{"id":394,"date":"2026-05-05T04:43:28","date_gmt":"2026-05-05T04:43:28","guid":{"rendered":"https:\/\/79.buffdemo.com\/index.php\/2026\/05\/05\/10-de-on-tap-kiem-tra-giua-ki-2-toan-12-ctm-theo-form-bgd-2025\/"},"modified":"2026-05-05T07:16:11","modified_gmt":"2026-05-05T07:16:11","slug":"10-de-on-tap-kiem-tra-giua-ki-2-toan-12-ctm-theo-form-bgd-2025","status":"publish","type":"post","link":"https:\/\/79.buffdemo.com\/index.php\/2026\/05\/05\/10-de-on-tap-kiem-tra-giua-ki-2-toan-12-ctm-theo-form-bgd-2025\/","title":{"rendered":"10 \u0111\u1ec1 \u00f4n t\u1eadp ki\u1ec3m tra gi\u1eefa k\u00ec 2 To\u00e1n 12 CTM theo form BGD 2025"},"content":{"rendered":"

Tuy\u1ec3n t\u1eadp 10 \u0111\u1ec1 \u00f4n t\u1eadp ki\u1ec3m tra gi\u1eefa k\u1ef3 2 To\u00e1n 12 CTM theo form B\u1ed9 GD 2025<\/strong>, \u0111\u01b0\u1ee3c bi\u00ean so\u1ea1n chi ti\u1ebft b\u1edfi th\u1ea7y Hu\u1ef3nh V\u0103n \u00c1nh. B\u1ed9 t\u00e0i li\u1ec7u g\u1ed3m 10 \u0111\u1ec1 theo c\u1ea5u tr\u00fac m\u1edbi (3-2-2-3), c\u00f3 \u0111\u1ea7y \u0111\u1ee7 \u0111\u00e1p \u00e1n v\u00e0 l\u1eddi gi\u1ea3i chi ti\u1ebft. Xem qua c\u1ea5u tr\u00fac chu\u1ea9n c\u1ee7a \u0111\u1ec1 thi c\u00f9ng T\u00e0i Li\u1ec7u \u00d4n Thi<\/a><\/strong> nh\u00e9!<\/p>\n

N\u1ed9i dung 10 \u0111\u1ec1 \u00f4n t\u1eadp ki\u1ec3m tra gi\u1eefa k\u00ec 2 To\u00e1n 12 CTM theo form BGD 2025<\/strong><\/h2>\n

N\u1ed9i dung c\u1ee7a b\u1ed9 \u0111\u1ec1 thi xoay quanh c\u00e1c ki\u1ebfn th\u1ee9c To\u00e1n 12<\/strong> tr\u1ecdng t\u00e2m, c\u00f9ng v\u1edbi c\u00e1c d\u1ea1ng b\u00e0i t\u1eadp chu\u1ea9n theo c\u1ea5u tr\u00fac m\u1edbi c\u1ee7a b\u1ed9 gi\u00e1o d\u1ee5c, gi\u00fap h\u1ecdc sinh d\u1ec5 d\u00e0ng \u00f4n t\u1eadp hi\u1ec7u qu\u1ea3. C\u1ea5u tr\u00fac chi ti\u1ebft 10 \u0111\u1ec1 \u00f4n t\u1eadp ki\u1ec3m tra gi\u1eefa k\u00ec 2 To\u00e1n 12 CTM theo form BGD 2025<\/strong> nh\u01b0 sau:<\/p>\n

Ph\u1ea7n I: Tr\u1eafc nghi\u1ec7m nhi\u1ec1u ph\u01b0\u01a1ng \u00e1n l\u1ef1a ch\u1ecdn<\/strong><\/span><\/p>\n

G\u1ed3m 12 c\u00e2u h\u1ecfi tr\u1eafc nghi\u1ec7m (chi\u1ebfm tr\u1ecdng s\u1ed1 30% c\u1ee7a t\u1ed5ng \u0111i\u1ec3m). Th\u00ed sinh l\u00e0m b\u00e0i b\u1eb1ng c\u00e1ch \u0111\u1ecdc \u0111\u1ec1 v\u00e0 ch\u1ecdn ph\u01b0\u01a1ng \u00e1n ch\u00ednh x\u00e1c nh\u1ea5t.<\/p>\n

C\u00e2u 1:<\/strong> H\u1ecd t\u1ea5t c\u1ea3 c\u00e1c nguy\u00ean h\u00e0m c\u1ee7a h\u00e0m s\u1ed1 f(x) = x\u00b3 l\u00e0<\/p>\n

A. 4x\u2074 + C .\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0B. 3x\u00b2 + C .\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0C. x\u2074 + C .\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0D. 1\/4x\u2074 + C .<\/p>\n

C\u00e2u 2<\/strong>: H\u1ecd t\u1ea5t c\u1ea3 c\u00e1c nguy\u00ean h\u00e0m c\u1ee7a h\u00e0m s\u1ed1 f(x) = 2sin x l\u00e0<\/p>\n

A. int 2sin x dx = -2cos x + C .\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 B. int 2sin x dx = 2cos x + C .<\/p>\n

C. int 2sin x dx = sin\u00b2 x + C .\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 D. int 2sin x dx = sin 2x + C .<\/p>\n

Ph\u1ea7n II: Tr\u1eafc nghi\u1ec7m \u0111\u00fang sai<\/strong><\/span><\/p>\n

M\u1ed7i c\u00e2u c\u00f3 c\u00e1c ph\u01b0\u01a1ng \u00e1n a), b), c), d), b\u1ea1n c\u1ea7n ch\u1ecdn m\u1ec7nh \u0111\u1ec1 \u0111\u00fang ho\u1eb7c sai cho m\u1ed7i c\u00e2u. Ph\u1ea7n n\u00e0y chi\u1ebfm 20% s\u1ed1 \u0111i\u1ec3m c\u1ee7a \u0111\u1ec1 thi v\u00e0 c\u00f3 c\u00e1c c\u00e2u h\u1ecfi sau \u0111\u00e2y:<\/p>\n

C\u00e2u 1:<\/strong> M\u1ed9t v\u1eadt \u0111ang chuy\u1ec3n \u0111\u1ed9ng v\u1edbi v\u1eadn t\u1ed1c v = 72 (km\/h) th\u00ec thay \u0111\u1ed5i v\u1eadn t\u1ed1c v\u1edbi gia t\u1ed1c \u0111\u01b0\u1ee3c t\u00ednh theo th\u1eddi gian t l\u00e0 a(t) = -4 + 2t (m\/s \u00b2 ).<\/p>\n

a) V\u1eadn t\u1ed1c c\u1ee7a v\u1eadt khi thay \u0111\u1ed5i l\u00e0 v(t) = t\u00b2 – 4t (m\/s).<\/p>\n

b) T\u1ea1i th\u1eddi \u0111i\u1ec3m t = 0 (khi v\u1eadt b\u1eaft \u0111\u1ea7u thay \u0111\u1ed5i v\u1eadn t\u1ed1c) ta c\u00f3 v\u2080 = 20 (m\/s). Suy ra v(t) = t\u00b2 – 4t + 72.\n<\/p>\n

c) Qu\u00e3ng \u0111\u01b0\u1eddng v\u1eadt \u0111i \u0111\u01b0\u1ee3c trong kho\u1ea3ng th\u1eddi gian 3 gi\u00e2y k\u1ec3 t\u1eeb khi b\u1eaft \u0111\u1ea7u t\u0103ng t\u1ed1c l\u00e0 9 (m).<\/p>\n

d) Qu\u00e3ng \u0111\u01b0\u1eddng v\u1eadt \u0111i \u0111\u01b0\u1ee3c k\u1ec3 t\u1eeb th\u1eddi \u0111i\u1ec3m thay \u0111\u1ed5i v\u1eadn t\u1ed1c \u0111\u1ebfn l\u00fac v\u1eadt \u0111\u1ea1t v\u1eadn t\u1ed1c b\u00e9 nh\u1ea5t l\u00e0 104\/3 (m).\n<\/p>\n

C\u00e2u 2<\/strong>: Trong kh\u00f4ng gian Oxyz , cho m\u1eb7t ph\u1eb3ng (P) : 2x – y + 2z – 8 = 0.\n<\/p>\n

a) \u0110i\u1ec3m M(2; -1; 2) thu\u1ed9c m\u1eb7t ph\u1eb3ng (P) .<\/p>\n

b) T\u1ecda \u0111\u1ed9 m\u1ed9t vect\u01a1 ph\u00e1p tuy\u1ebfn c\u1ee7a m\u1eb7t ph\u1eb3ng (P) l\u00e0 mathbfn = (2; -1; 2) .<\/p>\n

c) M\u1eb7t ph\u1eb3ng (Q) \u0111i qua \u0111i\u1ec3m A(3; 2; -1) v\u00e0 song song v\u1edbi m\u1eb7t ph\u1eb3ng (P) c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 2x – y + 2z = 0.\n<\/p>\n

d) G\u1ecdi (R) l\u00e0 m\u1eb7t ph\u1eb3ng vu\u00f4ng g\u00f3c v\u1edbi m\u1eb7t ph\u1eb3ng (P) v\u00e0 m\u1eb7t ph\u1eb3ng (Oxy) , \u0111\u1ed3ng th\u1eddi c\u00e1ch \u0111i\u1ec3m B(-3; 1; 5) m\u1ed9t kho\u1ea3ng b\u1eb1ng \u221a5 . Gi\u1ea3 s\u1eed (R) c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh ax + by + d = 0, (a > 0, d < 0), a, b in mathbbZ, (a, b) = 1.
\nGi\u00e1 tr\u1ecb c\u1ee7a bi\u1ec3u th\u1ee9c b + d = -4 .<\/p>\n

Ph\u1ea7n III: Tr\u1eafc nghi\u1ec7m tr\u1ea3 l\u1eddi ng\u1eafn<\/strong><\/span><\/p>\n

C\u00f3 4 c\u00e2u chi\u1ebfm 20% s\u1ed1 \u0111i\u1ec3m c\u1ee7a \u0111\u1ec1 thi. Tham kh\u1ea3o qua m\u1ed9t s\u1ed1 c\u00e2u h\u1ecfi nh\u01b0 sau \u0111\u1ec3 hi\u1ec3u r\u00f5 h\u01a1n nh\u00e9.<\/p>\n

C\u00e2u 3:<\/strong> Trong kh\u00f4ng gian v\u1edbi h\u1ec7 t\u1ecda \u0111\u1ed9 Oxyz , cho 3 \u0111i\u1ec3m A(0,0,3) , B(1,1,3) , C(0,1,1) . Kho\u1ea3ng c\u00e1ch t\u1eeb g\u1ed1c t\u1ecda \u0111\u1ed9 O \u0111\u1ebfn m\u1eb7t ph\u1eb3ng (ABC) b\u1eb1ng?<\/p>\n

C\u00e2u 4:<\/strong> Trong kh\u00f4ng gian Oxyz , cho m\u1eb7t ph\u1eb3ng (P): ax + by + cz – 27 = 0 qua hai \u0111i\u1ec3m A(3,2,1) v\u00e0 B(-3,5,2) v\u00e0 vu\u00f4ng g\u00f3c v\u1edbi m\u1eb7t ph\u1eb3ng (Q): 3x + y + z + 4 = 0 . T\u00ednh t\u1ed5ng S = a + b + c .<\/p>\n

Ph\u1ea7n IV: T\u1ef1 lu\u1eadn<\/strong><\/span><\/p>\n

G\u1ed3m 3 c\u00e2u h\u1ecfi t\u1ef1 lu\u1eadn chi\u1ebfm 30% t\u1ed5ng \u0111i\u1ec3m. \u0110\u1ec3 chi\u1ebfm tr\u1ecdn s\u1ed1 \u0111i\u1ec3m c\u1ee7a ph\u1ea7n n\u00e0y trong \u0111\u1ec1 thi gi\u1eefa k\u00ec 2 to\u00e1n 12<\/a>, b\u1ea1n n\u00ean xem qua c\u00e1c c\u00e2u h\u1ecfi sau \u0111\u00e2y:<\/p>\n

C\u00e2u 1:<\/strong> T\u00ednh t\u00edch ph\u00e2n int\u208b\u2081\u2075 x\u00b2 |x – 1| dx.\n<\/p>\n

C\u00e2u 2:<\/strong> T\u00ednh di\u1ec7n t\u00edch c\u1ee7a h\u00ecnh ph\u1eb3ng (H) gi\u1edbi h\u1ea1n b\u1edfi \u0111\u01b0\u1eddng cong y = x\u00b3 – x\u00b2 – 12x v\u00e0 tr\u1ee5c Ox .<\/p>\n

T\u1ea3i 10 \u0111\u1ec1 \u00f4n t\u1eadp ki\u1ec3m tra gi\u1eefa k\u00ec 2 To\u00e1n 12 CTM theo form BGD 2025<\/strong><\/h2>\n

\u0110\u1ec3 n\u1eafm v\u1eefng c\u1ea5u tr\u00fac \u0111\u1ec1 thi m\u1edbi v\u00e0 chu\u1ea9n b\u1ecb chi\u1ebfn l\u01b0\u1ee3c l\u00e0m b\u00e0i hi\u1ec7u qu\u1ea3 cho k\u1ef3 thi gi\u1eefa k\u1ef3 2 To\u00e1n 12, qu\u00fd th\u1ea7y c\u00f4 v\u00e0 c\u00e1c b\u1ea1n h\u00e3y t\u1ea3i ngay b\u1ed9 t\u00e0i li\u1ec7u 10 \u0111\u1ec1 \u00f4n t\u1eadp ki\u1ec3m tra gi\u1eefa k\u1ef3 2 To\u00e1n 12 CTM theo form B\u1ed9 GD 2025<\/strong> qua \u0111\u01b0\u1eddng link d\u01b0\u1edbi \u0111\u00e2y. B\u1eaft tay v\u00e0o luy\u1ec7n t\u1eadp ngay v\u00e0 \u0111\u1ed1i chi\u1ebfu \u0111\u00e1p \u00e1n sau m\u1ed7i \u0111\u1ec1 \u0111\u1ec3 c\u1ee7ng c\u1ed1 ki\u1ebfn th\u1ee9c nh\u00e9!<\/p>\n