{"id":476,"date":"2026-05-05T04:47:49","date_gmt":"2026-05-05T04:47:49","guid":{"rendered":"https:\/\/79.buffdemo.com\/index.php\/2026\/05\/05\/de-thi-thu-tot-nghiep-thpt-mon-toan-2024-2025-truong-thpt-chuyen-vinh-phuc\/"},"modified":"2026-05-05T07:20:52","modified_gmt":"2026-05-05T07:20:52","slug":"de-thi-thu-tot-nghiep-thpt-mon-toan-2024-2025-truong-thpt-chuyen-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-chuyen-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 Chuy\u00ean 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 Chuy\u00ean V\u0129nh Ph\u00fac<\/strong> l\u00e0 b\u00e0i ki\u1ec3m tra to\u00e0n di\u1ec7n, bao g\u1ed3m c\u00e1c d\u1ea1ng c\u00e2u h\u1ecfi tr\u1eafc nghi\u1ec7m v\u00e0 t\u1ef1 lu\u1eadn nh\u1eb1m \u0111\u00e1nh gi\u00e1 n\u0103ng l\u1ef1c t\u01b0 duy v\u00e0 ki\u1ebfn th\u1ee9c to\u00e1n h\u1ecdc c\u1ee7a h\u1ecdc sinh. \u0110\u1ec1 thi c\u00f3 c\u1ea5u tr\u00fac r\u00f5 r\u00e0ng, ph\u00e2n lo\u1ea1i c\u00e2u h\u1ecfi t\u1eeb c\u01a1 b\u1ea3n \u0111\u1ebfn n\u00e2ng cao, gi\u00fap th\u00ed sinh r\u00e8n luy\u1ec7n k\u1ef9 n\u0103ng gi\u1ea3i to\u00e1n v\u00e0 chu\u1ea9n b\u1ecb t\u1ed1t nh\u1ea5t cho k\u1ef3 thi ch\u00ednh th\u1ee9c. C\u00f9ng T\u00e0i Li\u1ec7u \u00d4n Thi<\/strong><\/a> l\u01b0u ngay b\u1ed9 \u0111\u1ec1 n\u00e0y \u0111\u1ec3 luy\u1ec7n t\u1eadp nh\u00e9!<\/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 Chuy\u00ean 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 Chuy\u00ean V\u0129nh Ph\u00fac<\/strong> g\u1ed3m 5 trang, v\u1edbi t\u1ed5ng c\u1ed9ng 22 c\u00e2u h\u1ecfi chia th\u00e0nh ba ph\u1ea7n: tr\u1eafc nghi\u1ec7m nhi\u1ec1u ph\u01b0\u01a1ng \u00e1n l\u1ef1a ch\u1ecdn, tr\u1eafc nghi\u1ec7m \u0111\u00fang sai, v\u00e0 tr\u1eafc nghi\u1ec7m tr\u1ea3 l\u1eddi ng\u1eafn. \u0110\u1ec1 thi th\u1eed to\u00e1n 12<\/a> c\u00f3 k\u00e8m \u0111\u00e1p \u00e1n chi ti\u1ebft, gi\u00fap h\u1ecdc sinh d\u1ec5 d\u00e0ng ki\u1ec3m tra v\u00e0 t\u1ef1 \u0111\u00e1nh gi\u00e1 sau khi l\u00e0m b\u00e0i. D\u01b0\u1edbi \u0111\u00e2y l\u00e0 m\u1ed9t s\u1ed1 c\u00e2u h\u1ecfi ti\u00eau bi\u1ec3u \u0111\u01b0\u1ee3c li\u1ec7t k\u00ea theo t\u1eebng d\u1ea1ng \u0111\u1ec3 b\u1ea1n tham kh\u1ea3o:<\/p>\n

Ph\u1ea7n I: 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 4 \u0111\u00e1p \u00e1n \u0111\u1ec3 l\u1ef1a ch\u1ecdn.<\/em><\/p>\n

C\u00e2u 1.<\/strong> Gi\u1ea3i b\u00f3ng \u0111\u00e1 Ngo\u1ea1i h\u1ea1ng Anh n\u0103m 2024 c\u00f3 20 \u0111\u1ed9i tham gia thi \u0111\u1ea5u. C\u1ee9 2 \u0111\u1ed9i th\u00ec ph\u1ea3i \u0111\u1ea5u v\u1edbi nhau 2 tr\u1eadn (l\u01b0\u1ee3t \u0111i v\u00e0 l\u01b0\u1ee3t v\u1ec1). H\u1ecfi gi\u1ea3i Ngo\u1ea1i h\u1ea1ng Anh c\u00f3 t\u1ea5t c\u1ea3 bao nhi\u00eau tr\u1eadn \u0111\u1ea5u?<\/p>\n

A. A\u2082\u2080\u00b2 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0B. C\u2082\u2080\u00b2 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0C. 20! \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0D. 40<\/p>\n

C\u00e2u 2.<\/strong> Nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh cos x = -1\/2 l\u00e0<\/p>\n

A. x = \u00b1 2\u03c0\/3 + k2\u03c0 (k in mathbbZ) \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0B. x = \u00b1 \u03c0\/6 + k\u03c0 (k in mathbbZ)
\nC. x = \u00b1 \u03c0\/3 + k2\u03c0 (k in mathbbZ) \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0D. x = \u00b1 \u03c0\/6 + k2\u03c0 (k in mathbbZ) <\/p>\n

Ph\u1ea7n II: Tr\u1eafc nghi\u1ec7m \u0111\u00fang sai<\/strong><\/p>\n

G\u1ed3m 4 c\u00e2u h\u1ecfi, m\u1ed7i c\u00e2u c\u00f3 c\u00e1c m\u1ec7nh \u0111\u1ec1 \u0111\u00fang ho\u1eb7c sai.<\/em><\/p>\n

C\u00e2u 1.<\/strong> Cho h\u00ecnh l\u0103ng tr\u1ee5 tam gi\u00e1c \u0111\u1ec1u ABC.A\u2019B\u2019C\u2019 c\u00f3 c\u1ea1nh \u0111\u00e1y b\u1eb1ng 2a , c\u1ea1nh b\u00ean AA\u2019 = a . Trong m\u1eb7t ph\u1eb3ng (A\u2019B\u2019C\u2019) , k\u1ebb A\u2019H perp B\u2019C\u2019 t\u1ea1i H .<\/p>\n

a) \u0110\u01b0\u1eddng th\u1eb3ng B\u2019C\u2019 vu\u00f4ng g\u00f3c v\u1edbi m\u1eb7t ph\u1eb3ng (AA\u2019H) .
\nb) Hai m\u1eb7t ph\u1eb3ng (ABC) v\u00e0 (AA\u2019H) vu\u00f4ng g\u00f3c.
\nc) Kho\u1ea3ng c\u00e1ch t\u1eeb \u0111i\u1ec3m A\u2019 \u0111\u1ebfn m\u1eb7t ph\u1eb3ng (AB\u2019C\u2019) b\u1eb1ng a\u221a3 .
\nd) Th\u1ec3 t\u00edch kh\u1ed1i t\u1ee9 di\u1ec7n BCB\u2019A\u2019 l\u00e0 a\u00b3 \u221a3 .<\/p>\n

C\u00e2u 2.<\/strong> Cho h\u00e0m s\u1ed1 f(x) = (1)\/(x\u00b2 – 4) <\/p>\n

a) Ta c\u00f3 int f(x) dx = 1\/4 int ( (1)\/(x – 2) – (1)\/(x + 2) ) dx
\nb) G\u1ecdi F(x) l\u00e0 nguy\u00ean h\u00e0m c\u1ee7a f(x) . Gi\u1ea3 s\u1eed F(3) = -1\/4 ln 5 , khi \u0111\u00f3 int f(x) dx = 1\/4 ln | (x – 2)\/(x + 2) | + 1
\nc) int\u2083\u2074 f(x) dx = 1\/4 ln a\/b v\u1edbi a\/b l\u00e0 ph\u00e2n s\u1ed1 t\u1ed1i gi\u1ea3n v\u00e0 a, b in mathbbN . Ta c\u00f3: ab = 15
\nd) int\u2083\u2074 [ (f(x) + f'(x))\/(f\u00b2(x)) ] dx = 1\/4 ln 5\/3 + 7 <\/p>\n

Ph\u1ea7n III: Tr\u1eafc nghi\u1ec7m tr\u1ea3 l\u1eddi ng\u1eafn<\/strong><\/p>\n

G\u1ed3m 6 c\u00e2u h\u1ecfi y\u00eau c\u1ea7u t\u00ednh to\u00e1n ho\u1eb7c gi\u1ea3i th\u00edch ng\u1eafn g\u1ecdn.<\/em><\/p>\n

C\u00e2u 1.<\/strong> Cho h\u00e0m s\u1ed1 y = f(x) l\u00e0 h\u00e0m \u0111a th\u1ee9c b\u1eadc b\u1ed1n, c\u00f3 \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1 y = f'(x) nh\u01b0 h\u00ecnh v\u1ebd. H\u00e0m s\u1ed1 y = f(5 – 2x) + 4x\u00b2 – 10x c\u00f3 bao nhi\u00eau \u0111i\u1ec3m c\u1ef1c ti\u1ec3u?<\/p>\n

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C\u00e2u 2.<\/strong> Hai th\u00e0nh ph\u1ed1 A v\u00e0 B c\u00e1ch nhau m\u1ed9t con s\u00f4ng. Ng\u01b0\u1eddi ta x\u00e2y d\u1ef1ng m\u1ed9t c\u00e2y c\u1ea7u EF b\u1eafc qua s\u00f4ng. Bi\u1ebft r\u1eb1ng th\u00e0nh ph\u1ed1 A c\u00e1ch con s\u00f4ng m\u1ed9t kho\u1ea3ng l\u00e0 4 km v\u00e0 th\u00e0nh ph\u1ed1 B c\u00e1ch con s\u00f4ng m\u1ed9t kho\u1ea3ng l\u00e0 6 km (h\u00ecnh v\u1ebd), bi\u1ebft HE + KF = 20 km v\u00e0 \u0111\u1ed9 d\u00e0i EF kh\u00f4ng \u0111\u1ed5i. H\u1ecfi x\u00e2y c\u00e2y c\u1ea7u c\u00e1ch th\u00e0nh ph\u1ed1 A l\u00e0 bao nhi\u00eau km \u0111\u1ec3 \u0111\u01b0\u1eddng \u0111i t\u1eeb th\u00e0nh ph\u1ed1 A \u0111\u1ebfn th\u00e0nh ph\u1ed1 B l\u00e0 ng\u1eafn nh\u1ea5t (\u0111i theo \u0111\u01b0\u1eddng AEF B )? (k\u1ebft qu\u1ea3 l\u00e0m tr\u00f2n \u0111\u1ebfn ph\u1ea7n tr\u0103m).<\/p>\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 Chuy\u00ean 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 Chuy\u00ean V\u0129nh Ph\u00fac<\/strong> <\/span>c\u00f3 s\u1ef1 k\u1ebft h\u1ee3p gi\u1eefa l\u00fd thuy\u1ebft v\u00e0 \u1ee9ng d\u1ee5ng th\u1ef1c t\u1ebf, gi\u00fap h\u1ecdc sinh ph\u00e1t tri\u1ec3n kh\u1ea3 n\u0103ng t\u01b0 duy logic v\u00e0 v\u1eadn d\u1ee5ng ki\u1ebfn th\u1ee9c v\u00e0o t\u00ecnh hu\u1ed1ng th\u1ef1c t\u1ebf. C\u00e1c c\u00e2u h\u1ecfi \u0111\u01b0\u1ee3c ph\u00e2n lo\u1ea1i t\u1eeb d\u1ec5 \u0111\u1ebfn kh\u00f3, ph\u00f9 h\u1ee3p v\u1edbi nhi\u1ec1u m\u1ee9c \u0111\u1ed9 h\u1ecdc l\u1ef1c. N\u1ebfu b\u1ea1n mu\u1ed1n luy\u1ec7n t\u1eadp h\u00e3y t\u1ea3i ngay file b\u00ean d\u01b0\u1edbi!<\/p>\n