{"id":428,"date":"2026-05-05T04:45:14","date_gmt":"2026-05-05T04:45:14","guid":{"rendered":"https:\/\/79.buffdemo.com\/index.php\/2026\/05\/05\/05-de-thi-thu-giua-hoc-ki-2-mon-toan-11-nam-hoc-2024-2025-co-dap-an\/"},"modified":"2026-05-05T07:18:00","modified_gmt":"2026-05-05T07:18:00","slug":"05-de-thi-thu-giua-hoc-ki-2-mon-toan-11-nam-hoc-2024-2025-co-dap-an","status":"publish","type":"post","link":"https:\/\/79.buffdemo.com\/index.php\/2026\/05\/05\/05-de-thi-thu-giua-hoc-ki-2-mon-toan-11-nam-hoc-2024-2025-co-dap-an\/","title":{"rendered":"05 \u0111\u1ec1 thi th\u1eed gi\u1eefa h\u1ecdc k\u00ec 2 m\u00f4n To\u00e1n 11 n\u0103m h\u1ecdc 2024 \u2013 2025 c\u00f3 \u0111\u00e1p \u00e1n"},"content":{"rendered":"

05 \u0111\u1ec1 thi th\u1eed gi\u1eefa h\u1ecdc k\u00ec 2 m\u00f4n To\u00e1n 11 n\u0103m h\u1ecdc 2024 \u2013 2025 c\u00f3 \u0111\u00e1p \u00e1n<\/strong> l\u00e0 t\u00e0i li\u1ec7u \u00f4n luy\u1ec7n ch\u1ea5t l\u01b0\u1ee3ng d\u00e0nh cho h\u1ecdc sinh l\u1edbp 11 trong giai \u0111o\u1ea1n n\u01b0\u1edbc r\u00fat tr\u01b0\u1edbc k\u1ef3 ki\u1ec3m tra quan tr\u1ecdng. B\u1ed9 \u0111\u1ec1 g\u1ed3m 05 \u0111\u1ec1 thi th\u1eed \u0111\u01b0\u1ee3c bi\u00ean so\u1ea1n k\u1ef9 l\u01b0\u1ee1ng, b\u00e1m s\u00e1t ch\u01b0\u01a1ng tr\u00ecnh s\u00e1ch gi\u00e1o khoa v\u00e0 c\u1ea5u tr\u00fac \u0111\u1ec1 thi gi\u1eefa k\u1ef3 m\u1edbi nh\u1ea5t. \u0110\u1eb7c bi\u1ec7t, m\u1ed7i \u0111\u1ec1 \u0111\u1ec1u c\u00f3 \u0111\u00e1p \u00e1n r\u00f5 r\u00e0ng, h\u1ed7 tr\u1ee3 h\u1ecdc sinh t\u1ef1 \u0111\u1ed1i chi\u1ebfu v\u00e0 c\u1ea3i thi\u1ec7n k\u1ef9 n\u0103ng gi\u1ea3i to\u00e1n. C\u00f9ng T\u00e0i Li\u1ec7u \u00d4n Thi<\/strong><\/a> kh\u00e1m ph\u00e1 xem b\u1ed9 \u0111\u1ec1 n\u00e0y c\u00f3 g\u00ec h\u1ea5p d\u1eabn v\u00e0 h\u1eefu \u00edch cho h\u00e0nh tr\u00ecnh chinh ph\u1ee5c \u0111i\u1ec3m cao c\u1ee7a b\u1ea1n nh\u00e9!<\/p>\n

N\u1ed9i dung 05 \u0111\u1ec1 thi th\u1eed gi\u1eefa h\u1ecdc k\u00ec 2 m\u00f4n To\u00e1n 11 n\u0103m h\u1ecdc 2024 \u2013 2025 c\u00f3 \u0111\u00e1p \u00e1n<\/span><\/strong><\/h2>\n

05 \u0111\u1ec1 thi th\u1eed gi\u1eefa h\u1ecdc k\u00ec 2 m\u00f4n To\u00e1n 11 n\u0103m h\u1ecdc 2024 \u2013 2025 c\u00f3 \u0111\u00e1p \u00e1n<\/strong> g\u1ed3m 21 trang, bao g\u1ed3m 5 \u0111\u1ec1 thi gi\u1eefa k\u00ec 2 to\u00e1n 11 v\u1edbi c\u1ea5u tr\u00fac 4 ph\u1ea7n nh\u01b0 sau:<\/p>\n

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

C\u00e2u 1:<\/strong> Cho c\u00e1c s\u1ed1 th\u1ef1c $a, b, m, n$ ($a, b > 0$). Kh\u1eb3ng \u0111\u1ecbnh n\u00e0o sau \u0111\u00e2y l\u00e0 \u0111\u00fang?<\/p>\n

A. $a^m \\cdot a^n = a^{m+n}$.\u00a0 \u00a0 \u00a0B. $(a^m)^n = a^{m+n}$.\u00a0 \u00a0 \u00a0C. $(a+b)^m = a^m + b^m$.\u00a0 \u00a0 \u00a0D. $\\frac{a^m}{a^n} = \\sqrt[n]{a^m}$.<\/p>\n

C\u00e2u 2:<\/strong> Cho $a$ l\u00e0 s\u1ed1 th\u1ef1c d\u01b0\u01a1ng. Gi\u00e1 tr\u1ecb c\u1ee7a bi\u1ec3u th\u1ee9c $P = a^{\\frac{2}{3}} \\cdot \\sqrt{a}$ b\u1eb1ng<\/p>\n

A. $a^{\\frac{2}{3}}$. \u00a0 \u00a0 B. $a^{\\frac{7}{6}}$. \u00a0 \u00a0 C. $a^5$. \u00a0 \u00a0 D. $a^{\\frac{5}{6}}$.<\/p>\n

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

C\u00e2u 1:<\/strong> Cho h\u00e0m s\u1ed1 $y = \\log_3 (5x \u2013 3)$. X\u00e9t t\u00ednh \u0111\u00fang sai c\u1ee7a c\u00e1c kh\u1eb3ng \u0111\u1ecbnh sau:<\/p>\n

a) T\u1eadp x\u00e1c \u0111\u1ecbnh c\u1ee7a h\u00e0m s\u1ed1 l\u00e0 $D = (0; +\\infty)$.
\nb) H\u00e0m s\u1ed1 \u0111\u1ed3ng bi\u1ebfn tr\u00ean $\\left(\\frac{3}{5}; +\\infty \\right)$.
\nc) \u0110\u1ed3 th\u1ecb h\u00e0m s\u1ed1 \u0111i qua \u0111i\u1ec3m $M(2;7)$.
\nd) T\u1ed5ng gi\u00e1 tr\u1ecb l\u1edbn nh\u1ea5t v\u00e0 gi\u00e1 tr\u1ecb nh\u1ecf nh\u1ea5t c\u1ee7a h\u00e0m s\u1ed1 $f(x)$ tr\u00ean $\\left[ \\frac{4}{5}; \\frac{12}{5} \\right]$ l\u00e0 $2$.<\/p>\n

C\u00e2u 2:<\/strong> Cho h\u00ecnh ch\u00f3p $S.ABCD$ c\u00f3 \u0111\u00e1y $ABCD$ l\u00e0 h\u00ecnh vu\u00f4ng c\u1ea1nh $a$ c\u00f3 $SA = a\\sqrt{6}$ v\u00e0 $SA \\perp (ABCD)$. X\u00e9t t\u00ednh \u0111\u00fang sai c\u1ee7a c\u00e1c kh\u1eb3ng \u0111\u1ecbnh sau:<\/p>\n

a) $AB \\perp (SAD)$.
\nb) G\u00f3c gi\u1eefa $SC$ v\u00e0 $(ABCD)$ b\u1eb1ng $45^\\circ$.
\nc) Sin c\u1ee7a g\u00f3c gi\u1eefa $SB$ v\u00e0 $(SAC)$ b\u1eb1ng $\\frac{\\sqrt{14}}{14}$.
\nd) Sin c\u1ee7a g\u00f3c gi\u1eefa $AC$ v\u00e0 $(SBC)$ b\u1eb1ng $\\frac{\\sqrt{7}}{7}$.<\/p>\n

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

C\u00e2u 1:<\/strong> Sau m\u1ed9t th\u1eddi gian l\u00e0m vi\u1ec7c, ch\u1ecb H\u1ea1nh c\u00f3 s\u1ed1 v\u1ed1n l\u00e0 450 tri\u1ec7u \u0111\u1ed3ng. Ch\u1ecb H\u1ea1nh chia s\u1ed1 ti\u1ec1n th\u00e0nh hai ph\u1ea7n v\u00e0 g\u1eedi \u1edf hai ng\u00e2n h\u00e0ng Agribank v\u00e0 Sacombank theo ph\u01b0\u01a1ng th\u1ee9c l\u00e3i k\u00e9p. S\u1ed1 ti\u1ec1n \u1edf ph\u1ea7n th\u1ee9 nh\u1ea5t ch\u1ecb H\u1ea1nh g\u1eedi \u1edf ng\u00e2n h\u00e0ng Agribank v\u1edbi l\u00e3i su\u1ea5t 2,1% m\u1ed9t qu\u00fd trong th\u1eddi gian 18 th\u00e1ng. S\u1ed1 ti\u1ec1n \u1edf ph\u1ea7n th\u1ee9 hai ch\u1ecb H\u1ea1nh g\u1eedi \u1edf ng\u00e2n h\u00e0ng Sacombank v\u1edbi l\u00e3i su\u1ea5t 0,73% m\u1ed9t th\u00e1ng trong th\u1eddi gian 10 th\u00e1ng. T\u1ed5ng s\u1ed1 ti\u1ec1n l\u00e3i thu \u0111\u01b0\u1ee3c \u1edf hai ng\u00e2n h\u00e0ng l\u00e0 500,010592 tri\u1ec7u \u0111\u1ed3ng. H\u1ecfi ch\u1ecb H\u1ea1nh \u0111\u00e3 g\u1eedi \u1edf ng\u00e2n h\u00e0ng Agribank l\u00e0 bao nhi\u00eau tri\u1ec7u \u0111\u1ed3ng?<\/p>\n

C\u00e2u 2:<\/strong> Cho $f(1) = 1$; $f(m+n) = f(m) + f(n) + mn$ v\u1edbi m\u1ecdi $m,n \\in \\mathbb{N}^*$. T\u00ednh gi\u00e1 tr\u1ecb c\u1ee7a bi\u1ec3u th\u1ee9c $T = \\log \\left[ \\frac{f(2019) \u2013 f(2009) \u2013 145}{2} \\right]$.<\/p>\n

PH\u1ea6N IV: T\u1ef1 lu\u1eadn<\/strong><\/p>\n

C\u00e2u 1:<\/strong> T\u00ecm $m$ \u0111\u1ec3 h\u00e0m s\u1ed1 $y = \\ln \\left( x^2 \u2013 2mx + 4 \\right)$ c\u00f3 t\u1eadp x\u00e1c \u0111\u1ecbnh l\u00e0 $\\mathbb{R}$.<\/p>\n

C\u00e2u 2:<\/strong> Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh: $\\log_{\\sqrt{3}} (x \u2013 2) + \\log_3 (x \u2013 4)^2 = 0.$<\/p>\n

T\u1ea3i 05 \u0111\u1ec1 thi th\u1eed gi\u1eefa h\u1ecdc k\u00ec 2 m\u00f4n To\u00e1n 11 n\u0103m h\u1ecdc 2024 \u2013 2025 c\u00f3 \u0111\u00e1p \u00e1n<\/span><\/strong><\/h2>\n

\u0110\u1ec3 d\u1ec5 d\u00e0ng \u00f4n t\u1eadp, b\u1ea1n c\u00f3 th\u1ec3 t\u1ea3i ngay 05 \u0111\u1ec1 thi th\u1eed gi\u1eefa h\u1ecdc k\u00ec 2 m\u00f4n To\u00e1n 11 n\u0103m h\u1ecdc 2024 \u2013 2025 c\u00f3 \u0111\u00e1p \u00e1n<\/strong> theo link d\u01b0\u1edbi \u0111\u00e2y v\u00e0 b\u1eaft \u0111\u1ea7u luy\u1ec7n t\u1eadp \u0111\u1ec3 \u0111\u1ea1t k\u1ebft qu\u1ea3 cao nh\u1ea5t nh\u00e9.<\/p>\n