{"id":524,"date":"2026-05-05T04:50:19","date_gmt":"2026-05-05T04:50:19","guid":{"rendered":"https:\/\/79.buffdemo.com\/index.php\/2026\/05\/05\/de-thi-thu-tot-nghiep-thpt-mon-toan-2024-2025-truong-thpt-sang-son-vinh-phuc\/"},"modified":"2026-05-05T07:23:41","modified_gmt":"2026-05-05T07:23:41","slug":"de-thi-thu-tot-nghiep-thpt-mon-toan-2024-2025-truong-thpt-sang-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-sang-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 S\u00e1ng 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 S\u00e1ng S\u01a1n V\u0129nh Ph\u00fac<\/strong> g\u1ed3m 5 trang v\u1edbi th\u1eddi gian l\u00e0m b\u00e0i 90 ph\u00fat. \u0110\u1ec1 b\u00e1m s\u00e1t c\u1ea5u tr\u00fac \u0111\u1ec1 thi ch\u00ednh th\u1ee9c v\u1edbi c\u00e1c d\u1ea1ng b\u00e0i quen thu\u1ed9c, gi\u00fap h\u1ecdc sinh \u00f4n t\u1eadp v\u00e0 l\u00e0m quen v\u1edbi format \u0111\u1ec1 thi t\u1ed1t nghi\u1ec7p. H\u00f4m nay, c\u00f9ng Tailieuonthi<\/strong><\/a> t\u00ecm hi\u1ec3u v\u1ec1 \u0111\u1ec1 thi n\u00e0y.<\/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 S\u00e1ng 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 S\u00e1ng S\u01a1n V\u0129nh Ph\u00fac<\/strong> <\/span>bao g\u1ed3m nhi\u1ec1u d\u1ea1ng b\u00e0i \u0111a d\u1ea1ng, t\u1eeb \u0111\u01a1n gi\u1ea3n \u0111\u1ebfn n\u00e2ng cao, gi\u00fap h\u1ecdc sinh \u00f4n t\u1eadp to\u00e0n di\u1ec7n c\u00e1c ki\u1ebfn th\u1ee9c to\u00e1n h\u1ecdc THPT. D\u01b0\u1edbi \u0111\u00e2y l\u00e0 m\u1ed9t s\u1ed1 c\u00e2u h\u1ecfi ti\u00eau bi\u1ec3u theo t\u1eebng d\u1ea1ng trong \u0111\u1ec1 thi.<\/p>\n

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

G\u1ed3m 12 c\u00e2u (t\u1eeb c\u00e2u 1 \u0111\u1ebfn c\u00e2u 12), m\u1ed7i c\u00e2u \u0111\u00fang \u0111\u01b0\u1ee3c 0,25 \u0111i\u1ec3m.<\/em><\/p>\n

C\u00e2u 1:<\/strong> Cho h\u00e0m s\u1ed1 $y = f(x) = ax^3 + bx^2 + cx + d$ c\u00f3 \u0111\u1ed3 th\u1ecb nh\u01b0 h\u00ecnh v\u1ebd.<\/p>\n

\"\"<\/p>\n

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

A. $(2;+\\infty)$\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 B. $(-\\infty;-1)$ \u00a0 \u00a0 \u00a0 \u00a0 C. $(0;2)$ \u00a0 \u00a0 \u00a0 \u00a0 D. $(-4;2)$<\/p>\n

C\u00e2u 2:<\/strong> \u0110\u1ed3 th\u1ecb h\u00e0m s\u1ed1 trong h\u00ecnh v\u1ebd l\u00e0 \u0111\u1ed3 th\u1ecb c\u1ee7a h\u00e0m s\u1ed1 n\u00e0o trong c\u00e1c h\u00e0m s\u1ed1 d\u01b0\u1edbi \u0111\u00e2y?<\/p>\n

\"\"<\/p>\n

A. $y = \\frac{2x-1}{x+1}$ \u00a0 \u00a0 \u00a0 \u00a0 B. $y = \\frac{2x-2}{x-1}$ \u00a0 \u00a0 \u00a0 \u00a0 C. $y = \\frac{2x-1}{x-1}$ \u00a0 \u00a0 \u00a0 \u00a0 D. $y = \\frac{2x+3}{x+1}$<\/p>\n

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

G\u1ed3m 4 c\u00e2u (t\u1eeb c\u00e2u 1 \u0111\u1ebfn c\u00e2u 4). V\u1edbi m\u1ed7i c\u00e2u, th\u00ed sinh ch\u1ecdn \u0111\u00fang ho\u1eb7c sai cho t\u1eebng m\u1ec7nh \u0111\u1ec1 a), b), c), d).<\/em><\/p>\n

C\u00e2u 1:<\/strong> Cho h\u00e0m s\u1ed1 $y = f(x) = x^3 \u2013 3x^2 \u2013 9x + 35$. X\u00e9t t\u00ednh \u0111\u00fang sai c\u1ee7a c\u00e1c m\u1ec7nh \u0111\u1ec1 sau:<\/p>\n

a) H\u00e0m s\u1ed1 c\u00f3 t\u1eadp x\u00e1c \u0111\u1ecbnh l\u00e0 $\\mathbb{R}$.<\/p>\n

b) H\u00e0m s\u1ed1 \u0111\u00e3 cho \u0111\u1ed3ng bi\u1ebfn tr\u00ean $(-\\infty;3)$.<\/p>\n

c) Gi\u00e1 tr\u1ecb c\u1ef1c ti\u1ec3u c\u1ee7a h\u00e0m s\u1ed1 b\u1eb1ng 8.<\/p>\n

d) $\\min_{[-4;4]} f(x) = 8$ \u0111\u1ea1t \u0111\u01b0\u1ee3c khi $x = 3$.<\/p>\n

C\u00e2u 2:<\/strong> Ch\u1ea5t \u0111i\u1ec3m chuy\u1ec3n \u0111\u1ed9ng theo quy lu\u1eadt v\u1eadn t\u1ed1c $v(t)(m\/s)$ c\u00f3 d\u1ea1ng \u0111\u01b0\u1eddng Parabol $(P)$ khi $0 \\leq t \\leq 5(s)$ v\u00e0 $v(t)$ c\u00f3 d\u1ea1ng \u0111\u01b0\u1eddng th\u1eb3ng khi $5 \\leq t \\leq 10(s)$ (H\u00ecnh v\u1ebd). Cho \u0111inh Parabol l\u00e0 $I(2,3)$. H\u1ecfi qu\u00e3ng \u0111\u01b0\u1eddng \u0111i \u0111\u01b0\u1ee3c ch\u1ea5t \u0111i\u1ec3m trong th\u1eddi gian $0 \\leq t \\leq 10(s)$ l\u00e0 bao nhi\u00eau m\u00e9t?<\/p>\n

\"\"<\/p>\n

a) Ph\u01b0\u01a1ng tr\u00ecnh Parabol $(P)$ c\u00f3 d\u1ea1ng: $y = 2x^2 \u2013 8x + 11$<\/p>\n

b) Qu\u00e3ng \u0111\u01b0\u1eddng v\u1eadt \u0111i chuy\u1ec3n trong kho\u1ea3ng th\u1eddi gian t\u1eeb $0 \\leq t \\leq 5(s)$ l\u00e0 $\\frac{112}{3}(m)$<\/p>\n

c) Qu\u00e3ng \u0111\u01b0\u1eddng v\u1eadt \u0111i chuy\u1ec3n trong kho\u1ea3ng th\u1eddi gian t\u1eeb $5 \\leq t \\leq 10(s)$ l\u00e0 $\\frac{105}{2}(m)$<\/p>\n

d) Qu\u00e3ng \u0111\u01b0\u1eddng \u0111i \u0111\u01b0\u1ee3c ch\u1ea5t \u0111i\u1ec3m trong th\u1eddi gian $0 \\leq t \\leq 10(s)$ l\u00e0 $\\frac{545}{6}(m)$<\/p>\n

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

G\u1ed3m 6 c\u00e2u (t\u1eeb c\u00e2u 1 \u0111\u1ebfn c\u00e2u 6). M\u1ed7i c\u00e2u tr\u1ea3 l\u1eddi \u0111\u00fang \u0111\u01b0\u1ee3c 0,5 \u0111i\u1ec3m.<\/em><\/p>\n

C\u00e2u 1:<\/strong> Cho m\u1ed9t t\u1ea5m nh\u00f4m h\u00ecnh vu\u00f4ng c\u00f3 c\u1ea1nh $1(m)$ nh\u01b0 h\u00ecnh v\u1ebd. Ng\u01b0\u1eddi ta c\u1eaft ph\u1ea7n t\u00f4 \u0111\u1eadm c\u1ee7a t\u1ea5m nh\u00f4m r\u1ed3i g\u1ea5p th\u00e0nh h\u00ecnh ch\u00f3p t\u1ee9 gi\u00e1c \u0111\u1ec1u c\u00f3 c\u1ea1nh \u0111\u00e1y b\u1eb1ng $\\frac{\\sqrt{2}}{3}(m)$. Kho\u1ea3ng c\u00e1ch t\u1eeb \u0111\u1ec9nh $S$ t\u1edbi \u0111\u00e1y $(ABCD)$ b\u1eb1ng $\\frac{a\\sqrt{6}}{b}(m)$, v\u1edbi $a, b$ l\u00e0 c\u00e1c s\u1ed1 nguy\u00ean d\u01b0\u01a1ng. T\u00ednh gi\u00e1 tr\u1ecb c\u1ee7a $b^2 \u2013 a^2$?<\/p>\n

\"\"<\/p>\n

C\u00e2u 2:<\/strong> Trung b\u00ecnh sau m\u1ed7i n\u0103m s\u1eed d\u1ee5ng, gi\u00e1 tr\u1ecb c\u00f2n l\u1ea1i c\u1ee7a m\u1ed9t chi\u1ebfc \u00f4 t\u00f4 gi\u1ea3m \u0111i 6% so v\u1edbi n\u0103m tr\u01b0\u1edbc \u0111\u00f3. Gi\u00e1 s\u1eed m\u1ed9t chi\u1ebfc \u00f4 t\u00f4 l\u00fac m\u1edbi mua l\u00e0 800 tri\u1ec7u \u0111\u1ed3ng. H\u1ecfi sau \u00edt nh\u1ea5t bao nhi\u00eau n\u0103m s\u1eed d\u1ee5ng th\u00ec gi\u00e1 tr\u1ecb c\u00f2n l\u1ea1i c\u1ee7a chi\u1ebfc \u00f4 t\u00f4 \u0111\u00f3 nh\u1ecf h\u01a1n 600 tri\u1ec7u \u0111\u1ed3ng (l\u00e0m tr\u00f2n k\u1ebft qu\u1ea3 \u0111\u1ebfn h\u00e0ng \u0111\u01a1n v\u1ecb)?<\/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 S\u00e1ng 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 S\u00e1ng S\u01a1n V\u0129nh Ph\u00fac <\/strong>c\u00f3 c\u1ea5u tr\u00fac b\u00e1m s\u00e1t \u0111\u1ec1 thi ch\u00ednh th\u1ee9c, gi\u00fap h\u1ecdc sinh r\u00e8n luy\u1ec7n k\u1ef9 n\u0103ng l\u00e0m b\u00e0i v\u00e0 c\u1ee7ng c\u1ed1 ki\u1ebfn th\u1ee9c tr\u01b0\u1edbc k\u1ef3 thi t\u1ed1t nghi\u1ec7p. C\u00e1c em h\u1ecdc sinh c\u00f3 th\u1ec3 t\u1ea3i file \u0111\u1ec1 thi th\u1eed to\u00e1n 12<\/a> v\u00e0 \u0111\u00e1p \u00e1n \u0111\u1ea7y \u0111\u1ee7 \u0111\u1ec3 t\u1ef1 luy\u1ec7n t\u1eadp, n\u00e2ng cao n\u0103ng l\u1ef1c gi\u1ea3i to\u00e1n v\u00e0 chu\u1ea9n b\u1ecb t\u1ed1t nh\u1ea5t cho k\u1ef3 thi s\u1eafp t\u1edbi.<\/p>\n