{"id":424,"date":"2026-05-05T04:45:01","date_gmt":"2026-05-05T04:45:01","guid":{"rendered":"https:\/\/79.buffdemo.com\/index.php\/2026\/05\/05\/de-minh-hoa-giua-ki-2-toan-12-nam-2024-2025-truong-ha-long-tinh-long-an\/"},"modified":"2026-05-05T07:17:47","modified_gmt":"2026-05-05T07:17:47","slug":"de-minh-hoa-giua-ki-2-toan-12-nam-2024-2025-truong-ha-long-tinh-long-an","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-ha-long-tinh-long-an\/","title":{"rendered":"\u0110\u1ec1 minh h\u1ecda gi\u1eefa k\u00ec 2 To\u00e1n 12 n\u0103m 2024 \u2013 2025 tr\u01b0\u1eddng H\u00e0 Long t\u1ec9nh Long An"},"content":{"rendered":"

C\u00e1c b\u1ea1n h\u1ecdc sinh 12 \u0111\u1eebng b\u1ecf qua c\u01a1 h\u1ed9i h\u1ec7 th\u1ed1ng l\u1ea1i ki\u1ebfn th\u1ee9c v\u1edbi \u0110\u1ec1 minh h\u1ecda gi\u1eefa k\u00ec 2 To\u00e1n 12 n\u0103m 2024 \u2013 2025 tr\u01b0\u1eddng H\u00e0 Long t\u1ec9nh Long An<\/strong>. V\u00ec \u0111\u00e2y ch\u00ednh l\u00e0 t\u1eadp t\u00e0i li\u1ec7u \u00f4n t\u1eadp c\u1ea7n thi\u1ebft, gi\u00fap c\u00e1c b\u1ea1n n\u1eafm ch\u1eafc ki\u1ebfn th\u1ee9c tr\u1ecdng t\u00e2m v\u00e0 s\u1eb5n s\u00e0ng cho k\u1ef3 thi. C\u00f9ng T\u00e0i Li\u1ec7u \u00d4n Thi<\/a><\/strong> t\u00ecm hi\u1ec3u b\u1ed9 \u0111\u1ec1 n\u00e0y c\u00f3 g\u00ec \u0111\u1eb7c bi\u1ec7t nh\u00e9!<\/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 H\u00e0 Long t\u1ec9nh Long An<\/strong><\/h2>\n

\u0110\u1ec1 minh h\u1ecda gi\u1eefa k\u00ec 2 To\u00e1n 12 n\u0103m 2024 \u2013 2025 tr\u01b0\u1eddng H\u00e0 Long t\u1ec9nh Long An<\/strong> l\u00e0 b\u1ed9 \u0111\u1ec1 cung c\u1ea5p c\u00e1i nh\u00ecn t\u1ed5ng quan cho c\u00e1c b\u1ea1n h\u1ecdc sinh v\u1ec1 ki\u1ebfn th\u1ee9c v\u00e0 k\u1ef9 n\u0103ng l\u00e0m b\u00e0i th\u1eadt t\u1ed1t cho k\u1ef3 thi gi\u1eefa k\u00ec 2. C\u1ea5u tr\u00fac \u0111\u1ec1 minh h\u1ecda g\u1ed3m 4 ph\u1ea7n theo nh\u01b0 quy \u0111\u1ecbnh chung v\u00e0 c\u00f3 n\u1ed9i dung nh\u01b0 sau:<\/p>\n

C\u00e1c n\u1ed9i dung l\u00fd thuy\u1ebft c\u1ea7n \u00f4n t\u1eadp:<\/strong><\/p>\n

PH\u1ea6N \u0110\u1ea0I S\u1ed0<\/strong><\/p>\n

\u2013 Kh\u00e1i ni\u1ec7m nguy\u00ean h\u00e0m v\u00e0 c\u00e1c ph\u01b0\u01a1ng ph\u00e1p t\u00ecm nguy\u00ean h\u00e0m.<\/p>\n

\u2013 \u0110\u1ecbnh ngh\u0129a t\u00edch ph\u00e2n x\u00e1c \u0111\u1ecbnh v\u00e0 c\u00e1c ph\u01b0\u01a1ng ph\u00e1p t\u00ednh t\u00edch ph\u00e2n.<\/p>\n

\u2013 \u1ee8ng d\u1ee5ng c\u1ee7a t\u00edch ph\u00e2n trong vi\u1ec7c t\u00ednh di\u1ec7n t\u00edch h\u00ecnh ph\u1eb3ng, th\u1ec3 t\u00edch v\u1eadt th\u1ec3 tr\u00f2n xoay, v\u00e0 c\u00e1c b\u00e0i to\u00e1n li\u00ean quan.<\/p>\n

\u2013 \u0110\u1ecbnh ngh\u0129a v\u00e0 c\u00e1c ph\u00e9p to\u00e1n c\u01a1 b\u1ea3n v\u1edbi s\u1ed1 ph\u1ee9c.<\/p>\n

\u2013 D\u1ea1ng l\u01b0\u1ee3ng gi\u00e1c c\u1ee7a s\u1ed1 ph\u1ee9c v\u00e0 \u1ee9ng d\u1ee5ng.<\/p>\n

PH\u1ea6N H\u00ccNH H\u1eccC<\/strong><\/p>\n

Ph\u01b0\u01a1ng tr\u00ecnh m\u1eb7t ph\u1eb3ng, \u0111\u01b0\u1eddng th\u1eb3ng v\u00e0 m\u1eb7t c\u1ea7u trong kh\u00f4ng gian.<\/p>\n

\u2013 T\u00ednh kho\u1ea3ng c\u00e1ch gi\u1eefa c\u00e1c \u0111\u1ed1i t\u01b0\u1ee3ng h\u00ecnh h\u1ecdc: \u0111i\u1ec3m \u0111\u1ebfn m\u1eb7t ph\u1eb3ng, \u0111\u01b0\u1eddng th\u1eb3ng \u0111\u1ebfn m\u1eb7t ph\u1eb3ng, hai \u0111\u01b0\u1eddng th\u1eb3ng ch\u00e9o nhau.<\/p>\n

\u2013 X\u00e1c \u0111\u1ecbnh g\u00f3c gi\u1eefa hai m\u1eb7t ph\u1eb3ng, gi\u1eefa \u0111\u01b0\u1eddng th\u1eb3ng v\u00e0 m\u1eb7t ph\u1eb3ng, gi\u1eefa hai \u0111\u01b0\u1eddng th\u1eb3ng.<\/p>\n

\u2013 X\u00e1c \u0111\u1ecbnh v\u1ecb tr\u00ed t\u01b0\u01a1ng \u0111\u1ed1i gi\u1eefa c\u00e1c \u0111\u1ed1i t\u01b0\u1ee3ng h\u00ecnh h\u1ecdc: \u0111i\u1ec3m, \u0111\u01b0\u1eddng th\u1eb3ng, m\u1eb7t ph\u1eb3ng v\u00e0 m\u1eb7t c\u1ea7u.<\/p>\n

C\u1ea5u tr\u00fac \u0111\u1ec1 thi gi\u1eefa k\u00ec 2 To\u00e1n 12<\/a> minh h\u1ecda c\u00f3:<\/strong><\/p>\n

Ph\u1ea7n I \u2013 C\u00e2u h\u1ecfi tr\u1eafc nghi\u1ec7m nhi\u1ec1u ph\u01b0\u01a1ng \u00e1n l\u1ef1a ch\u1ecdn ( 12 c\u00e2u )<\/strong><\/p>\n

H\u1ecdc sinh tr\u1ea3 l\u1eddi t\u1eeb c\u00e2u 1 \u0111\u1ebfn c\u00e2u 12 ch\u1ecdn ra \u0111\u00e1p \u00e1n ch\u00ednh x\u00e1c nh\u1ea5t v\u1edbi c\u00e2u h\u1ecfi c\u1ee7a \u0111\u1ec1:<\/p>\n

\\[
\n\\textbf{C\u00e2u 2:} \\text{ Trong c\u00e1c h\u00e0m s\u1ed1 sau, h\u00e0m s\u1ed1 n\u00e0o l\u00e0 m\u1ed9t nguy\u00ean h\u00e0m c\u1ee7a } f(x) = \\sin x?
\n\\]\n\\begin{align*}
\n\\text{A. } & F(x) = \\cos x. \\\\
\n\\text{B. } & G(x) = -\\cos x. \\\\
\n\\text{C. } & H(x) = \\sin x. \\\\
\n\\text{D. } & K(x) = -\\sin x. \\\\
\n\\end{align*}<\/p>\n

\\[
\n\\textbf{C\u00e2u 3:} \\text{ Cho } \\int_{2}^{5} f(x)dx = 10. \\text{ Khi \u0111\u00f3 } \\int_{5}^{2} f(x)dx \\text{ b\u1eb1ng:}
\n\\]\n\\begin{align*}
\n\\text{A. } & -30. \\\\
\n\\text{B. } & 30. \\\\
\n\\text{C. } & 10. \\\\
\n\\text{D. } & -10. \\\\
\n\\end{align*}<\/p>\n

Ph\u1ea7n II \u2013 C\u00e2u h\u1ecfi tr\u1eafc nghi\u1ec7m \u0111\u00fang sai ( 2 c\u00e2u )<\/strong><\/p>\n

Trong m\u1ed7i \u00fd a), b), c), d) m\u1ed7i c\u00e2u, th\u00ed sinh ch\u1ecdn m\u1ec7nh \u0111\u1ec1 \u0111\u00fang ho\u1eb7c sai, h\u00e3y tham kh\u1ea3o qua c\u00e1c c\u00e2u h\u1ecfi sau \u0111\u00e2y:<\/p>\n

C\u00e2u 2.<\/strong> C\u00e1c m\u1ec7nh \u0111\u1ec1 sau \u0111\u00fang hay sai?<\/p>\n

a\/ Trong kh\u00f4ng gian v\u1edbi h\u1ec7 tr\u1ee5c t\u1ecda \u0111\u1ed9 Oxyz , ph\u01b0\u01a1ng tr\u00ecnh m\u1eb7t ph\u1eb3ng (Oxy) l\u00e0: x + y = 0.\n<\/p>\n

b\/ Trong kh\u00f4ng gian v\u1edbi h\u1ec7 tr\u1ee5c t\u1ecda \u0111\u1ed9 Oxyz , m\u1eb7t ph\u1eb3ng (P) c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh: 3x – 2z + 2 = 0 \u0111i qua \u0111i\u1ec3m A(1;2;4) .<\/p>\n

c\/ Trong kh\u00f4ng gian v\u1edbi h\u1ec7 tr\u1ee5c t\u1ecda \u0111\u1ed9 Oxyz , \u0111\u01b0\u1eddng th\u1eb3ng \u0394 \u0111i qua \u0111i\u1ec3m A(2;-1;2) v\u00e0 nh\u1eadn vect\u01a1 ch\u1ec9 ph\u01b0\u01a1ng
\n u\u20d7 = (-1;2;-1) l\u00e0m vect\u01a1 ch\u1ec9 ph\u01b0\u01a1ng c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh ch\u00ednh t\u1eafc l\u00e0: (x + 2)\/(-1) = (y – 1)\/(2) = (z + 2)\/(-1).\n<\/p>\n

d\/ Trong kh\u00f4ng gian v\u1edbi h\u1ec7 tr\u1ee5c t\u1ecda \u0111\u1ed9 Oxyz , cho ph\u01b0\u01a1ng tr\u00ecnh m\u1eb7t c\u1ea7u (S) : x\u00b2 + y\u00b2 + z\u00b2 = 16. Khi \u0111\u00f3, t\u00e2m I v\u00e0 b\u00e1n k\u00ednh R c\u1ee7a m\u1eb7t c\u1ea7u (S) l\u1ea7n l\u01b0\u1ee3t l\u00e0 I(1;1;1) v\u00e0 R = 4 .<\/p>\n

Ph\u1ea7n III \u2013 C\u00e2u h\u1ecfi tr\u1ea3 l\u1eddi ng\u1eafn ( 4 c\u00e2u )<\/strong><\/p>\n

Tr\u1ea3 l\u1eddi theo d\u1ea1ng t\u1ef1 lu\u1eadn ng\u1eafn qua c\u00e1c c\u00e2u h\u1ecfi sau:<\/p>\n

C\u00e2u 1.<\/strong> Cho h\u00e0m s\u1ed1 f(x) \u0111\u01b0\u1ee3c x\u00e1c \u0111\u1ecbnh b\u1edfi:
\n\\[
\nf(x) =
\n\\begin{cases}
\n3x^2 & \\text{khi } 0 \\leq x \\leq 1 \\\\
\n4 \u2013 x & \\text{khi } 1 \\leq x \\leq 2
\n\\end{cases}
\n\\]T\u00ednh t\u00edch ph\u00e2n: int\u2080\u00b2 f(x) dx?\n<\/p>\n

C\u00e2u 2.<\/strong> M\u1ed9t bi\u1ec3n qu\u1ea3ng c\u00e1o d\u1ea1ng h\u00ecnh vu\u00f4ng ABCD c\u1ea1nh AB = 4m . Tr\u00ean t\u1ea5m bi\u1ec3n \u0111\u00f3 c\u00f3 c\u00e1c \u0111\u01b0\u1eddng tr\u00f2n t\u00e2m A v\u00e0 \u0111\u01b0\u1eddng tr\u00f2n t\u00e2m B c\u00f9ng b\u00e1n k\u00ednh R = 4m , hai \u0111\u01b0\u1eddng tr\u00f2n c\u1eaft nhau nh\u01b0 h\u00ecnh v\u1ebd. Chi ph\u00ed \u0111\u1ec3 s\u01a1n ph\u1ea7n g\u1ea1ch ch\u00e9o l\u00e0 150 000 \u0111\u1ed3ng\/m \u00b2 , chi ph\u00ed s\u01a1n ph\u1ea7n m\u00e0u \u0111en l\u00e0 100 000 \u0111\u1ed3ng\/m \u00b2 v\u00e0 chi ph\u00ed \u0111\u1ec3 s\u01a1n ph\u1ea7n c\u00f2n l\u1ea1i l\u00e0 250 000 \u0111\u1ed3ng\/m \u00b2 . Bi\u1ebft s\u1ed1 ti\u1ec1n \u0111\u1ec3 s\u01a1n bi\u1ec3n qu\u1ea3ng c\u00e1o theo c\u00e1ch tr\u00ean l\u00e0 a tri\u1ec7u \u0111\u1ed3ng. T\u00ecm gi\u00e1 tr\u1ecb c\u1ee7a a ? (K\u1ebft qu\u1ea3 l\u00e0m tr\u00f2n \u0111\u1ebfn h\u00e0ng ph\u1ea7n tr\u0103m).<\/p>\n

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

H\u1ecdc sinh c\u1ea7n v\u1eadn d\u1ee5ng c\u00e1c th\u1ee9c \u0111\u00e3 h\u1ecdc v\u00e0 k\u1ef9 n\u0103ng t\u01b0 duy \u0111\u1ec3 gi\u1ea3i nhanh 4 c\u00e2u h\u1ecfi trong ph\u1ea7n n\u00e0y nh\u00e9:<\/p>\n

C\u00e2u 3.<\/strong> C\u00f3 hai chi\u1ebfc h\u1ed9p, h\u1ed9p I c\u00f3 6 qu\u1ea3 b\u00f3ng m\u00e0u \u0111\u1ecf v\u00e0 4 qu\u1ea3 b\u00f3ng m\u00e0u v\u00e0ng, h\u1ed9p II c\u00f3 7 qu\u1ea3 b\u00f3ng m\u00e0u \u0111\u1ecf v\u00e0 3 qu\u1ea3 b\u00f3ng m\u00e0u v\u00e0ng, c\u00e1c qu\u1ea3 b\u00f3ng c\u00f3 c\u00f9ng k\u00edch th\u01b0\u1edbc v\u00e0 kh\u1ed1i l\u01b0\u1ee3ng. L\u1ea5y ng\u1eabu nhi\u00ean m\u1ed9t qu\u1ea3 b\u00f3ng t\u1eeb h\u1ed9p I b\u1ecf v\u00e0o h\u1ed9p II. Sau \u0111\u00f3, l\u1ea5y ra ng\u1eabu nhi\u00ean m\u1ed9t qu\u1ea3 b\u00f3ng t\u1eeb h\u1ed9p II. T\u00ednh x\u00e1c su\u1ea5t \u0111\u1ec3 qu\u1ea3 b\u00f3ng \u0111\u01b0\u1ee3c l\u1ea5y ra t\u1eeb h\u1ed9p II l\u00e0 qu\u1ea3 b\u00f3ng \u0111\u01b0\u1ee3c chuy\u1ec3n t\u1eeb h\u1ed9p I sang, bi\u1ebft r\u1eb1ng qu\u1ea3 b\u00f3ng \u0111\u00f3 c\u00f3 m\u00e0u \u0111\u1ecf. (L\u00e0m tr\u00f2n k\u1ebft qu\u1ea3 \u0111\u1ebfn h\u00e0ng ph\u1ea7n tr\u0103m).<\/p>\n

C\u00e2u 4.<\/strong> Trong kh\u00f4ng gian v\u1edbi h\u1ec7 tr\u1ee5c t\u1ecda \u0111\u1ed9 Oxyz , cho \u0111\u01b0\u1eddng th\u1eb3ng (d) c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh:
\n (x – 1)\/(1) = (y – 1)\/(-1) = z\/2 v\u00e0 \u0111\u01b0\u1eddng th\u1eb3ng (d\u2019) c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh: (x – 1)\/(5) = (y – 2)\/(1) = (z + 2)\/(-2). X\u00e9t v\u1ecb tr\u00ed t\u01b0\u01a1ng \u0111\u1ed1i c\u1ee7a hai \u0111\u01b0\u1eddng th\u1eb3ng \u0111\u00e3 cho.<\/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 H\u00e0 Long t\u1ec9nh Long An<\/strong><\/h2>\n

\u0110\u1ec1 minh h\u1ecda gi\u1eefa k\u00ec 2 To\u00e1n 12 n\u0103m 2024 \u2013 2025 tr\u01b0\u1eddng H\u00e0 Long t\u1ec9nh Long An<\/strong> cung c\u1ea5p \u0111a d\u1ea1ng b\u00e0i t\u1eadp t\u1eeb m\u1ee9c \u0111\u1ed9 nh\u1eadn bi\u1ebft \u0111\u1ebfn v\u1eadn d\u1ee5ng n\u00e2ng cao, gi\u00fap h\u1ecdc sinh r\u00e8n luy\u1ec7n t\u01b0 duy v\u00e0 l\u00e0m quen v\u1edbi c\u1ea5u tr\u00fac \u0111\u1ec1 thi chu\u1ea9n. \u0110\u00e2y l\u00e0 t\u00e0i li\u1ec7u quan tr\u1ecdng \u0111\u1ec3 \u00f4n t\u1eadp hi\u1ec7u qu\u1ea3 v\u00e0 \u0111\u1ea1t k\u1ebft qu\u1ea3 cao trong k\u1ef3 thi ch\u00ednh th\u1ee9c. H\u00e3y t\u1ea3i ngay t\u00e0i li\u1ec7u qua \u0111\u01b0\u1eddng link sau v\u00e0 b\u1eaft \u0111\u1ea7u \u00f4n luy\u1ec7n nh\u00e9!<\/p>\n