{"id":474,"date":"2026-05-05T04:47:43","date_gmt":"2026-05-05T04:47:43","guid":{"rendered":"https:\/\/79.buffdemo.com\/index.php\/2026\/05\/05\/de-thi-thu-tot-nghiep-thpt-mon-toan-2024-2025-truong-thpt-le-xoay-vinh-phuc\/"},"modified":"2026-05-05T07:20:46","modified_gmt":"2026-05-05T07:20:46","slug":"de-thi-thu-tot-nghiep-thpt-mon-toan-2024-2025-truong-thpt-le-xoay-vinh-phuc","status":"publish","type":"post","link":"https:\/\/79.buffdemo.com\/index.php\/2026\/05\/05\/de-thi-thu-tot-nghiep-thpt-mon-toan-2024-2025-truong-thpt-le-xoay-vinh-phuc\/","title":{"rendered":"\u0110\u1ec1 thi th\u1eed t\u1ed1t nghi\u1ec7p THPT m\u00f4n To\u00e1n 2024 \u2013 2025 tr\u01b0\u1eddng THPT L\u00ea Xoay V\u0129nh Ph\u00fac"},"content":{"rendered":"

\u0110\u1ec1 thi th\u1eed t\u1ed1t nghi\u1ec7p THPT m\u00f4n To\u00e1n 2024 \u2013 2025 tr\u01b0\u1eddng THPT L\u00ea Xoay V\u0129nh Ph\u00fac <\/strong>v\u1eeba \u0111\u01b0\u1ee3c c\u00f4ng b\u1ed1 b\u00e1m s\u00e1t chu\u1ea9n \u0111\u1ea7u ra v\u00e0 \u0111\u00e1p \u1ee9ng y\u00eau c\u1ea7u \u0111\u00e1nh gi\u00e1 to\u00e0n di\u1ec7n ki\u1ebfn th\u1ee9c, \u0111\u1ec1 thi bao g\u1ed3m nhi\u1ec1u d\u1ea1ng b\u00e0i t\u1eeb c\u01a1 b\u1ea3n \u0111\u1ebfn n\u00e2ng cao, gi\u00fap h\u1ecdc sinh r\u00e8n luy\u1ec7n k\u1ef9 n\u0103ng gi\u1ea3i quy\u1ebft v\u1ea5n \u0111\u1ec1 v\u00e0 t\u01b0 duy logic. C\u00f9ng Tailieuonthi<\/strong><\/a> tham kh\u1ea3o m\u1ed9t s\u1ed1 d\u1ea1ng b\u00e0i c\u00f3 trong \u0111\u1ec1 thi.<\/p>\n

N\u1ed9i dung \u0110\u1ec1 thi th\u1eed t\u1ed1t nghi\u1ec7p THPT m\u00f4n To\u00e1n 2024 \u2013 2025 tr\u01b0\u1eddng THPT L\u00ea Xoay V\u0129nh Ph\u00fac<\/strong><\/h2>\n

\u0110\u1ec1 thi th\u1eed t\u1ed1t nghi\u1ec7p THPT m\u00f4n To\u00e1n 2024 \u2013 2025 tr\u01b0\u1eddng THPT L\u00ea Xoay V\u0129nh Ph\u00fac<\/strong> c\u00f3 th\u1eddi gian l\u00e0m b\u00e0i l\u00e0 90 ph\u00fat, \u0111\u01b0\u1ee3c chia th\u00e0nh 3 ph\u1ea7n r\u00f5 r\u00e0ng, bao g\u1ed3m c\u1ea3 tr\u1eafc nghi\u1ec7m khoanh \u0111\u00e1p \u00e1n, \u0111\u00fang sai v\u00e0 tr\u1ea3 l\u1eddi ng\u1eafn. \u0110i\u1ec3m n\u1ed5i b\u1eadt c\u1ee7a \u0111\u1ec1 thi th\u1eed thpt qu\u1ed1c gia m\u00f4n to\u00e1n 2025<\/a> n\u00e0y l\u00e0 l\u1eddi gi\u1ea3i chi ti\u1ebft k\u00e8m \u0111\u00e1p \u00e1n, gi\u00fap h\u1ecdc sinh t\u1ef1 ki\u1ec3m tra, \u0111\u1ed1i chi\u1ebfu v\u00e0 kh\u1eafc ph\u1ee5c l\u1ed7 h\u1ed5ng ki\u1ebfn th\u1ee9c.<\/p>\n

PH\u1ea6N I: C\u00e2u tr\u1eafc nghi\u1ec7m nhi\u1ec1u ph\u01b0\u01a1ng \u00e1n l\u1ef1a ch\u1ecdn<\/strong><\/p>\n

G\u1ed3m 12 c\u00e2u h\u1ecfi, m\u1ed7i c\u00e2u c\u00f3 b\u1ed1n ph\u01b0\u01a1ng \u00e1n tr\u1ea3 l\u1eddi. Th\u00ed sinh ch\u1ec9 ch\u1ecdn m\u1ed9t ph\u01b0\u01a1ng \u00e1n \u0111\u00fang. N\u1ed9i dung c\u00e1c c\u00e2u h\u1ecfi bao g\u1ed3m:<\/p>\n

C\u00e2u 1.<\/strong> Cho h\u00e0m s\u1ed1 f(x) li\u00ean t\u1ee5c tr\u00ean [-1; 5] v\u00e0 c\u00f3 \u0111\u1ed3 th\u1ecb tr\u00ean \u0111o\u1ea1n [-1; 5] nh\u01b0 h\u00ecnh v\u1ebd b\u00ean d\u01b0\u1edbi. T\u1ed5ng gi\u00e1 tr\u1ecb l\u1edbn nh\u1ea5t v\u00e0 gi\u00e1 tr\u1ecb nh\u1ecf nh\u1ea5t c\u1ee7a h\u00e0m s\u1ed1 f(x) tr\u00ean \u0111o\u1ea1n [-1; 5] b\u1eb1ng<\/p>\n

\"\"<\/p>\n

A. -1 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0B. 4 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0C. 1 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0D. 2 <\/p>\n

C\u00e2u 2.<\/strong> \u0110\u01b0\u1eddng cong trong h\u00ecnh v\u1ebd b\u00ean l\u00e0 \u0111\u1ed3 th\u1ecb c\u1ee7a h\u00e0m s\u1ed1 n\u00e0o d\u01b0\u1edbi \u0111\u00e2y?<\/p>\n

\"\"<\/p>\n

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

PH\u1ea6N II: C\u00e2u tr\u1eafc nghi\u1ec7m \u0111\u00fang sai<\/strong><\/p>\n

G\u1ed3m 4 c\u00e2u, m\u1ed7i c\u00e2u c\u00f3 4 \u00fd nh\u1ecf (a, b, c, d). Th\u00ed sinh ch\u1ecdn \u0111\u00fang ho\u1eb7c sai cho t\u1eebng \u00fd. N\u1ed9i dung t\u1eadp trung v\u00e0o:<\/p>\n

C\u00e2u 1.<\/strong> Cho h\u00e0m s\u1ed1 y = f(x) = (ax\u00b2 + bx + c)\/(x + n) c\u00f3 \u0111\u1ed3 th\u1ecb (C) nh\u01b0 h\u00ecnh v\u1ebd b\u00ean.<\/p>\n

\"\"<\/p>\n

a) H\u00e0m s\u1ed1 \u0111\u1ed3ng bi\u1ebfn tr\u00ean kho\u1ea3ng (0;1) .<\/p>\n

b) H\u00e0m s\u1ed1 c\u00f3 hai \u0111i\u1ec3m c\u1ef1c tr\u1ecb.<\/p>\n

c) \u0110\u1ed3 th\u1ecb (C) c\u00f3 ti\u1ec7m c\u1eadn xi\u00ean \u0111i qua \u0111i\u1ec3m A(-1;2) .<\/p>\n

d) Ph\u01b0\u01a1ng tr\u00ecnh x \u00b7 | f(x) | = x – 4 c\u00f3 \u0111\u00fang 3 nghi\u1ec7m th\u1ef1c ph\u00e2n bi\u1ec7t.<\/p>\n

C\u00e2u 2.<\/strong> Cho h\u00e0m s\u1ed1 y = f(x) , f'(x) = 2 cos\u00b2 x\/2 + 3, forall x in mathbbR . C\u00e1c m\u1ec7nh \u0111\u1ec1 sau \u0111\u00fang hay sai?<\/p>\n

a) H\u00e0m s\u1ed1 y = f(x) c\u00f3 d\u1ea1ng f(x) = sin x + 4x + C v\u1edbi C l\u00e0 h\u1eb1ng s\u1ed1.<\/p>\n

b) N\u1ebfu f(0) = 4 th\u00ec f(\u03c0\/2) = 2\u03c0 + 5 .<\/p>\n

c) int\u2080^\u03c0\/2 f(x) dx = F( \u03c0\/2 ) – F(0) v\u1edbi F(x) l\u00e0 m\u1ed9t nguy\u00ean h\u00e0m c\u1ee7a f(x) .<\/p>\n

d) Di\u1ec7n t\u00edch h\u00ecnh ph\u1eb3ng gi\u1edbi h\u1ea1n b\u1edfi \u0111\u1ed3 th\u1ecb c\u1ee7a hai h\u00e0m s\u1ed1 y = f'(x); y = 6 v\u00e0 hai \u0111\u01b0\u1eddng th\u1eb3ng x = 0, x = \u03c0\/2 c\u00f3 d\u1ea1ng S = a + b\u03c0 th\u00ec a + 2b = -1 .<\/p>\n

PH\u1ea6N III: C\u00e2u tr\u1eafc nghi\u1ec7m tr\u1ea3 l\u1eddi ng\u1eafn<\/strong><\/p>\n

G\u1ed3m 6 c\u00e2u v\u1edbi y\u00eau c\u1ea7u t\u00ednh to\u00e1n c\u1ee5 th\u1ec3:<\/p>\n

C\u00e2u 1.<\/strong> Anh Vinh \u0111ang c\u1eafm tr\u1ea1i d\u01b0\u1edbi t\u00e1n c\u00e2y th\u00f4ng \u1edf \u0111i\u1ec3m X c\u00e1ch \u0111i\u1ec3m A m\u1ed9t kho\u1ea3ng 3 km . \u0110i\u1ec3m A n\u1eb1m tr\u00ean \u0111\u01b0\u1eddng b\u1edd bi\u1ec3n (\u0111\u01b0\u1eddng b\u1edd bi\u1ec3n l\u00e0 \u0111\u01b0\u1eddng th\u1eb3ng). \u00d4 t\u00f4 c\u1ee7a anh Vinh \u0111\u1ed7 \u1edf v\u1ecb tr\u00ed Y c\u00e1ch \u0111i\u1ec3m B m\u1ed9t kho\u1ea3ng 3 km . \u0110i\u1ec3m B c\u0169ng thu\u1ed9c \u0111\u01b0\u1eddng b\u1edd bi\u1ec3n. Bi\u1ebft r\u1eb1ng AB = 18 km, AM = NB = x km v\u00e0 AX = 3 km, BY = 3 km (minh ho\u1ea1 nh\u01b0 h\u00ecnh v\u1ebd).<\/p>\n

\"\"<\/p>\n

Khi \u0111ang d\u1ef1ng tr\u1ea1i t\u1ea1i v\u1ecb tr\u00ed X , anh Vinh kh\u00f4ng may b\u1ecb r\u1eafn c\u1eafn, ch\u1ea5t \u0111\u1ed9c lan v\u00e0o m\u00e1u. Sau khi b\u1ecb r\u1eafn c\u1eafn, n\u1ed3ng \u0111\u1ed9 ch\u1ea5t \u0111\u1ed9c trong m\u00e1u t\u0103ng theo th\u1eddi gian \u0111\u01b0\u1ee3c t\u00ednh theo ph\u01b0\u01a1ng tr\u00ecnh y = 50 log(t + 2) . Trong \u0111\u00f3, y l\u00e0 n\u1ed3ng \u0111\u1ed9 ch\u1ea5t \u0111\u1ed9c, t l\u00e0 th\u1eddi gian t\u00ednh b\u1eb1ng gi\u1edd sau khi b\u1ecb r\u1eafn c\u1eafn. Anh Vinh c\u1ea7n quay tr\u1edf l\u1ea1i \u00f4 t\u00f4 \u1edf v\u1ecb tr\u00ed Y \u0111\u1ec3 l\u1ea5y thu\u1ed1c gi\u1ea3i \u0111\u1ed9c. Anh ch\u1ea1y t\u1eeb ch\u1ed7 c\u00e2y th\u00f4ng \u1edf \u0111i\u1ec3m X ra th\u1eb3ng v\u1ecb tr\u00ed M v\u1edbi v\u1eadn t\u1ed1c l\u00e0 5 km\/h v\u00e0 ch\u1ea1y tr\u00ean b\u00e3i bi\u1ec3n t\u1eeb M t\u1edbi \u0111i\u1ec3m N v\u1edbi v\u1eadn t\u1ed1c l\u00e0 13 km\/h , sau \u0111\u00f3 ch\u1ea1y th\u1eb3ng \u0111\u1ebfn ch\u1ed7 \u00f4 t\u00f4 v\u1edbi v\u1eadn t\u1ed1c 5 km\/h . T\u00ednh n\u1ed3ng \u0111\u1ed9 ch\u1ea5t \u0111\u1ed9c trong m\u00e1u th\u1ea5p nh\u1ea5t khi anh Vinh v\u1ec1 \u0111\u1ebfn \u00f4 t\u00f4 (k\u1ebft qu\u1ea3 l\u00e0m tr\u00f2n \u0111\u1ebfn h\u00e0ng ph\u1ea7n ch\u1ee5c).<\/p>\n

C\u00e2u 2.<\/strong> Cho \u0111\u1ed3 th\u1ecb (C) c\u1ee7a h\u00e0m \u0111a th\u1ee9c b\u1eadc ba v\u00e0 parabola (P) c\u00f3 tr\u1ee5c \u0111\u1ed1i x\u1ee9ng vu\u00f4ng g\u00f3c v\u1edbi tr\u1ee5c ho\u00e0nh nh\u01b0 h\u00ecnh v\u1ebd.<\/p>\n

\"\"<\/p>\n

Bi\u1ebft ph\u1ea7n h\u00ecnh ph\u1eb3ng gi\u1edbi h\u1ea1n b\u1edfi (C) v\u00e0 (P) (ph\u1ea7n t\u00f4 \u0111\u1eadm c\u1ee7a h\u00ecnh v\u1ebd) c\u00f3 di\u1ec7n t\u00edch b\u1eb1ng m\/n (m, n in mathbbN; m\/n l\u00e0 ph\u00e2n s\u1ed1 t\u1ed1i gi\u1ea3n) . T\u00ednh m + n .<\/p>\n

T\u1ea3i \u0110\u1ec1 thi th\u1eed t\u1ed1t nghi\u1ec7p THPT m\u00f4n To\u00e1n 2024 \u2013 2025 tr\u01b0\u1eddng THPT L\u00ea Xoay V\u0129nh Ph\u00fac<\/span><\/strong><\/h2>\n

\u0110\u1ec1 thi th\u1eed t\u1ed1t nghi\u1ec7p THPT m\u00f4n To\u00e1n 2024 \u2013 2025 tr\u01b0\u1eddng THPT L\u00ea Xoay V\u0129nh Ph\u00fac <\/strong>bao ph\u1ee7 to\u00e0n di\u1ec7n ki\u1ebfn th\u1ee9c c\u1ee7a h\u1ecdc sinh v\u1ec1 c\u00e1c l\u0129nh v\u1ef1c: \u0111\u1ea1i s\u1ed1, gi\u1ea3i t\u00edch, h\u00ecnh h\u1ecdc kh\u00f4ng gian, x\u00e1c su\u1ea5t th\u1ed1ng k\u00ea v\u00e0 \u1ee9ng d\u1ee5ng th\u1ef1c t\u1ebf. C\u00e1c c\u00e2u h\u1ecfi c\u00f3 m\u1ee9c \u0111\u1ed9 kh\u00f3 \u0111a d\u1ea1ng t\u1eeb c\u01a1 b\u1ea3n \u0111\u1ebfn n\u00e2ng cao nh\u1eb1m \u0111\u00e1nh gi\u00e1 n\u0103ng l\u1ef1c t\u01b0 duy v\u00e0 kh\u1ea3 n\u0103ng v\u1eadn d\u1ee5ng ki\u1ebfn th\u1ee9c v\u00e0o th\u1ef1c ti\u1ec5n c\u1ee7a th\u00ed sinh. T\u1ea3i ngay<\/strong> \u0111\u1ec3 luy\u1ec7n t\u1eadp theo link b\u00ean d\u01b0\u1edbi!<\/p>\n