{"id":580,"date":"2026-05-05T04:53:12","date_gmt":"2026-05-05T04:53:12","guid":{"rendered":"https:\/\/79.buffdemo.com\/index.php\/2026\/05\/05\/de-khao-sat-chat-luong-mon-toan-12-nam-2024-2025-truong-thpt-thach-thanh-1-thanh-hoa\/"},"modified":"2026-05-05T07:27:07","modified_gmt":"2026-05-05T07:27:07","slug":"de-khao-sat-chat-luong-mon-toan-12-nam-2024-2025-truong-thpt-thach-thanh-1-thanh-hoa","status":"publish","type":"post","link":"https:\/\/79.buffdemo.com\/index.php\/2026\/05\/05\/de-khao-sat-chat-luong-mon-toan-12-nam-2024-2025-truong-thpt-thach-thanh-1-thanh-hoa\/","title":{"rendered":"\u0110\u1ec1 kh\u1ea3o s\u00e1t ch\u1ea5t l\u01b0\u1ee3ng m\u00f4n To\u00e1n 12 n\u0103m 2024 \u2013 2025 tr\u01b0\u1eddng THPT Th\u1ea1ch Th\u00e0nh 1 Thanh H\u00f3a"},"content":{"rendered":"

\u0110\u1ec1 kh\u1ea3o s\u00e1t ch\u1ea5t l\u01b0\u1ee3ng m\u00f4n To\u00e1n 12 n\u0103m 2024 \u2013 2025 tr\u01b0\u1eddng THPT Th\u1ea1ch Th\u00e0nh 1 Thanh H\u00f3a<\/strong> \u0111\u01b0\u1ee3c bi\u00ean so\u1ea1n nh\u1eb1m \u0111\u00e1nh gi\u00e1 m\u1ee9c \u0111\u1ed9 hi\u1ec3u bi\u1ebft v\u00e0 n\u0103ng l\u1ef1c h\u1ecdc sinh. Qua \u0111\u00f3 gi\u00fap c\u00e1c b\u1ea1n l\u00e0m quen v\u1edbi c\u1ea5u tr\u00fac \u0111\u1ec1 thi, c\u0169ng nh\u01b0 luy\u1ec7n t\u1eadp nh\u1eb1m t\u0103ng t\u1ed1c \u0111\u1ed9 gi\u1ea3i b\u00e0i v\u00e0 n\u00e2ng cao t\u01b0 duy x\u1eed l\u00fd \u0111\u1ec1. H\u00e3y c\u00f9ng T\u00e0i li\u1ec7u \u00d4n Thi<\/a><\/strong> t\u00ecm hi\u1ec3u qua n\u1ed9i dung c\u1ee7a \u0111\u1ec1 kh\u1ea3o s\u00e1t qua b\u00e0i vi\u1ebft sau \u0111\u00e2y.<\/p>\n

N\u1ed9i dung \u0110\u1ec1 kh\u1ea3o s\u00e1t ch\u1ea5t l\u01b0\u1ee3ng m\u00f4n To\u00e1n 12 n\u0103m 2024 \u2013 2025 tr\u01b0\u1eddng THPT Th\u1ea1ch Th\u00e0nh 1 Thanh H\u00f3a<\/strong><\/h2>\n

N\u1ed9i dung \u0110\u1ec1 kh\u1ea3o s\u00e1t ch\u1ea5t l\u01b0\u1ee3ng m\u00f4n To\u00e1n 12 n\u0103m 2024 \u2013 2025 tr\u01b0\u1eddng THPT Th\u1ea1ch Th\u00e0nh 1 Thanh H\u00f3a<\/strong> \u0111\u01b0\u1ee3c x\u00e2y d\u1ef1ng b\u00e1m s\u00e1t ki\u1ebfn th\u1ee9c ch\u01b0\u01a1ng tr\u00ecnh h\u1ecdc. G\u1ed3m 22 c\u00e2u h\u1ecfi \u0111\u01b0\u1ee3c ph\u00e2n chia theo t\u1eeb d\u1ea1ng b\u00e0i t\u1eadp nh\u1eadn bi\u1ebft \u0111\u1ebfn v\u1eadn d\u1ee5ng cao nh\u1eb1m \u0111\u00e1nh gi\u00e1 \u0111\u00fang n\u0103ng l\u1ef1c c\u1ee7a c\u00e1c b\u1ea1n h\u1ecdc sinh.
\n\u0110\u1eb7c bi\u1ec7t, t\u00e0i li\u1ec7u \u0111i k\u00e8m \u0111\u00e1p \u00e1n chi ti\u1ebft v\u00e0 h\u01b0\u1edbng d\u1eabn gi\u1ea3i c\u1ee5 th\u1ec3 cho t\u1eebng c\u00e2u, gi\u00fap b\u1ea1n d\u1ec5 d\u00e0ng ki\u1ec3m tra, ph\u00e1t hi\u1ec7n l\u1ed7i sai v\u00e0 r\u00fat kinh nghi\u1ec7m khi l\u00e0m b\u00e0i. \u0110\u1ec1 Kh\u1ea3o S\u00e1t Ch\u1ea5t L\u01b0\u1ee3ng To\u00e1n 12<\/a> c\u00f3 c\u1ea5u tr\u00fac v\u00e0 n\u1ed9i dung chi ti\u1ebft nh\u01b0 sau:<\/p>\n

Ph\u1ea7n 1. C\u00e2u h\u1ecfi tr\u1eafc nghi\u1ec7m ( 12 c\u00e2u )<\/strong><\/span><\/p>\n

Th\u00ed sinh tr\u1ea3 l\u1eddi b\u1eb1ng c\u00e1ch ch\u1ecdn ph\u01b0\u01a1ng \u00e1n ch\u00ednh x\u00e1c nh\u1ea5t t\u1eeb 4 \u0111\u00e1p \u00e1n A, B, C, D. H\u00e3y th\u1eed gi\u1ea3i ngay c\u00e1c c\u00e2u h\u1ecfi tr\u1eafc nghi\u1ec7m sau \u0111\u00e2y:<\/p>\n

C\u00e2u 1:<\/strong> Cho h\u00e0m s\u1ed1 y = f(x) li\u00ean t\u1ee5c v\u00e0 c\u00f3 b\u1ea3ng bi\u1ebfn thi\u00ean tr\u00ean \u0111o\u1ea1n [-1; 3] nh\u01b0 h\u00ecnh v\u1ebd b\u00ean.<\/p>\n

\"\"<\/p>\n

Kh\u1eb3ng \u0111\u1ecbnh n\u00e0o sau \u0111\u00e2y \u0111\u00fang?<\/p>\n

A. maxlimits_[-1;3] f(x) = 4 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0B. maxlimits_[-1;3] f(x) = 5 <\/p>\n

C. maxlimits_[-1;3] f(x) = 1 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0D. maxlimits_[-1;3] f(x) = 0 <\/p>\n

C\u00e2u 2<\/strong>: Ph\u01b0\u01a1ng tr\u00ecnh 4^x\u00b2 – 4 = 16 c\u00f3 nghi\u1ec7m l\u00e0:<\/p>\n

A. x = 2 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 B. x = 3 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 C. x = 4 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0D. x = 1 <\/p>\n

C\u00e2u 3<\/strong>: Gi\u00e1 tr\u1ecb l\u1edbn nh\u1ea5t c\u1ee7a h\u00e0m s\u1ed1 f(x) = x\u00b3 – 3x\u00b2 – 9x + 10 tr\u00ean \u0111o\u1ea1n [-2; 2] b\u1eb1ng:<\/p>\n

A. -12 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0B. 10 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 C. 15 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 D. -2 <\/p>\n

Ph\u1ea7n 2. C\u00e2u h\u1ecfi tr\u1eafc nghi\u1ec7m \u0111\u00fang sai ( 4 c\u00e2u )<\/strong><\/span><\/p>\n

Trong m\u1ed7i \u00fd a), b), c), d) \u1edf m\u1ed7i c\u00e2u h\u1ecfi, th\u00ed sinh ch\u1ecdn m\u1ec7nh \u0111\u1ec1 \u0111\u00fang h\u1ecdc sai cho m\u1ed7i \u0111\u00e1p \u00e1n. Xem qua c\u00e1c c\u00e2u h\u1ecfi sau \u0111\u00e2y \u0111\u1ec3 h\u00ecnh dung r\u00f5 h\u01a1n v\u1ec1 d\u1ea1ng b\u00e0i t\u1eadp n\u00e0y nh\u00e9:<\/p>\n

C\u00e2u 1:<\/strong> B\u00e1c L\u00e2m mu\u1ed1n g\u00f2 m\u1ed9t c\u00e1i th\u00f9ng b\u1eb1ng t\u00f4n d\u1ea1ng h\u00ecnh h\u1ed9p ch\u1eef nh\u1eadt kh\u00f4ng n\u1eafp c\u00f3 \u0111\u00e1y l\u00e0 h\u00ecnh vu\u00f4ng v\u00e0 \u0111\u1ef1ng \u0111\u1ea7y \u0111\u01b0\u1ee3c 32 l\u00edt n\u01b0\u1edbc. G\u1ecdi \u0111\u1ed9 d\u00e0i c\u1ea1nh \u0111\u00e1y c\u1ee7a th\u00f9ng l\u00e0 x (dm) , chi\u1ec1u cao c\u1ee7a th\u00f9ng l\u00e0 h (dm) .<\/p>\n

a) Th\u1ec3 t\u00edch c\u1ee7a th\u00f9ng l\u00e0 V = x\u00b2 h (dm\u00b3) .<\/p>\n

b) T\u1ed5ng di\u1ec7n t\u00edch xung quanh v\u00e0 di\u1ec7n t\u00edch \u0111\u00e1y c\u1ee7a th\u00f9ng l\u00e0: S = 4xh + x\u00b2 (dm\u00b2)\n<\/p>\n

c) Ta c\u00f3: x\u00b2 h = 32 \u21d2 h = (32)\/(x\u00b2) \u21d2 S(x) = 128\/x + x\u00b2 S'(x) = -(128)\/(x\u00b2) + 2x\n<\/p>\n

d) \u0110\u1ec3 l\u00e0m \u0111\u01b0\u1ee3c c\u00e1i th\u00f9ng m\u00e0 t\u1ed1n \u00edt nguy\u00ean li\u1ec7u nh\u1ea5t th\u00ec \u0111\u1ed9 d\u00e0i c\u1ea1nh \u0111\u00e1y c\u1ee7a th\u00f9ng l\u00e0 4 dm .<\/p>\n

C\u00e2u 2<\/strong>: K\u00ednh vi\u1ec5n v\u1ecdng kh\u00f4ng gian Hubble \u0111\u01b0\u1ee3c \u0111\u01b0a v\u00e0o v\u0169 tr\u1ee5 ng\u00e0y 24\/4\/1990 b\u1eb1ng t\u00e0u con thoi Discovery. V\u1eadn t\u1ed1c c\u1ee7a t\u00e0u con thoi trong s\u1ee9 m\u1ec7nh n\u00e0y, t\u1eeb l\u00fac c\u1ea5t c\u00e1nh t\u1ea1i th\u1eddi \u0111i\u1ec3m t = 0(s) cho \u0111\u1ebfn khi t\u00ean l\u1eeda \u0111\u1ea9y \u0111\u01b0\u1ee3c ph\u00f3ng \u0111i t\u1ea1i th\u1eddi \u0111i\u1ec3m t = 126(s) , cho b\u1edfi h\u00e0m s\u1ed1 sau: v(t) = 0,001302t\u00b3 – 0,09029t\u00b2 + 23,61t – 3,083 (v \u0111\u01b0\u1ee3c t\u00ednh b\u1eb1ng textitfeet\/s, 1 textitfeet = 0,3048m)\n<\/p>\n

\"\"<\/p>\n

X\u00e9t t\u00ednh \u0111\u00fang sai c\u1ee7a m\u1ec7nh \u0111\u1ec1 sau:<\/p>\n

a) V\u1eadn t\u1ed1c c\u1ee7a t\u00e0u con thoi lu\u00f4n t\u0103ng trong kho\u1ea3ng th\u1eddi gian t\u1eeb l\u00fac c\u1ea5t c\u00e1nh \u0111\u1ebfn khi t\u00ean l\u1eeda \u0111\u1ea9y \u0111\u01b0\u1ee3c ph\u00f3ng \u0111i.<\/p>\n

b) Gia t\u1ed1c l\u1edbn nh\u1ea5t m\u00e0 t\u00e0u con thoi c\u00f3 th\u1ec3 \u0111\u1ea1t \u0111\u01b0\u1ee3c trong l\u00fac th\u1ef1c hi\u1ec7n s\u1ee9 m\u1ec7nh tr\u00ean (l\u00e0m tr\u00f2n \u0111\u1ebfn h\u00e0ng ph\u1ea7n tr\u0103m) l\u00e0 62,87 textitfeet\/s\u00b2 .<\/p>\n

c) Gia t\u1ed1c c\u1ee7a t\u00e0u con thoi t\u0103ng trong kho\u1ea3ng th\u1eddi gian t\u1eeb th\u1eddi \u0111i\u1ec3m t = 23(s) .<\/p>\n

d) Gia t\u1ed1c c\u1ee7a t\u00e0u con thoi t\u0103ng trong kho\u1ea3ng th\u1eddi gian t\u1eeb t = 21,5(s) \u0111\u1ebfn t = 126(s) .<\/p>\n

Ph\u1ea7n 3. C\u00e2u h\u1ecfi tr\u1ea3 l\u1eddi ng\u1eafn ( 6 c\u00e2u )<\/strong><\/span><\/p>\n

M\u1ed7i \u0111\u1ec1 s\u1ebd c\u00f3 6 c\u00e2u h\u1ecfi t\u1ef1 lu\u1eadn \u0111\u01b0\u1ee3c ph\u00e2n b\u1ed5 t\u1eeb c\u00e2u d\u1ec5 \u0111\u1ebfn n\u00e2ng cao v\u00e0 c\u00f3 d\u1ea1ng:<\/p>\n

C\u00e2u 1<\/strong>: Cho h\u00ecnh ch\u00f3p t\u1ee9 gi\u00e1c \u0111\u1ec1u S.ABCD , c\u00f3 \u0111\u00e1y ABCD l\u00e0 h\u00ecnh vu\u00f4ng t\u00e2m O . C\u00e1c c\u1ea1nh b\u00ean v\u00e0 c\u00e1c c\u1ea1nh \u0111\u00e1y \u0111\u1ec1u b\u1eb1ng a . G\u1ecdi M l\u00e0 trung \u0111i\u1ec3m SC . T\u00ednh g\u00f3c gi\u1eefa hai m\u1eb7t ph\u1eb3ng (MBD) v\u00e0 (ABCD) .<\/p>\n

C\u00e2u 2<\/strong>: M\u1ed9t b\u1ec3 ch\u1ee9a 3000 l\u00edt n\u01b0\u1edbc tinh khi\u1ebft. Ng\u01b0\u1eddi ta b\u01a1m v\u00e0o b\u1ec3 \u0111\u00f3 n\u01b0\u1edbc mu\u1ed1i c\u00f3 n\u1ed3ng \u0111\u1ed9 25 gam mu\u1ed1i\/1 l\u00edt n\u01b0\u1edbc v\u1edbi t\u1ed1c \u0111\u1ed9 20 l\u00edt\/ph\u00fat. Gi\u1ea3 s\u1eed n\u1ed3ng \u0111\u1ed9 mu\u1ed1i trong n\u01b0\u1edbc b\u1ec3 sau t ph\u00fat \u0111\u01b0\u1ee3c x\u00e1c \u0111\u1ecbnh b\u1edfi m\u1ed9t h\u00e0m s\u1ed1 c(t) tr\u00ean [0; +\u221e) (gam\/l\u00edt). Khi t c\u00e0ng l\u1edbn th\u00ec n\u1ed3ng \u0111\u1ed9 mu\u1ed1i trong b\u1ec3 ti\u1ebfn g\u1ea7n \u0111\u1ebfn bao nhi\u00eau gam\/l\u00edt?<\/p>\n

T\u1ea3i \u0110\u1ec1 kh\u1ea3o s\u00e1t ch\u1ea5t l\u01b0\u1ee3ng m\u00f4n To\u00e1n 12 n\u0103m 2024 \u2013 2025 tr\u01b0\u1eddng THPT Th\u1ea1ch Th\u00e0nh 1 Thanh H\u00f3a<\/strong><\/h2>\n

H\u00e3y ki\u1ec3m tra n\u0103ng l\u1ef1c v\u00e0 c\u1ee7ng c\u1ed1 ki\u1ebfn th\u1ee9c c\u1ee7a m\u00ecnh b\u1eb1ng c\u00e1ch t\u1ea3i ngay \u0110\u1ec1 kh\u1ea3o s\u00e1t ch\u1ea5t l\u01b0\u1ee3ng m\u00f4n To\u00e1n 12 n\u0103m 2024 \u2013 2025 tr\u01b0\u1eddng THPT Th\u1ea1ch Th\u00e0nh 1 Thanh H\u00f3a<\/strong> qua \u0111\u01b0\u1eddng link sau \u0111\u00e2y. T\u00e0i li\u1ec7u n\u00e0y s\u1ebd gi\u00fap b\u1ea1n l\u00e0m quen v\u1edbi c\u1ea5u tr\u00fac \u0111\u1ec1 thi, \u0111\u00e1nh gi\u00e1 \u0111\u01b0\u1ee3c n\u0103ng l\u1ef1c c\u1ee7a m\u00ecnh \u0111\u1ec3 b\u1eaft \u0111\u1ea7u r\u00e8n luy\u1ec7n k\u1ef9 n\u0103ng l\u00e0m b\u00e0i v\u00e0 chu\u1ea9n b\u1ecb th\u1eadt t\u1ed1t cho k\u1ef3 thi s\u1eafp t\u1edbi nh\u00e9.<\/p>\n