{"id":534,"date":"2026-05-05T04:50:51","date_gmt":"2026-05-05T04:50:51","guid":{"rendered":"https:\/\/79.buffdemo.com\/index.php\/2026\/05\/05\/de-thi-thu-tot-nghiep-thpt-mon-toan-2024-2025-so-gd-dt-ha-tinh\/"},"modified":"2026-05-05T07:24:16","modified_gmt":"2026-05-05T07:24:16","slug":"de-thi-thu-tot-nghiep-thpt-mon-toan-2024-2025-so-gd-dt-ha-tinh","status":"publish","type":"post","link":"https:\/\/79.buffdemo.com\/index.php\/2026\/05\/05\/de-thi-thu-tot-nghiep-thpt-mon-toan-2024-2025-so-gd-dt-ha-tinh\/","title":{"rendered":"\u0110\u1ec1 thi th\u1eed t\u1ed1t nghi\u1ec7p THPT m\u00f4n To\u00e1n 2024 \u2013 2025 S\u1edf GD & \u0110T H\u00e0 T\u0129nh"},"content":{"rendered":"

\u0110\u1ec1 thi th\u1eed t\u1ed1t nghi\u1ec7p THPT m\u00f4n To\u00e1n 2024 \u2013 2025 S\u1edf GD & \u0110T H\u00e0 T\u0129nh<\/strong> l\u00e0 t\u00e0i li\u1ec7u \u00f4n luy\u1ec7n kh\u00f4ng th\u1ec3 thi\u1ebfu d\u00e0nh cho c\u00e1c b\u1ea1n h\u1ecdc sinh 12. C\u00e1c c\u00e2u h\u1ecfi b\u00e1m s\u00e1t v\u1edbi c\u1ea5u tr\u00fac \u0111\u1ec1 thi ch\u00ednh th\u1ee9c v\u00e0 c\u00f3 \u0111\u1ed9 ph\u00e2n h\u00f3a m\u1ee9c \u0111\u1ed9 r\u00f5 r\u00e0ng. Ph\u00f9 h\u1ee3p cho c\u00e1c b\u1ea1n h\u1ecdc sinh trong qu\u00e1 tr\u00ecnh \u00f4n t\u1eadp v\u00e0 r\u00e8n luy\u1ec7n k\u0129 n\u0103ng gi\u1ea3i \u0111\u1ec1. H\u00e3y c\u00f9ng T\u00e0i li\u1ec7u \u00d4n Thi<\/a> t\u00ecm hi\u1ec3u qua n\u1ed9i dung c\u1ee7a \u0111\u1ec1 c\u00f3 g\u00ec \u0111\u1eb7c bi\u1ec7t nh\u00e9!<\/p>\n

N\u1ed9i dung \u0110\u1ec1 thi th\u1eed t\u1ed1t nghi\u1ec7p THPT m\u00f4n To\u00e1n 2024 \u2013 2025 S\u1edf GD & \u0110T H\u00e0 T\u0129nh<\/strong><\/h2>\n

N\u1ed9i dung \u0110\u1ec1 thi th\u1eed t\u1ed1t nghi\u1ec7p THPT m\u00f4n To\u00e1n 2024 \u2013 2025 S\u1edf GD & \u0110T H\u00e0 T\u0129nh<\/strong> c\u00f3 t\u1ea5t c\u1ea3 22 c\u00e2u. G\u1ed3m c\u00e2u tr\u1eafc nghi\u1ec7m nhi\u1ec1u \u0111\u00e1p \u00e1n, c\u00e2u \u0111\u00fang sai, c\u00e2u tr\u1ea3 l\u1eddi ng\u1eafn. \u0110\u1eb7c bi\u1ec7t, \u0111\u1ec1 thi c\u00f3 k\u00e8m theo \u0111\u00e1p \u00e1n v\u00e0 c\u00e1ch gi\u1ea3i chi ti\u1ebft m\u1ed7i c\u00e2u. Gi\u00fap c\u00e1c b\u1ea1n h\u1ecdc sinh d\u1ec5 d\u00e0ng n\u1eafm v\u1eefng ki\u1ebfn th\u1ee9c v\u00e0 \u00f4n t\u1eadp d\u1ec5 d\u00e0ng h\u01a1n. C\u1ea5u tr\u00fac chi ti\u1ebft c\u00f3 d\u1ea1ng nh\u01b0 sau:<\/p>\n

Ph\u1ea7n 1 \u2013 C\u00e2u h\u1ecfi tr\u1eafc nghi\u1ec7m nhi\u1ec1u ph\u01b0\u01a1ng \u00e1n l\u1ef1a ch\u1ecdn
\n<\/strong>
\nC\u00f3 t\u1ea5t c\u1ea3 12 c\u00e2u h\u1ecfi, m\u1ed7i c\u00e2u th\u00ed sinh ch\u1ecdn ra \u0111\u00e1p \u00e1n ch\u00ednh x\u00e1c nh\u1ea5t cho m\u1ed7i c\u00e2u. Tham kh\u1ea3o c\u00e1c c\u00e2u h\u1ecfi sau:<\/p>\n

C\u00e2u 1<\/strong>: H\u00e0m s\u1ed1 n\u00e0o d\u01b0\u1edbi \u0111\u00e2y \u0111\u1ed3ng bi\u1ebfn tr\u00ean kho\u1ea3ng (-\u221e; +\u221e) ?<\/p>\n

A. y = -x\u00b2 – 2x + 1 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0B. y = (x – 2)\/(x + 1) \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0C. y = 3x\u00b3 + 3x – 2 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0D. y = 2x\u00b2 – 5x + 1 <\/p>\n

C\u00e2u 2<\/strong>: Cho h\u00e0m s\u1ed1 y = f(x) c\u00f3 \u0111\u1ea1o h\u00e0m f'(x) = (x\u00b2 – 4)(x + 2)(x – 3) v\u00e0 li\u00ean t\u1ee5c tr\u00ean mathbbR . S\u1ed1 \u0111i\u1ec3m c\u1ef1c tr\u1ecb c\u1ee7a h\u00e0m s\u1ed1 \u0111\u00e3 cho l\u00e0:<\/p>\n

A. 5\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0B. 2\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 C. 3\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0D. 1<\/p>\n

Ph\u1ea7n 2 \u2013 C\u00e2u tr\u1eafc nghi\u1ec7m \u0111\u00fang sai<\/strong><\/p>\n

H\u00e3y c\u00f9ng l\u00e0m quen c\u00e1c c\u00e2u h\u1ecfi trong \u0111\u1ec1 thi th\u1eed to\u00e1n 12<\/a> c\u00f3 d\u1ea1ng tr\u1eafc nghi\u1ec7m \u0110\u00daNG\/ SAI:<\/p>\n

C\u00e2u 1<\/strong>: M\u1ed9t lo\u1ea1i thu\u1ed1c \u0111\u01b0\u1ee3c ti\u00eam cho m\u1ed9t b\u1ec7nh nh\u00e2n v\u00e0 n\u1ed3ng \u0111\u1ed9 thu\u1ed1c trong m\u00e1u c\u1ee7a b\u1ec7nh nh\u00e2n \u0111\u01b0\u1ee3c gi\u1ea3m s\u00e1t b\u1edfi b\u00e1c s\u0129. Bi\u1ebft r\u1eb1ng n\u1ed3ng \u0111\u1ed9 thu\u1ed1c trong m\u00e1u c\u1ee7a b\u1ec7nh nh\u00e2n sau khi ti\u00eam v\u00e0o c\u01a1 th\u1ec3 trong t gi\u1edd \u0111\u01b0\u1ee3c cho b\u1edfi c\u00f4ng th\u1ee9c c(t) = (t)\/(t\u00b2 + 1) (mg\/l)\n<\/p>\n

a) Sau khi ti\u00eam thu\u1ed1c 2 gi\u1edd th\u00ec n\u1ed3ng \u0111\u1ed9 thu\u1ed1c trong m\u00e1u c\u1ee7a b\u1ec7nh nh\u00e2n b\u1eb1ng 0,4 (mg\/l) <\/p>\n

b) Sau khi ti\u00eam thu\u1ed1c th\u00ec n\u1ed3ng \u0111\u1ed9 thu\u1ed1c trong m\u00e1u c\u1ee7a b\u1ec7nh nh\u00e2n c\u00f3 th\u1ec3 v\u01b0\u1ee3t qu\u00e1 0,5 (mg\/l) <\/p>\n

c) Sau khi ti\u00eam thu\u1ed1c 1 gi\u1edd th\u00ec n\u1ed3ng \u0111\u1ed9 thu\u1ed1c trong m\u00e1u c\u1ee7a b\u1ec7nh nh\u00e2n nh\u1ecf nh\u1ea5t<\/p>\n

d) Sau khi ti\u00eam thu\u1ed1c th\u00ec n\u1ed3ng \u0111\u1ed9 thu\u1ed1c trong m\u00e1u c\u1ee7a b\u1ec7nh nh\u00e2n cao nh\u1ea5t b\u1eb1ng 0,5 (mg\/l) <\/p>\n

C\u00e2u 2<\/strong>: M\u1ed9t h\u1ed3 n\u01b0\u1edbc nh\u00e2n t\u1ea1o \u0111\u01b0\u1ee3c x\u00e2y d\u1ef1ng trong m\u1ed9t c\u00f4ng vi\u00ean gi\u1ea3i tr\u00ed. Trong m\u00f4 h\u00ecnh minh h\u1ecda, n\u01a1i \u0111\u01b0\u1ee3c gi\u1edbi h\u1ea1n b\u1edfi c\u00e1c tr\u1ee5c t\u1ecda \u0111\u1ed9 v\u00e0 \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1 y = f(x) = -0,1x\u2074 + 0,9x\u00b2 – 1,5x + 5,6 . \u0110\u01a1n v\u1ecb \u0111o \u0111\u1ed9 d\u00e0i tr\u00ean m\u1ed7i tr\u1ee5c t\u1ecda \u0111\u1ed9 l\u00e0 100 m.<\/p>\n

\"\"<\/p>\n

a) \u0110\u01b0\u1eddng d\u1ea1o ven h\u1ed3 ch\u1ea1y d\u1ecdc theo tr\u1ee5c Ox d\u00e0i 600 m<\/p>\n

b) Tr\u00ean \u0111\u01b0\u1eddng \u0111i d\u1ea1o ven h\u1ed3 ch\u1ea1y d\u1ecdc theo tr\u1ee5c Ox , \u0111i\u1ec3m c\u00e1ch g\u1ed1c O m\u1ed9t \u0111o\u1ea1n 500 m c\u00f3 kho\u1ea3ng c\u00e1ch theo ph\u01b0\u01a1ng \u0111\u1ee9ng \u0111\u1ebfn m\u1eb7t h\u1ed3 l\u1edbn nh\u1ea5t l\u00e0 16 m<\/p>\n

c) Kho\u1ea3ng c\u00e1ch nh\u1ecf nh\u1ea5t theo ph\u01b0\u01a1ng \u0111\u1ee9ng t\u1eeb m\u1ed9t \u0111i\u1ec3m tr\u00ean \u0111\u01b0\u1eddng \u0111i d\u1ea1o ven h\u1ed3 \u0111\u1ebfn b\u1edd h\u1ed3 c\u00f3 \u0111\u1ed9 d\u00e0i \u0111\u1ebfn 490 m<\/p>\n

d) X\u00e2y d\u1ef1ng th\u00eam m\u1ed9t con k\u00eanh n\u01b0\u1edbc ch\u1ea3y theo \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1 y = -1,5x + 1,8 . Ng\u01b0\u1eddi ta d\u1ef1 \u0111\u1ecbnh x\u00e2y d\u1ef1ng m\u1ed9t b\u1ebfn t\u00e0u \u0111\u1ec3 b\u1ebfn thuy\u1ec1n ra \u0111\u01b0\u1ee3c sao cho kho\u1ea3ng c\u00e1ch t\u1eeb b\u1ebfn thuy\u1ec1n \u0111\u1ebfn con \u0111\u01b0\u1eddng \u0111i d\u1ea1o l\u00e0 ng\u1eafn nh\u1ea5t. Bi\u1ebft t\u1ecda \u0111\u1ed9 c\u1ee7a \u0111i\u1ec3m \u0111\u1eb7t x\u00e2y b\u1ebfn thuy\u1ec1n n\u00e0y l\u00e0 M(a; b) . Gi\u00e1 tr\u1ecb c\u1ee7a a + b b\u1eb1ng 43<\/p>\n

Ph\u1ea7n 3 \u2013 C\u00e2u tr\u1eafc nghi\u1ec7m tr\u1ea3 l\u1eddi ng\u1eafn<\/strong><\/p>\n

Tr\u00edch m\u1ed9t s\u1ed1 c\u00e2u h\u1ecfi d\u1ea1ng tr\u1ea3 l\u1eddi ng\u1eafn trong \u0111\u1ec1 nh\u01b0 sau:<\/p>\n

C\u00e2u 1<\/strong>: Cho \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1 f(x) = 2sinx nh\u01b0 h\u00ecnh v\u1ebd b\u00ean. T\u00ednh di\u1ec7n t\u00edch tam gi\u00e1c ABC .<\/p>\n

\"\"<\/p>\n

C\u00e2u 2<\/strong>: Trong \u0111\u1ec1 ki\u1ec3m tra 15 ph\u00fat m\u00f4n To\u00e1n c\u00f3 20 c\u00e2u tr\u1eafc nghi\u1ec7m. M\u1ed7i c\u00e2u tr\u1eafc nghi\u1ec7m c\u00f3 4 ph\u01b0\u01a1ng \u00e1n tr\u1ea3 l\u1eddi, trong \u0111\u00f3 ch\u1ec9 c\u00f3 m\u1ed9t ph\u01b0\u01a1ng \u00e1n tr\u1ea3 l\u1eddi \u0111\u00fang. An gi\u1ea3i ch\u00ednh x\u00e1c \u0111\u00fang 10 c\u00e2u, 10 c\u00e2u c\u00f2n l\u1ea1i l\u1ef1a ch\u1ecdn ng\u1eabu nhi\u00ean \u0111\u1ec3 tr\u1ea3 l\u1eddi. Bi\u1ebft r\u1eb1ng m\u1ed7i c\u00e2u tr\u1ea3 l\u1eddi \u0111\u00fang \u0111\u01b0\u1ee3c 0{,}5 \u0111i\u1ec3m, tr\u1ea3 l\u1eddi sai kh\u00f4ng b\u1ecb tr\u1eeb \u0111i\u1ec3m. X\u00e1c su\u1ea5t \u0111\u1ec3 An \u0111\u1ea1t \u0111\u01b0\u1ee3c \u0111\u00fang 6 \u0111i\u1ec3m l\u00e0 P . Khi \u0111\u00f3, 100P b\u1eb1ng \\ldots<\/p>\n

T\u1ea3i \u0110\u1ec1 thi th\u1eed t\u1ed1t nghi\u1ec7p THPT m\u00f4n To\u00e1n 2024 \u2013 2025 S\u1edf GD & \u0110T H\u00e0 T\u0129nh<\/strong><\/h2>\n

\u0110\u1ec3 c\u00f3 m\u1ed9t k\u1ebft qu\u1ea3 t\u1ed1t trong k\u00ec thi t\u1ed1t nghi\u1ec7p THPT s\u1eafp t\u1edbi. C\u00e1c b\u1ea1n h\u1ecdc sinh t\u1ea3i ngay \u0110\u1ec1 thi th\u1eed t\u1ed1t nghi\u1ec7p THPT m\u00f4n To\u00e1n 2024 \u2013 2025 S\u1edf GD & \u0110T H\u00e0 T\u0129nh<\/strong> qua \u0111\u01b0\u1eddng link sau \u0111\u00e2y. V\u00e0 h\u00e3y b\u1eaft tay v\u00e0o luy\u1ec7n t\u1eadp ngay h\u00f4m nay \u0111\u1ec3 \u0111\u1ea1t th\u00e0nh t\u00edch cao nh\u1ea5t nh\u00e9.<\/p>\n