{"id":412,"date":"2026-05-05T04:44:25","date_gmt":"2026-05-05T04:44:25","guid":{"rendered":"https:\/\/79.buffdemo.com\/index.php\/2026\/05\/05\/de-minh-hoa-giua-ki-2-toan-11-nam-2024-2025-truong-ha-long-long-an\/"},"modified":"2026-05-05T07:17:10","modified_gmt":"2026-05-05T07:17:10","slug":"de-minh-hoa-giua-ki-2-toan-11-nam-2024-2025-truong-ha-long-long-an","status":"publish","type":"post","link":"https:\/\/79.buffdemo.com\/index.php\/2026\/05\/05\/de-minh-hoa-giua-ki-2-toan-11-nam-2024-2025-truong-ha-long-long-an\/","title":{"rendered":"\u0110\u1ec1 minh h\u1ecda gi\u1eefa k\u00ec 2 To\u00e1n 11 n\u0103m 2024 \u2013 2025 tr\u01b0\u1eddng H\u00e0 Long Long An"},"content":{"rendered":"

\u0110\u1ec1 minh h\u1ecda gi\u1eefa k\u00ec 2 To\u00e1n 11 n\u0103m 2024 \u2013 2025 tr\u01b0\u1eddng H\u00e0 Long Long An<\/strong> l\u00e0 t\u00e0i li\u1ec7u h\u1eefu \u00edch gi\u00fap h\u1ecdc sinh l\u1edbp 11 r\u00e8n luy\u1ec7n k\u1ef9 n\u0103ng l\u00e0m b\u00e0i, c\u1ee7ng c\u1ed1 ki\u1ebfn th\u1ee9c quan tr\u1ecdng \u0111\u1ec3 chu\u1ea9n b\u1ecb t\u1ed1t nh\u1ea5t cho k\u1ef3 ki\u1ec3m tra gi\u1eefa k\u1ef3 2. B\u1ed9 \u0111\u1ec1 \u0111\u01b0\u1ee3c bi\u00ean so\u1ea1n b\u1edfi \u0111\u1ed9i ng\u0169 gi\u00e1o vi\u00ean chuy\u00ean nghi\u1ec7p gi\u00fap h\u1ecdc sinh l\u00e0m quen v\u1edbi c\u1ea5u tr\u00fac \u0111\u1ec1 thi m\u00e0 c\u00f2n n\u00e2ng cao t\u01b0 duy gi\u1ea3i to\u00e1n v\u00e0 kh\u1ea3 n\u0103ng v\u1eadn d\u1ee5ng linh ho\u1ea1t c\u00e1c ph\u01b0\u01a1ng ph\u00e1p l\u00e0m b\u00e0i. C\u00f9ng T\u00e0i li\u1ec7u \u00d4n Thi<\/strong><\/a> kh\u00e1m ph\u00e1 chi ti\u1ebft v\u1ec1 b\u1ed9 \u0111\u1ec1 n\u00e0y qua b\u00e0i vi\u1ebft d\u01b0\u1edbi \u0111\u00e2y nh\u00e9!<\/p>\n

N\u1ed9i dung \u0110\u1ec1 minh h\u1ecda gi\u1eefa k\u00ec 2 To\u00e1n 11 n\u0103m 2024 \u2013 2025 tr\u01b0\u1eddng H\u00e0 Long Long An<\/span><\/strong><\/h2>\n

\u0110\u1ec1 minh h\u1ecda gi\u1eefa k\u00ec 2 To\u00e1n 11 n\u0103m 2024 \u2013 2025 tr\u01b0\u1eddng H\u00e0 Long Long An<\/strong> \u0111\u01b0\u1ee3c bi\u00ean so\u1ea1n nh\u1eb1m gi\u00fap h\u1ecdc sinh \u00f4n t\u1eadp v\u00e0 \u0111\u00e1nh gi\u00e1 ki\u1ebfn th\u1ee9c m\u1ed9t c\u00e1ch to\u00e0n di\u1ec7n. \u0110\u1ec1 g\u1ed3m b\u1ed1n ph\u1ea7n: 12 c\u00e2u tr\u1eafc nghi\u1ec7m nhi\u1ec1u ph\u01b0\u01a1ng \u00e1n l\u1ef1a ch\u1ecdn, 2 c\u00e2u tr\u1eafc nghi\u1ec7m \u0111\u00fang sai v\u1edbi 4 \u00fd nh\u1ecf, 4 c\u00e2u tr\u1eafc nghi\u1ec7m tr\u1ea3 l\u1eddi ng\u1eafn v\u00e0 4 c\u00e2u t\u1ef1 lu\u1eadn y\u00eau c\u1ea7u tr\u00ecnh b\u00e0y chi ti\u1ebft. Th\u1eddi gian l\u00e0m b\u00e0i 90 ph\u00fat gi\u00fap h\u1ecdc sinh r\u00e8n luy\u1ec7n t\u01b0 duy logic, ph\u1ea3n x\u1ea1 nhanh v\u00e0 l\u00e0m quen v\u1edbi c\u00e1c d\u1ea1ng b\u00e0i t\u1eadp kh\u00e1c nhau, t\u1ea1o n\u1ec1n t\u1ea3ng v\u1eefng ch\u1eafc cho c\u00e1c k\u1ef3 thi quan tr\u1ecdng.<\/p>\n

PH\u1ea6N 1: C\u00e2u tr\u1eafc nghi\u1ec7m nhi\u1ec1u ph\u01b0\u01a1ng \u00e1n l\u1ef1a ch\u1ecdn<\/strong><\/p>\n

C\u00e2u 1:<\/strong> Cho $a > 0, m, n \\in \\mathbb{R}$. Kh\u1eb3ng \u0111\u1ecbnh n\u00e0o sau \u0111\u00e2y \u0111\u00fang?<\/p>\n

A. $\\dfrac{a^m}{a^n} = a^{n \u2013 m}$. \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0B. $a^m + a^n = a^{m+n}$. \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0C. $a^m \\cdot a^n = a^{m-n}$. \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 D. $\\left(a^m\\right)^n = a^{m \\cdot n}$.<\/p>\n

C\u00e2u 2:<\/strong> V\u1edbi $a$ l\u00e0 s\u1ed1 th\u1ef1c d\u01b0\u01a1ng t\u00f9y \u00fd, $a^{\\frac{3}{5}}$ b\u1eb1ng?<\/p>\n

A. $\\sqrt[5]{a^3}$. \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0B. $a^2$. \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 C. $\\sqrt[3]{a^5}$. \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 D. $a^8$.<\/p>\n

PH\u1ea6N 2: C\u00e2u tr\u1eafc nghi\u1ec7m \u0111\u00fang sai<\/strong><\/p>\n

C\u00e2u 1:<\/strong><\/p>\n

a) Ph\u01b0\u01a1ng tr\u00ecnh logarit c\u01a1 b\u1ea3n $\\log_a x = b$ (v\u1edbi $0 < a \\neq 1$) c\u00f3 nghi\u1ec7m l\u00e0 $x = a^b$.
\nb) Ph\u01b0\u01a1ng tr\u00ecnh $2^x = 8$ c\u00f3 nghi\u1ec7m $x = 4$.
\nc) Ph\u01b0\u01a1ng tr\u00ecnh $\\log_3 x = 4$ v\u00f4 nghi\u1ec7m.
\nd) \u00d4ng A g\u1eedi 300 tri\u1ec7u \u0111\u1ed3ng v\u00e0o ng\u00e2n h\u00e0ng v\u1edbi l\u00e3i su\u1ea5t 7% tr\u00ean m\u1ed9t n\u0103m (l\u00e3i c\u1ed9ng d\u1ed3n v\u00e0o ti\u1ec1n g\u1ed1c). V\u1eady \u00edt nh\u1ea5t 9 n\u0103m, th\u00ec \u00f4ng A nh\u1eadn \u0111\u01b0\u1ee3c s\u1ed1 ti\u1ec1n nhi\u1ec1u h\u01a1n 600 tri\u1ec7u \u0111\u1ed3ng bao g\u1ed3m c\u1ea3 g\u1ed1c l\u1eabn l\u00e3i?<\/p>\n

C\u00e2u 2:<\/strong> Cho h\u00ecnh ch\u00f3p $S.ABCD$ c\u00f3 $ABCD$ l\u00e0 h\u00ecnh vu\u00f4ng t\u00e2m $O$ v\u00e0 $SA \\perp (ABCD)$. G\u1ecdi $H, I, K$ l\u1ea7n l\u01b0\u1ee3t l\u00e0 h\u00ecnh chi\u1ebfu vu\u00f4ng g\u00f3c c\u1ee7a \u0111i\u1ec3m $A$ tr\u00ean c\u00e1c c\u1ea1nh $SB, SC, SD$.<\/p>\n

\"\"<\/p>\n

a) \u0110\u01b0\u1eddng th\u1eb3ng $SA$ vu\u00f4ng g\u00f3c v\u1edbi \u0111\u01b0\u1eddng th\u1eb3ng $AB$.
\nb) $\\triangle SAD$ vu\u00f4ng t\u1ea1i $A$.
\nc) $AC \\perp (SBD)$ t\u1ea1i $O$.
\nd) $SC \\perp (AHK)$.<\/p>\n

PH\u1ea6N 3: C\u00e2u tr\u1eafc nghi\u1ec7m tr\u1ea3 l\u1eddi ng\u1eafn<\/strong><\/p>\n

C\u00e2u 1:<\/strong> T\u00ednh: $A = \\log_a \\frac{\\sqrt{a^3}}{a \\cdot \\sqrt[4]{a}}$<\/p>\n

C\u00e2u 2:<\/strong> Cho h\u00e0m s\u1ed1 y = log\u2082\u2080\u2082\u2085 ( -x\u00b2 + 6x – 5 ) . Bi\u1ebft h\u00e0m s\u1ed1 c\u00f3 t\u1eadp x\u00e1c \u0111\u1ecbnh l\u00e0 D = (a; b) . T\u00ednh: T = a – b .<\/p>\n

PH\u1ea6N 4: C\u00e2u t\u1ef1 lu\u1eadn<\/strong><\/p>\n

C\u00e2u 1:<\/strong> Th\u1ef1c hi\u1ec7n ph\u00e9p t\u00ednh: $A = \\frac{\\sqrt[3]{a^7} \\cdot a^{\\frac{11}{3}}}{a^4 \\cdot \\sqrt[4]{a^{-5}}}$<\/p>\n

C\u00e2u 2:<\/strong> N\u1ebfu D\u2080 l\u00e0 ch\u00eanh l\u1ec7ch nhi\u1ec7t \u0111\u1ed9 ban \u0111\u1ea7u gi\u1eefa m\u1ed9t v\u1eadt M v\u00e0 c\u00e1c v\u1eadt xung quanh, v\u00e0 n\u1ebfu c\u00e1c v\u1eadt xung quanh c\u00f3 nhi\u1ec7t \u0111\u1ed9 T_s , th\u00ec nhi\u1ec7t \u0111\u1ed9 c\u1ee7a v\u1eadt M t\u1ea1i th\u1eddi \u0111i\u1ec3m t \u0111\u01b0\u1ee3c m\u00f4 h\u00ecnh h\u00f3a b\u1edfi h\u00e0m s\u1ed1: $T(t) = T_s + D_0 \\cdot e^{-kt} \\quad (1)$ (trong \u0111\u00f3 k l\u00e0 h\u1eb1ng s\u1ed1 d\u01b0\u01a1ng ph\u1ee5 thu\u1ed9c v\u00e0o v\u1eadt M ). M\u1ed9t con g\u00e0 t\u00e2y n\u01b0\u1edbng \u0111\u01b0\u1ee3c l\u1ea5y t\u1eeb l\u00f2 n\u01b0\u1edbng khi nhi\u1ec7t \u0111\u1ed9 c\u1ee7a n\u00f3 \u0111\u00e3 \u0111\u1ea1t \u0111\u1ebfn 195^\u00b0 F v\u00e0 \u0111\u01b0\u1ee3c \u0111\u1eb7t tr\u00ean m\u1ed9t b\u00e0n trong m\u1ed9t c\u0103n ph\u00f2ng c\u00f3 nhi\u1ec7t \u0111\u1ed9 l\u00e0 65^\u00b0 F . N\u1ebfu nhi\u1ec7t \u0111\u1ed9 c\u1ee7a g\u00e0 t\u00e2y l\u00e0 150^\u00b0 F sau n\u1eeda gi\u1edd, nhi\u1ec7t \u0111\u1ed9 c\u1ee7a n\u00f3 sau 60 ph\u00fat l\u00e0 bao nhi\u00eau \u0111\u1ed9 F?<\/p>\n

T\u1ea3i \u0110\u1ec1 minh h\u1ecda gi\u1eefa k\u00ec 2 To\u00e1n 11 n\u0103m 2024 \u2013 2025 tr\u01b0\u1eddng H\u00e0 Long Long An<\/span><\/strong><\/h2>\n

\u0110\u1ec3 chu\u1ea9n b\u1ecb t\u1ed1t cho \u0111\u1ee3t kh\u1ea3o s\u00e1t ch\u1ea5t l\u01b0\u1ee3ng m\u00f4n To\u00e1n s\u1eafp t\u1edbi c\u0169ng nh\u01b0 n\u00e2ng cao ki\u1ebfn th\u1ee9c v\u1eefng ch\u1eafc cho k\u1ef3 thi cu\u1ed1i n\u0103m, c\u00e1c b\u1ea1n h\u1ecdc sinh v\u00e0 th\u1ea7y c\u00f4 h\u00e3y nhanh tay t\u1ea3i ngay \u0110\u1ec1 minh h\u1ecda gi\u1eefa k\u00ec 2 To\u00e1n 11 n\u0103m 2024 \u2013 2025 tr\u01b0\u1eddng H\u00e0 Long Long An<\/strong>. \u0110\u1ec1 ki\u1ec3m tra to\u00e1n l\u1edbp 11 gi\u1eefa h\u1ecdc k\u00ec 2 \u0111\u01b0\u1ee3c bi\u00ean so\u1ea1n s\u00e1t v\u1edbi ch\u01b0\u01a1ng tr\u00ecnh, gi\u00fap h\u1ecdc sinh r\u00e8n luy\u1ec7n k\u1ef9 n\u0103ng l\u00e0m b\u00e0i v\u00e0 c\u1ee7ng c\u1ed1 ki\u1ebfn th\u1ee9c m\u1ed9t c\u00e1ch hi\u1ec7u qu\u1ea3.<\/p>\n