{"id":396,"date":"2026-05-05T04:43:35","date_gmt":"2026-05-05T04:43:35","guid":{"rendered":"https:\/\/79.buffdemo.com\/index.php\/2026\/05\/05\/de-kiem-tra-danh-gia-giua-ki-2-toan-12-nam-hoc-2024-2025-truong-thpt-tay-thanh-tp-hcm\/"},"modified":"2026-05-05T07:16:18","modified_gmt":"2026-05-05T07:16:18","slug":"de-kiem-tra-danh-gia-giua-ki-2-toan-12-nam-hoc-2024-2025-truong-thpt-tay-thanh-tp-hcm","status":"publish","type":"post","link":"https:\/\/79.buffdemo.com\/index.php\/2026\/05\/05\/de-kiem-tra-danh-gia-giua-ki-2-toan-12-nam-hoc-2024-2025-truong-thpt-tay-thanh-tp-hcm\/","title":{"rendered":"\u0110\u1ec1 ki\u1ec3m tra \u0111\u00e1nh gi\u00e1 gi\u1eefa k\u00ec 2 To\u00e1n 12 n\u0103m h\u1ecdc 2024 \u2013 2025 Tr\u01b0\u1eddng THPT T\u00e2y Th\u1ea1nh TP HCM"},"content":{"rendered":"

\u0110\u1ec1 ki\u1ec3m tra \u0111\u00e1nh gi\u00e1 gi\u1eefa k\u00ec 2 To\u00e1n 12 n\u0103m h\u1ecdc 2024 \u2013 2025 Tr\u01b0\u1eddng THPT T\u00e2y Th\u1ea1nh TP HCM<\/strong> l\u00e0 t\u1eadp t\u00e0i li\u1ec7u v\u00f4 c\u00f9ng h\u1eefu \u00edch d\u00e0nh cho c\u00e1c h\u1ecdc sinh 12 \u00f4n t\u1eadp chu\u1ea9n b\u1ecb ki\u1ec3m tra \u0111\u00e1nh gi\u00e1 m\u00f4n To\u00e1n. C\u00f9ng T\u00e0i li\u1ec7u \u00d4n Thi<\/a> <\/strong>t\u00ecm hi\u1ec3u b\u1ed9 \u0111\u1ec1 n\u00e0y c\u00f3 \u0111i\u1ec3m g\u00ec \u0111\u1eb7c bi\u1ec7t nh\u00e9!<\/p>\n

N\u1ed9i dung \u0110\u1ec1 ki\u1ec3m tra \u0111\u00e1nh gi\u00e1 gi\u1eefa k\u00ec 2 To\u00e1n12 n\u0103m h\u1ecdc 2024 \u2013 2025 Tr\u01b0\u1eddng THPT T\u00e2y Th\u1ea1nh TP HCM<\/strong><\/h2>\n

\u0110\u1ec1 ki\u1ec3m tra \u0111\u00e1nh gi\u00e1 gi\u1eefa k\u00ec 2 To\u00e1n 12 n\u0103m h\u1ecdc 2024 \u2013 2025 Tr\u01b0\u1eddng THPT T\u00e2y Th\u1ea1nh TP HCM<\/strong> \u0111\u01b0\u1ee3c thi\u1ebft k\u1ebf v\u1edbi c\u1ea5u tr\u00fac 4 ph\u1ea7n, bao g\u1ed3m c\u00e1c d\u1ea1ng c\u00e2u h\u1ecfi b\u00e1m s\u00e1t l\u00fd thuy\u1ebft v\u1ec1 nguy\u00ean h\u00e0m, t\u00edch ph\u00e2n v\u00e0 h\u00ecnh h\u1ecdc kh\u00f4ng gian Oxyz. \u0110\u00e2y l\u00e0 t\u00e0i li\u1ec7u c\u1ea7n thi\u1ebft h\u1ed7 tr\u1ee3 h\u1ecdc sinh l\u00e0m quen v\u1edbi \u0111\u1ec1 thi \u0111\u00e1nh gi\u00e1, th\u1eed s\u1ee9c v\u00e0 r\u00fat kinh nghi\u1ec7m \u0111\u1ec3 chu\u1ea9n b\u1ecb t\u1ed1t cho k\u1ef3 thi ch\u00ednh th\u1ee9c.<\/p>\n

Ph\u1ea7n 1 \u2013 C\u00e2u h\u1ecfi tr\u1eafc nghi\u1ec7m nhi\u1ec1u \u0111\u00e1p \u00e1n (6 c\u00e2u)<\/b><\/span><\/p>\n

Tr\u00edch d\u1eabn m\u1ed9t s\u1ed1 c\u00e2u h\u1ecfi trong \u0111\u1ec1 nh\u01b0 sau, b\u1ea1n c\u1ea7n ph\u1ea3i \u0111\u1ecdc k\u0129 \u0111\u1ec1 v\u00e0 ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t.<\/p>\n

C\u00e2u 1:<\/strong> Trong kh\u00f4ng gian t\u1ecda \u0111\u1ed9 Oxyz , cho hai \u0111i\u1ec3m M(3;1;-2) v\u00e0 N(1;-3;4) . M\u1eb7t ph\u1eb3ng trung tr\u1ef1c c\u1ee7a \u0111o\u1ea1n th\u1eb3ng MN l\u00e0
\n A. x + 2y + 3z – 3 = 0. \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 B. x + 2y – 3x + 3 = 0.
\n C. x + 2y – 3z – 3 = 0. \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 D. x + 2y + 3z + 3 = 0.\n<\/p>\n

C\u00e2u 2<\/strong>: Trong kh\u00f4ng gian t\u1ecda \u0111\u1ed9 Oxyz , ph\u01b0\u01a1ng tr\u00ecnh m\u1eb7t c\u1ea7u t\u00e2m I(2;-3;-4) , b\u00e1n k\u00ednh b\u1eb1ng 4 c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh l\u00e0<\/p>\n

A. x\u00b2 + y\u00b2 + z\u00b2 + 4x – 6y – 8z + 13 = 0.] [B. (x + 2)\u00b2 + (y – 3)\u00b2 + (z – 4)\u00b2 = 4.<\/p>\n

C. (x – 2)\u00b2 + (y + 3)\u00b2 + (z + 4)\u00b2 = 4.] [D. x\u00b2 + y\u00b2 + z\u00b2 – 4x + 6y + 8z + 13 = 0.<\/p>\n

Ph\u1ea7n 2 \u2013 C\u00e2u h\u1ecfi \u0111\u00fang sai (2 c\u00e2u)<\/b><\/p>\n

C\u00f3 t\u1ea5t c\u1ea3 2 c\u00e2u h\u1ecfi \u1edf ph\u1ea7n n\u00e0y, th\u00ed sinh c\u1ea7n ch\u1ecdn m\u1ec7nh \u0111\u1ec1 \u0111\u00fang ho\u1eb7c sai cho m\u1ed7i c\u00e2u tr\u1ea3 l\u1eddi. C\u00f9ng xem qua c\u00e1c c\u00e2u h\u1ecfi sau \u0111\u00e2y<\/p>\n

C\u00e2u 1:<\/strong> Cho c\u00e1c h\u00e0m s\u1ed1 f(x) v\u00e0 F(x) li\u00ean t\u1ee5c tr\u00ean mathbbR th\u1ecfa F'(x) = f(x), forall x in mathbbR .<\/p>\n

a) N\u1ebfu F(0) = 2, F(1) = 8 . Khi \u0111\u00f3: int\u2080\u00b9 f(x)dx = -6.\n<\/p>\n

b) int_a^b f(x)dx = F(b) – F(a). <\/p>\n

c) Cho f(x) = 2^x v\u00e0 int\u2080^a f(x)dx = (a)\/(ln 2) . Khi \u0111\u00f3 a = 2 .<\/p>\n

d) int f(x)dx = F(x). <\/p>\n

C\u00e2u 2:<\/strong> Trong kh\u00f4ng gian h\u1ec7 t\u1ecda \u0111\u1ed9 Oxyz , cho m\u1eb7t ph\u1eb3ng (P): 2x – y – 2z + 7 = 0 , \u0111i\u1ec3m A(6;9;5) v\u00e0 \u0111\u01b0\u1eddng th\u1eb3ng \u0394 : (x – 2)\/(2) = (y + 1)\/(5) = (z + 5)\/(3).\n<\/p>\n

a) \u0110\u01b0\u1eddng th\u1eb3ng \u0394 v\u00e0 m\u1eb7t ph\u1eb3ng (P) c\u1eaft nhau t\u1ea1i A .<\/p>\n

b) Vecto k\u20d7 = (2;1;2) l\u00e0 m\u1ed9t vecto ph\u00e1p tuy\u1ebfn c\u1ee7a m\u1eb7t ph\u1eb3ng (P) .<\/p>\n

c) Ph\u01b0\u01a1ng tr\u00ecnh m\u1eb7t c\u1ea7u t\u00e2m I(1;-1;2) v\u00e0 ti\u1ebfp x\u00fac m\u1eb7t ph\u1eb3ng (P) c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh t\u1ed5ng qu\u00e1t: x\u00b2 + y\u00b2 + z\u00b2 – 2x + 2y – 4z + 2 = 0.\n<\/p>\n

d) G\u1ecdi (Q) l\u00e0 m\u1eb7t ph\u1eb3ng ch\u1ee9a \u0111i\u1ec3m A v\u00e0 \u0111\u01b0\u1eddng th\u1eb3ng \u0394 , m\u1eb7t ph\u1eb3ng (Q) c\u00f3 m\u1ed9t vecto ph\u00e1p tuy\u1ebfn l\u00e0 q\u20d7 = (1;4;-1) .<\/p>\n

Ph\u1ea7n 3 \u2013 C\u00e2u h\u1ecfi tr\u1ea3 l\u1eddi ng\u1eafn ( 2 c\u00e2u )<\/strong><\/span><\/p>\n

C\u00e1c c\u00e2u h\u1ecfi c\u1ee7a ph\u1ea7n 3 trong \u0111\u1ec1 thi nh\u01b0 sau:<\/p>\n

C\u00e2u 1:<\/strong> M\u1ef1c n\u01b0\u1edbc trong h\u1ed3 ch\u1ee9a c\u1ee7a nh\u00e0 m\u00e1y \u0111i\u1ec7n th\u1ee7y tri\u1ec1u thay \u0111\u1ed5i trong su\u1ed1t m\u1ed9t ng\u00e0y do n\u01b0\u1edbc ch\u1ea3y ra khi th\u1ee7y tri\u1ec1u xu\u1ed1ng v\u00e0 n\u01b0\u1edbc ch\u1ea3y v\u00e0o khi th\u1ee7y tri\u1ec1u l\u00ean. T\u1ed1c \u0111\u1ed9 thay \u0111\u1ed5i c\u1ee7a m\u1ef1c n\u01b0\u1edbc \u0111\u01b0\u1ee3c x\u00e1c \u0111\u1ecbnh b\u1edfi h\u00e0m s\u1ed1: h'(t) = 1\/90 (t\u00b2 – 17t + 60) trong \u0111\u00f3 t t\u00ednh b\u1eb1ng gi\u1edd (0 \u2264 t \u2264 24) v\u00e0 h(t) t\u00ednh b\u1eb1ng m\u00e9t\/gi\u1edd. T\u1ea1i th\u1eddi \u0111i\u1ec3m t = 0 , m\u1ef1c n\u01b0\u1edbc trong h\u1ed3 ch\u1ee9a cao 8 m\u00e9t. H\u1ecfi t\u1ea1i th\u1eddi \u0111i\u1ec3m t = 6 , m\u1ef1c n\u01b0\u1edbc trong h\u1ed3 ch\u1ee9a cao bao nhi\u00eau m\u00e9t?<\/p>\n

C\u00e2u 2:<\/strong> Khi g\u1eafn h\u1ec7 t\u1ecda \u0111\u1ed9 Oxyz (\u0111\u01a1n v\u1ecb tr\u00ean m\u1ed7i tr\u1ee5c l\u00e0 km) v\u00e0o m\u1ed9t s\u00e2n bay, m\u1eb7t ph\u1eb3ng (Oxy) tr\u00f9ng v\u1edbi m\u1eb7t s\u00e2n bay. M\u1ed9t m\u00e1y bay bay theo \u0111\u01b0\u1eddng th\u1eb3ng t\u1eeb v\u1ecb tr\u00ed: A(1;-5;7) \u0111\u1ebfn v\u1ecb tr\u00ed: B(6;5;4) v\u00e0 h\u1ea1 c\u00e1nh t\u1ea1i v\u1ecb tr\u00ed M(a;b;0) . Gi\u00e1 tr\u1ecb c\u1ee7a 5a + 5b l\u00e0 bao nhi\u00eau?<\/p>\n

Ph\u1ea7n 4 \u2013 C\u00e2u h\u1ecfi t\u1ef1 lu\u1eadn (3 c\u00e2u)<\/span><\/strong><\/p>\n

Th\u00ed sinh tr\u00ecnh b\u00e0y \u0111\u1ea7y \u0111\u1ee7 t\u1eeb c\u00e2u 1 \u0111\u1ebfn c\u00e2u 3 ra gi\u1ea5y v\u1edbi l\u1eadp lu\u1eadn, tr\u00ecnh b\u00e0y b\u00e0i gi\u1ea3i r\u00f5 r\u00e0ng.<\/p>\n

C\u00e2u 1<\/strong>: T\u00ednh t\u00edch ph\u00e2n: P = int\u2081\u00b2\u2075 ( 1\/x – (1)\/(x\u00b2) – (1)\/(x\u00b3) ) dx.\n<\/p>\n

C\u00e2u 2:<\/strong> Trong kh\u00f4ng gian v\u1edbi h\u1ec7 tr\u1ee5c Oxyz , cho \u0111i\u1ec3m I(3;-6;-2) . Vi\u1ebft ph\u01b0\u01a1ng tr\u00ecnh m\u1eb7t ph\u1eb3ng ch\u1ee9a \u0111i\u1ec3m I v\u00e0 vu\u00f4ng g\u00f3c v\u1edbi \u0111\u01b0\u1eddng th\u1eb3ng \u0394 : x+7\/15 = y+1\/-11 = z\/37.\n<\/p>\n

Ngo\u00e0i ra b\u1ea1n c\u0169ng c\u00f3 th\u1ec3 t\u00ecm hi\u1ec3u nhi\u1ec1u \u0111\u1ec1 thi gi\u1eefa k\u1ef3 2 to\u00e1n 12<\/a><\/strong> kh\u00e1c v\u00e0 th\u1eed s\u1ee9c v\u1edbi nhi\u1ec1u d\u1ea1ng b\u00e0i t\u1eadp nh\u00e9!<\/p>\n

T\u1ea3i \u0110\u1ec1 ki\u1ec3m tra \u0111\u00e1nh gi\u00e1 gi\u1eefa k\u00ec 2 To\u00e1n 12n\u0103m h\u1ecdc 2024 \u2013 2025 Tr\u01b0\u1eddng THPT T\u00e2y Th\u1ea1nh TP HCM<\/strong><\/h2>\n

\u0110\u1ec1 ki\u1ec3m tra \u0111\u00e1nh gi\u00e1 gi\u1eefa k\u00ec 2 To\u00e1n 12 n\u0103m h\u1ecdc 2024 \u2013 2025 Tr\u01b0\u1eddng THPT T\u00e2y Th\u1ea1nh TP HCM<\/strong> l\u00e0 tr\u1ee3 th\u1ee7 \u0111\u1eafc l\u1ef1c gi\u00fap c\u00e1c b\u1ea1n h\u1ecdc sinh \u00f4n t\u1eadp th\u1eadt v\u1eefng ki\u1ebfn th\u1ee9c tr\u01b0\u1edbc thi chu\u1ea9n b\u1ecb l\u00e0m b\u00e0i thi ch\u00ednh th\u1ee9c. H\u00e3y t\u1ea3i ngay t\u00e0i li\u1ec7u qua \u0111\u01b0\u1eddng link sau \u0111\u00e2y v\u00e0 luy\u1ec7n t\u1eadp ngay!<\/p>\n