{"id":576,"date":"2026-05-05T04:53:00","date_gmt":"2026-05-05T04:53:00","guid":{"rendered":"https:\/\/79.buffdemo.com\/index.php\/2026\/05\/05\/de-thi-thu-tot-nghiep-thpt-mon-toan-2024-2025-cum-cac-truong-thpt-hai-duong\/"},"modified":"2026-05-05T07:26:53","modified_gmt":"2026-05-05T07:26:53","slug":"de-thi-thu-tot-nghiep-thpt-mon-toan-2024-2025-cum-cac-truong-thpt-hai-duong","status":"publish","type":"post","link":"https:\/\/79.buffdemo.com\/index.php\/2026\/05\/05\/de-thi-thu-tot-nghiep-thpt-mon-toan-2024-2025-cum-cac-truong-thpt-hai-duong\/","title":{"rendered":"\u0110\u1ec1 thi th\u1eed t\u1ed1t nghi\u1ec7p THPT m\u00f4n To\u00e1n 2024 \u2013 2025 C\u1ee5m c\u00e1c tr\u01b0\u1eddng THPT H\u1ea3i D\u01b0\u01a1ng"},"content":{"rendered":"

\u0110\u1ec1 thi th\u1eed t\u1ed1t nghi\u1ec7p THPT m\u00f4n To\u00e1n 2024 \u2013 2025 C\u1ee5m c\u00e1c tr\u01b0\u1eddng THPT H\u1ea3i D\u01b0\u01a1ng<\/strong> l\u00e0 t\u00e0i li\u1ec7u tham kh\u1ea3o ch\u1ea5t l\u01b0\u1ee3ng, gi\u00fap h\u1ecdc sinh l\u1edbp 12 \u00f4n luy\u1ec7n hi\u1ec7u qu\u1ea3 tr\u01b0\u1edbc k\u1ef3 thi quan tr\u1ecdng. \u0110\u1ec1 thi \u0111\u01b0\u1ee3c x\u00e2y d\u1ef1ng b\u00e1m s\u00e1t c\u1ea5u tr\u00fac ch\u00ednh th\u1ee9c c\u1ee7a B\u1ed9 GD&\u0110T, ph\u00f9 h\u1ee3p \u0111\u1ec3 \u0111\u00e1nh gi\u00e1 n\u0103ng l\u1ef1c v\u00e0 l\u00e0m quen v\u1edbi \u0111\u1ec1 thi th\u1eadt. L\u00e0 m\u1ed9t trong nh\u1eefng b\u1ed9 \u0111\u1ec1 \u0111\u01b0\u1ee3c nhi\u1ec1u gi\u00e1o vi\u00ean v\u00e0 h\u1ecdc sinh \u0111\u00e1nh gi\u00e1 cao trong qu\u00e1 tr\u00ecnh luy\u1ec7n thi THPT Qu\u1ed1c gia. H\u00e3y c\u00f9ng T\u00e0i li\u1ec7u \u00d4n Thi<\/a> t\u00ecm hi\u1ec3u th\u00eam n\u1ed9i dung qua b\u00e0i vi\u1ebft sau.<\/p>\n

N\u1ed9i dung \u0110\u1ec1 thi th\u1eed t\u1ed1t nghi\u1ec7p THPT m\u00f4n To\u00e1n 2024 \u2013 2025 C\u1ee5m c\u00e1c tr\u01b0\u1eddng THPT H\u1ea3i D\u01b0\u01a1ng<\/strong><\/h2>\n

N\u1ed9i dung \u0110\u1ec1 thi th\u1eed t\u1ed1t nghi\u1ec7p THPT m\u00f4n To\u00e1n 2024 \u2013 2025 C\u1ee5m c\u00e1c tr\u01b0\u1eddng THPT H\u1ea3i D\u01b0\u01a1ng<\/strong> \u0111\u01b0\u1ee3c \u0111\u00e1nh gi\u00e1 cao nh\u1edd ch\u1ea5t l\u01b0\u1ee3ng c\u00e2u h\u1ecfi c\u00f3 \u0111\u1ed9 ph\u00e2n c\u1ea5p t\u1ed1t, gi\u00fap t\u1ea5t c\u1ea3 c\u00e1c b\u1ea1n \u0111\u1ec1u c\u00f3 th\u1ec3 ti\u1ebfp c\u1eadn v\u00e0 th\u1eed s\u1ee9c v\u1edbi \u0111\u1ec1 thi. V\u1edbi h\u00ecnh th\u1ee9c l\u00e0 tr\u1eafc nghi\u1ec7m c\u00f3 nhi\u1ec1u d\u1ea1ng, gi\u00fap h\u1ecdc sinh r\u00e8n luy\u1ec7n t\u01b0 duy ph\u1ea3n x\u1ea1 v\u1edbi \u0111\u1ec1 nhanh v\u00e0 ch\u00ednh x\u00e1c. Chi ti\u1ebft \u0111\u1ec1 thi nh\u01b0 sau:<\/p>\n

Ph\u1ea7n 1 \u2013 C\u00e2u h\u1ecfi tr\u1eafc nghi\u1ec7m nhi\u1ec1u ph\u01b0\u01a1ng \u00e1n<\/strong><\/p>\n

C\u00e1c b\u1ea1n \u0111\u1ecdc v\u00e0 tham kh\u1ea3o c\u00e1c c\u00e2u h\u1ecfi sau \u0111\u00e2y:<\/p>\n

C\u00e2u 1:<\/strong> Cho h\u00e0m s\u1ed1 y = f(x) li\u00ean t\u1ee5c tr\u00ean mathbbR v\u00e0 c\u00f3 b\u1ea3ng x\u00e9t d\u1ea5u f'(x) nh\u01b0 h\u00ecnh v\u1ebd d\u01b0\u1edbi \u0111\u00e2y:<\/p>\n

\"\"<\/p>\n

H\u00e0m s\u1ed1 y = f(x) \u0111\u1ed3ng bi\u1ebfn tr\u00ean kho\u1ea3ng n\u00e0o d\u01b0\u1edbi \u0111\u00e2y?<\/p>\n

(A) (2;3)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0(B) (-10; -5)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 (C) (0;1)\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 h\u00ecnh v\u1ebd d\u01b0\u1edbi \u0111\u00e2y.<\/p>\n

\"\"<\/p>\n

\u0110\u1ed3 th\u1ecb h\u00e0m s\u1ed1 y = f(x) c\u00f3 bao nhi\u00eau \u0111\u01b0\u1eddng ti\u1ec7m c\u1eadn?<\/p>\n

(A) 1\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 (B) 4\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0(C) 3\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0(D) 2<\/p>\n

Ph\u1ea7n 2 \u2013 C\u00e2u h\u1ecfi tr\u1eafc nghi\u1ec7m \u0111\u00fang sai<\/strong><\/p>\n

M\u1ed7i c\u00e2u tr\u1ea3 l\u1eddi, th\u00ed sinh ch\u1ec9 ch\u1ecdn gi\u00e1 tr\u1ecb \u0110\u00daNG ho\u1eb7c SAI cho m\u1ed7i \u00fd. V\u00ed d\u1ee5 c\u00e1c c\u00e2u sau:<\/p>\n

C\u00e2u 1:<\/strong> Trong m\u1ed9t tr\u00f2 ch\u01a1i th\u1eed th\u00e1ch, b\u1ea1n Gi\u00e1p \u0111ang \u1edf tr\u00ean thuy\u1ec1n (v\u1ecb tr\u00ed A) c\u00e1ch b\u1edd h\u1ed3 (v\u1ecb tr\u00ed C) 300 m v\u00e0 c\u1ea7n \u0111i \u0111\u1ebfn v\u1ecb tr\u00ed B tr\u00ean b\u1edd h\u1ed3 nh\u01b0 h\u00ecnh v\u1ebd d\u01b0\u1edbi \u0111\u00e2y, kho\u1ea3ng c\u00e1ch t\u1eeb C \u0111\u1ebfn B l\u00e0 400 m. L\u01b0u \u00fd l\u00e0 Gi\u00e1p c\u00f3 th\u1ec3 ch\u00e8o thuy\u1ec1n th\u1eb3ng t\u1eeb A \u0111\u1ebfn B ho\u1eb7c ch\u00e8o thuy\u1ec1n t\u1eeb A \u0111\u1ebfn m\u1ed9t \u0111i\u1ec3m n\u1eb1m gi\u1eefa C v\u00e0 B r\u1ed3i ch\u1ea1y b\u1ed9 \u0111\u1ebfn B. Bi\u1ebft r\u1eb1ng Gi\u00e1p ch\u00e8o thuy\u1ec1n v\u1edbi t\u1ed1c \u0111\u1ed9 50 m\/ph\u00fat v\u00e0 ch\u1ea1y b\u1ed9 v\u1edbi t\u1ed1c \u0111\u1ed9 100 m\/ph\u00fat.<\/p>\n

(a) Th\u1eddi gian Gi\u00e1p ch\u00e8o thuy\u1ec1n t\u1eeb A \u0111\u1ebfn C r\u1ed3i ch\u1ea1y b\u1ed9 t\u1eeb C \u0111\u1ebfn B l\u00e0 10 ph\u00fat.<\/p>\n

(b) Gi\u1ea3 s\u1eed Gi\u00e1p ch\u00e8o thuy\u1ec1n th\u1eb3ng \u0111\u1ebfn \u0111i\u1ec3m D n\u1eb1m gi\u1eefa B v\u00e0 C v\u00e0 c\u00e1ch C m\u1ed9t \u0111o\u1ea1n x (m) nh\u01b0 h\u00ecnh v\u1ebd d\u01b0\u1edbi \u0111\u00e2y, r\u1ed3i ch\u1ea1y b\u1ed9 \u0111\u1ebfn B th\u00ec th\u1eddi gian Gi\u00e1p \u0111i t\u1eeb A \u0111\u1ebfn B \u0111\u01b0\u1ee3c t\u00ednh b\u1eb1ng c\u00f4ng th\u1ee9c: f(x) = 1\/100 ( \u221ax\u00b2 + 90000 + 400 – x ) (ph\u00fat).\n<\/p>\n

\"\"<\/p>\n

(c) Th\u1eddi gian nhanh nh\u1ea5t \u0111\u1ec3 Gi\u00e1p \u0111i t\u1eeb A \u0111\u1ebfn B x\u1ea5p x\u1ec9 9,2 ph\u00fat (k\u1ebft qu\u1ea3 l\u00e0m tr\u00f2n \u0111\u1ebfn h\u00e0ng ph\u1ea7n ch\u1ee5c).<\/p>\n

(d) Th\u1eddi gian Gi\u00e1p ch\u00e8o thuy\u1ec1n th\u1eb3ng t\u1eeb A \u0111\u1ebfn B l\u00e0 10 ph\u00fat.<\/p>\n

C\u00e2u 2:<\/strong> Trong kh\u00f4ng gian Oxyz , cho h\u00ecnh l\u1eadp ph\u01b0\u01a1ng ABCD.A\u2019B\u2019C\u2019D\u2019 c\u00f3 c\u1ea1nh b\u1eb1ng 1 (tham kh\u1ea3o h\u00ecnh v\u1ebd).<\/p>\n

\"\"<\/p>\n

(a) N\u1ebfu A(0;0;0), B(1;0;0), D(0;1;0), A'(0;0;1) v\u00e0 \u0111i\u1ec3m M th\u1ecfa m\u00e3n 2MB\u20d7 – 3MC\u20d7 + 5MD\u2019\u20d7 = 0\u20d7 th\u00ec M = (-1;4;7) .<\/p>\n

(b) G\u1ecdi E, F l\u1ea7n l\u01b0\u1ee3t thu\u1ed9c c\u00e1c \u0111\u01b0\u1eddng th\u1eb3ng A\u2019A\u2019 v\u00e0 C\u2019D\u2019 sao cho \u0111\u01b0\u1eddng th\u1eb3ng EF vu\u00f4ng g\u00f3c v\u1edbi m\u1eb7t ph\u1eb3ng (A\u2019BC\u2019) . Khi \u0111\u00f3 EF = \u221a3 .<\/p>\n

(c) AC\u2019\u20d7 = AB\u20d7 + AD\u20d7 + AA\u2019\u20d7 <\/p>\n

(d) N\u1ebfu A(0;0;0), B(1;0;0), D(0;1;0), A'(0;0;1) th\u00ec C'(1;2;3) <\/p>\n

Ph\u1ea7n 3 \u2013 C\u00e2u h\u1ecfi tr\u1eafc nghi\u1ec7m tr\u1ea3 l\u1eddi ng\u1eafn<\/strong><\/p>\n

C\u00e2u 1:<\/strong> Cho hai s\u1ed1 th\u1ef1c d\u01b0\u01a1ng x, y th\u1ecfa m\u00e3n 2^x + y = 2x + log\u2082(x + 2^y). Gi\u00e1 tr\u1ecb nh\u1ecf nh\u1ea5t c\u1ee7a bi\u1ec3u th\u1ee9c T = x\/y b\u1eb1ng bao nhi\u00eau, k\u1ebft qu\u1ea3 l\u00e0m tr\u00f2n \u0111\u1ebfn h\u00e0ng ph\u1ea7n tr\u0103m?<\/p>\n

C\u00e2u 2:<\/strong> Trong kh\u00f4ng gian Oxyz , cho \u0111i\u1ec3m A(2;3;-1), B(-8;7;-3) v\u00e0 \u0111i\u1ec3m M(a;b;c) thu\u1ed9c m\u1eb7t ph\u1eb3ng (Oxy) . Bi\u1ebft r\u1eb1ng A, B, M th\u1eb3ng h\u00e0ng, h\u00e3y t\u00ednh 2a – b + 3c .<\/p>\n

T\u1ea3i \u0110\u1ec1 thi th\u1eed t\u1ed1t nghi\u1ec7p THPT m\u00f4n To\u00e1n 2024 \u2013 2025 C\u1ee5m c\u00e1c tr\u01b0\u1eddng THPT H\u1ea3i D\u01b0\u01a1ng<\/strong><\/h2>\n

Luy\u1ec7n t\u1eadp \u0110\u1ec1 thi th\u1eed t\u1ed1t nghi\u1ec7p THPT m\u00f4n To\u00e1n 2024 \u2013 2025 C\u1ee5m c\u00e1c tr\u01b0\u1eddng THPT H\u1ea3i D\u01b0\u01a1ng<\/strong> s\u1ebd l\u00e0 c\u01a1 h\u1ed9i gi\u00fap b\u1ea1n ki\u1ec3m tra n\u0103ng l\u1ef1c, r\u00e8n luy\u1ec7n t\u1ed1c \u0111\u1ed9 v\u00e0 chu\u1ea9n b\u1ecb t\u00e2m l\u00fd v\u1eefng v\u00e0ng tr\u01b0\u1edbc k\u1ef3 thi th\u1eadt. T\u1ea3i ngay \u0111\u1ec1 thi k\u00e8m \u0111\u00e1p \u00e1n qua \u0111\u01b0\u1eddng d\u1eabn d\u01b0\u1edbi \u0111\u00e2y. Ch\u00fac b\u1ea1n chinh ph\u1ee5c \u0111i\u1ec3m s\u1ed1 cao trong \u0111\u1ec1 thi t\u1ed1t nghi\u1ec7p thpt m\u00f4n to\u00e1n<\/a> s\u1eafp t\u1edbi.<\/p>\n