{"id":330,"date":"2026-05-05T04:39:58","date_gmt":"2026-05-05T04:39:58","guid":{"rendered":"https:\/\/79.buffdemo.com\/index.php\/2026\/05\/05\/bo-de-on-tap-kiem-tra-giua-hoc-ki-2-mon-toan-12-theo-cau-truc-moi\/"},"modified":"2026-05-05T07:12:41","modified_gmt":"2026-05-05T07:12:41","slug":"bo-de-on-tap-kiem-tra-giua-hoc-ki-2-mon-toan-12-theo-cau-truc-moi","status":"publish","type":"post","link":"https:\/\/79.buffdemo.com\/index.php\/2026\/05\/05\/bo-de-on-tap-kiem-tra-giua-hoc-ki-2-mon-toan-12-theo-cau-truc-moi\/","title":{"rendered":"B\u1ed9 \u0111\u1ec1 \u00f4n t\u1eadp ki\u1ec3m tra gi\u1eefa h\u1ecdc k\u00ec 2 m\u00f4n To\u00e1n 12 theo c\u1ea5u tr\u00fac m\u1edbi 3 b\u1ed9 s\u00e1ch"},"content":{"rendered":"

\u00d4n t\u1eadp ki\u1ec3m tra To\u00e1n 12 gi\u1eefa k\u00ec hi\u1ec7u qu\u1ea3 b\u1eb1ng B\u1ed9 \u0111\u1ec1 \u00f4n t\u1eadp ki\u1ec3m tra gi\u1eefa h\u1ecdc k\u00ec 2 m\u00f4n To\u00e1n 12 theo c\u1ea5u tr\u00fac m\u1edbi<\/strong>. T\u1eadp t\u00e0i li\u1ec7u g\u1ed3m c\u00e1c d\u1ea1ng \u0111\u1ec1 c\u1ee7a 3 \u0111\u1ea7u s\u00e1ch n\u1ed5i ti\u1ebfng nh\u01b0 S\u00e1ch C\u00e1nh Di\u1ec1u, S\u00e1ch Ch\u00e2n Tr\u1eddi S\u00e1ng T\u1ea1o, S\u00e1ch K\u1ebft N\u1ed1i Tri Th\u1ee9c & Cu\u1ed9c S\u1ed1ng do GV Ph\u1ea1m L\u00ea Duy bi\u00ean so\u1ea1n. H\u00e3y c\u00f9ng T\u00e0i Li\u1ec7u \u00d4n Thi<\/a> <\/strong>t\u00ecm hi\u1ec3u \u0111\u1eb7c \u0111i\u1ec3m n\u1ed5i b\u1eadt c\u1ee7a b\u1ed9 t\u00e0i li\u1ec7u n\u00e0y qua b\u00e0i vi\u1ebft sau \u0111\u00e2y nh\u00e9!<\/p>\n

N\u1ed9i dung B\u1ed9 \u0111\u1ec1 \u00f4n t\u1eadp ki\u1ec3m tra gi\u1eefa h\u1ecdc k\u00ec 2 m\u00f4n To\u00e1n 12 theo c\u1ea5u tr\u00fac m\u1edbi<\/strong><\/h2>\n

B\u1ed9 \u0111\u1ec1 \u00f4n t\u1eadp ki\u1ec3m tra gi\u1eefa h\u1ecdc k\u00ec 2 m\u00f4n To\u00e1n 12 theo c\u1ea5u tr\u00fac m\u1edbi<\/strong> c\u00f3 t\u1ea5t c\u1ea3 79 trang. M\u1ed7i b\u1ed9 s\u00e1ch c\u00f3 5 \u0111\u1ec1 ki\u1ec3m tra kh\u00e1c nhau v\u1edbi ki\u1ebfn th\u1ee9c v\u1ec1 \u0111\u1ea1o h\u00e0m, s\u1ed1 ph\u1ee9c, ph\u01b0\u01a1ng ph\u00e1p t\u1ecda \u0111\u1ed9 trong kh\u00f4ng gian, h\u00ecnh h\u1ecdc,\u2026.Gi\u00fap h\u1ecdc sinh l\u00e0m quen v\u1edbi c\u00e1c d\u1ea1ng \u0111\u1ec1 v\u00e0 n\u00e2ng cao \u0111i\u1ec3m s\u1ed1 trong k\u00ec thi s\u1eafp \u0111\u1ebfn.<\/p>\n

C\u1ea5u tr\u00fac \u0111\u1ec1 thi g\u1ed3m 3 ph\u1ea7n ch\u00ednh nh\u01b0 sau:<\/strong><\/p>\n

Ph\u1ea7n 1 \u2013 C\u00e2u h\u1ecfi tr\u1eafc nghi\u1ec7m<\/strong><\/span><\/p>\n

Ph\u1ea7n 2 \u2013 C\u00e2u tr\u1eafc nghi\u1ec7m \u0111\u00fang sai<\/strong><\/span><\/p>\n

Ph\u1ea7n 3 \u2013 C\u00e2u tr\u1eafc nghi\u1ec7m tr\u1ea3 l\u1eddi ng\u1eafn<\/strong><\/span><\/p>\n

(***) Tr\u00edch m\u1ed9t s\u1ed1 c\u00e2u h\u1ecfi trong B\u1ed9 \u0111\u1ec1 \u00f4n t\u1eadp ki\u1ec3m tra gi\u1eefa h\u1ecdc k\u00ec 2 m\u00f4n To\u00e1n 12 theo c\u1ea5u tr\u00fac m\u1edbi<\/strong><\/p>\n

C\u00e2u 1<\/strong>. Cho F(x) l\u00e0 m\u1ed9t nguy\u00ean h\u00e0m c\u1ee7a h\u00e0m s\u1ed1 f(x) = 2x – 2 , bi\u1ebft r\u1eb1ng F(1) = 1 . X\u00e9t t\u00ednh \u0111\u00fang sai c\u1ee7a c\u00e1c kh\u1eb3ng \u0111\u1ecbnh sau:<\/p>\n

a) F(x) lu\u00f4n x\u00e1c \u0111\u1ecbnh tr\u00ean mathbbR .<\/p>\n

b) F(x) = x\u00b2 – 2x .<\/p>\n

c) F(x) > 0 , v\u1edbi m\u1ecdi x in mathbbR .<\/p>\n

d) F(-2) = 2 .<\/p>\n

C\u00e2u 2<\/strong>. Cho h\u00e0m s\u1ed1 f(x) = 5x\u2074 + 4x – 2 . X\u00e9t t\u00ednh \u0111\u00fang \u2013 sai c\u1ee7a c\u00e1c kh\u1eb3ng \u0111\u1ecbnh sau:<\/p>\n

a) int\u2080\u00b9 f(x) dx = 1.\n<\/p>\n

b) H\u1ecd nguy\u00ean h\u00e0m c\u1ee7a f(x) l\u00e0 F(x) = x\u2075 + 2x\u00b2 – 2x + C .<\/p>\n

c) Nguy\u00ean h\u00e0m F(x) c\u1ee7a f(x) v\u00e0 th\u1ecfa F(0) = 2 l\u00e0 F(x) = x\u2075 + 2x\u00b2 – 2x + 4.\n<\/p>\n

d) F(x) l\u00e0 m\u1ed9t nguy\u00ean h\u00e0m c\u1ee7a f(x) th\u1ecfa F(1) = 4 th\u00ec F(3) = 258 .<\/p>\n

C\u00e2u 5<\/strong>. Hai \u00f4 t\u00f4 xu\u1ea5t ph\u00e1t t\u1ea1i c\u00f9ng m\u1ed9t th\u1eddi \u0111i\u1ec3m tr\u00ean c\u00f9ng \u0111o\u1ea1n \u0111\u01b0\u1eddng th\u1eb3ng AB , \u00f4 t\u00f4 th\u1ee9 nh\u1ea5t b\u1eaft \u0111\u1ea7u xu\u1ea5t ph\u00e1t t\u1eeb A v\u00e0 \u0111i theo h\u01b0\u1edbng t\u1eeb A \u0111\u1ebfn B v\u1edbi v\u1eadn t\u1ed1c v_a(t) = 2t + 1 (km\/h); \u00f4 t\u00f4 th\u1ee9 hai xu\u1ea5t ph\u00e1t t\u1eeb O c\u00e1ch A m\u1ed9t kho\u1ea3ng 22 km v\u00e0 \u0111i theo h\u01b0\u1edbng t\u1eeb A \u0111\u1ebfn B v\u1edbi v\u1eadn t\u1ed1c 10 km\/h, sau m\u1ed9t kho\u1ea3ng th\u1eddi gian ng\u01b0\u1eddi l\u00e1i \u0111\u1ea1p phanh; t\u1eeb th\u1eddi \u0111i\u1ec3m \u0111\u00f3, \u00f4 t\u00f4 th\u1ee9 hai chuy\u1ec3n \u0111\u1ed9ng ch\u1eadm d\u1ea7n \u0111\u1ec1u v\u1edbi v\u1eadn t\u1ed1c v_o(t) = -5t + 20 (km\/h). H\u1ecfi sau kho\u1ea3ng th\u1eddi gian bao l\u00e2u k\u1ec3 t\u1eeb khi xu\u1ea5t ph\u00e1t hai \u00f4 t\u00f4 s\u1ebd g\u1eb7p nhau.<\/p>\n

C\u00e2u 6<\/strong>. Trong kh\u00f4ng gian Oxyz , cho ba \u0111i\u1ec3m A(1;1;1) , B(-1;2;0) , C(3;-1;2) v\u00e0 M l\u00e0 \u0111i\u1ec3m thu\u1ed9c m\u1eb7t ph\u1eb3ng (\u03b1): 2x – y + 2z + 7 = 0 . T\u00ednh gi\u00e1 tr\u1ecb nh\u1ecf nh\u1ea5t c\u1ee7a P = | 3MA\u20d7 + 5MB\u20d7 – 7MC\u20d7 |.\n<\/p>\n

T\u1ea3i B\u1ed9 \u0111\u1ec1 \u00f4n t\u1eadp ki\u1ec3m tra gi\u1eefa h\u1ecdc k\u00ec 2 m\u00f4n To\u00e1n 12 theo c\u1ea5u tr\u00fac m\u1edbi<\/strong><\/h2>\n

\u0110\u00e2y l\u00e0 b\u1ed9 t\u00e0i li\u1ec7u chu\u1ea9n ch\u1ec9nh v\u1edbi t\u1ea5t c\u1ea3 c\u00e1c d\u1ea1ng \u0111\u1ec1 thi \u0111\u01b0\u1ee3c bi\u00ean so\u1ea1n t\u1eeb c\u00e1c \u0111\u1ea7u s\u00e1ch uy t\u00edn. T\u1ea3i ngay B\u1ed9 \u0111\u1ec1 \u00f4n t\u1eadp ki\u1ec3m tra gi\u1eefa h\u1ecdc k\u00ec 2 m\u00f4n To\u00e1n 12 theo c\u1ea5u tr\u00fac m\u1edbi<\/strong> qua \u0111\u01b0\u1eddng link d\u01b0\u1edbi \u0111\u00e2y \u0111\u1ec3 luy\u1ec7n t\u1eadp m\u1ed7i ng\u00e0y v\u00e0 chinh ph\u1ee5c \u0111i\u1ec3m cao nh\u00e9.<\/p>\n