{"id":508,"date":"2026-05-05T04:49:30","date_gmt":"2026-05-05T04:49:30","guid":{"rendered":"https:\/\/79.buffdemo.com\/index.php\/2026\/05\/05\/de-thi-thu-tot-nghiep-thpt-mon-toan-2024-2025-truong-thpt-lien-son-vinh-phuc\/"},"modified":"2026-05-05T07:22:48","modified_gmt":"2026-05-05T07:22:48","slug":"de-thi-thu-tot-nghiep-thpt-mon-toan-2024-2025-truong-thpt-lien-son-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-lien-son-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 Li\u1ec5n S\u01a1n 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 Li\u1ec5n S\u01a1n V\u0129nh Ph\u00fac<\/strong> l\u00e0 t\u00e0i li\u1ec7u tham kh\u1ea3o h\u1eefu \u00edch d\u00e0nh cho h\u1ecdc sinh l\u1edbp 12. \u0110\u1ec1 thi c\u00f3 c\u1ea5u tr\u00fac chu\u1ea9n g\u1ed3m 5 trang gi\u00fap h\u1ecdc sinh r\u00e8n luy\u1ec7n k\u1ef9 n\u0103ng gi\u1ea3i to\u00e1n m\u00e0 c\u00f2n l\u00e0m quen v\u1edbi c\u00e1c d\u1ea1ng b\u00e0i th\u01b0\u1eddng xu\u1ea5t hi\u1ec7n trong k\u1ef3 thi ch\u00ednh th\u1ee9c. \u0110\u1eb7c bi\u1ec7t, \u0111\u1ec1 thi c\u00f3 k\u00e8m \u0111\u00e1p \u00e1n chi ti\u1ebft, h\u1ed7 tr\u1ee3 h\u1ecdc sinh t\u1ef1 \u0111\u00e1nh gi\u00e1 v\u00e0 c\u1ea3i thi\u1ec7n hi\u1ec7u qu\u1ea3 h\u1ecdc t\u1eadp. T\u1ea3i ngay \u0111\u1ec3 luy\u1ec7n t\u1eadp c\u00f9ng T\u00e0i Li\u1ec7u \u00d4n Thi<\/a><\/strong> 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 Li\u1ec5n S\u01a1n 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 Li\u1ec5n S\u01a1n V\u0129nh Ph\u00fac<\/strong> \u0111\u01b0\u1ee3c ph\u00e2n chia th\u00e0nh nhi\u1ec1u d\u1ea1ng b\u00e0i t\u1eadp t\u1eeb c\u01a1 b\u1ea3n \u0111\u1ebfn n\u00e2ng cao. \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 \u0111\u1ed1i chi\u1ebfu v\u00e0 \u0111\u00e1nh gi\u00e1 k\u1ebft qu\u1ea3 sau khi l\u00e0m b\u00e0i. Sau \u0111\u00e2y l\u00e0 m\u1ed9t s\u1ed1 d\u1ea1ng c\u00e2u h\u1ecfi ti\u00eau bi\u1ec3u trong \u0111\u1ec1 thi:<\/p>\n

Ph\u1ea7n I: Tr\u1eafc nghi\u1ec7m nhi\u1ec1u ph\u01b0\u01a1ng \u00e1n l\u1ef1a ch\u1ecdn<\/strong><\/span><\/p>\n

C\u00e2u 1:<\/strong> Cho h\u00e0m s\u1ed1 b\u1eadc ba y = f(x) c\u00f3 \u0111\u1ed3 th\u1ecb l\u00e0 \u0111\u01b0\u1eddng cong trong h\u00ecnh v\u1ebd. H\u00e0m s\u1ed1 \u0111\u00e3 cho \u0111\u1ed3ng bi\u1ebfn tr\u00ean kho\u1ea3ng n\u00e0o sau \u0111\u00e2y?<\/p>\n

\"\"<\/p>\n

A. (2; +\u221e) \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 B. (-\u221e; 0) \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 C. (1; 3) \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 D. (0; 2) <\/p>\n

C\u00e2u 2:<\/strong> Cho h\u00e0m s\u1ed1 y = f(x) c\u00f3 b\u1ea3ng bi\u1ebfn thi\u00ean nh\u01b0 sau:<\/p>\n

\"\"<\/p>\n

\u0110\u1ed3 th\u1ecb h\u00e0m s\u1ed1 \u0111\u00e3 cho c\u00f3 bao nhi\u00eau \u0111\u01b0\u1eddng ti\u1ec7m c\u1eadn?<\/p>\n

A. 4\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0B. 2\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0C. 1\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0D. 3<\/p>\n

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

C\u00e2u 1. <\/strong>Cho h\u00e0m s\u1ed1 y = x + 4\/x .<\/p>\n

a) Ti\u1ec7m c\u1eadn \u0111\u1ee9ng c\u1ee7a \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1 \u0111\u00e3 cho l\u00e0 tr\u1ee5c tung.<\/p>\n

b) \u0110\u1ea1o h\u00e0m c\u1ee7a h\u00e0m s\u1ed1 \u0111\u00e3 cho l\u00e0 y\u2019 = 1 – (4)\/(x\u00b2) .<\/p>\n

c) H\u00e0m s\u1ed1 \u0111\u00e3 cho c\u00f3 b\u1ea3ng bi\u1ebfn thi\u00ean nh\u01b0 sau:<\/p>\n

\"\"<\/p>\n

d) Gi\u00e1 tr\u1ecb l\u1edbn nh\u1ea5t c\u1ee7a h\u00e0m s\u1ed1 \u0111\u00e3 cho b\u1eb1ng 4.<\/p>\n

C\u00e2u 2.<\/strong> M\u1ed9t xe \u00f4 t\u00f4 \u0111ang ch\u1ea1y v\u1edbi v\u1eadn t\u1ed1c 65 km\/h th\u00ec ng\u01b0\u1eddi l\u00e1i xe b\u1ea5t ng\u1edd ph\u00e1t hi\u1ec7n ch\u01b0\u1edbng ng\u1ea1i v\u1eadt tr\u00ean \u0111\u01b0\u1eddng c\u00e1ch \u0111\u00f3 50 m . Ng\u01b0\u1eddi l\u00e1i xe ph\u1ea3n \u1ee9ng m\u1ed9t gi\u00e2y, sau \u0111\u00f3 \u0111\u1ea1p phanh kh\u1ea9n c\u1ea5p. K\u1ec3 t\u1eeb th\u1eddi \u0111i\u1ec3m n\u00e0y, \u00f4 t\u00f4 chuy\u1ec3n \u0111\u1ed9ng ch\u1eadm d\u1ea7n \u0111\u1ec1u v\u1edbi t\u1ed1c \u0111\u1ed9 v(t) = -10t + 20 (m\/s) , trong \u0111\u00f3 t l\u00e0 th\u1eddi gian t\u00ednh b\u1eb1ng gi\u00e2y k\u1ec3 t\u1eeb l\u00fac \u0111\u1ea1p phanh. G\u1ecdi s(t) l\u00e0 qu\u00e3ng \u0111\u01b0\u1eddng xe \u00f4 t\u00f4 \u0111i \u0111\u01b0\u1ee3c trong t (s) k\u1ec3 t\u1eeb l\u00fac \u0111\u1ea1p phanh.<\/p>\n

a) Qu\u00e3ng \u0111\u01b0\u1eddng s(t) m\u00e0 xe \u00f4 t\u00f4 \u0111i \u0111\u01b0\u1ee3c trong th\u1eddi gian t (s) l\u00e0 m\u1ed9t nguy\u00ean h\u00e0m c\u1ee7a h\u00e0m s\u1ed1 v(t) .<\/p>\n

b) s(t) = -5t\u00b2 + 20t .<\/p>\n

c) Th\u1eddi gian k\u1ec3 t\u1eeb l\u00fac \u0111\u1ea1p phanh \u0111\u1ebfn khi xe \u00f4 t\u00f4 d\u1eebng h\u1eb3n l\u00e0 20 s .<\/p>\n

d) Xe \u00f4 t\u00f4 \u0111\u00f3 kh\u00f4ng va v\u00e0o ch\u01b0\u1edbng ng\u1ea1i v\u1eadt \u1edf tr\u00ean \u0111\u01b0\u1eddng.<\/p>\n

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

C\u00e2u 1.<\/strong>\u00a0Trong m\u1ed9t ph\u00f2ng h\u1ecdc \u0111\u01b0\u1ee3c thi\u1ebft k\u1ebf d\u1ea1ng h\u00ecnh h\u1ed9p ch\u1eef nh\u1eadt v\u1edbi chi\u1ec1u d\u00e0i 8 m , chi\u1ec1u r\u1ed9ng 6 m v\u00e0 chi\u1ec1u cao 3 m . Hai b\u1ea1n An v\u00e0 B\u00ecnh l\u00e0m nhi\u1ec7m v\u1ee5 tr\u1ef1c nh\u1eadt, m\u1ea1ng nh\u1ec7n c\u1ea7n qu\u00e9t \u1edf g\u00f3c ngo\u00e0i c\u00f9ng tr\u00ean tr\u1ea7n nh\u00e0. An b\u1ea3o kh\u00f4ng n\u00ean \u0111\u1ee9ng ngay v\u1ecb tr\u00ed \u0111\u00f3 \u1edf n\u1ec1n nh\u00e0 qu\u00e9t v\u00ec s\u1ebd b\u1ee5i r\u01a1i xu\u1ed1ng ng\u01b0\u1eddi m\u00ecnh, An l\u1ea1i \u0111\u1ed1 b\u1ea1n B\u00ecnh: n\u1ebfu m\u00ecnh \u0111\u1ee9ng \u1edf gi\u1eefa nh\u00e0 qu\u00e9t th\u00ec m\u00ecnh ph\u1ea3i k\u00e9o ch\u1ed5i qu\u00e9t nh\u00e0 d\u00e0i ra m\u1ea5y m\u00e9t (l\u00e0m tr\u00f2n \u0111\u1ebfn h\u00e0ng ph\u1ea7n tr\u0103m) \u0111\u1ec3 qu\u00e9t \u0111\u01b0\u1ee3c v\u1ecb tr\u00ed m\u1ea1ng nh\u1ec7n, bi\u1ebft An c\u1ea7m ch\u1ed5i cao 1,5 m . B\u00ecnh tr\u1ea3 l\u1eddi \u0111\u1ee9ng v\u1ecb tr\u00ed \u0111\u00f3 ch\u1ed5i d\u00e0i 5 m c\u0169ng kh\u00f4ng t\u1edbi. H\u1ecfi B\u00ecnh \u0111\u00e3 t\u00ednh \u0111\u01b0\u1ee3c bao nhi\u00eau?<\/p>\n

C\u00e2u 2.<\/strong> Gi\u1ea3 s\u1eed gi\u00e1 tr\u1ecb c\u00f2n l\u1ea1i (t\u00ednh theo tri\u1ec7u \u0111\u1ed3ng) c\u1ee7a m\u1ed9t chi\u1ebfc \u00f4 t\u00f4 sau t n\u0103m s\u1eed d\u1ee5ng \u0111\u01b0\u1ee3c m\u00f4 h\u00ecnh ho\u00e1 b\u1eb1ng c\u00f4ng th\u1ee9c: V(t) = A \u00b7 (0,905)^t , trong \u0111\u00f3 A l\u00e0 gi\u00e1 xe (t\u00ednh theo tri\u1ec7u \u0111\u1ed3ng) l\u00fac m\u1edbi mua. H\u1ecfi n\u1ebfu theo m\u00f4 h\u00ecnh n\u00e0y, sau bao nhi\u00eau n\u0103m s\u1eed d\u1ee5ng th\u00ec gi\u00e1 tr\u1ecb c\u1ee7a chi\u1ebfc xe \u0111\u00f3 c\u00f2n l\u1ea1i kh\u00f4ng qu\u00e1 300 tri\u1ec7u \u0111\u1ed3ng? (K\u1ebft qu\u1ea3 l\u00e0m tr\u00f2n \u0111\u1ebfn h\u00e0ng \u0111\u01a1n v\u1ecb). Bi\u1ebft A = 780 (tri\u1ec7u \u0111\u1ed3ng).<\/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 Li\u1ec5n S\u01a1n 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 Li\u1ec5n S\u01a1n V\u0129nh Ph\u00fac <\/strong>gi\u00fap h\u1ecdc sinh l\u00e0m quen v\u1edbi c\u00e1c d\u1ea1ng b\u00e0i tr\u1ecdng t\u00e2m, t\u1eeb \u0111\u00f3 n\u00e2ng cao k\u1ef9 n\u0103ng gi\u1ea3i to\u00e1n v\u00e0 t\u1ef1 tin h\u01a1n khi b\u01b0\u1edbc v\u00e0o k\u1ef3 thi ch\u00ednh th\u1ee9c. H\u00e3y t\u1ea3i ngay file \u0111\u1ec1 thi \u0111\u1ec3 b\u1eaft \u0111\u1ea7u luy\u1ec7n t\u1eadp v\u00e0 \u0111\u1ea1t k\u1ebft qu\u1ea3 cao trong k\u1ef3 thi s\u1eafp t\u1edbi!<\/p>\n