{"id":374,"date":"2026-05-05T04:42:24","date_gmt":"2026-05-05T04:42:24","guid":{"rendered":"https:\/\/79.buffdemo.com\/index.php\/2026\/05\/05\/de-thi-thu-tn-thpt-2025-mon-toan-truong-thpt-anh-son-3-nghe-an\/"},"modified":"2026-05-05T07:15:05","modified_gmt":"2026-05-05T07:15:05","slug":"de-thi-thu-tn-thpt-2025-mon-toan-truong-thpt-anh-son-3-nghe-an","status":"publish","type":"post","link":"https:\/\/79.buffdemo.com\/index.php\/2026\/05\/05\/de-thi-thu-tn-thpt-2025-mon-toan-truong-thpt-anh-son-3-nghe-an\/","title":{"rendered":"\u0110\u1ec1 thi th\u1eed TN THPT 2025 m\u00f4n To\u00e1n tr\u01b0\u1eddng THPT Anh S\u01a1n 3 Ngh\u1ec7 An c\u00f3 \u0111\u00e1p \u00e1n"},"content":{"rendered":"

\u0110\u1ec1 thi th\u1eed TN THPT 2025 m\u00f4n To\u00e1n tr\u01b0\u1eddng THPT Anh S\u01a1n 3 Ngh\u1ec7 An<\/strong> l\u00e0 t\u1eadp t\u00e0i li\u1ec7u r\u1ea5t c\u1ea7n thi\u1ebft d\u00e0nh cho c\u00e1c b\u1ea1n h\u1ecdc sinh 12 \u0111ang chu\u1ea9n b\u1ecb \u00f4n t\u1eadp thi THPT Qu\u1ed1c gia \u2013 k\u1ef3 thi v\u00f4 c\u00f9ng quan tr\u1ecdng. \u0110\u1ec1 thi th\u1eed n\u00e0y t\u1ed5ng h\u1ee3p \u0111\u1ea7y \u0111\u1ee7 c\u00e1c d\u1ea1ng b\u00e0i t\u1eadp c\u00f3 trong \u0111\u1ec1 thi ch\u00ednh th\u1ee9c, gi\u00fap c\u00e1c b\u1ea1n l\u00e0m quen v\u1edbi c\u1ea5u tr\u00fac \u0111\u1ec1 v\u00e0 r\u00e8n luy\u1ec7n k\u1ef9 n\u0103ng gi\u1ea3i b\u00e0i hi\u1ec7u qu\u1ea3 nh\u1ea5t. C\u00f9ng T\u00e0i li\u1ec7u \u00d4n Thi<\/a><\/strong> t\u00ecm hi\u1ec3u qua n\u1ed9i dung \u0111\u1ec1 thi trong b\u00e0i vi\u1ebft sau \u0111\u00e2y nh\u00e9.<\/p>\n

N\u1ed9i dung \u0110\u1ec1 thi th\u1eed TN THPT 2025 m\u00f4n To\u00e1n tr\u01b0\u1eddng THPT Anh S\u01a1n 3 Ngh\u1ec7 An<\/strong><\/h2>\n

\u0110\u1ec1 thi th\u1eed TN THPT 2025 m\u00f4n To\u00e1n tr\u01b0\u1eddng THPT Anh S\u01a1n 3 Ngh\u1ec7 An<\/strong> g\u1ed3m 2 \u0111\u1ec1 thi th\u1eed k\u00e8m \u0111\u00e1p \u00e1n chi ti\u1ebft v\u00e0 \u0111\u00e2y c\u0169ng ch\u00ednh l\u00e0 ch\u00eca kh\u00f3a gi\u00fap c\u00e1c b\u1ea1n \u00f4n t\u1eadp d\u1ec5 d\u00e0ng h\u01a1n v\u00e0 l\u00e0m quen \u0111\u01b0\u1ee3c ma tr\u1eadn \u0111\u1ec1 chu\u1ea9n b\u1ecb th\u1eadt t\u1ed1t cho k\u1ef3 thi THPT Qu\u1ed1c gia s\u1eafp t\u1edbi. T\u00e0i li\u1ec7u g\u1ed3m \u0111\u1ec1 thi v\u00e0 \u0111\u00e1p \u00e1n c\u1ee7a m\u00e3 \u0111\u1ec1 101 v\u00e0 102 v\u1edbi c\u1ea5u tr\u00fac \u0111\u1ec1 c\u00f3 3 ph\u1ea7n ch\u00ednh sau \u0111\u00e2y:<\/p>\n

PH\u1ea6N I \u2013 C\u00c2U H\u1eceI TR\u1eaeC NGHI\u1ec6M ( 12 c\u00e2u)<\/strong><\/p>\n

M\u1ed7i \u0111\u1ec1 s\u1ebd c\u00f3 12 c\u00e2u tr\u1eafc nghi\u1ec7m, c\u00e1c b\u1ea1n h\u1ecdc sinh s\u1ebd ch\u1ecdn \u0111\u00e1p \u00e1n ch\u00ednh x\u00e1c nh\u1ea5t trong 4 \u0111\u00e1p \u00e1n.<\/p>\n

C\u00e2u 5:<\/strong> T\u00ecm s\u1ed1 m\u1eb7t v\u00e0 s\u1ed1 c\u1ea1nh c\u1ee7a m\u1ed9t h\u00ecnh ch\u00f3p c\u00f3 \u0111\u00e1y l\u00e0 ng\u0169 gi\u00e1c.<\/p>\n

mathbfA. 6 m\u1eb7t, 10 c\u1ea1nh. mathbfB. 5 m\u1eb7t, 10 c\u1ea1nh. mathbfC. 6 m\u1eb7t, 5 c\u1ea1nh. mathbfD. 5 m\u1eb7t, 5 c\u1ea1nh.<\/p>\n

C\u00e2u 6:<\/strong> Cho h\u00ecnh h\u1ed9p ch\u1eef nh\u1eadt ABCD.A\u2019B\u2019C\u2019D\u2019 . Vector n\u00e0o sau \u0111\u00e2y b\u1eb1ng vector AB\u20d7 ?<\/p>\n

mathbfA. B\u2019A\u2019\u20d7. mathbfB. CD\u20d7. mathbfC. BA\u20d7. mathbfD. D\u2019C\u2019\u20d7.<\/p>\n

PH\u1ea6N II \u2013 C\u00c2U H\u1eceI \u0110\u00daNG SAI ( 4 c\u00e2u )<\/strong><\/p>\n

C\u00e2u 1<\/strong>: Cho h\u00e0m s\u1ed1 f(x) = log_1\/2(x\u00b2 – 5x + 4)\n<\/p>\n

a) D = (-\u221e;1] cup [4;+\u221e) .<\/p>\n

b) f'(x) = (5 – 2x)\/((x\u00b2 – 5x + 4) ln 2) .<\/p>\n

c) \u0110\u1ed3ng bi\u1ebfn tr\u00ean ( -\u221e; 5\/2 ) .<\/p>\n

d) B\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh f(x) > 0 c\u00f3 \u0111\u00fang 4 nghi\u1ec7m nguy\u00ean.<\/p>\n

PH\u1ea6N III \u2013 C\u00c2U H\u1eceI T\u1ef0 LU\u1eacN ( 6 c\u00e2u )<\/strong><\/p>\n

Ph\u1ea7n t\u1ef1 lu\u1eadn g\u1ed3m 6 c\u00e2u g\u1ed3m c\u00e1c c\u00e2u v\u1ec1 h\u00ecnh h\u1ecdc kh\u00f4ng gian, ph\u01b0\u01a1ng tr\u00ecnh, to\u00e1n n\u00e2ng cao,\u2026 M\u1ed9t s\u1ed1 c\u00e2u h\u1ecfi trong \u0111\u1ec1 thi th\u1eed nh\u01b0 sau:<\/p>\n

C\u00e2u 6.<\/strong> G\u1eafn h\u1ec7 tr\u1ee5c t\u1ecda \u0111\u1ed9 Oxyz sao cho O n\u1eb1m tr\u00ean m\u1eb7t n\u01b0\u1edbc, m\u1eb7t ph\u1eb3ng (Oxy) l\u00e0 m\u1eb7t n\u01b0\u1edbc, tr\u1ee5c Oz h\u01b0\u1edbng l\u00ean tr\u00ean (\u0111\u01a1n v\u1ecb \u0111o: m\u00e9t), m\u1ed9t con chim b\u00f3i c\u00e1 \u0111ang \u1edf v\u1ecb tr\u00ed (c\u00f3 ho\u00e0nh \u0111\u1ed9, tung \u0111\u1ed9 v\u00e0 cao \u0111\u1ed9 l\u00e0 c\u00e1c s\u1ed1 th\u1ef1c d\u01b0\u01a1ng) c\u00e1ch m\u1eb7t n\u01b0\u1edbc 2m , c\u00e1ch m\u1eb7t ph\u1eb3ng (Oxz) , (Oyz) l\u1ea7n l\u01b0\u1ee3t l\u00e0 3m v\u00e0 1m . Ph\u00f3ng th\u1eb3ng xu\u1ed1ng v\u1ecb tr\u00ed con c\u00e1, bi\u1ebft con c\u00e1 (c\u00f3 ho\u00e0nh \u0111\u1ed9 v\u00e0 tung \u0111\u1ed9 l\u00e0 c\u00e1c s\u1ed1 th\u1ef1c d\u01b0\u01a1ng) c\u00e1ch m\u1eb7t n\u01b0\u1edbc 50cm , c\u00e1ch m\u1eb7t ph\u1eb3ng (Oxz) , (Oyz) l\u1ea7n l\u01b0\u1ee3t l\u00e0 1m v\u00e0 1.5m .T\u1ecda \u0111\u1ed9 \u0111i\u1ec3m B l\u00fac chim b\u00f3i c\u00e1 v\u1eeba ti\u1ebfp x\u00fac v\u1edbi m\u1eb7t n\u01b0\u1edbc l\u00e0 (a;b;c) . T\u00ednh T = 3a + 5b + c .<\/p>\n

T\u1ea3i \u0110\u1ec1 thi th\u1eed TN THPT 2025 m\u00f4n To\u00e1n tr\u01b0\u1eddng THPT Anh S\u01a1n 3 Ngh\u1ec7 An<\/strong><\/h2>\n

H\u00e3y t\u1ea3i ngay b\u1ed9 \u0111\u1ec1 thi th\u1eed h\u1eefu \u00edch n\u00e0y v\u00e0 luy\u1ec7n t\u1eadp m\u1ed7i ng\u00e0y tr\u01b0\u1edbc khi k\u1ef3 thi b\u1eaft \u0111\u1ea7u nh\u00e9. Tin ch\u1eafc r\u1eb1ng, n\u1ebfu b\u1ea1n luy\u1ec7n t\u1eadp \u0110\u1ec1 thi th\u1eed TN THPT 2025 m\u00f4n To\u00e1n tr\u01b0\u1eddng THPT Anh S\u01a1n 3 Ngh\u1ec7 An<\/strong> th\u1eadt t\u1ed1t m\u1ed7i ng\u00e0y th\u00ec s\u1ebd \u0111\u1ea1t \u0111\u01b0\u1ee3c k\u1ebft qu\u1ea3 cao trong k\u1ef3 thi THPT Qu\u1ed1c gia s\u1eafp \u0111\u1ebfn.
\n