{"id":546,"date":"2026-05-05T04:51:28","date_gmt":"2026-05-05T04:51:28","guid":{"rendered":"https:\/\/79.buffdemo.com\/index.php\/2026\/05\/05\/de-thi-thu-tot-nghiep-thpt-mon-toan-2024-2025-truong-thpt-mai-thuc-loan-ha-tinh\/"},"modified":"2026-05-05T07:25:00","modified_gmt":"2026-05-05T07:25:00","slug":"de-thi-thu-tot-nghiep-thpt-mon-toan-2024-2025-truong-thpt-mai-thuc-loan-ha-tinh","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-mai-thuc-loan-ha-tinh\/","title":{"rendered":"\u0110\u1ec1 thi th\u1eed t\u1ed1t nghi\u1ec7p THPT m\u00f4n To\u00e1n 2024 \u2013 2025 tr\u01b0\u1eddng THPT Mai Th\u00fac Loan H\u00e0 T\u0129nh"},"content":{"rendered":"

\u0110\u1ec1 thi th\u1eed t\u1ed1t nghi\u1ec7p THPT m\u00f4n To\u00e1n 2024 \u2013 2025 tr\u01b0\u1eddng THPT Mai Th\u00fac Loan H\u00e0 T\u0129nh<\/strong> l\u00e0 t\u00e0i li\u1ec7u m\u00f4n To\u00e1n d\u00e0nh cho th\u1ea7y c\u00f4 v\u00e0 c\u00e1c b\u1ea1n h\u1ecdc sinh 12 c\u00f3 th\u00eam t\u00e0i nguy\u00ean trong qu\u00e1 tr\u00ecnh \u00f4n t\u1eadp v\u00e0 gi\u1ea3i \u0111\u1ec1 chu\u1ea9n b\u1ecb cho k\u00ec thi THPT m\u00f4n To\u00e1n ch\u00ednh th\u1ee9c. V\u1edbi c\u1ea5u tr\u00fac chu\u1ea9n b\u00e1m s\u00e1t theo quy \u0111\u1ecbnh c\u1ee7a b\u1ed9 GD&DT gi\u00fap c\u00e1c b\u1ea1n l\u00e0m quen d\u1ea7n v\u1edbi t\u01b0 duy ra \u0111\u1ec1 v\u00e0 c\u00e1ch gi\u1ea3i b\u00e0i thi. H\u00e3y c\u00f9ng T\u00e0i li\u1ec7u \u00d4n Thi<\/a> t\u00ecm hi\u1ec3u th\u00eam v\u1ec1 n\u1ed9i dung qua b\u00e0i vi\u1ebft d\u01b0\u1edbi \u0111\u00e2y.<\/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 Mai Th\u00fac Loan H\u00e0 T\u0129nh<\/strong><\/h2>\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 Mai Th\u00fac Loan H\u00e0 T\u0129nh<\/strong> c\u00f3 h\u00ecnh th\u1ee9c l\u00e0 c\u00e2u h\u1ecfi tr\u1eafc nghi\u1ec7m v\u1edbi c\u00e1c d\u1ea1ng kh\u00e1c nhau. Th\u00ed sinh c\u00f3 th\u1eddi gian 90 ph\u00fat \u0111\u1ec3 ho\u00e0n th\u00e0nh b\u00e0i thi v\u1edbi 22 c\u00e2u h\u1ecfi c\u00f3 \u0111\u1ed9 ph\u00e2n h\u00f3a r\u00f5 r\u00e0ng, nh\u1eb1m \u0111\u00e1nh gi\u00e1 \u0111\u00fang n\u0103ng l\u1ef1c c\u1ee7a c\u00e1c b\u1ea1n v\u00e0 chu\u1ea9n b\u1ecb qu\u00e1 tr\u00ecnh \u00f4n luy\u1ec7n t\u1ed1t nh\u1ea5t cho k\u00ec thi ch\u00ednh th\u1ee9c. D\u01b0\u1edbi \u0111\u00e2y l\u00e0 chi ti\u1ebft n\u1ed9i dung c\u1ee7a \u0111\u1ec1 thi, m\u1eddi c\u00e1c b\u1ea1n tham kh\u1ea3o qua:<\/p>\n

Ph\u1ea7n 1 \u2013 C\u00e2u tr\u1eafc nghi\u1ec7m nhi\u1ec1u ph\u01b0\u01a1ng \u00e1n l\u1ef1a ch\u1ecdn (12 c\u00e2u)<\/strong><\/p>\n

Chi ti\u1ebft c\u00e1c c\u00e2u h\u1ecfi nh\u01b0 sau:<\/p>\n

C\u00e2u 1<\/strong>: T\u1eadp nghi\u1ec7m c\u1ee7a b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh log\u2083(x – 2) – 1 > 0 l\u00e0:<\/p>\n

A. $(4; +\\infty)$\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 B. $(3; +\\infty)$\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0C. $(5; +\\infty)$\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0D. $(6; +\\infty)$<\/p>\n

C\u00e2u 2<\/strong>: Trong kh\u00f4ng gian v\u1edbi h\u1ec7 t\u1ecda \u0111\u1ed9 Oxyz, cho vect\u01a1 u\u20d7 = (3;0;1), v\u20d7 = (2;1;0). T\u00ednh t\u00edch v\u00f4 h\u01b0\u1edbng $\\vec{u} \\cdot \\vec{v}$:<\/p>\n

A. $\\vec{u} \\cdot \\vec{v} = 8$\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 B. $\\vec{u} \\cdot \\vec{v} = 6$<\/p>\n

C. $\\vec{u} \\cdot \\vec{v} = -6$\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 D. $\\vec{u} \\cdot \\vec{v} = 0$<\/p>\n

Ph\u1ea7n 2 \u2013 C\u00e2u tr\u1eafc nghi\u1ec7m \u0111\u00fang sai (4 c\u00e2u)
\n<\/strong>
\n\u1ede ph\u1ea7n n\u00e0y, th\u00ed sinh c\u1ea7n n\u1eafm v\u1eefng ki\u1ebfn th\u1ee9c \u0111\u1ec3 ph\u00e2n bi\u1ec7t \u0111\u01b0\u1ee3c \u0111\u00e1p \u00e1n n\u00e0o l\u00e0 gi\u00e1 tr\u1ecb \u0110\u00daNG\/SAI<\/strong>. Tham kh\u1ea3o ngay c\u00e1c c\u00e2u h\u1ecfi \u0111\u1ec3 hi\u1ec3u r\u00f5 h\u01a1n nh\u00e9:<\/p>\n

C\u00e2u 1<\/strong>: B\u1ea1n Tu\u1ea5n gieo m\u1ed9t con x\u00fac x\u1eafc \u0111\u1ec3 ghi l\u1ea1i s\u1ed1 ch\u1ea5m xu\u1ea5t hi\u1ec7n v\u00e0 b\u1ea1n Minh ch\u1ecdn ng\u1eabu nhi\u00ean m\u1ed9t l\u00e1 b\u00e0i t\u1eeb m\u1ed9t b\u1ed9 b\u00e0i t\u00fa l\u01a1 kh\u01a1 c\u00f3 52 l\u00e1 b\u00e0i.<\/p>\n

a) X\u00e1c su\u1ea5t \u0111\u1ec3 Tu\u1ea5n gieo \u0111\u01b0\u1ee3c con x\u00fac x\u1eafc c\u00f3 m\u1eb7t s\u00e1u ch\u1ea5m b\u1eb1ng 1\/6.\n<\/p>\n

b) X\u00e1c su\u1ea5t \u0111\u1ec3 Minh ch\u1ecdn \u0111\u01b0\u1ee3c m\u1ed9t l\u00e1 b\u00e0i \u00c1t b\u1eb1ng 1\/13.\n<\/p>\n

c) X\u00e1c su\u1ea5t \u0111\u1ec3 Tu\u1ea5n gieo \u0111\u01b0\u1ee3c con x\u00fac x\u1eafc c\u00f3 m\u1eb7t s\u00e1u ch\u1ea5m v\u00e0 Minh ch\u1ecdn \u0111\u01b0\u1ee3c m\u1ed9t l\u00e1 b\u00e0i \u00c1t b\u1eb1ng 1\/78.\n<\/p>\n

d) X\u00e1c su\u1ea5t \u0111\u1ec3 s\u1ed1 ch\u1ea5m tr\u00ean con x\u00fac x\u1eafc v\u00e0 s\u1ed1 c\u1ee7a l\u00e1 b\u00e0i l\u00e0 gi\u1ed1ng nhau b\u1eb1ng 1\/16.\n<\/p>\n

C\u00e2u 2<\/strong>: X\u00e9t m\u1ed9t ch\u1ea5t \u0111i\u1ec3m chuy\u1ec3n \u0111\u1ed9ng tr\u00ean m\u1ed9t tr\u1ee5c th\u1eb3ng \u0111\u1ee9ng, chi\u1ec1u d\u01b0\u01a1ng h\u01b0\u1edbng l\u00ean tr\u00ean. Gi\u1ea3 s\u1eed v\u1ecb tr\u00ed S(t) (m\u00e9t) c\u1ee7a ch\u1ea5t \u0111i\u1ec3m tr\u00ean tr\u1ee5c s\u1ed1 \u0111\u00e3 ch\u1ecdn t\u1ea1i th\u1eddi \u0111i\u1ec3m t (gi\u00e2y) \u0111\u01b0\u1ee3c cho b\u1edfi c\u00f4ng th\u1ee9c: S(t) = -t\u00b3 + 9t\u00b2 – 15t + 2, t \u2265 0.\n<\/p>\n

a) H\u00e0m v\u1eadn t\u1ed1c c\u1ee7a ch\u1ea5t \u0111i\u1ec3m l\u00e0 v(t) = -3t\u00b2 + 18t – 15.\n<\/p>\n

b) V\u1eadn t\u1ed1c c\u1ee7a ch\u1ea5t \u0111i\u1ec3m sau 2 gi\u00e2y l\u00e0 v(2) = 9 (m\/s).\n<\/p>\n

c) V\u1eadn t\u1ed1c c\u1ee7a ch\u1ea5t \u0111i\u1ec3m l\u1edbn nh\u1ea5t t\u1ea1i th\u1eddi \u0111i\u1ec3m t = 3 gi\u00e2y.\n<\/p>\n

d) Trong 6 gi\u00e2y \u0111\u1ea7u ti\u00ean, ch\u1ea5t \u0111i\u1ec3m \u0111i \u0111\u01b0\u1ee3c qu\u00e3ng \u0111\u01b0\u1eddng l\u00e0 20 m\u00e9t.\n<\/p>\n

Ph\u1ea7n 3 \u2013 C\u00e2u tr\u1eafc nghi\u1ec7m t\u1ef1 lu\u1eadn (6 c\u00e2u)
\n<\/strong>
\nV\u00ed d\u1ee5 c\u00e1c c\u00e2u h\u1ecfi minh h\u1ecda d\u01b0\u1edbi \u0111\u00e2y:<\/p>\n

C\u00e2u 1<\/strong>: Trong tr\u1eadn thi \u0111\u1ea5u b\u00f3ng b\u00e0n \u0111\u01a1n nam gi\u1eefa v\u1eadn \u0111\u1ed9ng vi\u00ean Nguy\u1ec5n \u0110\u1ee9c Tu\u1ea5n (ng\u01b0\u1eddi t\u1eebng \u0111o\u1ea1t huy ch\u01b0\u01a1ng v\u00e0ng \u0111\u01a1n nam m\u00f4n b\u00f3ng b\u00e0n t\u1ea1i Seagames 31) v\u1edbi m\u1ed9t v\u1eadn \u0111\u1ed9ng vi\u00ean n\u01b0\u1edbc ngo\u00e0i, tr\u1eadn \u0111\u1ea5u g\u1ed3m t\u1ed1i \u0111a 5 set (s\u00e9c), ng\u01b0\u1eddi n\u00e0o th\u1eafng tr\u01b0\u1edbc 3 set s\u1ebd gi\u00e0nh chi\u1ebfn th\u1eafng chung cu\u1ed9c. X\u00e1c su\u1ea5t \u0111\u1ec3 v\u1eadn \u0111\u1ed9ng vi\u00ean Tu\u1ea5n th\u1eafng m\u1ed7i set l\u00e0 0,6 . T\u00ednh x\u00e1c su\u1ea5t \u0111\u1ec3 v\u1eadn \u0111\u1ed9ng vi\u00ean Tu\u1ea5n gi\u00e0nh chi\u1ebfn th\u1eafng trong tr\u1eadn \u0111\u1ea5u (l\u00e0m tr\u00f2n k\u1ebft qu\u1ea3 \u0111\u1ebfn h\u00e0ng ph\u1ea7n tr\u0103m).<\/p>\n

C\u00e2u 2<\/strong>: Cho h\u00ecnh ch\u00f3p S.ABCD c\u00f3 \u0111\u00e1y ABCD l\u00e0 h\u00ecnh vu\u00f4ng, SA vu\u00f4ng g\u00f3c v\u1edbi \u0111\u00e1y v\u00e0 SA = 3 . Bi\u1ebft r\u1eb1ng kho\u1ea3ng c\u00e1ch gi\u1eefa hai \u0111\u01b0\u1eddng th\u1eb3ng AB v\u00e0 SD b\u1eb1ng 12\/5. Th\u1ec3 t\u00edch c\u1ee7a kh\u1ed1i ch\u00f3p \u0111\u00e3 cho b\u1eb1ng bao nhi\u00eau?<\/p>\n

Ngo\u00e0i ra, c\u00e1c b\u1ea1n h\u1ecdc sinh 12 c\u00f3 th\u1ec3 tham kh\u1ea3o qua c\u00e1c \u0111\u1ec1 thi th\u1eed thpt qu\u1ed1c gia m\u00f4n to\u00e1n<\/a> kh\u00e1c \u0111\u1ec3 n\u00e2ng cao t\u01b0 duy gi\u1ea3i \u0111\u1ec1 v\u00e0 l\u00e0m quen v\u1edbi nhi\u1ec1u \u0111\u1ec1 thi, chu\u1ea9n b\u1ecb t\u00e2m l\u00fd v\u1eefng tr\u01b0\u1edbc khi v\u00e0o k\u00ec thi ch\u00ednh th\u1ee9c nh\u00e9.<\/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 Mai Th\u00fac Loan H\u00e0 T\u0129nh<\/strong><\/h2>\n

\u0110\u1ec1 thi th\u1eed t\u1ed1t nghi\u1ec7p THPT m\u00f4n To\u00e1n 2024 \u2013 2025 tr\u01b0\u1eddng THPT Mai Th\u00fac Loan H\u00e0 T\u0129nh<\/strong> l\u00e0 t\u00e0i li\u1ec7u h\u1eefu \u00edch gi\u00fap c\u00e1c b\u1ea1n \u00f4n luy\u1ec7n hi\u1ec7u qu\u1ea3 h\u01a1n. H\u00e3y th\u00eam t\u00e0i nguy\u00ean n\u00e0y v\u00e0o b\u1ed9 \u0111\u1ec1 thi c\u1ee7a b\u1ea1n v\u00e0 \u00f4n t\u1eadp b\u1eb1ng c\u00e1ch t\u1ea3i \u0111\u1ec1 thi k\u00e8m \u0111\u00e1p \u00e1n chi ti\u1ebft qua \u0111\u01b0\u1eddng link d\u01b0\u1edbi \u0111\u00e2y. \u0110\u1eb7c bi\u1ec7t, t\u00e0i li\u1ec7u \u0111\u00e3 c\u00f3 \u0111\u00e1p \u00e1n v\u00e0 c\u00e1ch gi\u1ea3i c\u1ee5 th\u1ec3 n\u00ean b\u1ea1n s\u1ebd ti\u1ebft ki\u1ec7m th\u1eddi gian trong vi\u1ec7c tra c\u1ee9u \u0111\u00e1p \u00e1n v\u00e0 gi\u1ea3i \u0111\u1ec1.<\/p>\n