{"id":526,"date":"2026-05-05T04:50:26","date_gmt":"2026-05-05T04:50:26","guid":{"rendered":"https:\/\/79.buffdemo.com\/index.php\/2026\/05\/05\/de-thi-thu-tot-nghiep-thpt-mon-toan-2024-2025-truong-thpt-tam-dao-vinh-phuc\/"},"modified":"2026-05-05T07:23:47","modified_gmt":"2026-05-05T07:23:47","slug":"de-thi-thu-tot-nghiep-thpt-mon-toan-2024-2025-truong-thpt-tam-dao-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-tam-dao-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 Tam \u0110\u1ea3o 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 Tam \u0110\u1ea3o V\u0129nh Ph\u00fac<\/strong> \u2013 t\u1ed5 To\u00e1n Tin g\u1ed3m 4 trang v\u1edbi th\u1eddi gian l\u00e0m b\u00e0i 90 ph\u00fat. \u0110\u1ec1 thi c\u00f3 c\u1ea5u tr\u00fac \u0111a d\u1ea1ng v\u1edbi nhi\u1ec1u d\u1ea1ng b\u00e0i t\u1eadp t\u1eeb c\u01a1 b\u1ea3n \u0111\u1ebfn n\u00e2ng cao, gi\u00fap h\u1ecdc sinh r\u00e8n luy\u1ec7n k\u1ef9 n\u0103ng gi\u1ea3i to\u00e1n v\u00e0 chu\u1ea9n b\u1ecb t\u1ed1t cho k\u1ef3 thi t\u1ed1t nghi\u1ec7p THPT s\u1eafp t\u1edbi. Tham kh\u1ea3o nh\u1eefng ch\u1ee7 \u0111\u1ec1 to\u00e1n h\u1ecdc c\u00f3 trong \u0111\u1ec1 thi c\u00f9ng Tailieuonthi<\/strong><\/a>.<\/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 Tam \u0110\u1ea3o 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 Tam \u0110\u1ea3o V\u0129nh Ph\u00fac<\/strong> c\u00f3 nhi\u1ec1u d\u1ea1ng to\u00e1n ph\u1ed5 bi\u1ebfn t\u1eeb \u0111\u1ea1i s\u1ed1, gi\u1ea3i t\u00edch \u0111\u1ebfn h\u00ecnh h\u1ecdc kh\u00f4ng gian v\u00e0 x\u00e1c su\u1ea5t, gi\u00fap h\u1ecdc sinh c\u00f3 c\u00e1i nh\u00ecn t\u1ed5ng quan v\u1ec1 c\u00e1c ch\u1ee7 \u0111\u1ec1 s\u1ebd xu\u1ea5t hi\u1ec7n trong k\u1ef3 thi t\u1ed1t nghi\u1ec7p. H\u00e3y c\u00f9ng \u0111i\u1ec3m qua m\u1ed9t s\u1ed1 c\u00e2u h\u1ecfi \u0111\u1eb7c tr\u01b0ng trong t\u1eebng ph\u1ea7n c\u1ee7a \u0111\u1ec1 thi t\u1ed1t nghi\u1ec7p thpt m\u00f4n to\u00e1n<\/a> n\u00e0y.<\/p>\n

PH\u1ea6N I. C\u00e2u tr\u1eafc nghi\u1ec7m nhi\u1ec1u ph\u01b0\u01a1ng \u00e1n l\u1ef1a ch\u1ecdn (3 \u0111i\u1ec3m)<\/strong><\/span><\/p>\n

C\u00e2u 1:<\/strong> Cho h\u00e0m s\u1ed1 $y = f(x)$ li\u00ean t\u1ee5c tr\u00ean $\\mathbb{R}$ v\u00e0 c\u00f3 b\u1ea3ng bi\u1ebfn thi\u00ean nh\u01b0 h\u00ecnh v\u1ebd b\u00ean d\u01b0\u1edbi.<\/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. $(-1;0)$\u00a0 \u00a0 \u00a0 \u00a0 B. $(0;+\\infty)$ \u00a0 \u00a0 \u00a0 C. $(-\\infty;0)$ \u00a0 \u00a0 \u00a0 D. $(0;1)$<\/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

\\begin{array}{|c|c|c|c|c|c|c|c|}
\n\\hline
\nx & -\\infty & & & 2 & & & +\\infty \\\\
\n\\hline
\nf(x) & -5 & \\searrow & -\\infty & & 1 & \\searrow & -5 \\\\
\n\\hline
\n\\end{array}<\/p>\n

T\u1ed5ng s\u1ed1 ti\u1ec7m c\u1eadn ngang v\u00e0 ti\u1ec7m c\u1eadn \u0111\u1ee9ng c\u1ee7a \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1 \u0111\u00e3 cho l\u00e0<\/p>\n

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

PH\u1ea6N II. C\u00e2u tr\u1eafc nghi\u1ec7m \u0111\u00fang sai (4 \u0111i\u1ec3m)<\/strong><\/span><\/p>\n

C\u00e2u 1:<\/strong> Cho h\u00e0m s\u1ed1 $y = \\frac{x^2 \u2013 2x + 2024}{x \u2013 1}$ c\u00f3 \u0111\u1ed3 th\u1ecb $(C)$.<\/p>\n

a) $(C)$ c\u00f3 \u0111\u01b0\u1eddng ti\u1ec7m c\u1eadn \u0111\u1ee9ng l\u00e0 $x = 1$.<\/p>\n

b) $(C)$ c\u00f3 \u0111\u01b0\u1eddng ti\u1ec7m c\u1eadn xi\u00ean l\u00e0 $y = -x + 1$.<\/p>\n

c) $(C)$ H\u00e0m s\u1ed1 c\u00f3 gi\u00e1 tr\u1ecb l\u1edbn nh\u1ea5t v\u00e0 gi\u00e1 tr\u1ecb nh\u1ecf nh\u1ea5t.<\/p>\n

d) Tr\u00ean $(C)$ c\u00f3 \u0111\u00fang 4 \u0111i\u1ec3m c\u00f3 t\u1ecda \u0111\u1ed9 nguy\u00ean.<\/p>\n

C\u00e2u 2:<\/strong> Cho $I = \\int 2x(3x-2)^6 dx = A(3x-2)^8 + B(3x-2)^7 + C$ v\u1edbi $A, B, C \\in \\mathbb{Q}$.<\/p>\n

a) \u0110\u1eb7t $t = 3x -2$ th\u00ec $dt = 3 dx$<\/p>\n

b) Khi \u0111\u00f3 $I = \\int 2(t+2)t^6 dt$<\/p>\n

c) $I = \\frac{1}{4}(3x-2)^8 + \\frac{4}{7}(3x-2)^7 + C$<\/p>\n

d) Gi\u00e1 tr\u1ecb c\u1ee7a bi\u1ec3u th\u1ee9c $12A + 7B$ l\u00e0 $\\frac{7}{9}$<\/p>\n

PH\u1ea6N III. C\u00e2u tr\u1eafc nghi\u1ec7m tr\u1ea3 l\u1eddi ng\u1eafn (3 \u0111i\u1ec3m)<\/strong><\/span><\/p>\n

C\u00e2u 1:<\/strong> M\u1ed9t v\u1eadt chuy\u1ec3n \u0111\u1ed9ng theo quy lu\u1eadt $s = \\frac{1}{2}t^3 + 9t^2$ v\u1edbi $t$ (gi\u00e2y) l\u00e0 kho\u1ea3ng th\u1eddi gian t\u00ednh t\u1eeb l\u00fac b\u1eaft \u0111\u1ea7u chuy\u1ec3n \u0111\u1ed9ng v\u00e0 $s$ (m\u00e9t) l\u00e0 qu\u00e3ng \u0111\u01b0\u1eddng v\u1eadt \u0111i \u0111\u01b0\u1ee3c trong kho\u1ea3ng th\u1eddi gian \u0111\u00f3. H\u1ecfi trong kho\u1ea3ng th\u1eddi gian 10 gi\u00e2y, k\u1ec3 t\u1eeb l\u00fac b\u1eaft \u0111\u1ea7u chuy\u1ec3n \u0111\u1ed9ng, v\u1eadn t\u1ed1c l\u1edbn nh\u1ea5t c\u1ee7a v\u1eadt \u0111\u1ea1t \u0111\u01b0\u1ee3c b\u1eb1ng bao nhi\u00eau?<\/p>\n

A. $216 (m\/s)$\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 B. $30 (m\/s)$ \u00a0 \u00a0 \u00a0 \u00a0 C. $400 (m\/s)$ \u00a0 \u00a0 \u00a0 \u00a0 D. $54 (m\/s)$<\/p>\n

C\u00e2u 2:<\/strong> Anh Phong c\u00f3 m\u1ed9t c\u00e1i ao v\u1edbi di\u1ec7n t\u00edch $50m^2$ \u0111\u1ec3 nu\u00f4i c\u00e1 \u0111i\u00eau h\u1ed3ng. V\u1ee5 v\u1eeba qua, anh nu\u00f4i v\u1edbi m\u1eadt \u0111\u1ed9 $20con\/m^2$ v\u00e0 thu \u0111\u01b0\u1ee3c 1,5 t\u1ea5n c\u00e1 th\u00e0nh ph\u1ea9m. Theo kinh nghi\u1ec7m nu\u00f4i c\u00e1 c\u1ee7a m\u00ecnh anh th\u1ea5y c\u1ee9 th\u1ea3 gi\u1ea3m \u0111i $8con\/m^2$ th\u00ec m\u1ed7i con c\u00e1 th\u00e0nh ph\u1ea9m thu \u0111\u01b0\u1ee3c t\u0103ng th\u00eam $0,5kg$. \u0110\u1ec3 t\u1ed5ng n\u0103ng su\u1ea5t cao nh\u1ea5t th\u00ec v\u1ee5 t\u1edbi anh n\u00ean mua bao nhi\u00eau c\u00e1 gi\u1ed1ng \u0111\u1ec3 th\u1ea3? (gi\u1ea3 s\u1eed kh\u00f4ng c\u00f3 hao h\u1ee5t trong qu\u00e1 tr\u00ecnh nu\u00f4i)<\/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 Tam \u0110\u1ea3o V\u0129nh Ph\u00fac<\/span><\/strong><\/h2>\n

C\u00e1c b\u1ea1n c\u00f3 th\u1ec3 t\u1ea3i b\u1ea3n \u0111\u1ea7y \u0111\u1ee7 c\u1ee7a \u0110\u1ec1 thi th\u1eed t\u1ed1t nghi\u1ec7p THPT m\u00f4n To\u00e1n 2024 \u2013 2025 tr\u01b0\u1eddng THPT Tam \u0110\u1ea3o V\u0129nh Ph\u00fac <\/strong>k\u00e8m \u0111\u00e1p \u00e1n chi ti\u1ebft v\u1ec1 \u0111\u1ec3 th\u1ef1c h\u00e0nh, t\u1eeb \u0111\u00f3 n\u00e2ng cao kh\u1ea3 n\u0103ng gi\u1ea3i quy\u1ebft v\u1ea5n \u0111\u1ec1 v\u00e0 t\u1ed1c \u0111\u1ed9 l\u00e0m b\u00e0i tr\u01b0\u1edbc k\u1ef3 thi ch\u00ednh th\u1ee9c.<\/p>\n