{"id":386,"date":"2026-05-05T04:43:01","date_gmt":"2026-05-05T04:43:01","guid":{"rendered":"https:\/\/79.buffdemo.com\/index.php\/2026\/05\/05\/de-minh-hoa-giua-ki-2-toan-12-nam-2024-2025-truong-thpt-huong-hoa-quang-tri\/"},"modified":"2026-05-05T07:15:45","modified_gmt":"2026-05-05T07:15:45","slug":"de-minh-hoa-giua-ki-2-toan-12-nam-2024-2025-truong-thpt-huong-hoa-quang-tri","status":"publish","type":"post","link":"https:\/\/79.buffdemo.com\/index.php\/2026\/05\/05\/de-minh-hoa-giua-ki-2-toan-12-nam-2024-2025-truong-thpt-huong-hoa-quang-tri\/","title":{"rendered":"\u0110\u1ec1 minh h\u1ecda gi\u1eefa k\u00ec 2 To\u00e1n 12 n\u0103m 2024 \u2013 2025 tr\u01b0\u1eddng THPT H\u01b0\u1edbng H\u00f3a \u2013 Qu\u1ea3ng Tr\u1ecb"},"content":{"rendered":"

\u0110\u1ec1 minh h\u1ecda gi\u1eefa k\u00ec 2 To\u00e1n 12 n\u0103m 2024 \u2013 2025 tr\u01b0\u1eddng THPT H\u01b0\u1edbng H\u00f3a \u2013 Qu\u1ea3ng Tr\u1ecb\u00a0<\/strong>l\u00e0 t\u00e0i li\u1ec7u t\u1ed5ng h\u1ee3p ki\u1ebfn th\u1ee9c quan tr\u1ecdng v\u1ec1 nguy\u00ean h\u00e0m, t\u00edch ph\u00e2n v\u00e0 ph\u01b0\u01a1ng tr\u00ecnh m\u1eb7t ph\u1eb3ng. C\u00f9ng c\u1ea5u tr\u00fac \u0111\u1ec1 thi v\u1edbi c\u00e1c d\u1ea1ng b\u00e0i t\u1eadp v\u1eadn d\u1ee5ng theo t\u1eebng m\u1ee9c \u0111\u1ed9 nh\u1eadn bi\u1ebft, gi\u00fap h\u1ecdc sinh l\u1edbp 12 r\u00e8n luy\u1ec7n v\u00e0 c\u1ee7ng c\u1ed1 ki\u1ebfn th\u1ee9c hi\u1ec7u qu\u1ea3. \u0110\u1ec3 n\u1eafm v\u1eefng t\u00e0i li\u1ec7u n\u00e0y h\u00e3y c\u00f9ng \u00d4n thi<\/a><\/strong> xem qua n\u1ed9i dung c\u1ee7a t\u1eadp \u0111\u1ec1 minh h\u1ecda c\u00f3 g\u00ec \u0111\u1eb7c bi\u1ec7t ngay sau \u0111\u00e2y.<\/p>\n

N\u1ed9i dung \u0110\u1ec1 minh h\u1ecda gi\u1eefa k\u00ec 2 To\u00e1n 12 n\u0103m 2024 \u2013 2025 tr\u01b0\u1eddng THPT H\u01b0\u1edbng H\u00f3a \u2013 Qu\u1ea3ng Tr\u1ecb<\/strong><\/h2>\n

\u0110\u1ec1 minh h\u1ecda gi\u1eefa k\u00ec 2 To\u00e1n 12 n\u0103m 2024 \u2013 2025 tr\u01b0\u1eddng THPT H\u01b0\u1edbng H\u00f3a \u2013 Qu\u1ea3ng Tr\u1ecb<\/strong> g\u1ed3m 16 c\u00e2u tr\u1eafc nghi\u1ec7m v\u1edbi nhi\u1ec1u ph\u01b0\u01a1ng \u00e1n l\u1ef1a ch\u1ecdn, 04 c\u00e2u tr\u1eafc nghi\u1ec7m \u0111\u00fang sai v\u00e0 04 c\u00e2u tr\u1eafc nghi\u1ec7m tr\u1ea3 l\u1eddi ng\u1eafn, v\u1edbi th\u1eddi gian l\u00e0m b\u00e0i l\u00e0 90 ph\u00fat. D\u01b0\u1edbi \u0111\u00e2y l\u00e0 ma tr\u1eadn \u0111\u1ec1 minh h\u1ecda b\u1ea1n c\u00f3 th\u1ec3 tham kh\u1ea3o qua:<\/p>\n

N\u1ed9i dung \u00f4n t\u1eadp<\/strong><\/span><\/p>\n

Ph\u1ea7n I \u2013 Nguy\u00ean h\u00e0m, t\u00edch ph\u00e2n v\u00e0 \u1ee9ng d\u1ee5ng (15 ti\u1ebft)<\/strong><\/span><\/p>\n

+ Nguy\u00ean h\u00e0m (5 ti\u1ebft).<\/strong><\/p>\n

+ T\u00edch ph\u00e2n (4 ti\u1ebft).<\/strong><\/p>\n

+ \u1ee8ng d\u1ee5ng c\u1ee7a t\u00edch ph\u00e2n (4 ti\u1ebft).<\/strong><\/p>\n

Ph\u1ea7n II \u2013 Ph\u01b0\u01a1ng tr\u00ecnh m\u1eb7t ph\u1eb3ng (6 ti\u1ebft)<\/strong><\/span><\/p>\n

+ Ph\u01b0\u01a1ng tr\u00ecnh m\u1eb7t ph\u1eb3ng (6 ti\u1ebft).<\/strong><\/p>\n

C\u1ea5u tr\u00fac \u0111\u1ec1 thi<\/strong><\/span><\/p>\n

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

\u2013 G\u1ed3m 16 c\u00e2u h\u1ecfi tr\u1eafc nghi\u1ec7m v\u1edbi 4 ph\u01b0\u01a1ng \u00e1n a), b), c), d). M\u1ed7i c\u00e2u ch\u1ec9 \u0111\u01b0\u1ee3c ch\u1ecdn m\u1ed9t \u0111\u00e1p \u00e1n ch\u00ednh x\u00e1c nh\u1ea5t, h\u00e3y xem qua c\u00e1c c\u00e2u h\u1ecfi sau \u0111\u00e2y:<\/p>\n

C\u00e2u 1:<\/strong> Cho h\u00e0m s\u1ed1 f(x) x\u00e1c \u0111\u1ecbnh tr\u00ean m\u1ed9t kho\u1ea3ng K . H\u00e0m s\u1ed1 F(x) \u0111\u01b0\u1ee3c g\u1ecdi l\u00e0 m\u1ed9t nguy\u00ean h\u00e0m c\u1ee7a h\u00e0m s\u1ed1 f(x) tr\u00ean K n\u1ebfu<\/p>\n

A. F'(x) = f(x), forall x in K. \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0B. F\u201d(x) = f(x), forall x in K.\n<\/p>\n

C. F(x) = f'(x), forall x in K. \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0D. F(x) = f\u201d(x), forall x in K.\n<\/p>\n

C\u00e2u 2:<\/strong> Cho h\u00e0m s\u1ed1 f(x) li\u00ean t\u1ee5c tr\u00ean [a;b] v\u00e0 F(x) l\u00e0 m\u1ed9t nguy\u00ean h\u00e0m c\u1ee7a f(x) . M\u1ec7nh \u0111\u1ec1 n\u00e0o sau \u0111\u00e2y \u0111\u00fang?<\/p>\n

A. int_a^b f(x)dx = F(b) – F(a). \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 B. int_a^b f(x)dx = F(a) – F(b).\n<\/p>\n

C. int_a^b f(x)dx = F(a) + F(b). \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0D. int_a^b f(x)dx = f(b) – f(a).\n<\/p>\n

Ph\u1ea7n II \u2013 C\u00e2u h\u1ecfi tr\u1eafc nghi\u1ec7m \u0111\u00fang sai<\/strong><\/span><\/p>\n

C\u00e1c b\u1ea1n h\u1ecdc sinh h\u00e3y tr\u1ea3 l\u1eddi nhanh c\u00e1c c\u00e2u sau v\u00e0 ch\u1ecdn m\u1ec7nh \u0111\u1ec1 \u0111\u00fang ho\u1eb7c sai cho m\u1ed7i c\u00e2u l\u1eddi:<\/p>\n

C\u00e2u 1:<\/strong> Cho h\u00e0m s\u1ed1 f(x) = 4x\u00b3 – 2 .<\/p>\n

a) N\u1ebfu F(x) l\u00e0 m\u1ed9t nguy\u00ean h\u00e0m c\u1ee7a f(x) th\u00ec F(x) = f'(x) .<\/p>\n

b) f(x) c\u00f3 m\u1ed9t nguy\u00ean h\u00e0m l\u00e0 F(x) = x\u2074 – 2x + 5 .<\/p>\n

c) int f(x)dx = int 4x\u00b3 dx – int 2dx\n<\/p>\n

d) int f(x)dx = x\u2074 – 2x + C.\n<\/p>\n

C\u00e2u 2:<\/strong> Cho h\u00e0m s\u1ed1 f(x) li\u00ean t\u1ee5c tr\u00ean mathbbR , F(x) l\u00e0 m\u1ed9t nguy\u00ean h\u00e0m c\u1ee7a f(x) , int\u2080\u2079 f(x)dx = 9 v\u00e0 h\u00e0m s\u1ed1 g(x) = 2cos x – sin x .<\/p>\n

a) Ta c\u00f3 int\u2080\u2079 f(x)dx = F(9) – F(0).\n<\/p>\n

b) int\u2080\u2076 f(x)dx + int\u2086\u2079 f(x)dx = 18.\n<\/p>\n

c) g(x) c\u00f3 m\u1ed9t nguy\u00ean h\u00e0m l\u00e0 G(x) = -2sin x + cos x + 3 v\u00e0 G(0) = 4 .<\/p>\n

d) G\u1ecdi G(x) l\u00e0 m\u1ed9t nguy\u00ean h\u00e0m c\u1ee7a h\u00e0m s\u1ed1 g(x) th\u1ecfa m\u00e3n G(\u03c0\/2) = 4 th\u00ec G(\u03c0\/3) = 4.\n<\/p>\n

Ph\u1ea7n III \u2013 C\u00e2u h\u1ecfi tr\u1ea3 l\u1eddi ng\u1eafn<\/strong><\/span><\/p>\n

M\u1ed9t s\u1ed1 c\u00e2u h\u1ecfi tr\u1ea3 l\u1eddi ng\u1eafn tham kh\u1ea3o \u0111\u1ec3 chu\u1ea9n b\u1ecb t\u1ed1t cho \u0111\u1ec1 thi gi\u1eefa h\u1ecdc k\u00ec 2 to\u00e1n 12<\/a> nh\u01b0 sau:<\/p>\n

C\u00e2u 1:<\/strong> M\u1ed9t qu\u1ea7n th\u1ec3 vi sinh v\u1eadt c\u00f3 t\u1ed1c \u0111\u1ed9 t\u0103ng s\u1ed1 l\u01b0\u1ee3ng c\u00e1 th\u1ec3 \u0111\u01b0\u1ee3c \u01b0\u1edbc l\u01b0\u1ee3ng b\u1edfi P'(t) = 150\u221at (c\u00e1 th\u1ec3\/ng\u00e0y)
\nv\u1edbi 0 \u2264 t \u2264 10 , trong \u0111\u00f3 P(t) l\u00e0 s\u1ed1 l\u01b0\u1ee3ng c\u00e1 th\u1ec3 vi sinh v\u1eadt t\u1ea1i th\u1eddi \u0111i\u1ec3m t ng\u00e0y k\u1ec3 t\u1eeb th\u1eddi \u0111i\u1ec3m ban \u0111\u1ea7u. Bi\u1ebft r\u1eb1ng ban \u0111\u1ea7u qu\u1ea7n th\u1ec3 c\u00f3 1000 c\u00e1 th\u1ec3. \u01af\u1edbc l\u01b0\u1ee3ng s\u1ed1 c\u00e1 th\u1ec3 c\u1ee7a qu\u1ea7n th\u1ec3 sau 5 ng\u00e0y k\u1ec3 t\u1eeb th\u1eddi \u0111i\u1ec3m ban \u0111\u1ea7u (k\u1ebft qu\u1ea3 l\u00e0m tr\u00f2n \u0111\u1ebfn h\u00e0ng tr\u0103m).<\/p>\n

C\u00e2u 2:<\/strong> Gi\u1ea3 s\u1eed anh Nam nh\u1ea3y d\u00f9 t\u1eeb m\u1ed9t chi\u1ebfc tr\u1ef1c th\u0103ng. V\u00e0o th\u1eddi \u0111i\u1ec3m 19 gi\u00e2y sau khi r\u1eddi kh\u1ecfi tr\u1ef1c th\u0103ng, anh Nam m\u1edf chi\u1ebfc d\u00f9 c\u1ee7a m\u00ecnh trong 2 gi\u00e2y, anh Nam ch\u1ea1m \u0111\u1ea5t sau 19 gi\u00e2y k\u1ec3 t\u1eeb l\u00fac bung d\u00f9.T\u1ea1i th\u1eddi \u0111i\u1ec3m t (gi\u00e2y), v\u1ecb tr\u00ed c\u1ee7a anh Nam c\u00e1ch m\u1eb7t \u0111\u1ea5t m\u1ed9t kho\u1ea3ng h(t) m\u00e9t v\u00e0 v\u1eadn t\u1ed1c r\u01a1i c\u1ee7a anh Nam (t\u00ednh b\u1eb1ng m\/s) l\u00e0 m\u1ed9t h\u00e0m s\u1ed1 \u0111\u01b0\u1ee3c cho b\u1edfi c\u00f4ng th\u1ee9c: v(t) = h'(t) =begincases-80 & khi 0 \u2264 t < 19 37t – 783 & khi 19 \u2264 t < 21 -6 & khi 21 \u2264 t \u2264 40.endcases \u0110\u1ed9 cao v\u1ecb tr\u00ed c\u1ee7a anh Nam khi b\u1eaft \u0111\u1ea7u nh\u1ea3y ra kh\u1ecfi tr\u1ef1c th\u0103ng b\u1eb1ng bao nhi\u00eau m\u00e9t?<\/p>\n

T\u1ea3i \u0110\u1ec1 minh h\u1ecda gi\u1eefa k\u00ec 2 To\u00e1n 12 n\u0103m 2024 \u2013 2025 tr\u01b0\u1eddng THPT H\u01b0\u1edbng H\u00f3a \u2013 Qu\u1ea3ng Tr\u1ecb<\/strong><\/h2>\n

\u0110\u1ec3 thu\u1eadn ti\u1ec7n cho vi\u1ec7c \u00f4n t\u1eadp k\u1ef3 thi gi\u1eefa h\u1ecdc k\u00ec 2 m\u00f4n To\u00e1n, c\u00e1c b\u1ea1n h\u1ecdc sinh v\u00e0 th\u1ea7y c\u00f4 h\u00e3y t\u1ea3i ngay \u0110\u1ec1 minh h\u1ecda gi\u1eefa k\u00ec 2 To\u00e1n 12 n\u0103m 2024 \u2013 2025 tr\u01b0\u1eddng THPT H\u01b0\u1edbng H\u00f3a \u2013 Qu\u1ea3ng Tr\u1ecb<\/strong> qua \u0111\u01b0\u1eddng d\u1eabn sau \u0111\u00e2y. Ch\u00fac c\u00e1c b\u1ea1n c\u00f3 m\u1ed9t l\u1ed9 tr\u00ecnh \u00f4n t\u1eadp th\u1eadt hi\u1ec7u qu\u1ea3 v\u00e0 \u0111\u1ea1t \u0111i\u1ec3m s\u1ed1 cao.<\/p>\n