{"id":504,"date":"2026-05-05T04:49:17","date_gmt":"2026-05-05T04:49:17","guid":{"rendered":"https:\/\/79.buffdemo.com\/index.php\/2026\/05\/05\/de-thi-thu-tot-nghiep-thpt-mon-toan-2024-2025-truong-thpt-hai-ba-trung-vinh-phuc\/"},"modified":"2026-05-05T07:22:33","modified_gmt":"2026-05-05T07:22:33","slug":"de-thi-thu-tot-nghiep-thpt-mon-toan-2024-2025-truong-thpt-hai-ba-trung-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-hai-ba-trung-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 Hai B\u00e0 Tr\u01b0ng 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 Hai B\u00e0 Tr\u01b0ng V\u0129nh Ph\u00fac <\/strong>s\u1ebd gi\u00fap h\u1ecdc sinh l\u00e0m quen v\u1edbi c\u1ea5u tr\u00fac \u0111\u1ec1 thi ch\u00ednh th\u1ee9c, r\u00e8n luy\u1ec7n k\u1ef9 n\u0103ng gi\u1ea3i to\u00e1n v\u00e0 t\u01b0 duy logic. \u0110\u1ec1 thi bao g\u1ed3m c\u00e1c d\u1ea1ng c\u00e2u h\u1ecfi \u0111a d\u1ea1ng, t\u1eeb tr\u1eafc nghi\u1ec7m \u0111\u1ebfn b\u00e0i to\u00e1n tr\u1ea3 l\u1eddi ng\u1eafn, b\u00e1m s\u00e1t n\u1ed9i dung ch\u01b0\u01a1ng tr\u00ecnh THPT, h\u1ed7 tr\u1ee3 h\u1ecdc sinh chu\u1ea9n b\u1ecb t\u1ed1t nh\u1ea5t cho k\u1ef3 thi s\u1eafp t\u1edbi. T\u00ecm hi\u1ec3u k\u1ef9 \u0111\u1ec1 thi c\u00f9ng Tailieuonthi<\/strong><\/a> qua b\u00e0i vi\u1ebft d\u01b0\u1edbi \u0111\u00e2y.<\/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 Hai B\u00e0 Tr\u01b0ng 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 Hai B\u00e0 Tr\u01b0ng V\u0129nh Ph\u00fac<\/span> g\u1ed3m 4 trang, v\u1edbi t\u1ed5ng c\u1ed9ng 22 c\u00e2u h\u1ecfi tr\u1eafc nghi\u1ec7m v\u00e0 c\u00f3 k\u00e8m \u0111\u00e1p \u00e1n chi ti\u1ebft. \u0110\u1ec1 thi \u0111\u01b0\u1ee3c chia th\u00e0nh nhi\u1ec1u d\u1ea1ng b\u00e0i t\u1eadp, gi\u00fap h\u1ecdc sinh r\u00e8n luy\u1ec7n to\u00e0n di\u1ec7n c\u00e1c k\u1ef9 n\u0103ng to\u00e1n h\u1ecdc.<\/p>\n

Ph\u1ea7n I: Tr\u1eafc nghi\u1ec7m nhi\u1ec1u l\u1ef1a ch\u1ecdn<\/strong><\/span><\/p>\n

G\u1ed3m 12 c\u00e2u h\u1ecfi t\u1eadp trung v\u00e0o c\u00e1c ch\u1ee7 \u0111\u1ec1 c\u01a1 b\u1ea3n nh\u01b0 c\u1ea5p s\u1ed1 nh\u00e2n, h\u00ecnh h\u1ecdc kh\u00f4ng gian v\u00e0 h\u00e0m s\u1ed1. V\u00ed d\u1ee5:<\/p>\n

C\u00e2u 1.<\/strong> Cho c\u1ea5p s\u1ed1 c\u1ed9ng (u_n) v\u1edbi s\u1ed1 h\u1ea1ng \u0111\u1ea7u u\u2081 = -6 v\u00e0 c\u00f4ng sai d = 4 . T\u00ednh t\u1ed5ng S c\u1ee7a 14 s\u1ed1 h\u1ea1ng \u0111\u1ea7u ti\u00ean c\u1ee7a c\u1ea5p s\u1ed1 c\u1ed9ng \u0111\u00f3.<\/p>\n

A. S = 46\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 B. S = 308 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 C. S = 644 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 D. S = 280<\/p>\n

C\u00e2u 2.<\/strong> Nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh log\u2081\u2086(x + 5) = 1\/2 l\u00e0:<\/p>\n

A. 3 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 B. -1\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 C. -3 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 D. 27<\/p>\n

Ph\u1ea7n II: C\u00e2u h\u1ecfi \u0111\u00fang\/sai<\/strong><\/span><\/p>\n

G\u1ed3m 4 c\u00e2u h\u1ecfi ki\u1ec3m tra kh\u1ea3 n\u0103ng nh\u1eadn bi\u1ebft v\u00e0 ph\u00e2n t\u00edch ki\u1ebfn th\u1ee9c to\u00e1n h\u1ecdc.<\/p>\n

C\u00e2u 1.<\/strong> Cho h\u00e0m s\u1ed1 y = f(x) = x\u00b3 – 3x\u00b2 + 2 c\u00f3 \u0111\u1ed3 th\u1ecb nh\u01b0 h\u00ecnh b\u00ean d\u01b0\u1edbi<\/p>\n

\"\"<\/p>\n

a) H\u00e0m s\u1ed1 f(x) \u0111\u1ed3ng bi\u1ebfn tr\u00ean kho\u1ea3ng (-\u221e; 2) .<\/p>\n

b) H\u00e0m s\u1ed1 f(x) \u0111\u1ea1t c\u1ef1c ti\u1ec3u t\u1ea1i x = 2 .<\/p>\n

c) \u0110\u1ed3 th\u1ecb h\u00e0m s\u1ed1 f(x) c\u00f3 hai \u0111i\u1ec3m c\u1ef1c tr\u1ecb thu\u1ed9c \u0111\u01b0\u1eddng th\u1eb3ng y = -2x + 2 .<\/p>\n

d) C\u00f3 1 gi\u00e1 tr\u1ecb nguy\u00ean m \u0111\u1ec3 ph\u01b0\u01a1ng tr\u00ecnh x\u00b3 – 3x\u00b2 + 2 – 2m = 0 c\u00f3 3 nghi\u1ec7m ph\u00e2n bi\u1ec7t.<\/p>\n

C\u00e2u 2.<\/strong> Cho parabol (P): y = x\u00b2 v\u00e0 \u0111\u01b0\u1eddng th\u1eb3ng d: y = 2x <\/p>\n

a) \u0110\u01b0\u1eddng th\u1eb3ng d c\u1eaft parabol (P) t\u1ea1i hai \u0111i\u1ec3m c\u00f3 ho\u00e0nh \u0111\u1ed9 x = 0 v\u00e0 x = 2 .<\/p>\n

b) Di\u1ec7n t\u00edch h\u00ecnh ph\u1eb3ng gi\u1edbi h\u1ea1n b\u1edfi \u0111\u01b0\u1eddng th\u1eb3ng d v\u00e0 parabol (P) l\u00e0 4\/3 .<\/p>\n

c) C\u00f4ng th\u1ee9c t\u00ednh th\u1ec3 t\u00edch c\u1ee7a kh\u1ed1i tr\u00f2n xoay khi cho h\u00ecnh ph\u1eb3ng gi\u1edbi h\u1ea1n b\u1edfi parabol (P) v\u00e0 \u0111\u01b0\u1eddng th\u1eb3ng d quay xung quanh tr\u1ee5c Ox l\u00e0: \\[
\nV = \\pi \\int_{0}^{2} \\left(x^4 \u2013 2x^2\\right) dx
\n\\]\n<\/p>\n

d) Th\u1ec3 t\u00edch c\u1ee7a kh\u1ed1i tr\u00f2n xoay khi cho h\u00ecnh ph\u1eb3ng gi\u1edbi h\u1ea1n b\u1edfi parabol (P) v\u00e0 \u0111\u01b0\u1eddng th\u1eb3ng d quay xung quanh tr\u1ee5c Ox l\u00e0 16\u03c0\/15 .<\/p>\n

Ph\u1ea7n III: B\u00e0i to\u00e1n tr\u1ea3 l\u1eddi ng\u1eafn<\/strong><\/span><\/p>\n

G\u1ed3m 6 b\u00e0i to\u00e1n y\u00eau c\u1ea7u h\u1ecdc sinh v\u1eadn d\u1ee5ng ki\u1ebfn th\u1ee9c \u0111\u1ec3 gi\u1ea3i quy\u1ebft v\u1ea5n \u0111\u1ec1 th\u1ef1c ti\u1ec5n. V\u00ed d\u1ee5:<\/p>\n

C\u00e2u 1.<\/strong> M\u1ed9t \u0111o\u00e0n c\u1ee9u tr\u1ee3 l\u0169 l\u1ee5t \u0111ang \u1edf v\u1ecb tr\u00ed A c\u1ee7a m\u1ed9t t\u1ec9nh mi\u1ec1n trung mu\u1ed1n \u0111\u1ebfn x\u00e3 C \u0111\u1ec3 ti\u1ebfp t\u1ebf l\u01b0\u01a1ng th\u1ef1c v\u00e0 thu\u1ed1c men. \u0110\u1ec3 \u0111i \u0111\u1ebfn C , \u0111o\u00e0n c\u1ee9u tr\u1ee3 ph\u1ea3i ch\u00e8o thuy\u1ec1n t\u1eeb A \u0111\u1ebfn v\u1ecb tr\u00ed D v\u1edbi v\u1eadn t\u1ed1c 4 (km\/h) , r\u1ed3i \u0111i b\u1ed9 \u0111\u1ebfn v\u1ecb tr\u00ed C v\u1edbi v\u1eadn t\u1ed1c 6 (km\/h) . Bi\u1ebft A c\u00e1ch B m\u1ed9t kho\u1ea3ng 5 km , B c\u00e1ch C m\u1ed9t kho\u1ea3ng 7 km (h\u00ecnh v\u1ebd). G\u1ecdi D\u2080 l\u00e0 v\u1ecb tr\u00ed \u0111\u1ec3 \u0111o\u00e0n c\u1ee9u tr\u1ee3 \u0111i \u0111\u1ebfn x\u00e3 C nhanh nh\u1ea5t. T\u00ednh S = 100 \u221a5AD\u2080? <\/p>\n

\"\"<\/p>\n

Tr\u1ea3 l\u1eddi: \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026<\/p>\n

C\u00e2u 2.<\/strong> M\u1ed9t c\u00e1i tr\u1ed1ng tr\u01b0\u1eddng c\u00f3 b\u00e1n k\u00ednh c\u00e1c \u0111\u00e1y l\u00e0 30 cm , thi\u1ebft di\u1ec7n vu\u00f4ng g\u00f3c v\u1edbi tr\u1ee5c v\u00e0 c\u00e1ch \u0111\u1ec1u hai \u0111\u00e1y c\u00f3 di\u1ec7n t\u00edch l\u00e0 1600\u03c0 (cm\u00b2) , chi\u1ec1u d\u00e0i c\u1ee7a tr\u1ed1ng l\u00e0 1 m . Bi\u1ebft r\u1eb1ng m\u1eb7t ph\u1eb3ng ch\u1ee9a tr\u1ee5c c\u1eaft m\u1eb7t xung quanh c\u1ee7a tr\u1ed1ng l\u00e0 c\u00e1c \u0111\u01b0\u1eddng Parabol. G\u1ecdi V l\u00e0 th\u1ec3 t\u00edch c\u1ee7a c\u00e1i tr\u1ed1ng. T\u00ednh 10V ?<\/p>\n

\"\"<\/p>\n

Tr\u1ea3 l\u1eddi: \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026<\/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 Hai B\u00e0 Tr\u01b0ng 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 Hai B\u00e0 Tr\u01b0ng V\u0129nh Ph\u00fac <\/strong>gi\u00fap h\u1ecdc sinh l\u00e0m quen v\u1edbi c\u1ea5u tr\u00fac v\u00e0 d\u1ea1ng b\u00e0i c\u1ee7a k\u1ef3 thi ch\u00ednh th\u1ee9c v\u00e0 c\u00f2n r\u1ea5t l\u00e0 h\u1eefu \u00edch \u0111\u1ec3 t\u1ef1 \u00f4n luy\u1ec7n, n\u00e2ng cao t\u01b0 duy v\u00e0 k\u1ef9 n\u0103ng gi\u1ea3i to\u00e1n. H\u00e3y t\u1ea3i ngay \u0111\u1ec1 thi th\u1eed to\u00e1n 12<\/a> \u0111\u1ec3 b\u1eaft \u0111\u1ea7u luy\u1ec7n t\u1eadp hi\u1ec7u qu\u1ea3!<\/p>\n