 Ti\u1ec1n b\u00e9o ph\u00ec 23,0 \u2212 24,9 B\u00e9o ph\u00ec \u0111\u1ed9 I 25,0 \u2212 29,9 B\u00e9o ph\u00ec \u0111\u1ed9 II \uf0b3 30,0 B\u1ea1n H\u1eb1ng n\u1eb7ng 50 kg ; cao 160cm th\u00ec thu\u1ed9c lo\u1ea1i th\u1ec3 tr\u1ea1ng n\u00e0o? c) BFP (Body Fat Percentage) \u2013 Ph\u1ea7n tr\u0103m M\u1ee1 c\u01a1 th\u1ec3 l\u00e0 t\u1ef7 l\u1ec7 m\u1ee1 so v\u1edbi kh\u1ed1i l\u01b0\u1ee3ng c\u01a1 th\u1ec3. C\u00f4ng th\u1ee9c ph\u1ea7n tr\u0103m m\u1ee1 c\u01a1 th\u1ec3 (BFP) \u0111\u1ed1i v\u1edbi nam gi\u1edbi tr\u01b0\u1edfng th\u00e0nh: BFP(%) = 1,20 \uf0b4 BMI + 0,23\uf0b4Tuoi \u2013 16,2 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 C\u00f4ng th\u1ee9c t\u1ef7 l\u1ec7 ph\u1ea7n tr\u0103m m\u1ee1 c\u01a1 th\u1ec3 (BFP) \u0111\u1ed1i v\u1edbi ph\u1ee5 n\u1eef tr\u01b0\u1edfng th\u00e0nh: BFP(%) = 1,20 \uf0b4 BMI + 0,23\uf0b4Tuoi \u2013 5,4 M\u1ed9t b\u1ea1n nam 20 tu\u1ed5i; cao 162cm ; c\u00f3 t\u1ec9 l\u1ec7 m\u1ee1 l\u00e0 25% . H\u1ecfi b\u1ea1n n\u00e0y c\u1ea7n n\u1ea1p v\u00e0o c\u01a1 th\u1ec3 bao nhi\u00eau l\u00edt n\u01b0\u1edbc m\u1ed7i ng\u00e0y (k\u1ebft qu\u1ea3 l\u00e0m tr\u00f2n 1 s\u1ed1 th\u1eadp ph\u00e2n). L\u1eddi gi\u1ea3i a) B\u1ea1n Nga n\u1eb7ng 50 kg th\u00ec ph\u1ea3i n\u1ea1p bao nhi\u00eau l\u00edt n\u01b0\u1edbc m\u1ed7i ng\u00e0y? S\u1ed1 l\u00edt n\u01b0\u1edbc Nga ph\u1ea3i n\u1ea1p m\u1ed7i ng\u00e0y: 0,033.50 = 1,65(l) . b) B\u1ea1n H\u1eb1ng n\u1eb7ng 50 kg ; cao 160cm th\u00ec thu\u1ed9c lo\u1ea1i th\u1ec3 tr\u1ea1ng n\u00e0o? Ch\u1ec9 s\u1ed1 kh\u1ed1i c\u01a1 th\u1ec3 c\u1ee7a b\u1ea1n H\u1eb1ng l\u00e0: 50 : 1,602 \uf0bb 19,5 V\u1eady b\u1ea1n H\u1eb1ng thu\u1ed9c t\u00ecnh tr\u1ea1ng dinh d\u01b0\u1ee1ng b\u00ecnh th\u01b0\u1eddng. c) M\u1ed9t b\u1ea1n nam 20 tu\u1ed5i; cao 162cm ; c\u00f3 t\u1ec9 l\u1ec7 m\u1ee1 l\u00e0 25% . H\u1ecfi b\u1ea1n n\u00e0y c\u1ea7n n\u1ea1p v\u00e0o c\u01a1 th\u1ec3 bao nhi\u00eau l\u00edt n\u01b0\u1edbc m\u1ed7i ng\u00e0y (k\u1ebft qu\u1ea3 l\u00e0m tr\u00f2n 1 s\u1ed1 th\u1eadp ph\u00e2n). Ch\u1ec9 s\u1ed1 BMI c\u1ee7a b\u1ea1n nam l\u00e0: BMI = 25 \u2212 0,23.20 + 16,2 = 30,5 1, 20 C\u00e2n n\u1eb7ng c\u1ee7a b\u1ea1n nam l\u00e0: 30,5.1,622 = 80,0442 S\u1ed1 l\u00edt n\u01b0\u1edbc b\u1ea1n nam n\u00e0y c\u1ea7n n\u1ea1p v\u00e0o c\u01a1 th\u1ec3 m\u1ed7i ng\u00e0y l\u00e0: 0,033.80,0442 \uf0bb 2,6(l) C\u00e2u 4. (0,75 \u0111i\u1ec3m). V\u00e0o \u0111\u1ea7u n\u0103m 2021, anh Ho\u00e0ng chia kho\u1ea3n ti\u1ec1n 1 t\u1ec9 500 tri\u1ec7u \u0111\u1ed3ng m\u00e0 anh \u0111ang c\u00f3 th\u00e0nh hai th\u00e0nh ph\u1ea7n: m\u1ed9t ph\u1ea7n anh g\u1eedi ti\u1ebft ki\u1ec7m k\u00ec h\u1ea1n 1 n\u0103m v\u00e0o ng\u00e2n h\u00e0ng v\u1edbi m\u1ee9c l\u00e3i su\u1ea5t 7 % \/n\u0103m (theo h\u00ecnh th\u1ee9c l\u00e3i k\u00e9p); m\u1ed9t ph\u1ea7n anh \u0111\u1ea7u t\u01b0 v\u00e0o m\u1ed9t c\u00f4ng ty th\u01b0\u01a1ng m\u1ea1i. Bi\u1ebft r\u1eb1ng sau \u0111\u00fang 1 n\u0103m, d\u01b0\u1edbi s\u1ef1 \u1ea3nh h\u01b0\u1edfng c\u1ee7a d\u1ecbch COVID-19 th\u00ec t\u00ecnh h\u00ecnh kinh doanh kh\u00f3 kh\u0103n, kho\u1ea3n \u0111\u1ea7u t\u01b0 v\u00e0o c\u00f4ng ty \u0111\u00e3 b\u1ecb l\u1ed7 10,5% . Anh Ho\u00e0ng r\u00fat kho\u1ea3n ti\u1ec1n l\u00e3i t\u1eeb ng\u00e2n h\u00e0ng ra th\u00ec v\u1eeba \u0111\u1ee7 \u0111\u1ec3 b\u00f9 l\u1ed7. T\u00ednh s\u1ed1 ti\u1ec1n anh Ho\u00e0ng \u0111\u00e3 \u0111\u1ea7u t\u01b0 v\u00e0o c\u00f4ng ty th\u01b0\u01a1ng m\u1ea1i. L\u1eddi gi\u1ea3i G\u1ecdi x (tri\u1ec7u \u0111\u1ed3ng) l\u00e0 s\u1ed1 ti\u1ec1n anh Ho\u00e0ng \u0111\u00e3 \u0111\u1ea7u t\u01b0 v\u00e0o c\u00f4ng ty th\u01b0\u01a1ng m\u1ea1i (x \uf03e 0) S\u1ed1 ti\u1ec1n anh Ho\u00e0ng g\u1eedi ti\u1ebft ki\u1ec7m k\u00ec h\u1ea1n 1 n\u0103m: 1 500 \u2212 x ( tri\u1ec7u \u0111\u1ed3ng) S\u1ed1 ti\u1ec1n anh Ho\u00e0ng b\u1ecb l\u1ed7: 10,5% x (\u0111\u1ed3ng) S\u1ed1 ti\u1ec1n l\u1eddi anh Ho\u00e0ng r\u00fat ra t\u1eeb vi\u1ec7c g\u1edfi ng\u00e2n h\u00e0ng: 7%(1 500 \u2212 x) (tri\u1ec7u \u0111\u1ed3ng)\u2019 Do Anh Ho\u00e0ng r\u00fat kho\u1ea3n ti\u1ec1n l\u00e3i t\u1eeb ng\u00e2n h\u00e0ng ra th\u00ec v\u1eeba \u0111\u1ee7 \u0111\u1ec3 b\u00f9 l\u1ed7 n\u00ean ta c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh: 7%(1 500 \u2212 x) = 10,5%x \uf0db x = 600(N) T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N 6","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH V\u1eady anh Ho\u00e0ng \u0111\u00e3 \u0111\u1ea7u t\u01b0 v\u00e0o c\u00f4ng ty th\u01b0\u01a1ng m\u1ea1i 600 tri\u1ec7u \u0111\u1ed3ng. C\u00e2u 5. (1 \u0111i\u1ec3m) Trong ti\u1ebft th\u1ef1c h\u00e0nh v\u1eadt l\u00fd; nh\u00f3m b\u1ea1n An \u0111\u01b0\u1ee3c c\u00f4 giao ghi l\u1ea1i th\u1eddi gian \u0111un s\u00f4i c\u1ee7a n\u01b0\u1edbc \u0111\u00e1 l\u00e0m t\u1eeb n\u01b0\u1edbc c\u1ea5t (b\u1ecf qua s\u1ef1 ph\u1ee5 thu\u1ed9c \u0111\u1ed9 cao). Nh\u00f3m b\u1ea1n ghi l\u1ea1i nh\u01b0 sau: T\u1ea1i ph\u00fat th\u1ee9 10 n\u01b0\u1edbc \u0111\u00e1 \u0111\u00e3 chuy\u1ec3n ho\u00e0n to\u00e0n t\u1eeb th\u1ec3 r\u1eafn sang th\u1ec3 l\u1ecfng v\u00e0 nhi\u1ec7t \u0111\u1ed9 \u0111o \u0111\u01b0\u1ee3c t\u1eeb nhi\u1ec7t k\u1ebf l\u00e0 0\uf0b0C . C\u1ee9 m\u1ed7i m\u1ed9t ph\u00fat \u0111un ti\u1ebfp theo v\u1edbi c\u00f9ng nhi\u1ec7t \u0111\u1ed9 l\u1eeda th\u00ec nh\u00f3m b\u1ea1n ghi nh\u1eadn nhi\u1ec7t \u0111\u1ed9 c\u1ee7a n\u01b0\u1edbc t\u0103ng th\u00eam 10\uf0b0C . G\u1ecdi h(\uf0b0C) l\u00e0 nhi\u1ec7t \u0111\u1ed9 n\u01b0\u1edbc \u0111o \u0111\u01b0\u1ee3c t\u1ea1i t (ph\u00fat) t\u1eeb l\u00fac n\u01b0\u1edbc \u1edf 0\uf0b0C \u0111\u1ebfn khi n\u01b0\u1edbc s\u00f4i c\u00f3 li\u00ean h\u1ec7 b\u1edfi h\u00e0m s\u1ed1 h = at + b(t \uf0b3 10) . a) X\u00e1c \u0111\u1ecbnh h\u1ec7 s\u1ed1 a, b c\u1ee7a h\u00e0m s\u1ed1 n\u00e0y. b) \u0110\u1ed9 F \u0111\u01b0\u1ee3c ra \u0111\u1eddi v\u00e0o n\u0103m 1724 b\u1edfi nh\u00e0 v\u1eadt l\u00fd h\u1ecdc ng\u01b0\u1eddi \u0110\u1ee9c Daniel Gabriel Fahrenheit ( )1686 \u2212 1736 \u0111\u01b0\u1ee3c k\u00fd hi\u1ec7u l\u00e0 \uf0b0F . G\u1ecdi TC l\u00e0 nhi\u1ec7t \u0111\u1ed9 C ; TF l\u00e0 nhi\u1ec7t \u0111\u1ed9 F c\u00f3 c\u00f4ng th\u1ee9c chuy\u1ec3n \u0111\u1ed5i nh\u01b0 sau: TC = 5 (TF \u2212 32) . H\u1ecfi sau khi \u0111un 18 ph\u00fat th\u00ec n\u01b0\u1edbc \u0111\u01b0\u1ee3c bao nhi\u00eau \u0111\u1ed9 9 F. 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 \uf0ech = 0 \uf0de 0 = 10a + b . (1) \uf0ed\uf0eet = 10 V\u1edbi \uf0ech = 50 \uf0de 50 = 15a + b . (2) \uf0ee\uf0edt = 15 T\u1eeb (1) v\u00e0 (2) ta c\u00f3 h\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh: \uf0ec10a + b = 0 \uf0db \uf0eca = 10 . \uf0ed\uf0ee15a + b = 50 \uf0ed\uf0eeb = \u2212100 V\u1eady: a = 10 , b = \u2212100 v\u00e0 h = 10t \u2212100 . b) H\u1ecfi sau khi \u0111un 18 ph\u00fat th\u00ec n\u01b0\u1edbc \u0111\u01b0\u1ee3c bao nhi\u00eau \u0111\u1ed9 F ? 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 Nhi\u1ec7t \u0111\u1ed9 F sau khi \u0111un 18 ph\u00fat: h = 10.18 \u2212 100 = 80\uf0b0F . C\u00e2u 6. (1 \u0111i\u1ec3m) C\u1eeda h\u00e0ng ABC nh\u1eadp v\u1ec1 m\u1ed9t s\u1ed1 \u00e1o v\u1edbi gi\u00e1 v\u1ed1n l\u00e0 300 000 \u0111\u1ed3ng\/c\u00e1i. Tu\u1ea7n th\u1ee9 nh\u1ea5t c\u1eeda h\u00e0ng b\u00e1n \u0111\u01b0\u1ee3c m\u1ed9t n\u1eeda s\u1ed1 l\u01b0\u1ee3ng \u00e1o th\u00ec l\u1eddi \u0111\u01b0\u1ee3c 40% gi\u00e1 v\u1ed1n. Tu\u1ea7n th\u1ee9 hai c\u1eeda h\u00e0ng b\u00e1n 3 s\u1ed1 \u00e1o c\u00f2n l\u1ea1i v\u1edbi gi\u00e1 v\u1ed1n. Tu\u1ea7n th\u1ee9 ba c\u1eeda h\u00e0ng b\u00e1n h\u1ebft s\u1ed1 \u00e1o c\u00f2n l\u1ea1i th\u00ec l\u1ed7 20% gi\u00e1 v\u1ed1n. 5 Sau khi th\u1ed1ng k\u1ebf th\u00ec c\u1ee7a h\u00e0ng thu l\u1ee3i nhu\u1eadn t\u1eeb vi\u1ec7c b\u00e1n h\u1ebft s\u1ed1 \u00e1o tr\u00ean l\u00e0 4 800 000 \u0111\u1ed3ng. H\u1ecfi c\u1ee7a h\u00e0ng \u0111\u00e3 nh\u1eadp v\u1ec1 bao nhi\u00eau c\u00e1i \u00e1o? L\u1eddi gi\u1ea3i G\u1ecdi x (c\u00e1i \u00e1o) l\u00e0 s\u1ed1 c\u00e1i \u00e1o c\u1eeda h\u00e0ng nh\u1eadp v\u1ec1 (x \uf03e 0) S\u1ed1 ti\u1ec1n c\u1eeda h\u00e0ng l\u1eddi \u0111\u01b0\u1ee3c khi b\u00e1n n\u1eeda s\u1ed1 \u00e1o: 300 000.40% x = 120 000x (\u0111\u1ed3ng) S\u1ed1 ti\u1ec1n c\u1eeda h\u00e0ng l\u1ed7 khi b\u00e1n h\u1ebft s\u1ed1 \u00e1o c\u00f2n l\u1ea1i: 300 000.20%.\uf0e6\uf0e7 x \u2212 3 x \uf0f6 = 24 000x (\u0111\u1ed3ng) \uf0e8 5 \uf0f7 \uf0f8 Do c\u1ee7a h\u00e0ng thu l\u1ee3i nhu\u1eadn t\u1eeb vi\u1ec7c b\u00e1n h\u1ebft s\u1ed1 \u00e1o tr\u00ean l\u00e0 4 800 000 \u0111\u1ed3ng n\u00ean ta c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh: 120 000x \u2212 24 000x = 4 800 000 \uf0db x = 50(N) V\u1eady c\u1eeda h\u00e0ng nh\u1eadp v\u1ec1 50 c\u00e1i \u00e1o. C\u00e2u 7. (1 \u0111i\u1ec3m) \u0110\u1ec3 l\u00e0m th\u00ed nghi\u1ec7m v\u1ec1 s\u1ef1 n\u1ed5i c\u1ee7a c\u00e1c v\u1eadt th\u1ec3, Minh chu\u1ea9n b\u1ecb m\u1ed9t c\u00e1i c\u1ed1c th\u1ee7y tinh c\u00f3 d\u1ea1ng l\u00f2ng trong h\u00ecnh tr\u1ee5 c\u00f3 \u0111\u01b0\u1eddng k\u00ednh \u0111\u00e1y 6 cm v\u00e0 chi\u1ec1u cao l\u00e0 10cm ; m\u1ed9t qu\u1ea3 b\u00f3ng b\u00e0n ti\u00eau chu\u1ea9n c\u1ee7a c\u00e1c gi\u1ea3i \u0111\u1ea5u qu\u1ed1c t\u1ebf c\u00f3 d\u1ea1ng h\u00ecnh c\u1ea7u \u0111\u01b0\u1eddng k\u00ednh 40 mm . Minh b\u1ecf qu\u1ea3 b\u00f3ng b\u00e0n v\u00e0o trong c\u1ed1c, r\u00f3t t\u1eeb t\u1eeb 200cm3 n\u01b0\u1edbc v\u00e0 \u0111o \u0111\u01b0\u1ee3c m\u1ef1c n\u01b0\u1edbc d\u00e2ng l\u00ean cao 7,2cm . a) T\u00ednh th\u1ec3 t\u00edch c\u1ee7a qu\u1ea3 b\u00f3ng b\u00e0n. b) T\u00ednh t\u1ec9 l\u1ec7 ph\u1ea7n tr\u0103m th\u1ec3 t\u00edch ph\u1ea7n n\u1ed5i c\u1ee7a qu\u1ea3 b\u00f3ng b\u00e0n trong th\u00ed nghi\u1ec7m tr\u00ean. 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 Bi\u1ebft c\u00f4ng th\u1ee9c x\u00e1c \u0111\u1ecbnh th\u1ec3 t\u00edch c\u1ee7a kh\u1ed1i c\u1ea7u b\u00e1n k\u00ednh R l\u00e0 V = 4 \uf070 R3 v\u00e0 c\u00f4ng th\u1ee9c t\u00ednh th\u1ec3 3 t\u00edch h\u00ecnh tr\u1ee5 b\u00e1n k\u00ednh r ,chi\u1ec1u cao h l\u00e0 V = \uf070 r2h . L\u1ea5y \uf070 \uf0bb 3,14 v\u00e0 c\u00e1c k\u1ebft qu\u1ea3 l\u00e0m tr\u00f2n \u0111\u1ebfn ch\u1eef s\u1ed1 th\u1eadp ph\u00e2n th\u1ee9 hai. L\u1eddi gi\u1ea3i a) T\u00ednh th\u1ec3 t\u00edch c\u1ee7a qu\u1ea3 b\u00f3ng b\u00e0n. = 4 \uf070 R3 = 4 \uf0d7\uf070 \uf0d7 23 = 32 \uf070 3 33 ( )Th\u1ec3 t\u00edch qu\u1ea3 b\u00f3ng b\u00e0n l\u00e0: cm3 V \uf0bb 33,51 b) T\u00ednh t\u1ec9 l\u1ec7 ph\u1ea7n tr\u0103m th\u1ec3 t\u00edch ph\u1ea7n n\u1ed5i c\u1ee7a qu\u1ea3 b\u00f3ng b\u00e0n trong th\u00ed nghi\u1ec7m tr\u00ean. ( )Th\u1ec3 t\u00edch n\u01b0\u1edbc v\u00e0 ph\u1ea7n ch\u00ecm trong c\u1ed1c l\u00e0: V = \uf070 r2h = 32.7,2\uf070 = 324 \uf070 cm3 5 ( )Th\u1ec3 t\u00edch ph\u1ea7n ch\u00ecm l\u00e0: 324 \uf070 \u2212 200 \uf0bb 3,58 cm3 5 ( )Th\u1ec3 t\u00edch ph\u1ea7n n\u1ed5i l\u00e0:32 \uf0e6 324 \uf0f6 3 \uf070 \u2212 \uf0e7 5 \uf070 \u2212 200 \uf0f7 \uf0bb 16, 53 cm3 \uf0e8 \uf0f8 T\u1ec9 l\u1ec7 ph\u1ea7n tr\u0103m th\u1ec3 t\u00edch ph\u1ea7n n\u1ed5i c\u1ee7a qu\u1ea3 b\u00f3ng b\u00e0n trong th\u00ed nghi\u1ec7m tr\u00ean l\u00e0: 16,53 \uf0bb 49,33% 33, 51 C\u00e2u 9. (3 \u0111i\u1ec3m) Cho tam gi\u00e1c ABC ( AB \uf03c AC) c\u00f3 ba g\u00f3c nh\u1ecdn n\u1ed9i ti\u1ebfp \u0111\u01b0\u1eddng tr\u00f2n (O) . \u0110\u01b0\u1eddng cao AD, BE, CF c\u1eaft nhau t\u1ea1i H . L\u1ea5y M t\u00f9y \u00fd thu\u1ed9c cung nh\u1ecf BC . a) Ch\u1ee9ng minh t\u1ee9 gi\u00e1c CDHE n\u1ed9i ti\u1ebfp v\u00e0 AMB = BHD . b) G\u1ecdi I l\u00e0 \u0111i\u1ec3m \u0111\u1ed1i x\u1ee9ng cu\u1ea3 M qua AB . Ch\u1ee9ng minh t\u1ee9 gi\u00e1c AHBI n\u1ed9i ti\u1ebfp v\u00e0 MAB = BHI . c) G\u1ecdi K l\u00e0 \u0111i\u1ec3m \u0111\u1ed1i x\u1ee9ng c\u1ee7a M qua \u0111\u01b0\u1eddng th\u1eb3ng AC . Ch\u1ee9ng minh ba \u0111i\u1ec3m I, H, K th\u1eb3ng h\u00e0ng. L\u1eddi gi\u1ea3i T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N 9","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 CDHE n\u1ed9i ti\u1ebfp v\u00e0 AMB = BHD . X\u00e9t t\u1ee9 gi\u00e1c CHDE , c\u00f3: \uf0ec\uf0efHDC = 90\uf0b0( AD \u22a5 BC) \uf0ed \uf0ef\uf0eeHEC = 90\uf0b0 ( BE \u22a5 AC ) \uf0de HDC + HEC = 180\uf0b0 \uf0de T\u1ee9 gi\u00e1c CHDE n\u1ed9i ti\u1ebfp v\u00ec c\u00f3 hai g\u00f3c \u0111\u1ed1i b\u00f9 nhau. \uf0de BHD = ECD (g\u00f3c ngo\u00e0i b\u1eb1ng g\u00f3c \u0111\u1ed1i trong). M\u00e0: AMB = ECD (hai g\u00f3c n\u1ed9i ti\u1ebfp c\u00f9ng ch\u1eafn AB ) N\u00ean: AMB = BHD . b) G\u1ecdi I l\u00e0 \u0111i\u1ec3m \u0111\u1ed1i x\u1ee9ng cu\u1ea3 M qua AB . Ch\u1ee9ng minh t\u1ee9 gi\u00e1c AHBI n\u1ed9i ti\u1ebfp v\u00e0 MAB = BHI . Ta c\u00f3 I l\u00e0 \u0111i\u1ec3m \u0111\u1ed1i x\u1ee9ng cu\u1ea3 M qua AB \uf0de AIB = AMB (t\u00ednh ch\u1ea5t \u0111\u1ed1i x\u1ee9ng) M\u00e0 BHD = AMB (cmt) N\u00ean BHD = AIB \uf0de AHBI n\u1ed9i ti\u1ebfp \uf0de IAB = IHB M\u00e0 IAB = MAB ( I l\u00e0 \u0111i\u1ec3m \u0111\u1ed1i x\u1ee9ng cu\u1ea3 M qua AB ) N\u00ean IHB = MAB T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N 10","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH c) G\u1ecdi K l\u00e0 \u0111i\u1ec3m \u0111\u1ed1i x\u1ee9ng c\u1ee7a M qua \u0111\u01b0\u1eddng th\u1eb3ng AC . Ch\u1ee9ng minh ba \u0111i\u1ec3m I, H, K th\u1eb3ng h\u00e0ng. X\u00e9t t\u1ee9 gi\u00e1c BFHD c\u00f3: BFH + BDH = 180\uf0b0 \uf0de BFHD n\u1ed9i ti\u1ebfp \uf0de DHC = FBD = AMC M\u00e0 AKC = AMC ( K l\u00e0 \u0111i\u1ec3m \u0111\u1ed1i x\u1ee9ng c\u1ee7a M qua \u0111\u01b0\u1eddng th\u1eb3ng AC ) N\u00ean DHC = AKC \uf0de AHCK n\u1ed9i ti\u1ebfp \uf0de CHK = CAK M\u00e0 CAM = CAK ( K l\u00e0 \u0111i\u1ec3m \u0111\u1ed1i x\u1ee9ng c\u1ee7a M qua \u0111\u01b0\u1eddng th\u1eb3ng AC ) N\u00ean CAM = CHK X\u00e9t t\u1ee9 gi\u00e1c AFHE c\u00f3: BAC + FHE = 360\uf0b0 \u2212 90\uf0b0 \u2212 90\uf0b0 = 180\uf0b0 \uf0de BAM + MAC + FHE = 180\uf0b0 M\u00e0 \uf0ec\uf0ef\uf0ef\uf0edCBAAMM = IHB (cmt) = CHK (cmt) \uf0ef = BHC ( 2 goc doi dinh) \uf0ef\uf0eeFHE \uf0de IHK = 180\uf0b0 \uf0de I, H, K th\u1eb3ng h\u00e0ng. ----H\u1ebeT--- T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N 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 HUY\u1ec6N NH\u00c0 B\u00c8 NA\u00caM HO\u00cfC: 2023 - 2024 M\u00d4N: TO\u00c1N 9 \u0110\u1ec0 THAM KH\u1ea2O M\u00c3 \u0110\u1ec0: Huy\u1ec7n Nh\u00e0 B\u00e8 - 1 \u0110\u00ea thi g\u1ed3m 8 c\u00e2u ho\u0309i t\u01b0\u0323 lu\u00e2\u0323n. Th\u01a1\u0300i gian: 120 phu\u0301t (kh\u00f4ng k\u00ea\u0309 th\u01a1\u0300i gian pha\u0301t \u0111\u00ea\u0300) C\u00e2u 1. (1,5 \u0111i\u1ec3m). a) V\u1ebd hai \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1 (d) : y = 1 x \u2212 2 v\u00e0 (P) : y = \u22121 x2 tr\u00ean c\u00f9ng m\u1ed9t m\u1eb7t ph\u1eb3ng t\u1ecda \u0111\u1ed9. 24 b) T\u00ecm t\u1ecda \u0111\u1ed9 giao \u0111i\u1ec3m c\u1ee7a hai \u0111\u1ed3 th\u1ecb b\u1eb1ng ph\u00e9p to\u00e1n. C\u00e2u 2. (1 \u0111i\u1ec3m). Cho ph\u01b0\u01a1ng tr\u00ecnh 2x2 \u2212 9x + 4 = 0 . Kh\u00f4ng gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh, h\u00e3y t\u00ednh gi\u00e1 tr\u1ecb c\u1ee7a bi\u1ec3u th\u1ee9c C = 49 . ( ) ( )x1 x1 \u2212 x2 + x2 x2 \u2212 x1 C\u00e2u 3. (1 \u0111i\u1ec3m). M\u1ed9t c\u1eeda h\u00e0ng t\u1ea1p ho\u00e1 nh\u1eadp v\u1ec1 1 th\u00f9ng coca v\u1edbi gi\u00e1 g\u1ed1c ph\u00e2n ph\u1ed1i t\u1eeb \u0111\u1ea1i l\u00fd l\u00e0 192000 \u0111\u1ed3ng\/ 1 th\u00f9ng. Sau \u0111\u00f3 b\u00e1n l\u1ebb cho kh\u00e1ch v\u1edbi gi\u00e1 10000 \u0111\u1ed3ng\/ 1 lon. a) H\u1ecfi v\u1edbi vi\u1ec7c mua v\u00e0 b\u00e1n nh\u01b0 th\u1ebf th\u00ec c\u1eeda h\u00e0ng \u0111\u00e3 thu l\u1eddi bao nhi\u00eau ph\u1ea7n tr\u0103m so v\u1edbi gi\u00e1 g\u1ed1c? (Bi\u1ebft m\u1ed9t th\u00f9ng coca c\u00f3 24 lon) b) \u0110\u1ec3 thu l\u1eddi l\u00e0 50% th\u00ec c\u1eeda h\u00e0ng c\u1ea7n b\u00e1n l\u1ebb cho kh\u00e1ch v\u1edbi gi\u00e1 l\u00e0 bao nhi\u00eau tr\u00ean 1 lon. C\u00e2u 4. (0,75 \u0111i\u1ec3m). C\u00e0ng l\u00ean cao kh\u00f4ng kh\u00ed c\u00e0ng lo\u00e3ng n\u00ean \u00e1p su\u1ea5t kh\u00ed quy\u1ec3n c\u00e0ng gi\u1ea3m. G\u1ecdi y l\u00e0 \u0111\u1ea1i l\u01b0\u1ee3ng bi\u1ec3u th\u1ecb cho \u00e1p su\u1ea5t c\u1ee7a kh\u00ed quy\u1ec3n (t\u00ednh b\u1eb1ng mmHg ) v\u00e0 x l\u00e0 \u0111\u1ea1i l\u01b0\u1ee3ng bi\u1ec3u th\u1ecb cho \u0111\u1ed9 cao so v\u1edbi m\u1eb7t n\u01b0\u1edbc bi\u1ec3n (t\u00ednh b\u1eb1ng m\u00e9t). Ng\u01b0\u1eddi ta th\u1ea5y v\u1edbi nh\u1eefng \u0111\u1ed9 cao kh\u00f4ng l\u1edbn l\u1eafm th\u00ec 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 c\u00f3 \u0111\u1ed3 th\u1ecb nh\u01b0 h\u00ecnh v\u1ebd sau: a) H\u00e3y x\u00e1c \u0111\u1ecbnh c\u00e1c h\u1ec7 s\u1ed1 a v\u00e0 b b) T\u1ea1i c\u1ef1c b\u1eafc c\u1ee7a Vi\u1ec7t Nam c\u00f3 m\u1ed9t \u0111\u1ecba danh kh\u00e1 n\u1ed5i ti\u1ebfng l\u00e0 C\u1ed9t c\u1edd L\u0169ng C\u00fa \u1edf x\u00e3 \u0110\u1ed3ng V\u0103n, t\u1ec9nh H\u00e0 Giang n\u1eb1m \u1edf \u0111\u1ed9 cao kho\u1ea3ng 1470m so v\u1edbi m\u1ef1c n\u01b0\u1edbc bi\u1ec3n. H\u1ecfi \u00e1p su\u1ea5t kh\u00ed quy\u1ec3n t\u1ea1i \u0111\u00e2y l\u00e0 bao nhi\u00eau? C\u00e2u 5. (0,75 \u0111i\u1ec3m). M\u1ed9t doanh nghi\u1ec7p v\u1eadn t\u1ea3i d\u1ef1 \u0111\u1ecbnh s\u1ebd ch\u1edf 40 t\u1ea5n g\u1ea1o trong m\u1ed9t ng\u00e0y \u0111\u1ec3 ph\u00e2n ph\u1ed1i \u0111\u1ebfn c\u00e1c \u0111\u1ea1i l\u00fd t\u1eeb m\u1ed9t kho h\u00e0ng l\u01b0\u01a1ng th\u1ef1c. Nh\u01b0ng tr\u00ean th\u1ef1c t\u1ebf, doanh nghi\u1ec7p v\u1eadn t\u1ea3i \u0111\u00e3 ch\u1edf \u0111\u01b0\u1ee3c v\u01b0\u1ee3t m\u1ee9c 25% , v\u00ec v\u1eady m\u00e0 \u0111\u00e3 th\u1ef1c hi\u1ec7n \u0111\u01b0\u1ee3c s\u1edbm h\u01a1n 4 ng\u00e0y so v\u1edbi d\u1ef1 \u0111\u1ecbnh. H\u1ecfi ban \u0111\u1ea7u trong kho c\u00f3 bao nhi\u00eau t\u1ea5n g\u1ea1o? C\u00e2u 6. (1 \u0111i\u1ec3m). Hai c\u1eeda h\u00e0ng A v\u00e0 B \u0111\u1ec1u nh\u1eadp v\u1ec1 (gi\u00e1 g\u1ed1c) m\u1ed9t nh\u00e3n h\u00e0ng ti vi v\u1edbi gi\u00e1 l\u00e0 10000000 \u0111\u1ed3ng. C\u1eeda h\u00e0ng A ni\u00eam y\u1ebft s\u1ea3n ph\u1ea9m \u0111\u00f3 v\u1edbi gi\u00e1 t\u0103ng 40% so v\u1edbi gi\u00e1 nh\u1eadp v\u1ec1, nh\u01b0ng l\u1ea1i b\u00e1n v\u1edbi gi\u00e1 gi\u1ea3m 20% so v\u1edbi gi\u00e1 ni\u00eam y\u1ebft. C\u1eeda h\u00e0ng B ni\u00eam y\u1ebft s\u1ea3n ph\u1ea9m \u0111\u00f3 v\u1edbi gi\u00e1 t\u0103ng 20% so v\u1edbi gi\u00e1 nh\u1eadp v\u1ec1, nh\u01b0ng l\u1ea1i b\u00e1n v\u1edbi gi\u00e1 gi\u1ea3m 5% so v\u1edbi gi\u00e1 ni\u00eam y\u1ebft. Bi\u1ebft gi\u00e1 ni\u00eam y\u1ebft l\u00e0 gi\u00e1 m\u00e0 c\u1eeda h\u00e0ng \u0111\u1ec1 xu\u1ea5t v\u1edbi ng\u01b0\u1eddi ti\u00eau d\u00f9ng. Theo em, ng\u01b0\u1eddi ti\u00eau d\u00f9ng ch\u1ecdn mua ti vi t\u1eeb c\u1eeda h\u00e0ng n\u00e0o s\u1ebd c\u00f3 l\u1ee3i h\u01a1n? Em h\u00e3y gi\u1ea3i th\u00edch? C\u00e2u 7. (1 \u0111i\u1ec3m). N\u00f3n l\u00e1 l\u00e0 bi\u1ec3u t\u01b0\u1ee3ng cho s\u1ef1 d\u1ecbu d\u00e0ng, b\u00ecnh d\u1ecb, th\u00e2n thi\u1ec7n c\u1ee7a ng\u01b0\u1eddi ph\u1ee5 n\u1eef Vi\u1ec7t Nam t\u1eeb ng\u00e0n \u0111\u1eddi nay; n\u00f3n l\u00e1 b\u00e0i th\u01a1 l\u00e0 m\u1ed9t \u0111\u1eb7c tr\u01b0ng c\u1ee7a x\u1ee9 Hu\u1ebf. M\u1ed9t chi\u1ebfc n\u00f3n l\u00e1 ho\u00e0n thi\u1ec7n c\u1ea7n qua nhi\u1ec1u c\u00f4ng \u0111o\u1ea1n t\u1eeb l\u00ean r\u1eebng h\u00e1i l\u00e1, r\u1ed3i s\u1ea5y l\u00e1, m\u1edf, \u1ee7i, ch\u1ecdn l\u00e1, x\u00e2y \u0111\u1ed9n v\u00e0nh, 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 ch\u1eb1m, c\u1eaft l\u00e1, n\u1ee9c v\u00e0nh, c\u1eaft ch\u1ec9, \u2026 Nh\u1eb1m l\u00e0m \u0111\u1eb9p v\u00e0 t\u00f4n vinh th\u00eam cho chi\u1ebfc n\u00f3n l\u00e1 x\u1ee9 Hu\u1ebf, c\u00e1c ngh\u1ec7 nh\u00e2n c\u00f2n \u00e9p tranh v\u00e0 v\u00e0i d\u00f2ng th\u01a1 v\u00e0o gi\u1eefa hai l\u1edbp l\u00e1: Khung c\u1ee7a n\u00f3n l\u00e1 c\u00f3 d\u1ea1ng h\u00ecnh n\u00f3n \u0111\u01b0\u1ee3c l\u00e0m b\u1edfi c\u00e1c thanh g\u1ed7 n\u1ed1i t\u1eeb \u0111\u1ec9nh t\u1edbi \u0111\u00e1y nh\u01b0 c\u00e1c \u0111\u01b0\u1eddng sinh ( ) , 16 v\u00e0nh n\u00f3n \u0111\u01b0\u1ee3c l\u00e0m t\u1eeb nh\u1eefng thanh tre m\u1ea3nh nh\u1ecf, d\u1ebbo dai u\u1ed1n th\u00e0nh nh\u1eefng v\u00f2ng tr\u00f2n c\u00f3 \u0111\u01b0\u1eddng k\u00ednh to, nh\u1ecf kh\u00e1c nhau, c\u00e1i nh\u1ecf nh\u1ea5t to b\u1eb1ng \u0111\u1ed3ng xu. \u2022 \u0110\u01b0\u1eddng k\u00ednh (d = 2r) c\u1ee7a v\u00e0nh n\u00f3n l\u1edbn nh\u1ea5t kho\u1ea3ng 40(cm) . \u2022 Chi\u1ec1u cao (h) c\u1ee7a chi\u1ebfc n\u00f3n l\u00e1 kho\u1ea3ng 18(cm) . a) T\u00ednh \u0111\u1ed9 d\u00e0i c\u1ee7a thanh tre u\u1ed1n th\u00e0nh v\u00f2ng tr\u00f2n l\u1edbn nh\u1ea5t c\u1ee7a v\u00e0nh chi\u1ebfc n\u00f3n l\u00e1. (Kh\u00f4ng k\u1ec3 ph\u1ea7n ch\u1eafp n\u1ed1i, bi\u1ebft \uf070 \uf0bb 3,14 ). b) T\u00ednh di\u1ec7n t\u00edch ph\u1ea7n l\u00e1 ph\u1ee7 xung quanh c\u1ee7a chi\u1ebfc n\u00f3n l\u00e1. (Kh\u00f4ng k\u1ec3 ph\u1ea7n ch\u1eafp n\u1ed1i t\u00ednh g\u1ea7n \u0111\u00fang \u0111\u1ebfn h\u00e0ng \u0111\u01a1n v\u1ecb). Bi\u1ebft di\u1ec7n t\u00edch xung quanh c\u1ee7a h\u00ecnh n\u00f3n l\u00e0 S = \uf070.r. . C\u00e2u 8. (3 \u0111i\u1ec3m) Cho \uf044ABC nh\u1ecdn ( AB \uf03c AC) n\u1ed9i ti\u1ebfp \u0111\u01b0\u1eddng tr\u00f2n (O) c\u00f3 2 \u0111\u01b0\u1eddng cao BE v\u00e0 CF c\u1eaft nhau t\u1ea1i H . EF k\u00e9o d\u00e0i c\u1eaft BC t\u1ea1i M , v\u00e0 MD l\u00e0 ti\u1ebfp tuy\u1ebfn c\u1ee7a \u0111\u01b0\u1eddng tr\u00f2n (O) ( D thu\u1ed9c cung nh\u1ecf BC ) a) Ch\u1ee9ng minh: AEHF n\u1ed9i ti\u1ebfp, v\u00e0 BFEC n\u1ed9i ti\u1ebfp. b) Ch\u1ee9ng minh: OA vu\u00f4ng g\u00f3c v\u1edbi EF v\u00e0 MD2 = ME.MF c) Ch\u1ee9ng minh: DA l\u00e0 ph\u00e2n gi\u00e1c c\u1ee7a g\u00f3c EDF . ----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). a) V\u1ebd hai \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1 (d) : y = 1 x \u2212 2 v\u00e0 (P) : y = \u22121 x2 tr\u00ean c\u00f9ng m\u1ed9t m\u1eb7t ph\u1eb3ng t\u1ecda \u0111\u1ed9. 24 b) T\u00ecm t\u1ecda \u0111\u1ed9 giao \u0111i\u1ec3m c\u1ee7a hai \u0111\u1ed3 th\u1ecb b\u1eb1ng ph\u00e9p to\u00e1n. L\u1eddi gi\u1ea3i a) V\u1ebd hai \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1 (d) : y = 1 x \u2212 2 v\u00e0 (P) : y = \u22121 x2 tr\u00ean c\u00f9ng m\u1ed9t m\u1eb7t ph\u1eb3ng t\u1ecda \u0111\u1ed9. 24 BGT: x \u22124 \u22122 0 2 4 y = \u22121 x2 \u22124 \u22121 0 \u22121 \u22124 4 x 02 y= 1x\u22122 \u22122 \u22121 2 b) T\u00ecm t\u1ecda \u0111\u1ed9 giao \u0111i\u1ec3m c\u1ee7a c\u1ee7a hai \u0111\u1ed3 th\u1ecb b\u1eb1ng ph\u00e9p to\u00e1n. Ph\u01b0\u01a1ng tr\u00ecnh ho\u00e0nh \u0111\u1ed9 giao \u0111i\u1ec3m c\u1ee7a (P) v\u00e0 (d) : \u2212 1 x2 = 1 x \u2212 2 42 \uf0db x2 + 2x \u2212 8 = 0 \uf0db \uf0e9x = 2 \uf0ea\uf0ebx = \u22124 Thay x = 2 v\u00e0o y = 1 x \u2212 2 , ta \u0111\u01b0\u1ee3c: y = 1 .2 \u2212 2 = \u22121 . 22 Thay x = \u22124 v\u00e0o y = 1 x \u2212 2 , ta \u0111\u01b0\u1ee3c: y = 1 .(\u22124) \u2212 2 = \u22124 . 22 V\u1eady (2; \u2212 1) , (\u22124; \u2212 4) l\u00e0 hai giao \u0111i\u1ec3m c\u1ea7n t\u00ecm. T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N 3","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH C\u00e2u 2. (1 \u0111i\u1ec3m) Cho ph\u01b0\u01a1ng tr\u00ecnh 2x2 \u2212 9x + 4 = 0 . Kh\u00f4ng gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh, h\u00e3y t\u00ednh gi\u00e1 tr\u1ecb c\u1ee7a bi\u1ec3u th\u1ee9c C = 49 ( ) ( )x1 x1 \u2212 x2 + x2 x2 \u2212 x1 L\u1eddi gi\u1ea3i V\u00ec \uf044 = b2 \u2212 4ac = (\u22129)2 \u2212 4.2.4 = 49 \uf03e 0 N\u00ean ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m ph\u00e2n bi\u1ec7t x1 , x2 . \uf0ec\uf0ef\uf0efS = x1 + x2 = \u2212b = 9 \uf0ed = x1 a 2 Theo \u0111\u1ecbnh l\u00ed Vi-et, ta c\u00f3: \uf0ef\uf0ef\uf0eeP =2 c .x2 = a 49 ( ) ( )Ta c\u00f3: C = x1 x1 \u2212 x2 + x2 x2 \u2212 x1 C = x12 \u2212 49 \u2212 x1x2 x1x2 + x22 C = S2 \u2212 49 \u2212 2P 2P C = S2 49 \u2212 4P C = 49 \uf0e6 9 \uf0f62 \u2212 4.2 \uf0e7\uf0e8 2 \uf0f7 \uf0f8 C=4 C\u00e2u 3. (1 \u0111i\u1ec3m) M\u1ed9t c\u1eeda h\u00e0ng t\u1ea1p ho\u00e1 nh\u1eadp v\u1ec1 1 th\u00f9ng coca v\u1edbi gi\u00e1 g\u1ed1c ph\u00e2n ph\u1ed1i t\u1eeb \u0111\u1ea1i l\u00fd l\u00e0 192 000\u0111\u1ed3ng\/ 1 th\u00f9ng. Sau \u0111\u00f3 b\u00e1n l\u1ebb cho kh\u00e1ch v\u1edbi gi\u00e1 10 000 \u0111\u1ed3ng\/ 1 lon. a) H\u1ecfi v\u1edbi vi\u1ec7c mua v\u00e0 b\u00e1n nh\u01b0 th\u1ebf th\u00ec c\u1eeda h\u00e0ng \u0111\u00e3 thu l\u1eddi bao nhi\u00eau ph\u1ea7n tr\u0103m so v\u1edbi gi\u00e1 g\u1ed1c? (Bi\u1ebft m\u1ed9t th\u00f9ng coca c\u00f3 24 lon) b) \u0110\u1ec3 thu l\u1eddi l\u00e0 50% th\u00ec c\u1eeda h\u00e0ng c\u1ea7n b\u00e1n l\u1ebb cho kh\u00e1ch v\u1edbi gi\u00e1 l\u00e0 bao nhi\u00eau tr\u00ean 1 lon. L\u1eddi gi\u1ea3i a) H\u1ecfi v\u1edbi vi\u1ec7c mua v\u00e0 b\u00e1n nh\u01b0 th\u1ebf th\u00ec c\u1eeda h\u00e0ng \u0111\u00e3 thu l\u1eddi bao nhi\u00eau ph\u1ea7n tr\u0103m so v\u1edbi gi\u00e1 g\u1ed1c? (Bi\u1ebft m\u1ed9t th\u00f9ng coca c\u00f3 24 lon) Ph\u1ea7n tr\u0103m l\u00e3i so v\u1edbi gi\u00e1 g\u1ed1c : 24.10 000 192 000 .100% 25% 192 000 b) \u0110\u1ec3 thu l\u1eddi l\u00e0 50% th\u00ec c\u1eeda h\u00e0ng c\u1ea7n b\u00e1n l\u1ebb cho kh\u00e1ch v\u1edbi gi\u00e1 l\u00e0 bao nhi\u00eau tr\u00ean 1 lon. G\u1ecdi x (\u0111\u1ed3ng) l\u00e0 gi\u00e1 ti\u1ec1n 1 lon n\u01b0\u1edbc (x 0) T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N 4","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH 24.x 192 000 .100% 50% 192 000 24.x 192 000 0, 5.192 000 x 12 000 n V\u1eady gi\u00e1 ti\u1ec1n 1 lon n\u01b0\u1edbc l\u00e0 12 000 \u0111\u1ed3ng \u0111\u1ec3 thu l\u1eddi l\u00e0 50% C\u00e2u 4. (0,75 \u0111i\u1ec3m) C\u00e0ng l\u00ean cao kh\u00f4ng kh\u00ed c\u00e0ng lo\u00e3ng n\u00ean \u00e1p su\u1ea5t kh\u00ed quy\u1ec3n c\u00e0ng gi\u1ea3m. G\u1ecdi y l\u00e0 \u0111\u1ea1i l\u01b0\u1ee3ng bi\u1ec3u th\u1ecb cho \u00e1p su\u1ea5t c\u1ee7a kh\u00ed quy\u1ec3n (t\u00ednh b\u1eb1ng mmHg ) v\u00e0 x l\u00e0 \u0111\u1ea1i l\u01b0\u1ee3ng bi\u1ec3u th\u1ecb cho \u0111\u1ed9 cao so v\u1edbi m\u1eb7t n\u01b0\u1edbc bi\u1ec3n (t\u00ednh b\u1eb1ng m\u00e9t). Ng\u01b0\u1eddi ta th\u1ea5y v\u1edbi nh\u1eefng \u0111\u1ed9 cao kh\u00f4ng l\u1edbn l\u1eafm th\u00ec 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 c\u00f3 \u0111\u1ed3 th\u1ecb nh\u01b0 h\u00ecnh v\u1ebd sau: a) H\u00e3y x\u00e1c \u0111\u1ecbnh c\u00e1c h\u1ec7 s\u1ed1 a v\u00e0 b b) T\u1ea1i c\u1ef1c b\u1eafc c\u1ee7a Vi\u1ec7t Nam c\u00f3 m\u1ed9t \u0111\u1ecba danh kh\u00e1 n\u1ed5i ti\u1ebfng l\u00e0 C\u1ed9t c\u1edd L\u0169ng C\u00fa \u1edf x\u00e3 \u0110\u1ed3ng V\u0103n, t\u1ec9nh H\u00e0 Giang n\u1eb1m \u1edf \u0111\u1ed9 cao kho\u1ea3ng 1470m so v\u1edbi m\u1ef1c n\u01b0\u1edbc bi\u1ec3n. H\u1ecfi \u00e1p su\u1ea5t kh\u00ed quy\u1ec3n t\u1ea1i \u0111\u00e2y l\u00e0 bao nhi\u00eau? L\u1eddi gi\u1ea3i a) H\u00e3y x\u00e1c \u0111\u1ecbnh c\u00e1c h\u1ec7 s\u1ed1 a v\u00e0 b Theo \u0111\u1ec1 b\u00e0i, ta c\u00f3: V\u1edbi \uf0ecx = 1600 \uf0de 632 = 1600.a + b . (1) \uf0ed\uf0eey = 632 V\u1edbi \uf0ecx =0 \uf0de 760 = 0.a + b . (2) \uf0ed\uf0eey = 760 T\u1eeb (1) v\u00e0 (2) ta c\u00f3 h\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh: \uf0ec0a + b = 760 \uf0db \uf0ec\uf0efa = \u2212 2 . \uf0ed\uf0ee1600a + b = 632 \uf0ed 25 \uf0ef\uf0eeb = 760 V\u1eady: a = \u2212 2 , b = 760 v\u00e0 y = \u2212 2 x + 760 . 25 25 b) T\u1ea1i c\u1ef1c b\u1eafc c\u1ee7a Vi\u1ec7t Nam c\u00f3 m\u1ed9t \u0111\u1ecba danh kh\u00e1 n\u1ed5i ti\u1ebfng l\u00e0 C\u1ed9t c\u1edd L\u0169ng C\u00fa \u1edf x\u00e3 \u0110\u1ed3ng V\u0103n, t\u1ec9nh H\u00e0 Giang n\u1eb1m \u1edf \u0111\u1ed9 cao kho\u1ea3ng 1470m so v\u1edbi m\u1ef1c n\u01b0\u1edbc bi\u1ec3n. H\u1ecfi \u00e1p su\u1ea5t kh\u00ed quy\u1ec3n t\u1ea1i \u0111\u00e2y l\u00e0 bao nhi\u00eau? Thay x 1470 v\u00e0o y = \u2212 2 x + 760 ta \u0111\u01b0\u1ee3c: 25 y = \u2212 2 .1470 + 760 25 \uf0db y = 642,4 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 V\u1eady \u00e1p su\u1ea5t kh\u00ed quy\u1ec3n t\u1ea1i \u0111\u00e2y l\u00e0 642, 4mmHg C\u00e2u 5. (0,75 \u0111i\u1ec3m). M\u1ed9t doanh nghi\u1ec7p v\u1eadn t\u1ea3i d\u1ef1 \u0111\u1ecbnh s\u1ebd ch\u1edf 40 t\u1ea5n g\u1ea1o trong m\u1ed9t ng\u00e0y \u0111\u1ec3 ph\u00e2n ph\u1ed1i \u0111\u1ebfn c\u00e1c \u0111\u1ea1i l\u00fd t\u1eeb m\u1ed9t kho h\u00e0ng l\u01b0\u01a1ng th\u1ef1c. Nh\u01b0ng tr\u00ean th\u1ef1c t\u1ebf, doanh nghi\u1ec7p v\u1eadn t\u1ea3i \u0111\u00e3 ch\u1edf \u0111\u01b0\u1ee3c v\u01b0\u1ee3t m\u1ee9c 25% , v\u00ec v\u1eady m\u00e0 \u0111\u00e3 th\u1ef1c hi\u1ec7n \u0111\u01b0\u1ee3c s\u1edbm h\u01a1n 4 ng\u00e0y so v\u1edbi d\u1ef1 \u0111\u1ecbnh. H\u1ecfi ban \u0111\u1ea7u trong kho c\u00f3 bao nhi\u00eau t\u1ea5n g\u1ea1o? L\u1eddi gi\u1ea3i G\u1ecdi x (t\u1ea5n) l\u00e0 s\u1ed1 t\u1ea5n g\u1ea1o ban \u0111\u1ea7u x 0 x 40 (ng\u00e0y) l\u00e0 s\u1ed1 ng\u00e0y giao h\u00e0ng d\u1ef1 \u0111\u1ecbnh \u0111\u1ec3 giao h\u1ebft s\u1ed1 t\u1ea5n g\u1ea1o. x 50 (ng\u00e0y) l\u00e0 s\u1ed1 ng\u00e0y giao h\u00e0ng th\u1ef1c t\u1ebf \u0111\u1ec3 giao h\u1ebft s\u1ed1 t\u1ea5n g\u1ea1o. Ta c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh: x x 4 40 50 50x 40x 4 2000 10x 8000 x 800 n V\u1eady ban \u0111\u1ea7u c\u00f3 800 t\u1ea5n g\u1ea1o. C\u00e2u 6. (1 \u0111i\u1ec3m). Hai c\u1eeda h\u00e0ng A v\u00e0 B \u0111\u1ec1u nh\u1eadp v\u1ec1 (gi\u00e1 g\u1ed1c) m\u1ed9t nh\u00e3n h\u00e0ng ti vi v\u1edbi gi\u00e1 l\u00e0 10 000 000 \u0111\u1ed3ng. C\u1eeda h\u00e0ng A ni\u00eam y\u1ebft s\u1ea3n ph\u1ea9m \u0111\u00f3 v\u1edbi gi\u00e1 t\u0103ng 40% so v\u1edbi gi\u00e1 nh\u1eadp v\u1ec1, nh\u01b0ng l\u1ea1i b\u00e1n v\u1edbi gi\u00e1 gi\u1ea3m 20% so v\u1edbi gi\u00e1 ni\u00eam y\u1ebft. C\u1eeda h\u00e0ng B ni\u00eam y\u1ebft s\u1ea3n ph\u1ea9m \u0111\u00f3 v\u1edbi gi\u00e1 t\u0103ng 20% so v\u1edbi gi\u00e1 nh\u1eadp v\u1ec1, nh\u01b0ng l\u1ea1i b\u00e1n v\u1edbi gi\u00e1 gi\u1ea3m 5% so v\u1edbi gi\u00e1 ni\u00eam y\u1ebft. Bi\u1ebft gi\u00e1 ni\u00eam y\u1ebft l\u00e0 gi\u00e1 m\u00e0 c\u1eeda h\u00e0ng \u0111\u1ec1 xu\u1ea5t v\u1edbi ng\u01b0\u1eddi ti\u00eau d\u00f9ng. Theo em, ng\u01b0\u1eddi ti\u00eau d\u00f9ng ch\u1ecdn mua ti vi t\u1eeb c\u1eeda h\u00e0ng n\u00e0o s\u1ebd c\u00f3 l\u1ee3i h\u01a1n? Em h\u00e3y gi\u1ea3i th\u00edch? L\u1eddi gi\u1ea3i Gi\u00e1 ni\u00eam y\u1ebft ti vi \u1edf c\u1eeda h\u00e0ng A : 10 000 000. 1 40% . 1 20% 11200 000 (\u0111\u1ed3ng) Gi\u00e1 ni\u00eam y\u1ebft ti vi \u1edf c\u1eeda h\u00e0ng B : 10 000 000. 1 20% . 1 5% 11400 000 (\u0111\u1ed3ng) V\u00ec 11200 000 11400 000 n\u00ean mua \u1edf c\u1eeda h\u00e0ng A s\u1ebd l\u1ee3i h\u01a1n. C\u00e2u 7. (1 \u0111i\u1ec3m). N\u00f3n l\u00e1 l\u00e0 bi\u1ec3u t\u01b0\u1ee3ng cho s\u1ef1 d\u1ecbu d\u00e0ng, b\u00ecnh d\u1ecb, th\u00e2n thi\u1ec7n c\u1ee7a ng\u01b0\u1eddi ph\u1ee5 n\u1eef Vi\u1ec7t Nam t\u1eeb ng\u00e0n \u0111\u1eddi nay; n\u00f3n l\u00e1 b\u00e0i th\u01a1 l\u00e0 m\u1ed9t \u0111\u1eb7c tr\u01b0ng c\u1ee7a x\u1ee9 Hu\u1ebf. M\u1ed9t chi\u1ebfc n\u00f3n l\u00e1 ho\u00e0n thi\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\u1ea7n qua nhi\u1ec1u c\u00f4ng \u0111o\u1ea1n t\u1eeb l\u00ean r\u1eebng h\u00e1i l\u00e1, r\u1ed3i s\u1ea5y l\u00e1, m\u1edf, \u1ee7i, ch\u1ecdn l\u00e1, x\u00e2y \u0111\u1ed9n v\u00e0nh, ch\u1eb1m, c\u1eaft l\u00e1, n\u1ee9c v\u00e0nh, c\u1eaft ch\u1ec9, \u2026 Nh\u1eb1m l\u00e0m \u0111\u1eb9p v\u00e0 t\u00f4n vinh th\u00eam cho chi\u1ebfc n\u00f3n l\u00e1 x\u1ee9 Hu\u1ebf, c\u00e1c ngh\u1ec7 nh\u00e2n c\u00f2n \u00e9p tranh v\u00e0 v\u00e0i d\u00f2ng th\u01a1 v\u00e0o gi\u1eefa hai l\u1edbp l\u00e1: Khung c\u1ee7a n\u00f3n l\u00e1 c\u00f3 d\u1ea1ng h\u00ecnh n\u00f3n \u0111\u01b0\u1ee3c l\u00e0m b\u1edfi c\u00e1c thanh g\u1ed7 n\u1ed1i t\u1eeb \u0111\u1ec9nh t\u1edbi \u0111\u00e1y nh\u01b0 c\u00e1c \u0111\u01b0\u1eddng sinh , 16 v\u00e0nh n\u00f3n \u0111\u01b0\u1ee3c l\u00e0m t\u1eeb nh\u1eefng thanh tre m\u1ea3nh nh\u1ecf, d\u1ebbo dai u\u1ed1n th\u00e0nh nh\u1eefng v\u00f2ng tr\u00f2n c\u00f3 \u0111\u01b0\u1eddng k\u00ednh to, nh\u1ecf kh\u00e1c nhau, c\u00e1i nh\u1ecf nh\u1ea5t to b\u1eb1ng \u0111\u1ed3ng xu. \u2022 \u0110\u01b0\u1eddng k\u00ednh d 2r c\u1ee7a v\u00e0nh n\u00f3n l\u1edbn nh\u1ea5t kho\u1ea3ng 40(cm) \u2022 Chi\u1ec1u cao h c\u1ee7a chi\u1ebfc n\u00f3n l\u00e1 kho\u1ea3ng 18(cm) a) T\u00ednh \u0111\u1ed9 d\u00e0i c\u1ee7a thanh tre u\u1ed1n th\u00e0nh v\u00f2ng tr\u00f2n l\u1edbn nh\u1ea5t c\u1ee7a v\u00e0nh chi\u1ebfc n\u00f3n l\u00e1. (Kh\u00f4ng k\u1ec3 ph\u1ea7n ch\u1eafp n\u1ed1i, bi\u1ebft 3,14 ) b) T\u00ednh di\u1ec7n t\u00edch ph\u1ea7n l\u00e1 ph\u1ee7 xung quanh c\u1ee7a chi\u1ebfc n\u00f3n l\u00e1. (Kh\u00f4ng k\u1ec3 ph\u1ea7n ch\u1eafp n\u1ed1i t\u00ednh g\u1ea7n \u0111\u00fang \u0111\u1ebfn h\u00e0ng \u0111\u01a1n v\u1ecb). Bi\u1ebft di\u1ec7n t\u00edch xung quanh c\u1ee7a h\u00ecnh n\u00f3n l\u00e0 S .r. L\u1eddi gi\u1ea3i a) T\u00ednh \u0111\u1ed9 d\u00e0i c\u1ee7a thanh tre u\u1ed1n th\u00e0nh v\u00f2ng tr\u00f2n l\u1edbn nh\u1ea5t c\u1ee7a v\u00e0nh chi\u1ebfc n\u00f3n l\u00e1. (Kh\u00f4ng k\u1ec3 ph\u1ea7n ch\u1eafp n\u1ed1i, bi\u1ebft 3,14 ) \u0110\u1ed9 d\u00e0i c\u1ee7a thanh tre u\u1ed1n th\u00e0nh v\u00f2ng tr\u00f2n l\u1edbn l\u00e0: 20 20.3,14 62, 83 cm b. T\u00ednh di\u1ec7n t\u00edch ph\u1ea7n l\u00e1 ph\u1ee7 xung quanh c\u1ee7a chi\u1ebfc n\u00f3n l\u00e1. (Kh\u00f4ng k\u1ec3 ph\u1ea7n ch\u1eafp n\u1ed1i t\u00ednh g\u1ea7n \u0111\u00fang \u0111\u1ebfn h\u00e0ng \u0111\u01a1n v\u1ecb). Bi\u1ebft di\u1ec7n t\u00edch xung quanh c\u1ee7a h\u00ecnh n\u00f3n l\u00e0 S .r. X\u00e9t ABC vu\u00f4ng t\u1ea1i B AB2 BC 2 AC 2 182 202 AC 2 724 AC 2 AC 2 181 cm Di\u1ec7n t\u00edch ph\u1ea7n l\u00e1 ph\u1ee7 xung quanh: S .r. 3,14.20.2 181 1690, 63 cm2 T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N 7","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH C\u00e2u 8. (3 \u0111i\u1ec3m) Cho \uf044ABC nh\u1ecdn ( AB \uf03c AC) n\u1ed9i ti\u1ebfp \u0111\u01b0\u1eddng tr\u00f2n (O) c\u00f3 2 \u0111\u01b0\u1eddng cao BE v\u00e0 CF c\u1eaft nhau t\u1ea1i H . EF k\u00e9o d\u00e0i c\u1eaft BC t\u1ea1i M , v\u00e0 MD l\u00e0 ti\u1ebfp tuy\u1ebfn c\u1ee7a \u0111\u01b0\u1eddng tr\u00f2n O ( D thu\u1ed9c cung nh\u1ecf BC ) a) Ch\u1ee9ng minh: AEHF n\u1ed9i ti\u1ebfp, v\u00e0 BFEC n\u1ed9i ti\u1ebfp. b) Ch\u1ee9ng minh: OA vu\u00f4ng g\u00f3c v\u1edbi EF v\u00e0 MD2 ME.MF c) Ch\u1ee9ng minh: DA l\u00e0 ph\u00e2n gi\u00e1c c\u1ee7a g\u00f3c EDF. L\u1eddi gi\u1ea3i x A Q E I S F O H MB C D a) Ch\u1ee9ng minh: AEHF n\u1ed9i ti\u1ebfp, v\u00e0 BFEC n\u1ed9i ti\u1ebfp. X\u00e9t t\u1ee9 gi\u00e1c AEHF , c\u00f3: \uf0ec\uf0efAEH = 90\uf0b0(BE \u22a5 AC) \uf0ed \uf0ef\uf0ee AFH = 90\uf0b0(CF \u22a5 AB) \uf0de AEH + AFH = 180\uf0b0 \uf0de T\u1ee9 gi\u00e1c AEHF n\u1ed9i ti\u1ebfp v\u00ec c\u00f3 hai g\u00f3c \u0111\u1ed1i b\u00f9 nhau. X\u00e9t t\u1ee9 gi\u00e1c BFEC , c\u00f3: \uf0ef\uf0ecBFC = 90\uf0b0(CF \u22a5 AB) \uf0ed \uf0ee\uf0efBEC = 90\uf0b0( BE \u22a5 AC ) \uf0de BFC = BEC = 90\uf0b0 T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N 8","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH \uf0de T\u1ee9 gi\u00e1c BFEC n\u1ed9i ti\u1ebfp v\u00ec c\u00f3 hai \u0111\u1ec9nh k\u1ec1 c\u00f9ng nh\u00ecn m\u1ed9t c\u1ea1nh d\u01b0\u1edbi hai g\u00f3c b\u1eb1ng nhau. b) Ch\u1ee9ng minh: OA vu\u00f4ng g\u00f3c v\u1edbi EF v\u00e0 MD2 ME.MF V\u1ebd Ax l\u00e0 ti\u1ebfp tuy\u1ebfn c\u1ee7a O xAE ABC (c\u00f9ng ch\u1eafn AC ) AEF ABC ( BFEC n\u1ed9i ti\u1ebfp, g\u00f3c ngo\u00e0i b\u1eb1ng g\u00f3c \u0111\u1ed1i trong) xAE AEF M\u00e0 hai g\u00f3c n\u00e0y \u1edf v\u1ecb tr\u00ed so le trong \uf0de Ax \/\/ EF M\u00e0 OA Ax OA EF Ch\u1ee9ng minh MBD \u0111\u1ed3ng d\u1ea1ng MDC (g-g) Suy ra MD2 MB.MC Ch\u1ee9ng minh MFB \u0111\u1ed3ng d\u1ea1ng MEC (g-g) Suy ra MF.ME MB.MC Do \u0111\u00f3 MD2 ME.MF c) Ch\u1ee9ng minh: DA l\u00e0 ph\u00e2n gi\u00e1c c\u1ee7a g\u00f3c EDF. K\u00e9o d\u00e0i DF , DE c\u1eaft O t\u1ea1i I v\u00e0 Q Ch\u1ee9ng minh \u0111\u01b0\u1ee3c MFD \u0111\u1ed3ng d\u1ea1ng MDE (c-g-c) MDF MED M\u00e0 MDF MDI IQD (c\u00f9ng ch\u1eafn DI ) EFD IQD m\u00e0 hai g\u00f3c n\u00e0y \u1edf v\u1ecb tr\u00ed \u0111\u1ed3ng v\u1ecb \uf0de IQ \/\/ EF IQ OA t\u1ea1i S AIQ c\u00f3 AS l\u00e0 \u0111\u01b0\u1eddng trung tr\u1ef1c AIQ c\u00e2n t\u1ea1i A IA IQ s\u0111IA = s\u0111 AQ IDA ADQ hay FDA ADE DA l\u00e0 ph\u00e2n gi\u00e1c c\u1ee7a g\u00f3c DEF . 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 ----H\u1ebeT--- T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N 10","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH S\u00d4\u00db GD&\u00d1T TP HO\u00c0 CH\u00cd MINH \u00d1E\u00c0 THAM KHA\u00dbO TUYE\u00c5N SINH 10 PHO\u00d8NG G\u00d1&\u00d1T HUY\u1ec6N NH\u00c0 B\u00c8 NA\u00caM HO\u00cfC: 2023 - 2024 M\u00d4N: TO\u00c1N 9 \u0110\u1ec0 THAM KH\u1ea2O M\u00c3 \u0110\u1ec0: Huy\u1ec7n Nh\u00e0 B\u00e8 - 2 \u0110\u00ea thi g\u1ed3m 8 c\u00e2u ho\u0309i t\u01b0\u0323 lu\u00e2\u0323n. Th\u01a1\u0300i gian: 120 phu\u0301t (kh\u00f4ng k\u00ea\u0309 th\u01a1\u0300i gian pha\u0301t \u0111\u00ea\u0300) B\u00e0i 1. (1,5 \u0111i\u1ec3m) a) V\u1ebd hai \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1 (d ) : y = x + 4 v\u00e0 ( P) : y = 1 x2 tr\u00ean c\u00f9ng m\u1ed9t m\u1eb7t ph\u1eb3ng t\u1ecda \u0111\u1ed9 2 b) T\u00ecm t\u1ecda \u0111\u1ed9 giao \u0111i\u1ec3m c\u1ee7a hai \u0111\u1ed3 th\u1ecb b\u1eb1ng ph\u00e9p to\u00e1n. B\u00e0i 2. (1 \u0111i\u1ec3m) Cho ph\u01b0\u01a1ng tr\u00ecnh: 2x2 \u2212 7x + 6 = 0 . B\u00e0i 3. Kh\u00f4ng gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh h\u00e3y t\u00ednh gi\u00e1 tr\u1ecb bi\u1ec3u th\u1ee9c: A = 2x1 + 2x2 3x1 \u2212 x2 3x2 \u2212 x1 (0,75 \u0111i\u1ec3m) Theo quy \u0111\u1ecbnh c\u1ee7a c\u00f4ng ty, m\u1ed9t c\u00f4ng nh\u00e2n \u0111\u01b0\u1ee3c tr\u1ea3 l\u01b0\u01a1ng nh\u01b0 sau: 400 000 \u0111\u1ed3ng cho m\u1ed9t ng\u00e0y l\u00e0m vi\u1ec7c b\u00ecnh th\u01b0\u1eddng (t\u1eeb th\u1ee9 hai \u0111\u1ebfn th\u1ee9 s\u00e1u h\u00e0ng tu\u1ea7n, m\u1ed9t ng\u00e0y l\u00e0m 8 gi\u1edd). N\u1ebfu l\u00e0m t\u0103ng ca v\u00e0o c\u00e1c ng\u00e0y th\u1ee9 7 - ch\u1ee7 nh\u1eadt ho\u1eb7c l\u1ec5 th\u00ec \u0111\u01b0\u1ee3c h\u01b0\u1edfng l\u01b0\u01a1ng b\u1eb1ng 150% ti\u1ec1n l\u01b0\u01a1ng c\u1ee7a m\u1ed9t ng\u00e0y l\u00e0m vi\u1ec7c b\u00ecnh th\u01b0\u1eddng Anh Th\u1eafng l\u00e0 nh\u00e2n vi\u00ean c\u1ee7a c\u00f4ng ty tr\u00ean v\u00e0 trong th\u00e1ng 5 \/ 2022 v\u1eeba qua anh \u0111\u00e3 \u0111\u01b0\u1ee3c tr\u1ea3 l\u01b0\u01a1ng l\u00e0 10 400 000 . H\u1ecfi anh \u0111\u00e3 l\u00e0m vi\u1ec7c bao nhi\u00eau ng\u00e0y t\u0103ng ca? (bi\u1ebft s\u1ed1 ng\u00e0y l\u00e0m vi\u1ec7c b\u00ecnh th\u01b0\u1eddng trong th\u00e1ng 5 c\u1ee7a anh l\u00e0 20 ng\u00e0y). B\u00e0i 4. (1 \u0111i\u1ec3m)M\u1ed9t h\u1ed9 gia \u0111\u00ecnh c\u00f3 \u00fd \u0111\u1ecbnh mua m\u1ed9t c\u00e1i m\u00e1y b\u01a1m \u0111\u1ec3 ph\u1ee5c v\u1ee5 cho vi\u1ec7c t\u01b0\u1edbi ti\u00eau. Khi \u0111\u1ebfn c\u1eeda h\u00e0ng th\u00ec \u0111\u01b0\u1ee3c nh\u00e2n vi\u00ean gi\u1edbi thi\u1ec7u 2 lo\u1ea1i m\u00e1y b\u01a1m c\u00f3 l\u01b0u l\u01b0\u1ee3ng n\u01b0\u1edbc trong m\u1ed9t gi\u1edd v\u00e0 ch\u1ea5t l\u01b0\u1ee3ng m\u00e1y l\u00e0 nh\u01b0 nhau. Gi\u00e1 b\u00e1n v\u00e0 hao ph\u00ed \u0111i\u1ec7n n\u0103ng c\u1ee7a m\u1ed7i m\u00e1y nh\u01b0 sau : M\u00e1y th\u1ee9 nh\u1ea5t : gi\u00e1 3 tri\u1ec7u v\u00e0 trong m\u1ed9t gi\u1edd ti\u00eau th\u1ee5 h\u1ebft 1,5 kWh. M\u00e1y th\u1ee9 hai : gi\u00e1 2 tri\u1ec7u v\u00e0 trong m\u1ed9t gi\u1edd ti\u00eau th\u1ee5 h\u1ebft 2 kWh. Bi\u1ebft gi\u00e1 1 kWh l\u00e0 2 000 \u0111\u1ed3ng v\u00e0 m\u1ed9t a) G\u1ecdi y l\u00e0 t\u1ed5ng chi ph\u00ed (bao g\u1ed3m ti\u1ec1n mua m\u00e1y b\u01a1m v\u00e0 ti\u1ec1n \u0111i\u1ec7n ph\u1ea3i tr\u1ea3) khi mua m\u1ed7i lo\u1ea1i m\u00e1y b\u01a1m v\u00e0 s\u1eed d\u1ee5ng trong x gi\u1edd. H\u00e3y l\u1eadp c\u00f4ng th\u1ee9c bi\u1ec3u di\u1ec5n y theo x c\u1ee7a t\u1eebng lo\u1ea1i m\u00e1y b\u01a1m. b) N\u1ebfu ng\u01b0\u1eddi n\u00f4ng d\u00e2n ch\u1ec9 s\u1eed d\u1ee5ng trong hai n\u0103m v\u00e0 m\u1ed7i ng\u00e0y ch\u1ec9 s\u1eed d\u1ee5ng 3 gi\u1edd th\u00ec n\u00ean ch\u1ecdn mua lo\u1ea1i m\u00e1y n\u00e0o c\u00f3 l\u1ee3i h\u01a1n. B\u00e0i 5. (1 \u0111i\u1ec3m)M\u1ed9t n\u0103m b\u00ecnh th\u01b0\u1eddng s\u1ebd c\u00f3 12 th\u00e1ng v\u00e0 365 ng\u00e0y. Khi m\u1ed9t n\u0103m c\u00f3 s\u1ed1 ng\u00e0y ho\u1eb7c s\u1ed1 th\u00e1ng t\u0103ng l\u00ean (theo D\u01b0\u01a1ng l\u1ecbch ho\u1eb7c \u00c2m l\u1ecbch) th\u00ec s\u1ebd \u0111\u01b0\u1ee3c g\u1ecdi l\u00e0 n\u0103m nhu\u1eadn, trong \u0111\u00f3 c\u00f3 nh\u1eefng ng\u00e0y nhu\u1eadn v\u00e0 th\u00e1ng nhu\u1eadn. N\u0103m nhu\u1eadn l\u00e0 n\u0103m c\u00f3 ng\u00e0y 29 th\u00e1ng 2 D\u01b0\u01a1ng l\u1ecbch (kh\u00f4ng nhu\u1eadn l\u00e0 28 ng\u00e0y). C\u00e1ch t\u00ednh n\u0103m nhu\u1eadn theo d\u01b0\u01a1ng l\u1ecbch l\u00e0 nh\u1eefng n\u0103m d\u01b0\u01a1ng l\u1ecbch n\u00e0o chia cho 4 th\u00ec \u0111\u00f3 s\u1ebd l\u00e0 n\u0103m nhu\u1eadn. V\u00ed d\u1ee5: 2016 chia h\u1ebft cho 4 n\u00ean n\u0103m 2016 l\u00e0 n\u0103m nhu\u1eadn. Ngo\u00e0i ra, \u0111\u1ed1i v\u1edbi nh\u1eefng tr\u00f2n th\u1ebf k\u1ef7 (nh\u1eefng n\u0103m c\u00f3 hai s\u1ed1 cu\u1ed1i l\u00e0 s\u1ed1 0 ) th\u00ec ch\u00fang ta s\u1ebd l\u1ea5y s\u1ed1 n\u0103m \u0111em chia cho 400 , n\u1ebfu nh\u01b0 chia h\u1ebft th\u00ec \u0111\u00f3 s\u1ebd l\u00e0 n\u0103m nhu\u1eadn. V\u00ed d\u1ee5: 1600 v\u00e0 2000 l\u00e0 c\u00e1c n\u0103m nhu\u1eadn nh\u01b0ng 1700 ,1800 v\u00e0 1900 kh\u00f4ng ph\u1ea3i n\u0103m nhu\u1eadn. 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) Em h\u00e3y d\u00f9ng quy t\u1eafc tr\u00ean \u0111\u1ec3 x\u00e1c \u0111\u1ecbnh n\u0103m 2022 , n\u0103m 2024 c\u00f3 ph\u1ea3i l\u00e0 n\u0103m nhu\u1eadn d\u01b0\u01a1ng l\u1ecbch kh\u00f4ng? b) Bi\u1ebft r\u1eb1ng ng\u00e0y 30 \/ 04 \/ 2022 l\u00e0 r\u01a1i v\u00e0o th\u1ee9 b\u1ea3y. H\u1ecfi ng\u00e0y 30 \/ 04 \/1975 l\u00e0 r\u01a1i v\u00e0o th\u1ee9 m\u1ea5y? Em h\u00e3y gi\u1ea3i th\u00edch. B\u00e0i 6. (1 \u0111i\u1ec3m)Trong k\u00ec thi HK2 m\u00f4n To\u00e1n l\u1edbp \u00a09 , m\u1ed9t ph\u00f2ng thi c\u00f3 24 th\u00ed sinh d\u1ef1 thi. C\u00e1c th\u00ed sinh \u0111\u1ec1u ph\u1ea3i l\u00e0m b\u00e0i tr\u00ean gi\u1ea5y thi c\u1ee7a tr\u01b0\u1eddng ph\u00e1t cho. Cu\u1ed1i bu\u1ed5i thi, sau khi thu b\u00e0i, gi\u00e1m th\u1ecb coi thi \u0111\u1ebfm \u0111\u01b0\u1ee3c t\u1ed5ng s\u1ed1 t\u1edd gi\u1ea5y thi l\u00e0 49 t\u1edd. H\u1ecfi trong ph\u00f2ng thi \u0111\u00f3 c\u00f3 bao nhi\u00eau th\u00ed sinh l\u00e0m b\u00e0i 2 t\u1edd gi\u1ea5y thi, bao nhi\u00eau th\u00ed sinh l\u00e0m b\u00e0i 3 t\u1edd gi\u1ea5y thi? Bi\u1ebft r\u1eb1ng c\u00f3 5 th\u00ed sinh ch\u1ec9 l\u00e0m 1 t\u1edd gi\u1ea5y thi, v\u00e0 kh\u00f4ng c\u00f3 th\u00ed sinh n\u00e0o l\u00e0m tr\u00ean 3 t\u1edd gi\u1ea5y thi. B\u00e0i 7. (0,75 \u0111i\u1ec3m) Ng\u01b0\u1eddi ta thi\u1ebft k\u1ebf m\u1ed9t h\u1ed3 b\u01a1i c\u00f3 d\u1ea1ng l\u00e0 m\u1ed9t l\u0103ng tr\u1ee5 \u0111\u1ee9ng t\u1ee9 gi\u00e1c v\u1edbi \u0111\u00e1y l\u00e0 h\u00ecnh thang vu\u00f4ng (m\u1eb7t s\u1ed1 \u00a01) c\u1ee7a h\u1ed3 b\u01a1i, c\u00f9ng c\u00e1c k\u00edch th\u01b0\u1edbc nh\u01b0 \u0111\u00e3 cho (xem h\u00ecnh v\u1ebd). Bi\u1ebft r\u1eb1ng ng\u01b0\u1eddi ta d\u00f9ng m\u1ed9t m\u00e1y b\u01a1m v\u1edbi l\u01b0u l\u01b0\u1ee3ng l\u00e0 6m3 \/ ph\u00fat v\u00e0 s\u1ebd b\u01a1m \u0111\u1ea7y h\u1ed3 m\u1ea5t 35 ph\u00fat. Em h\u00e3y t\u00ednh chi\u1ec1u d\u00e0i c\u1ee7a h\u1ed3. B\u00e0i 8. (3 \u0111i\u1ec3m) T\u1eeb \u0111i\u1ec3m A \u1edf ngo\u00e0i \u0111\u01b0\u1eddng tr\u00f2n t\u00e2m O , v\u1ebd hai ti\u1ebfp tuy\u1ebfn AB, AC v\u1edbi (O) (B , C l\u00e0 hai ti\u1ebfp \u0111i\u1ec3m). V\u1ebd c\u00e1t tuy\u1ebfn AMN v\u1edbi (O) sao cho AM \uf03c AN v\u00e0 tia AM n\u1eb1m gi\u1eefa tia OA v\u00e0 tia OC . G\u1ecdi E l\u00e0 trung \u0111i\u1ec3m c\u1ee7a MN . a) Ch\u1ee9ng minh t\u1ee9 gi\u00e1c ABOE n\u1ed9i ti\u1ebfp, v\u00e0 AB2 = AM .AN b) \u0110o\u1ea1n th\u1eb3ng BC c\u1eaft OA v\u00e0 MN l\u1ea7n l\u01b0\u1ee3t t\u1ea1i H v\u00e0 K . Ch\u1ee9ng minh: OA vu\u00f4ng g\u00f3c v\u1edbi BC t\u1ea1i H v\u00e0 AE.AK = AM.AN . c) Cho bi\u1ebft OA = 2R . Tr\u00ean \u0111o\u1ea1n th\u1eb3ng BC l\u1ea5y m\u1ed9t \u0111i\u1ec3m F b\u1ea5t k\u00ec, qua F v\u1ebd \u0111\u01b0\u1eddng th\u1eb3ng vu\u00f4ng g\u00f3c v\u1edbi OF t\u1ea1i F c\u1eaft AB v\u00e0 AC t\u1ea1i P v\u00e0 Q . Ch\u1ee9ng minh g\u00f3c POQ lu\u00f4n kh\u00f4ng \u0111\u1ed5i khi F di chuy\u1ec3n tr\u00ean \u0111o\u1ea1n BC . ----H\u1ebeT--- T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N 2","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH H\u01af\u1edaNG D\u1eaaN GI\u1ea2I B\u00e0i 1. (1,5 \u0111i\u1ec3m) a) V\u1ebd hai \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1 (d) : y = x + 4 v\u00e0 (P) : y = 1 x2 tr\u00ean c\u00f9ng m\u1ed9t m\u1eb7t ph\u1eb3ng t\u1ecda \u0111\u1ed9. 2 b) T\u00ecm t\u1ecda \u0111\u1ed9 giao \u0111i\u1ec3m c\u1ee7a hai \u0111\u1ed3 th\u1ecb 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: \u22122 \u22121 0 1 2 y x y = 1 x2 2 1 0 12 1 2 22 y = \u2219x2 2 x 0 \u22122 y=x+4 4 y = x+4 4 2 3 2 1 1\/2 -4 -3 -2 -1 O 12 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) : 1 x2 = x + 4 2 \uf0db x2 \u2212 2x \u2212 8 = 0 \uf0db \uf0e9x = 4 \uf0ea\uf0ebx = \u22122 Thay x = 4 v\u00e0o y = x + 4 , ta \u0111\u01b0\u1ee3c: y = 4 + 4 = 8 . Thay x = \u22122 v\u00e0o y = x + 4 , ta \u0111\u01b0\u1ee3c: y = \u22122 + 4 = 2 . V\u1eady (4; 8) , (\u22122; 2) l\u00e0 hai giao \u0111i\u1ec3m c\u1ea7n t\u00ecm. B\u00e0i 2. (1 \u0111i\u1ec3m) Cho ph\u01b0\u01a1ng tr\u00ecnh: 2x2 \u2212 7x + 6 = 0 T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N 3","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH Kh\u00f4ng gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh h\u00e3y t\u00ednh gi\u00e1 tr\u1ecb bi\u1ec3u th\u1ee9c: A = 2x1 + 2x2 3x1 \u2212 x2 3x2 \u2212 x1 L\u1eddi gi\u1ea3i V\u00ec \uf044 = b2 \u2212 4ac = (\u22127 )2 \u2212 4.2.6 = 1 \uf03e 0 N\u00ean ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m ph\u00e2n bi\u1ec7t x1 , x2 . \uf0ef\uf0ef\uf0ecS = x1 + x2 = \u2212b = \u2212 \u22127 = 7 \uf0ed = x1 a 2 2 Theo \u0111\u1ecbnh l\u00ed Vi-et, ta c\u00f3: \uf0ef\uf0ee\uf0efP c = 6 =3 .x2 = a 2 Ta c\u00f3: A = 2x1 + 2x2 3x1 \u2212 x2 3x2 \u2212 x1 A = 2x1 (3x2 \u2212 x1 ) + 2x2 (3x1 \u2212 x2 ) (3x1 \u2212 x2 )(3x2 \u2212 x1 ) A = 16x1x2 \u2212 2( x1 + x2 )2 = 16.3 \u2212 2.\uf0e6\uf0e7\uf0e8 7 \uf0f62 = 94 16x1x2 \u2212 3( x1 + x2 )2 16.3 \u2212 3.\uf0e6\uf0e7\uf0e8 2 \uf0f7\uf0f8 45 7 \uf0f62 2 \uf0f7\uf0f8 B\u00e0i 3. (0,75 \u0111i\u1ec3m) Theo quy \u0111\u1ecbnh c\u1ee7a c\u00f4ng ty, m\u1ed9t c\u00f4ng nh\u00e2n \u0111\u01b0\u1ee3c tr\u1ea3 l\u01b0\u01a1ng nh\u01b0 sau: 400 000 \u0111\u1ed3ng cho m\u1ed9t ng\u00e0y l\u00e0m vi\u1ec7c b\u00ecnh th\u01b0\u1eddng (t\u1eeb th\u1ee9 hai \u0111\u1ebfn th\u1ee9 s\u00e1u h\u00e0ng tu\u1ea7n, m\u1ed9t ng\u00e0y l\u00e0m 8 gi\u1edd). N\u1ebfu l\u00e0m t\u0103ng ca v\u00e0o c\u00e1c ng\u00e0y th\u1ee9 7 - ch\u1ee7 nh\u1eadt ho\u1eb7c l\u1ec5 th\u00ec \u0111\u01b0\u1ee3c h\u01b0\u1edfng l\u01b0\u01a1ng b\u1eb1ng 150% ti\u1ec1n l\u01b0\u01a1ng c\u1ee7a m\u1ed9t ng\u00e0y l\u00e0m vi\u1ec7c b\u00ecnh th\u01b0\u1eddng Anh Th\u1eafng l\u00e0 nh\u00e2n vi\u00ean c\u1ee7a c\u00f4ng ty tr\u00ean v\u00e0 trong th\u00e1ng 5 \/ 2022 v\u1eeba qua anh \u0111\u00e3 \u0111\u01b0\u1ee3c tr\u1ea3 l\u01b0\u01a1ng l\u00e0 10 400 000 . H\u1ecfi anh \u0111\u00e3 l\u00e0m vi\u1ec7c bao nhi\u00eau ng\u00e0y t\u0103ng ca? (bi\u1ebft s\u1ed1 ng\u00e0y l\u00e0m vi\u1ec7c b\u00ecnh th\u01b0\u1eddng trong th\u00e1ng 5 c\u1ee7a anh l\u00e0 20 ng\u00e0y). L\u1eddi gi\u1ea3i M\u1ed9t ng\u00e0y t\u0103ng ca anh Th\u1eafng \u0111\u01b0\u1ee3c tr\u1ea3 l\u01b0\u01a1ng l\u00e0: 400000.150% = 600000 ( \u0111\u1ed3ng) Ti\u1ec1n l\u01b0\u01a1ng 1 th\u00e1ng anh Th\u1eafng \u0111\u01b0\u1ee3c tr\u1ea3( kh\u00f4ng t\u0103ng ca) l\u00e0: 400000.20 = 8000000 ( \u0111\u1ed3ng) S\u1ed1 ti\u1ec1n anh Th\u1eafng \u0111\u01b0\u1ee3c nh\u1eadn khi t\u0103ng ca l\u00e0: 10400000 \u22128000000 = 2400000( \u0111\u1ed3ng) S\u1ed1 ng\u00e0y anh Th\u1eafng \u0111\u00e3 l\u00e0m t\u0103ng ca trong th\u00e1ng 5 l\u00e0: 2400000 : 600000 = 4 ( ng\u00e0y) B\u00e0i 4. (1 \u0111i\u1ec3m) M\u1ed9t h\u1ed9 gia \u0111\u00ecnh c\u00f3 \u00fd \u0111\u1ecbnh mua m\u1ed9t c\u00e1i m\u00e1y b\u01a1m \u0111\u1ec3 ph\u1ee5c v\u1ee5 cho vi\u1ec7c t\u01b0\u1edbi ti\u00eau. Khi \u0111\u1ebfn c\u1eeda h\u00e0ng th\u00ec \u0111\u01b0\u1ee3c nh\u00e2n vi\u00ean gi\u1edbi thi\u1ec7u 2 lo\u1ea1i m\u00e1y b\u01a1m c\u00f3 l\u01b0u l\u01b0\u1ee3ng n\u01b0\u1edbc trong m\u1ed9t gi\u1edd v\u00e0 ch\u1ea5t l\u01b0\u1ee3ng m\u00e1y l\u00e0 nh\u01b0 nhau. Gi\u00e1 b\u00e1n v\u00e0 hao ph\u00ed \u0111i\u1ec7n n\u0103ng c\u1ee7a m\u1ed7i m\u00e1y nh\u01b0 sau : 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 M\u00e1y th\u1ee9 nh\u1ea5t : gi\u00e1 3 tri\u1ec7u v\u00e0 trong m\u1ed9t gi\u1edd ti\u00eau th\u1ee5 h\u1ebft 1,5 kWh. a) M\u00e1y th\u1ee9 hai : gi\u00e1 2 tri\u1ec7u v\u00e0 trong m\u1ed9t gi\u1edd ti\u00eau th\u1ee5 h\u1ebft 2 kWh. Bi\u1ebft gi\u00e1 1 kWh l\u00e0 2 000 \u0111\u1ed3ng v\u00e0 m\u1ed9t G\u1ecdi y l\u00e0 t\u1ed5ng chi ph\u00ed (bao g\u1ed3m ti\u1ec1n mua m\u00e1y b\u01a1m v\u00e0 ti\u1ec1n \u0111i\u1ec7n ph\u1ea3i tr\u1ea3) khi mua m\u1ed7i lo\u1ea1i m\u00e1y b\u01a1m v\u00e0 s\u1eed d\u1ee5ng trong x gi\u1edd. H\u00e3y l\u1eadp c\u00f4ng th\u1ee9c bi\u1ec3u di\u1ec5n y theo x c\u1ee7a t\u1eebng lo\u1ea1i m\u00e1y b\u01a1m. b) N\u1ebfu ng\u01b0\u1eddi n\u00f4ng d\u00e2n ch\u1ec9 s\u1eed d\u1ee5ng trong hai n\u0103m v\u00e0 m\u1ed7i ng\u00e0y ch\u1ec9 s\u1eed d\u1ee5ng 3 gi\u1edd th\u00ec n\u00ean ch\u1ecdn mua lo\u1ea1i m\u00e1y n\u00e0o c\u00f3 l\u1ee3i h\u01a1n. L\u1eddi gi\u1ea3i a) C\u00f4ng th\u1ee9c bi\u1ec3u di\u1ec5n y theo x c\u1ee7a t\u1eebng lo\u1ea1i m\u00e1y b\u01a1m - M\u00e1y b\u01a1m I : y = 3000000 +1,5x \uf0d71500 = 3000000 + 2250x (\u0111\u1ed3ng) - M\u00e1y b\u01a1m II : y = 2000000 + 2x \uf0d71500 = 2000000 + 3000x (\u0111\u1ed3ng) b) S\u1ed1 gi\u1edd s\u1eed d\u1ee5ng: 2\uf0d7365\uf0d73 = 2190 (gi\u1edd) Thay x = 2190 , ta c\u00f3: - S\u1ed1 ti\u1ec1n ph\u1ea3i tr\u1ea3 n\u1ebfu s\u1eed d\u1ee5ng m\u00e1y b\u01a1m I l\u00e0: y = 3000000 + 2250 \uf0d7 2190 = 7927500 (\u0111\u1ed3ng) - S\u1ed1 ti\u1ec1n ph\u1ea3i tr\u1ea3 n\u1ebfu s\u1eed d\u1ee5ng m\u00e1y b\u01a1m II l\u00e0: y = 2000000 + 3000\uf0d7 2190 = 8570000 (\u0111\u1ed3ng) V\u1eady n\u00ean s\u1eed d\u1ee5ng m\u00e1y I c\u00f3 l\u1ee3i h\u01a1n. B\u00e0i 5. (1 \u0111i\u1ec3m) M\u1ed9t n\u0103m b\u00ecnh th\u01b0\u1eddng s\u1ebd c\u00f3 12 th\u00e1ng v\u00e0 365 ng\u00e0y. Khi m\u1ed9t n\u0103m c\u00f3 s\u1ed1 ng\u00e0y ho\u1eb7c s\u1ed1 th\u00e1ng t\u0103ng l\u00ean (theo D\u01b0\u01a1ng l\u1ecbch ho\u1eb7c \u00c2m l\u1ecbch) th\u00ec s\u1ebd \u0111\u01b0\u1ee3c g\u1ecdi l\u00e0 n\u0103m nhu\u1eadn, trong \u0111\u00f3 c\u00f3 nh\u1eefng ng\u00e0y nhu\u1eadn v\u00e0 th\u00e1ng nhu\u1eadn. N\u0103m nhu\u1eadn l\u00e0 n\u0103m c\u00f3 ng\u00e0y 29 th\u00e1ng 2 D\u01b0\u01a1ng l\u1ecbch (kh\u00f4ng nhu\u1eadn l\u00e0 28 ng\u00e0y). C\u00e1ch t\u00ednh n\u0103m nhu\u1eadn theo d\u01b0\u01a1ng l\u1ecbch l\u00e0 nh\u1eefng n\u0103m d\u01b0\u01a1ng l\u1ecbch n\u00e0o chia cho 4 th\u00ec \u0111\u00f3 s\u1ebd l\u00e0 n\u0103m nhu\u1eadn. V\u00ed d\u1ee5: 2016 chia h\u1ebft cho 4 n\u00ean n\u0103m 2016 l\u00e0 n\u0103m nhu\u1eadn. Ngo\u00e0i ra, \u0111\u1ed1i v\u1edbi nh\u1eefng tr\u00f2n th\u1ebf k\u1ef7 (nh\u1eefng n\u0103m c\u00f3 hai s\u1ed1 cu\u1ed1i l\u00e0 s\u1ed1 0 ) th\u00ec ch\u00fang ta s\u1ebd l\u1ea5y s\u1ed1 n\u0103m \u0111em chia cho 400 , n\u1ebfu nh\u01b0 chia h\u1ebft th\u00ec \u0111\u00f3 s\u1ebd l\u00e0 n\u0103m nhu\u1eadn. V\u00ed d\u1ee5: 1600 v\u00e0 2000 l\u00e0 c\u00e1c n\u0103m nhu\u1eadn nh\u01b0ng 1700 ,1800 v\u00e0 1900 kh\u00f4ng ph\u1ea3i n\u0103m nhu\u1eadn. a) Em h\u00e3y d\u00f9ng quy t\u1eafc tr\u00ean \u0111\u1ec3 x\u00e1c \u0111\u1ecbnh n\u0103m \u00a02022 , n\u0103m 2024 c\u00f3 ph\u1ea3i l\u00e0 n\u0103m nhu\u1eadn d\u01b0\u01a1ng l\u1ecbch kh\u00f4ng? T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N 5","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH b) Bi\u1ebft r\u1eb1ng ng\u00e0y 30 \/ 04 \/ 2022 l\u00e0 r\u01a1i v\u00e0o th\u1ee9 b\u1ea3y. H\u1ecfi ng\u00e0y 30 \/ 04 \/ 1975 l\u00e0 r\u01a1i v\u00e0o th\u1ee9 m\u1ea5y? Em h\u00e3y gi\u1ea3i th\u00edch. L\u1eddi gi\u1ea3i a) 2022 4 n\u00ean n\u0103m 2022 kh\u00f4ng l\u00e0 n\u0103m nhu\u1eadn 2022 4 n\u00ean n\u0103m 2022 l\u00e0 n\u0103m nhu\u1eadn b) S\u1ed1 n\u0103m nhu\u1eadn l\u00e0: (2020 \u22121976) : 4 +1 = 12 (n\u0103m) T\u1eeb 30 \/ 04 \/ 1975 \u0111\u1ebfn 30 \/ 04 \/ 2022 c\u00f3 s\u1ed1 n\u0103m l\u00e0 (2022 \u22121975) +1 = 48 (n\u0103m) S\u1ed1 ng\u00e0y t\u1eeb 30 \/ 04 \/ 1975 \u0111\u1ebfn 30 \/ 04 \/ 2022 l\u00e0: 48.365 12.1 1\u00a0 17533 (ng\u00e0y) 17533 chia \u00a07 d\u01b0 3 V\u00ec 30 \/ 04 \/ 2022 l\u00e0 r\u01a1i v\u00e0o th\u1ee9 b\u1ea3y n\u00ean ng\u00e0y 30 \/ 04 \/ 1975 tr\u01b0\u1edbc ba ng\u00e0y k\u1ec3 t\u1eeb ng\u00e0y th\u1ee9 B\u1ea3y v\u00e0 l\u00e0 ng\u00e0y Th\u1ee9 T\u01b0. B\u00e0i 6. (1 \u0111i\u1ec3m) Trong k\u00ec thi HK2 m\u00f4n To\u00e1n l\u1edbp 9, m\u1ed9t ph\u00f2ng thi c\u00f3 24 th\u00ed sinh d\u1ef1 thi. C\u00e1c th\u00ed sinh \u0111\u1ec1u ph\u1ea3i l\u00e0m b\u00e0i tr\u00ean gi\u1ea5y thi c\u1ee7a tr\u01b0\u1eddng ph\u00e1t cho. Cu\u1ed1i bu\u1ed5i thi, sau khi thu b\u00e0i, gi\u00e1m th\u1ecb coi thi \u0111\u1ebfm \u0111\u01b0\u1ee3c t\u1ed5ng s\u1ed1 t\u1edd gi\u1ea5y thi l\u00e0 49 t\u1edd. H\u1ecfi trong ph\u00f2ng thi \u0111\u00f3 c\u00f3 bao nhi\u00eau th\u00ed sinh l\u00e0m b\u00e0i 2 t\u1edd gi\u1ea5y thi, bao nhi\u00eau th\u00ed sinh l\u00e0m b\u00e0i 3 t\u1edd gi\u1ea5y thi? Bi\u1ebft r\u1eb1ng c\u00f3 5 th\u00ed sinh ch\u1ec9 l\u00e0m 1 t\u1edd gi\u1ea5y thi, v\u00e0 kh\u00f4ng c\u00f3 th\u00ed sinh n\u00e0o l\u00e0m tr\u00ean 3 t\u1edd gi\u1ea5y thi L\u1eddi gi\u1ea3i V\u00ec ph\u00f2ng thi 24 th\u00ed sinh c\u00f3 5 th\u00ed sinh ch\u1ec9 l\u00e0m 1 t\u1edd gi\u1ea5y thi v\u00e0 kh\u00f4ng c\u00f3 th\u00ed sinh n\u00e0o l\u00e0m tr\u00ean 3 t\u1edd gi\u1ea5y thi n\u00ean t\u1ed5ng s\u1ed1 th\u00ed sinh l\u00e0m 2 t\u1edd ho\u1eb7c 3 t\u1edd gi\u1ea5y thi l\u00e0 24 \u2212 5 = 19 th\u00ed sinh v\u00e0 t\u1ed5ng s\u1ed1 t\u1edd l\u00e0 49 \u2212 5 = 44 G\u1ecdi x , y (th\u00ed sinh) l\u1ea7n l\u01b0\u1ee3t l\u00e0 s\u1ed1 th\u00ed sinh l\u00e0m 2 t\u1edd ho\u1eb7c 3 t\u1edd gi\u1ea5y thi. (\u0110i\u1ec1u ki\u1ec7n: 0 \uf03c x, y \uf03c 19 ) Theo \u0111\u1ec1 b\u00e0i ta c\u00f3 h\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh: \uf0ecx + y = 19 Gi\u1ea3i h\u1ec7 ta \u0111\u01b0\u1ee3c: \uf0ecx = 13 (Th\u1ea3o m\u00e3n \u0111i\u1ec1u ki\u1ec7n) \uf0ed\uf0ee2x + 3y = 44 \uf0ed\uf0eey =6 V\u1eady c\u00f3 13 th\u00ed sinh l\u00e0m 2 t\u1edd v\u00e0 6 th\u00ed sinh l\u00e0m 3 t\u1edd. B\u00e0i 7. (0,75 \u0111i\u1ec3m) Ng\u01b0\u1eddi ta thi\u1ebft k\u1ebf m\u1ed9t h\u1ed3 b\u01a1i c\u00f3 d\u1ea1ng l\u00e0 m\u1ed9t l\u0103ng tr\u1ee5 \u0111\u1ee9ng t\u1ee9 gi\u00e1c v\u1edbi \u0111\u00e1y l\u00e0 h\u00ecnh thang vu\u00f4ng (m\u1eb7t s\u1ed1 (1)) c\u1ee7a h\u1ed3 b\u01a1i, c\u00f9ng c\u00e1c k\u00edch th\u01b0\u1edbc nh\u01b0 \u0111\u00e3 cho (xem h\u00ecnh v\u1ebd). Bi\u1ebft r\u1eb1ng ng\u01b0\u1eddi ta d\u00f9ng m\u1ed9t m\u00e1y b\u01a1m v\u1edbi l\u01b0u l\u01b0\u1ee3ng l\u00e0 6m3 \/ ph\u00fat v\u00e0 s\u1ebd b\u01a1m \u0111\u1ea7y h\u1ed3 m\u1ea5t 35 ph\u00fat. Em h\u00e3y t\u00ednh chi\u1ec1u d\u00e0i c\u1ee7a h\u1ed3. 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 2,5m 0,5m 3m Gh\u00e9p th\u00eam v\u00e0o l\u0103ng tr\u1ee5 t\u1ee9 gi\u00e1c m\u1ed9t l\u0103ng tr\u1ee5 \u0111\u1ee9ng c\u00f3 \u0111\u00e1y l\u00e0 tam gi\u00e1c vu\u00f4ng (nh\u01b0 h\u00ecnh v\u1ebd) sao cho t\u1ea1o th\u00e0nh m\u1ed9t h\u00ecnh h\u1ed9p ch\u1eef nh\u1eadt. G\u1ecdi x l\u00e0 \u0111\u1ed9 d\u00e0i c\u1ea1nh c\u00f2n l\u1ea1i c\u1ee7a h\u00ecnh h\u1ed9p ch\u1eef nh\u1eadt \u0111\u00f3. Ta c\u00f3: ( )Vhhcn = 6.3.x = 18x m3 ( )Vhoboi = 6.35 = 210 m3 2,5.x.6. 1 2 =( )Vlangtrudungtamgiac = m3 3.2, 5.x Khi \u0111\u00f3 Vhoboi = Vhhcn \u2212Vlangtrudungtamgiac \uf0de 18x \u2212 3.2, 5x = 210 \uf0de x = 20(m) Chi\u1ec1u d\u00e0i c\u1ee7a h\u1ed3 l\u00e0: 2,52 + 202 \uf0bb 20,16(m) B\u00e0i 8. (3 \u0111i\u1ec3m) T\u1eeb \u0111i\u1ec3m A \u1edf ngo\u00e0i \u0111\u01b0\u1eddng tr\u00f2n t\u00e2m O , v\u1ebd hai ti\u1ebfp tuy\u1ebfn AB, AC v\u1edbi (O) (B , C l\u00e0 hai ti\u1ebfp \u0111i\u1ec3m). V\u1ebd c\u00e1t tuy\u1ebfn AMN v\u1edbi (O) sao cho AM \uf03c AN v\u00e0 tia AM n\u1eb1m gi\u1eefa tia OA v\u00e0 tia OC . G\u1ecdi E l\u00e0 trung \u0111i\u1ec3m c\u1ee7a MN . a) Ch\u1ee9ng minh t\u1ee9 gi\u00e1c ABOE n\u1ed9i ti\u1ebfp, v\u00e0 AB2 = AM .AN b) \u0110o\u1ea1n th\u1eb3ng BC c\u1eaft OA v\u00e0 MN l\u1ea7n l\u01b0\u1ee3t t\u1ea1i H v\u00e0 K . Ch\u1ee9ng minh: OA vu\u00f4ng g\u00f3c v\u1edbi BC t\u1ea1i H v\u00e0 AE.AK = AM.AN . 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) Cho bi\u1ebft OA = 2R . Tr\u00ean \u0111o\u1ea1n th\u1eb3ng BC l\u1ea5y m\u1ed9t \u0111i\u1ec3m F b\u1ea5t k\u00ec, qua F v\u1ebd \u0111\u01b0\u1eddng th\u1eb3ng vu\u00f4ng g\u00f3c v\u1edbi OF t\u1ea1i F c\u1eaft AB v\u00e0 AC t\u1ea1i P v\u00e0 Q . Ch\u1ee9ng minh g\u00f3c POQ lu\u00f4n kh\u00f4ng \u0111\u1ed5i khi F di chuy\u1ec3n tr\u00ean \u0111o\u1ea1n BC . L\u1eddi gi\u1ea3i C M KE N A HO B a) X\u00e9t (O) c\u00f3 AB l\u00e0 ti\u1ebfp tuy\u1ebfn t\u1ea1i B .Suy ra ABO = 90\uf0b0 . L\u1ea1i c\u00f3 EM = EN n\u00ean OE \u22a5 MN hay AEO = 90\uf0b0 X\u00e9t t\u1ee9 gi\u00e1c ABOE c\u00f3 ABO + AEO = 90\uf0b0 + 90\uf0b0 = 180\uf0b0 . Suy ra t\u1ee9 gi\u00e1c ABOE n\u1ed9i ti\u1ebfp \u0111\u01b0\u1eddng tr\u00f2n \u0111\u01b0\u1eddng k\u00ednh AO . X\u00e9t \uf044AMC v\u00e0 \uf044ACN c\u00f3 : AMC g\u00f3c chung ; ANC = ACM ( g\u00f3c n\u1ed9i ti\u1ebfp v\u00e0 g\u00f3c t\u1ea1o b\u1edfi tia ti\u1ebfp tuy\u1ebfn v\u00e0 d\u00e2y cung c\u00f9ng ch\u1eafn cung MC ) \uf0de \uf044AMC\u223d\uf044ACN (g-g) Suy ra: AM = AC \uf0de AC2 = AM .AN 1 AC AN AB, AC l\u00e0 hai ti\u1ebfp tuy\u1ebfn c\u1eaft nhau t\u1ea1i A . Suy ra AB = AC hay AB2 = AM .AN b) V\u00ec AB = AC ; OB = OC = R n\u00ean OA l\u00e0 trung tr\u1ef1c c\u1ee7a BC \uf0de OA \u22a5 BC t\u1ea1i H X\u00e9t \uf044ACO vu\u00f4ng t\u1ea1i C c\u00f3 CH \u22a5 OA \uf0de AC2 = AH.AO 2 ( h\u1ec7 th\u1ee9c l\u01b0\u1ee3ng trong tam gi\u00e1c vu\u00f4ng). X\u00e9t \uf044AHK v\u00e0 \uf044AEO c\u00f3 AHK = AEO = 90\uf0b0 EOA g\u00f3c chung \uf0de \uf044AMC\u223d\uf044ACN (g-g) AH = AK \uf0de AK.AE = AH.AO 3 AE AO T\u1eeb (1),(2),(3) ta c\u00f3 AK.AE = AM.AN . 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 Q E N M K A I HO F B P c) X\u00e9t t\u1ee9 gi\u00e1c OFQC c\u00f3 OFC + OQC = 90\uf0b0 + 90\uf0b0 = 180\uf0b0 . Suy ra t\u1ee9 gi\u00e1c OFQC n\u1ed9i ti\u1ebfp \u0111\u01b0\u1eddng tr\u00f2n. COQ = CFQ (*) hai g\u00f3c n\u1ed9i ti\u1ebfp c\u00f9ng ch\u1eafn cung CQ . X\u00e9t t\u1ee9 gi\u00e1c OFBP c\u00f3 OFP = OBP = 90\uf0b0 . Hai \u0111i\u1ec3m B, F c\u00f9ng nh\u00ecn \u0111o\u1ea1n OP d\u01b0\u1edbi 1 g\u00f3c b\u1eb1ng nhau 90\uf0b0 . Suy ra t\u1ee9 gi\u00e1c OFBP n\u1ed9i ti\u1ebfp \u0111\u01b0\u1eddng tr\u00f2n \u0111\u01b0\u1eddng k\u00ednh OP . Suy ra BFP = BOQ (**) hai g\u00f3c n\u1ed9i ti\u1ebfp c\u00f9ng ch\u1eafn cung BP . M\u00e0 CFQ = BFP (***) (\u0111\u1ed1i \u0111\u1ec9nh). T\u1eeb (*), (**), (***) ta c\u00f3: CFQ = BOP \uf0de COQ + QOB = QOB + BOP \uf0de COB = QOP X\u00e9t (O) c\u00f3 AB , AC l\u00e0 hai ti\u1ebfp tuy\u1ebfn t\u1ea1i B v\u00e0 C \uf0de COA = AOB . V\u00ec OA = 2R . X\u00e9t \uf044ACO vu\u00f4ng t\u1ea1i C , c\u00f3 CI = 1 AO = R ( trung tuy\u1ebfn thu\u1ed9c c\u1ea1nh huy\u1ec1n) \uf0de \uf044COI 2 \u0111\u1ec1u ( OI = CI = OC = R ) . Suy ra COI = 60\uf0b0 hay \uf0de COA = AOB = 90\uf0b0 \uf0de COB = 120\uf0b0 \uf0de COB = POQ = 120\uf0b0 . 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 HUY\u1ec6N NH\u00c0 B\u00c8 NA\u00caM HO\u00cfC: 2023 - 2024 M\u00d4N: TO\u00c1N 9 PHU\u00d9 \u0110\u1ec0 THAM KH\u1ea2O \u0110\u00ea thi g\u1ed3m 8 c\u00e2u ho\u0309i t\u01b0\u0323 lu\u00e2\u0323n. Th\u01a1\u0300i gian: 120 phu\u0301t (kh\u00f4ng k\u00ea\u0309 th\u01a1\u0300i gian pha\u0301t \u0111\u00ea\u0300) M\u00c3 \u0110\u1ec0: Huy\u1ec7n Nh\u00e0 B\u00e8 - 3 C\u00e2u 1. (1,5 \u0111i\u1ec3m). a) V\u1ebd \u0111\u1ed3 th\u1ecb (P) : y = 1 x2 v\u00e0 \u0111\u01b0\u1eddng th\u1eb3ng (d) : y = \u2212 3 x + 2 tr\u00ean c\u00f9ng h\u1ec7 tr\u1ee5c t\u1ecda \u0111\u1ed9. 22 b) T\u00ecm t\u1ecda \u0111\u1ed9 giao \u0111i\u1ec3m c\u1ee7a (P) v\u00e0 (d) \u1edf c\u00e2u tr\u00ean b\u1eb1ng ph\u00e9p t\u00ednh. C\u00e2u 2. (1 \u0111i\u1ec3m). Cho ph\u01b0\u01a1ng tr\u00ecnh 2x2 \u2212 x \u2212 6 = 0 c\u00f3 2 nghi\u1ec7m l\u00e0 x1 , x2 . Kh\u00f4ng gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh, ( )h\u00e3y t\u00ednh gi\u00e1 tr\u1ecb c\u1ee7a bi\u1ec3u th\u1ee9c x1 \u2212 x2 2 C\u00e2u 3. (1 \u0111i\u1ec3m). Nh\u00e2n d\u1ecbp khai tr\u01b0\u01a1ng, c\u1eeda h\u00e0ng ch\u1ecb H\u01b0\u01a1ng \u0111\u00e3 quy\u1ebft \u0111\u1ecbnh gi\u1ea3m gi\u00e1 20% cho m\u1ed9t quy\u1ec3n t\u1eadp v\u00e0 n\u1ebfu kh\u00e1ch h\u00e0ng mua 10 quy\u1ec3n t\u1eadp tr\u1edf l\u00ean th\u00ec t\u1eeb quy\u1ec3n th\u1ee9 11 tr\u1edf \u0111i, kh\u00e1ch h\u00e0ng ch\u1ec9 ph\u1ea3i tr\u1ea3 90% gi\u00e1 \u0111\u00e3 gi\u1ea3m. Bi\u1ebft r\u1eb1ng ban \u0111\u1ea7u gi\u00e1 m\u1ed9t quy\u1ec3n t\u1eadp l\u00e0 8 000 \u0111\u1ed3ng. a) M\u1eb9 b\u1ea1n An \u0111\u1eb7t mua cho hai ch\u1ecb em b\u1ea1n An 60 quy\u1ec3n t\u1eadp. T\u00ednh s\u1ed1 ti\u1ec1n m\u1eb9 b\u1ea1n An ph\u1ea3i tr\u1ea3. b) M\u1ed9t kh\u00e1ch h\u00e0ng \u0111\u00e3 mua t\u1eadp \u1edf c\u1eeda h\u00e0ng ch\u1ecb H\u01b0\u01a1ng v\u00e0 t\u1ed5ng s\u1ed1 ti\u1ec1n m\u00e0 v\u1ecb kh\u00e1ch n\u00e0y ph\u1ea3i tr\u1ea3 l\u00e0 928 000 \u0111\u1ed3ng. H\u1ecfi kh\u00e1ch h\u00e0ng n\u00e0y \u0111\u00e3 mua bao nhi\u00eau quy\u1ec3n t\u1eadp? C\u00e2u 4. (1 \u0111i\u1ec3m). Hai b\u1ea1n Thanh v\u00e0 Li\u00ean \u0111i c\u00f9ng tr\u00ean m\u1ed9t con \u0111\u01b0\u1eddng. L\u00fac \u0111\u1ea7u hai b\u1ea1n \u1edf c\u00f9ng m\u1ed9t ph\u00eda \u0111\u1ed1i v\u1edbi tr\u1ea1m xe bus v\u00e0 c\u00e1ch tr\u1ea1m xe bus l\u1ea7n l\u01b0\u1ee3t l\u00e0 200m v\u00e0 500m; hai b\u1ea1n c\u00f9ng \u0111i ng\u01b0\u1ee3c h\u01b0\u1edbng v\u1edbi tr\u1ea1m xe bus . Trung b\u00ecnh m\u1ed7i gi\u1edd Thanh \u0111i \u0111\u01b0\u1ee3c 3km v\u00e0 Li\u00ean \u0111i \u0111\u01b0\u1ee3c 1km . G\u1ecdi d l\u00e0 kho\u1ea3ng c\u00e1ch c\u1ee7a Thanh, Li\u00ean v\u1edbi tr\u1ea1m xe bus sau khi \u0111i \u0111\u01b0\u1ee3c t gi\u1edd. a) L\u1eadp h\u00e0m s\u1ed1 c\u1ee7a d theo t \u0111\u1ed1i v\u1edbi m\u1ed7i b\u1ea1n b) Sau bao l\u00e2u th\u00ec hai b\u1ea1n g\u1eb7p nhau? C\u00e2u 5. (0,75 \u0111i\u1ec3m). V\u00e0o d\u1ecbp li\u00ean hoan cu\u1ed1i n\u0103m gia \u0111\u00ecnh b\u1ea1n Ph\u01b0\u01a1ng d\u1ef1 \u0111\u1ecbnh mua m\u1ed9t s\u1ed1 b\u00e1nh pizza \u0111\u1ec3 d\u00f9ng. Bi\u1ebft b\u00e1nh pizza c\u1ee1 7 inch c\u00f3 gi\u00e1 99 000 \u0111\u1ed3ng v\u00e0 c\u1ee1 9 inch c\u00f3 gi\u00e1 189 000 \u0111\u1ed3ng (ch\u1ea5t l\u01b0\u1ee3ng v\u00e0 b\u1ec1 d\u00e0y c\u1ee7a hai c\u1ee1 b\u00e1nh l\u00e0 nh\u01b0 nhau). Em h\u00e3y t\u00ednh xem mua b\u00e1nh c\u1ee1 n\u00e0o s\u1ebd c\u00f3 l\u1ee3i h\u01a1n? C\u00e2u 6. (1 \u0111i\u1ec3m). \u201cV\u00e0ng 24K c\u00f2n \u0111\u01b0\u1ee3c g\u1ecdi l\u00e0 v\u00e0ng r\u00f2ng (l\u00e0 lo\u1ea1i v\u00e0ng tinh khi\u1ebft nh\u1ea5t, g\u1ea7n nh\u01b0 kh\u00f4ng c\u00f3 pha l\u1eabn t\u1ea1p ch\u1ea5t, c\u00f3 gi\u00e1 tr\u1ecb cao nh\u1ea5t trong c\u00e1c lo\u1ea1i v\u00e0ng) l\u00e0 m\u1ed9t kim lo\u1ea1i c\u00f3 \u00e1nh kim \u0111\u1eadm nh\u1ea5t nh\u01b0ng kh\u00e1 m\u1ec1m. Trong ng\u00e0nh c\u00f4ng ngh\u1ec7 ch\u1ebf t\u1ea1o trang s\u1ee9c, ng\u01b0\u1eddi ta \u00edt d\u00f9ng v\u00e0ng 24K m\u00e0 thay th\u1ebf b\u1eb1ng v\u00e0ng 14K l\u00e0 h\u1ee3p kim c\u1ee7a v\u00e0ng v\u00e0 \u0111\u1ed3ng \u0111\u1ec3 d\u1ec5 \u0111\u00e1nh b\u00f3ng v\u00e0 t\u1ea1o ra nhi\u1ec1u 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 ki\u1ec3u d\u00e1ng \u0111a d\u1ea1ng\u201d. M\u1ed9t m\u00f3n trang s\u1ee9c \u0111\u01b0\u1ee3c l\u00e0m t\u1eeb v\u00e0ng 14K c\u00f3 th\u1ec3 t\u00edch 10cm3 v\u00e0 n\u1eb7ng 151,8g . H\u00e3y t\u00ednh th\u1ec3 t\u00edch v\u00e0ng nguy\u00ean ch\u1ea5t v\u00e0 \u0111\u1ed3ng \u0111\u01b0\u1ee3c d\u00f9ng \u0111\u1ec3 l\u00e0m ra m\u00f3n trang s\u1ee9c; bi\u1ebft kh\u1ed1i l\u01b0\u1ee3ng ri\u00eang c\u1ee7a v\u00e0ng nguy\u00ean ch\u1ea5t l\u00e0 19, 3g \/ cm3 , kh\u1ed1i l\u01b0\u1ee3ng ri\u00eang c\u1ee7a \u0111\u1ed3ng l\u00e0 9g \/ cm3 v\u00e0 c\u00f4ng th\u1ee9c li\u00ean h\u1ec7 gi\u1eefa kh\u1ed1i l\u01b0\u1ee3ng ri\u00eang v\u00e0 th\u1ec3 t\u00edch l\u00e0 m = D.V . C\u00e2u 7. (1 \u0111i\u1ec3m). V\u00e0o kho\u1ea3ng n\u0103m 200 tr\u01b0\u1edbc C\u00f4ng Nguy\u00ean, \u01a0rat\u00f4xten, m\u1ed9t nh\u00e0 to\u00e1n h\u1ecdc v\u00e0 thi\u00ean v\u0103n h\u1ecdc Hi L\u1ea1p, \u0111\u00e3 \u01b0\u1edbc l\u01b0\u1ee3ng \u0111\u01b0\u1ee3c \\\"chu vi\\\" c\u1ee7a Tr\u00e1i \u0110\u1ea5t (chu vi \u0111\u01b0\u1eddng X\u00edch \u0110\u1ea1o) nh\u1edd hai quan s\u00e1t sau: 1) M\u1ed9t ng\u00e0y trong n\u0103m, \u00f4ng ta \u0111\u1ec3 \u00fd th\u1ea5y M\u1eb7t Tr\u1eddi chi\u1ebfu th\u1eb3ng c\u00e1c \u0111\u00e1y gi\u1ebfng \u1edf th\u00e0nh ph\u1ed1 Xy-en (nay g\u1ecdi l\u00e0 At-xu-an), t\u1ee9c l\u00e0 tia s\u00e1ng chi\u1ebfu th\u1eb3ng \u0111\u1ee9ng 2) C\u00f9ng l\u00fac \u0111\u00f3 \u1edf th\u00e0nh ph\u1ed1 A-l\u1ebfch-x\u0103ng-\u0111ria c\u00e1ch Xy-en 800km, m\u1ed9t th\u00e1p cao 25m c\u00f3 b\u00f3ng tr\u00ean m\u1eb7t \u0111\u1ea5t d\u00e0i 3,1m . T\u1eeb hai quan s\u00e1t tr\u00ean, em h\u00e3y t\u00ednh x\u1ea5p x\u1ec9 \\\"chu vi\\\" c\u1ee7a Tr\u00e1i \u0110\u1ea5t. (Tr\u00ean h\u00ecnh \u0111i\u1ec3m S t\u01b0\u1ee3ng tr\u01b0ng cho th\u00e0nh ph\u1ed1 Xy-en, \u0111i\u1ec3m A t\u01b0\u1ee3ng tr\u01b0ng cho th\u00e0nh ph\u1ed1 A- l\u1ebfch-x\u0103ng-\u0111ri-a, b\u00f3ng c\u1ee7a th\u00e1p tr\u00ean m\u1eb7t \u0111\u1ea5t \u0111\u01b0\u1ee3c coi l\u00e0 \u0111o\u1ea1n th\u1eb3ng AB ) C\u00e2u 8. (3 \u0111i\u1ec3m) T\u1eeb \u0111i\u1ec3m A n\u1eb1m ngo\u00e0i \u0111\u01b0\u1eddng tr\u00f2n (O; R) , v\u1ebd hai ti\u1ebfp tuy\u1ebfn AM, AN ( M, N l\u00e0 hai ti\u1ebfp \u0111i\u1ec3m) v\u00e0 c\u00e1t tuy\u1ebfn ABC ( B n\u1eb1m gi\u1eefa A v\u00e0 C , tia AC n\u1eb1m gi\u1eefa hai tia AO, AN ). G\u1ecdi I l\u00e0 trung \u0111i\u1ec3m BC . a) Ch\u1ee9ng minh n\u0103m \u0111i\u1ec3m A, M, N, O, I c\u00f9ng thu\u1ed9c m\u1ed9t \u0111\u01b0\u1eddng tr\u00f2n. b) MN c\u1eaft BC t\u1ea1i K . Ch\u1ee9ng minh KB.KC = KM.KN v\u00e0 BC = 2 KC . BA KA c) MN c\u1eaft OA t\u1ea1i H . G\u1ecdi S l\u00e0 trung \u0111i\u1ec3m c\u1ee7a AH ; MS c\u1eaft \u0111\u01b0\u1eddng tr\u00f2n (O) t\u1ea1i E . V\u1ebd \u0111\u01b0\u1eddng k\u00ednh MF c\u1ee7a \u0111\u01b0\u1eddng tr\u00f2n (O) . Ch\u1ee9ng minh F, H, E th\u1eb3ng h\u00e0ng. 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) a) V\u1ebd \u0111\u1ed3 th\u1ecb (P) : y = 1 x2 v\u00e0 \u0111\u01b0\u1eddng th\u1eb3ng (d) : y = \u2212 3 x + 2 tr\u00ean c\u00f9ng h\u1ec7 tr\u1ee5c t\u1ecda \u0111\u1ed9. 22 b) T\u00ecm t\u1ecda \u0111\u1ed9 giao \u0111i\u1ec3m c\u1ee7a (P) v\u00e0 (d) \u1edf c\u00e2u tr\u00ean 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 = 1 x2 2 1 0 1 2 2 22 x 02 y = \u22123x+2 2 \u22121 2 b) T\u00ecm t\u1ecda \u0111\u1ed9 giao \u0111i\u1ec3m c\u1ee7a (P) v\u00e0 (d) b\u1eb1ng ph\u00e9p t\u00ednh. Ph\u01b0\u01a1ng tr\u00ecnh ho\u00e0nh \u0111\u1ed9 giao \u0111i\u1ec3m c\u1ee7a (P) v\u00e0 (d) : 1 x2 = \u2212 3 x + 2 22 \uf0db 1 x2 + 3 x \u2212 2 = 0 22 \uf0db \uf0e9x = 1 \uf0ea\uf0ebx = \u22124 Thay x = 1 v\u00e0o y = 1 x2 , ta \u0111\u01b0\u1ee3c: y = 1 .(1)2 = 1 . 2 22 T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N 3","TUY\u1ec2N T\u1eacP \u0110\u1ec0 THAM KH\u1ea2O TUY\u1ec2N SINH 10 TP H\u1ed2 CH\u00cd MINH Thay x = \u22124 v\u00e0o y = 1 x2 , ta \u0111\u01b0\u1ee3c: y = 1 .(\u22124)2 = 8 . 22 V\u1eady \uf0e6\uf0e7 1; 1\uf0f6 , (\u22124; 8) l\u00e0 hai giao \u0111i\u1ec3m c\u1ea7n t\u00ecm. \uf0e8 \uf0f7 2 \uf0f8 C\u00e2u 2. (1 \u0111i\u1ec3m) Cho ph\u01b0\u01a1ng tr\u00ecnh 2x2 \u2212 x \u2212 6 = 0 c\u00f3 2 nghi\u1ec7m l\u00e0 x1 , x2 . Kh\u00f4ng gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh, ( )h\u00e3y t\u00ednh gi\u00e1 tr\u1ecb c\u1ee7a bi\u1ec3u th\u1ee9c x1 \u2212 x2 2 L\u1eddi gi\u1ea3i V\u00ec \uf044 = b2 \u2212 4ac = 49 \uf03e 0 N\u00ean ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m ph\u00e2n bi\u1ec7t x1 ,x2 . \uf0ec\uf0ef\uf0efS = x1 + x2 = \u2212b = 1 \uf0ed = x1 a 2 Theo \u0111\u1ecbnh l\u00ed Vi-et, ta c\u00f3: \uf0ef\uf0ef\uf0eeP = \u22123 c .x2 = a ( )Ta c\u00f3: x1 \u2212 x2 2 = x12 \u2212 2x1x2 + x22 = x12 + x22 \u2212 2x1x2 = S2 \u2212 2P \u2212 2P = S2 \u2212 4P = \uf0e6 1 \uf0f62 \u2212 4. ( \u22123) = 49 \uf0e7\uf0e8 2 \uf0f7\uf0f8 4 C\u00e2u 3. (1 \u0111i\u1ec3m) Nh\u00e2n d\u1ecbp khai tr\u01b0\u01a1ng, c\u1eeda h\u00e0ng ch\u1ecb H\u01b0\u01a1ng \u0111\u00e3 quy\u1ebft \u0111\u1ecbnh gi\u1ea3m gi\u00e1 20% cho m\u1ed9t quy\u1ec3n t\u1eadp v\u00e0 n\u1ebfu kh\u00e1ch h\u00e0ng mua 10 quy\u1ec3n t\u1eadp tr\u1edf l\u00ean th\u00ec t\u1eeb quy\u1ec3n th\u1ee9 11 tr\u1edf \u0111i, kh\u00e1ch h\u00e0ng ch\u1ec9 ph\u1ea3i tr\u1ea3 90% gi\u00e1 \u0111\u00e3 gi\u1ea3m. Bi\u1ebft r\u1eb1ng ban \u0111\u1ea7u gi\u00e1 m\u1ed9t quy\u1ec3n t\u1eadp l\u00e0 8 000 \u0111\u1ed3ng. a) M\u1eb9 b\u1ea1n An \u0111\u1eb7t mua cho hai ch\u1ecb em b\u1ea1n An 60 quy\u1ec3n t\u1eadp. T\u00ednh s\u1ed1 ti\u1ec1n m\u1eb9 b\u1ea1n An ph\u1ea3i tr\u1ea3. b) M\u1ed9t kh\u00e1ch h\u00e0ng \u0111\u00e3 mua t\u1eadp \u1edf c\u1eeda h\u00e0ng ch\u1ecb H\u01b0\u01a1ng v\u00e0 t\u1ed5ng s\u1ed1 ti\u1ec1n m\u00e0 v\u1ecb kh\u00e1ch n\u00e0y ph\u1ea3i tr\u1ea3 l\u00e0 928 000 \u0111\u1ed3ng. H\u1ecfi kh\u00e1ch h\u00e0ng n\u00e0y \u0111\u00e3 mua bao nhi\u00eau quy\u1ec3n t\u1eadp? L\u1eddi gi\u1ea3i ( )a) Gi\u00e1 ti\u1ec1n m\u1ed9t quy\u1ec3n t\u1eadp sau khi gi\u1ea3m 20% l\u00e0: 8 000 1 \u2212 20% = 6 400 (\u0111\u1ed3ng) Do m\u1eb9 b\u1ea1n An mua 60 quy\u1ec3n t\u1eadp n\u00ean s\u1ebd ch\u1ec9 ph\u1ea3i tr\u1ea3 90% gi\u00e1 \u0111\u00e3 gi\u1ea3m. V\u00ec v\u1eady gi\u00e1 ti\u1ec1n m\u1ed9t quy\u1ec3n t\u1eadp ph\u1ea3i tr\u1ea3 l\u00e0: 6 400\uf0b490% = 5 760 (\u0111\u1ed3ng) S\u1ed1 ti\u1ec1n m\u1eb9 An ph\u1ea3i tr\u1ea3: 6 400 \uf0b410 + 5 760 \uf0b4 50 = 352 000 (\u0111\u1ed3ng). b) G\u1ecdi x l\u00e0 s\u1ed1 quy\u1ec3n t\u1eadp m\u00e0 v\u1ecb kh\u00e1ch mua. Do ph\u1ea3i tr\u1ea3 t\u1edbi 928 000 \u0111\u1ed3ng n\u00ean s\u1ed1 quy\u1ec3n t\u1eadp ph\u1ea3i nhi\u1ec1u h\u01a1n 10 quy\u1ec3n. T\u00c0I LI\u1ec6U \u0110\u01af\u1ee2C NH\u00d3M TO\u00c1N THCS TP HCM BI\u00caN SO\u1ea0N 4"]