{"id":460,"date":"2026-05-05T04:46:58","date_gmt":"2026-05-05T04:46:58","guid":{"rendered":"https:\/\/79.buffdemo.com\/index.php\/2026\/05\/05\/de-thi-thu-tot-nghiep-thpt-mon-toan-2024-2025-truong-thpt-yen-lac-2-vinh-phuc\/"},"modified":"2026-05-05T07:19:58","modified_gmt":"2026-05-05T07:19:58","slug":"de-thi-thu-tot-nghiep-thpt-mon-toan-2024-2025-truong-thpt-yen-lac-2-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-yen-lac-2-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 Y\u00ean L\u1ea1c 2 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 Y\u00ean L\u1ea1c 2 V\u0129nh Ph\u00fac<\/strong> l\u00e0 b\u1ed9 t\u00e0i li\u1ec7u m\u00f4n To\u00e1n v\u00f4 c\u00f9ng h\u1eefu \u00edch d\u00e0nh cho c\u00e1c b\u1ea1n h\u1ecdc sinh 12 chu\u1ea9n b\u1ecb \u00f4n t\u1eadp b\u01b0\u1edbc v\u00e0o k\u00ec thi t\u1ed1t nghi\u1ec7p THPT. \u0110\u1ec1 thi th\u1eed n\u00e0y t\u1eadp h\u1ee3p c\u00e1c d\u1ea1ng c\u00e2u h\u1ecfi t\u1eeb ki\u1ebfn th\u1ee9c nh\u1eadn bi\u1ebft \u0111\u1ebfn n\u00e2ng cao v\u00e0 k\u00e8m l\u1eddi gi\u1ea3i chi ti\u1ebft, gi\u00fap c\u00e1c b\u1ea1n r\u00e8n luy\u1ec7n t\u01b0 duy gi\u1ea3i \u0111\u1ec1 t\u1ed1t nh\u1ea5t. H\u00e3y c\u00f9ng T\u00e0i li\u1ec7u \u00d4n Thi<\/a><\/strong> t\u00ecm hi\u1ec3u qua n\u1ed9i dung c\u1ee7a \u0111\u1ec1 c\u00f3 g\u00ec \u0111\u1eb7c bi\u1ec7t 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 Y\u00ean L\u1ea1c 2 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 Y\u00ean L\u1ea1c 2 V\u0129nh Ph\u00fac<\/strong> \u0111\u01b0\u1ee3c bi\u00ean so\u1ea1n theo \u0111\u1ecbnh h\u01b0\u1edbng thi Tr\u1eafc nghi\u1ec7m n\u0103m 2025 v\u1edbi th\u1eddi gian l\u00e0m b\u00e0i 90 ph\u00fat. B\u1ed9 \u0111\u1ec1 thi th\u1eed c\u00f3 t\u1ea5t c\u1ea3 22 c\u00e2u \u0111\u01b0\u1ee3c chia l\u00e0m 3 ph\u1ea7n ch\u00ednh. N\u1ed9i dung chi ti\u1ebft c\u1ee7a \u0111\u1ec1 thi nh\u01b0 sau:<\/p>\n

Ph\u1ea7n I \u2013 C\u00e2u h\u1ecfi tr\u1eafc nghi\u1ec7m nhi\u1ec1u ph\u01b0\u01a1ng \u00e1n ( 12 c\u00e2u )<\/strong><\/span><\/p>\n

Ph\u1ea7n n\u00e0y c\u00f3 t\u1ea5t c\u1ea3 12 c\u00e2u tr\u1eafc nghi\u1ec7m v\u1edbi 4 ph\u01b0\u01a1ng \u00e1n A, B, C, D. C\u00e1c b\u1ea1n h\u1ecdc sinh c\u1ea7n l\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t. Tham kh\u1ea3o c\u00e1c c\u00e2u h\u1ecfi trong \u0111\u1ec1 thi th\u1eed sau \u0111\u00e2y:<\/p>\n

C\u00e2u 1. Cho h\u00e0m s\u1ed1 y = f(x) c\u00f3 \u0111\u1ed3 th\u1ecb nh\u01b0 h\u00ecnh v\u1ebd. H\u00e0m s\u1ed1 \u0111\u1ed3ng bi\u1ebfn tr\u00ean kho\u1ea3ng n\u00e0o d\u01b0\u1edbi \u0111\u00e2y?<\/p>\n

\"\"<\/p>\n

A. (0; +\u221e) \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 B. (-2; 2) \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0C. (-1; 1) \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 D. (-2; 1) <\/p>\n

C\u00e2u 2. Cho h\u00e0m s\u1ed1 y = f(x) c\u00f3 b\u1ea3ng bi\u1ebfn thi\u00ean nh\u01b0 sau:\\[
\n\\begin{array}{c|cccccc}
\nx & -\\infty & 0 & & 3 & & +\\infty \\\\
\n\\hline
\nf'(x) & & + & 0 & \u2013 & 0 & + \\\\
\n\\hline
\nf(x) & \\nearrow & 2 & \\searrow & -4 & \\nearrow & +\\infty
\n\\end{array}
\n\\]\n<\/p>\n

Gi\u00e1 tr\u1ecb c\u1ef1c ti\u1ec3u c\u1ee7a h\u00e0m s\u1ed1 \u0111\u00e3 cho b\u1eb1ng:<\/p>\n

A. -4 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 B. 3 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0C. 0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0D. 2 <\/p>\n

Ph\u1ea7n II \u2013 C\u00e2u h\u1ecfi tr\u1eafc nghi\u1ec7m \u0111\u00fang sai ( 4 c\u00e2u )<\/strong><\/span><\/p>\n

M\u1ed7i c\u00e2u h\u1ecfi tr\u1eafc nghi\u1ec7m s\u1ebd c\u00f3 4 c\u00e2u tr\u1ea3 l\u1eddi, c\u00e1c b\u1ea1n h\u1ecdc sinh c\u1ea7n \u0111\u1ecdc k\u0129 \u0111\u1ec1 v\u00e0 ch\u1ecdn m\u1ec7nh \u0111\u1ec1 \u0111\u00fang ho\u1eb7c sai cho m\u1ed7i c\u00e2u tr\u1ea3 l\u1eddi.<\/p>\n

C\u00e2u 1<\/strong>. Cho h\u00e0m s\u1ed1 y = x\u00b3 – 3x + 1.\n<\/p>\n

a) \u0110i\u1ec3m c\u1ef1c ti\u1ec3u c\u1ee7a h\u00e0m s\u1ed1 l\u00e0\u00a0 x = 1<\/p>\n

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

c) G\u1ecdi hai \u0111i\u1ec3m c\u1ef1c tr\u1ecb c\u1ee7a h\u00e0m s\u1ed1 l\u00e0 x\u2081, x\u2082 . Khi \u0111\u00f3 gi\u00e1 tr\u1ecb x\u2081 \u00b7 x\u2082 = -1.\n<\/p>\n

d) G\u1ecdi A, B l\u1ea7n l\u01b0\u1ee3t l\u00e0 \u0111i\u1ec3m c\u1ef1c \u0111\u1ea1i v\u00e0 \u0111i\u1ec3m c\u1ef1c ti\u1ec3u c\u1ee7a \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1. Khi \u0111\u00f3, di\u1ec7n t\u00edch tam gi\u00e1c ABC l\u00e0 12 v\u1edbi C(-1; 2).\n<\/p>\n

Ph\u1ea7n III \u2013 C\u00e2u h\u1ecfi tr\u1eafc nghi\u1ec7m tr\u1ea3 l\u1eddi ng\u1eafn ( 6 c\u00e2u )<\/span><\/strong><\/p>\n

C\u00e1c d\u1ea1ng c\u00e2u h\u1ecfi trong \u0111\u1ec1 c\u1ee7a ph\u1ea7n 3 xoay quanh c\u00e1c ki\u1ebfn th\u1ee9c v\u1ec1 h\u00ecnh h\u1ecdc kh\u00f4ng gian, ph\u01b0\u01a1ng tr\u00ecnh, t\u1ecda \u0111\u1ed9,\u2026 ph\u00e2n c\u1ea5p t\u1eeb c\u00e2u d\u1ec5 \u0111\u1ebfn n\u00e2ng cao. \u0110\u1ec3 chinh ph\u1ee5c tr\u1ecdn s\u1ed1 \u0111i\u1ec3m c\u1ee7a ph\u1ea7n 3 trong \u0111\u1ec1 thi thpt to\u00e1n<\/a> v\u00e0 h\u00e3y gi\u1ea3i th\u1eed c\u00e1c c\u00e2u h\u1ecfi sau \u0111\u00e2y nh\u00e9:<\/p>\n

C\u00e2u 1<\/strong>. Trong kh\u00f4ng gian v\u1edbi h\u1ec7 t\u1ecda \u0111\u1ed9 Oxyz , cho tam gi\u00e1c ABC c\u00f3 A(1; 2; -1), B(2; -1; 3), C(-4; 7; 5). G\u1ecdi D l\u00e0 ch\u00e2n \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c trong g\u00f3c B c\u1ee7a tam gi\u00e1c ABC . \u0110\u1ed9 d\u00e0i \u0111o\u1ea1n AD = (\u221aa)\/(3), gi\u00e1 tr\u1ecb c\u1ee7a a l\u00e0?<\/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 t\u1ed1i thi\u1ec3u 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? (L\u00e0m tr\u00f2n k\u1ebft qu\u1ea3 \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 Y\u00ean L\u1ea1c 2 V\u0129nh Ph\u00fac<\/strong><\/h2>\n

\u0110\u1ec3 \u0111\u1ea1t k\u1ebft qu\u1ea3 cao trong k\u00ec thi t\u1ed1t nghi\u1ec7p THPT s\u1eafp \u0111\u1ebfn, c\u00e1c b\u1ea1n h\u00e3y t\u1ea3i ngay \u0110\u1ec1 thi th\u1eed t\u1ed1t nghi\u1ec7p THPT m\u00f4n To\u00e1n 2024 \u2013 2025 tr\u01b0\u1eddng THPT Y\u00ean L\u1ea1c 2 V\u0129nh Ph\u00fac<\/strong> v\u00e0 b\u1eaft tay v\u00e0o luy\u1ec7n t\u1eadp m\u1ed7i ng\u00e0y nh\u00e9. B\u1ed9 \u0111\u1ec1 \u0111\u01b0\u1ee3c bi\u00ean so\u1ea1n chi ti\u1ebft, d\u1ec5 hi\u1ec3u, k\u00e8m \u0111\u00e1p \u00e1n v\u00e0 l\u1eddi gi\u1ea3i c\u1ee5 th\u1ec3 t\u1eebng ph\u1ea7n, gi\u00fap b\u1ea1n x\u00e1c \u0111\u1ecbnh l\u1ed7 h\u1ed5ng ki\u1ebfn th\u1ee9c v\u00e0 chu\u1ea9n b\u1ecb v\u1eefng v\u00e0ng cho k\u1ef3 thi.<\/p>\n