{"id":414,"date":"2026-05-05T04:44:30","date_gmt":"2026-05-05T04:44:30","guid":{"rendered":"https:\/\/79.buffdemo.com\/index.php\/2026\/05\/05\/de-kiem-tra-giua-ky-2-toan-11-nam-2024-2025-truong-thpt-nguyen-tat-thanh-tp-hcm\/"},"modified":"2026-05-05T07:17:16","modified_gmt":"2026-05-05T07:17:16","slug":"de-kiem-tra-giua-ky-2-toan-11-nam-2024-2025-truong-thpt-nguyen-tat-thanh-tp-hcm","status":"publish","type":"post","link":"https:\/\/79.buffdemo.com\/index.php\/2026\/05\/05\/de-kiem-tra-giua-ky-2-toan-11-nam-2024-2025-truong-thpt-nguyen-tat-thanh-tp-hcm\/","title":{"rendered":"\u0110\u1ec1 ki\u1ec3m tra gi\u1eefa k\u1ef3 2 To\u00e1n 11 n\u0103m 2024 \u2013 2025 tr\u01b0\u1eddng THPT Nguy\u1ec5n T\u1ea5t Th\u00e0nh TP HCM"},"content":{"rendered":"

\u0110\u1ec1 ki\u1ec3m tra gi\u1eefa k\u1ef3 2 To\u00e1n 11 n\u0103m 2024 \u2013 2025 tr\u01b0\u1eddng THPT Nguy\u1ec5n T\u1ea5t Th\u00e0nh TP HCM<\/strong> l\u00e0 t\u00e0i li\u1ec7u t\u1ed5ng h\u1ee3p \u0111\u1ea7y \u0111\u1ee7 c\u00e1c d\u1ea1ng b\u00e0i t\u1eadp quan tr\u1ecdng, bao g\u1ed3m c\u1ea5p s\u1ed1 c\u1ed9ng \u2013 c\u1ea5p s\u1ed1 nh\u00e2n, ph\u01b0\u01a1ng tr\u00ecnh l\u01b0\u1ee3ng gi\u00e1c, t\u1ed5 h\u1ee3p \u2013 x\u00e1c su\u1ea5t, v\u00e0 h\u00ecnh h\u1ecdc kh\u00f4ng gian. N\u1ed9i dung b\u00e1m s\u00e1t ch\u01b0\u01a1ng tr\u00ecnh gi\u00fap h\u1ecdc sinh l\u1edbp 11 \u00f4n t\u1eadp hi\u1ec7u qu\u1ea3, n\u1eafm v\u1eefng ki\u1ebfn th\u1ee9c n\u1ec1n t\u1ea3ng \u0111\u1ec3 \u0111\u1ea1t \u0111i\u1ec3m cao trong k\u1ef3 thi. C\u00f9ng T\u00e0i Li\u1ec7u \u00d4n Thi<\/a><\/strong> kh\u00e1m ph\u00e1 chi ti\u1ebft \u0111\u1ec1 ki\u1ec3m tra n\u00e0y qua b\u00e0i vi\u1ebft d\u01b0\u1edbi \u0111\u00e2y!<\/p>\n

N\u1ed9i dung \u0110\u1ec1 ki\u1ec3m tra gi\u1eefa k\u1ef3 2 To\u00e1n 11 n\u0103m 2024 \u2013 2025 tr\u01b0\u1eddng THPT Nguy\u1ec5n T\u1ea5t Th\u00e0nh TP HCM<\/span><\/strong><\/h2>\n

\u0110\u1ec1 ki\u1ec3m tra gi\u1eefa h\u1ecdc k\u00ec 2 To\u00e1n 11 n\u0103m 2024 \u2013 2025 tr\u01b0\u1eddng THPT H\u00f9ng V\u01b0\u01a1ng TP HCM<\/strong> g\u1ed3m 2 m\u00e3 \u0111\u1ec1 135 v\u00e0 246 (m\u1ed7i \u0111\u1ec1 2 trang). T\u00e0i li\u1ec7u \u0111\u1ec1 to\u00e1n gi\u1eefa k\u00ec 2 l\u1edbp 11 gi\u00fap c\u00e1c b\u1ea1n h\u1ecdc sinh d\u1ec5 d\u00e0ng \u00f4n t\u1eadp 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 c\u00e1c b\u1ea1n chu\u1ea9n b\u1ecb t\u1ed1t nh\u1ea5t cho k\u1ef3 thi gi\u1eefa k\u1ef3 2, n\u1eafm v\u1eefng ki\u1ebfn th\u1ee9c v\u00e0 r\u00e8n luy\u1ec7n k\u1ef9 n\u0103ng l\u00e0m b\u00e0i.<\/p>\n

Ph\u1ea7n 1 (3 \u0111i\u1ec3m): Tr\u1eafc nghi\u1ec7m nhi\u1ec1u l\u1ef1a ch\u1ecdn<\/strong><\/p>\n

C\u00e2u 1.<\/strong> H\u00e0m s\u1ed1 n\u00e0o sau \u0111\u00e2y c\u00f3 t\u1eadp x\u00e1c \u0111\u1ecbnh l\u00e0 $\\mathbb{R}$?<\/p>\n

A. $y = \\log_2{x}$ \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0B. $y = \\log_3{(x+2)}$ \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 C. $y = \\log_3{(1 \u2013 x^2)}$\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 D. $y = \\log_2{(x^2+1)}$<\/p>\n

C\u00e2u 2.<\/strong> Cho $x$ v\u00e0 $y$ l\u00e0 s\u1ed1 th\u1ef1c d\u01b0\u01a1ng. Kh\u1eb3ng \u0111\u1ecbnh n\u00e0o d\u01b0\u1edbi \u0111\u00e2y \u0111\u00fang?<\/p>\n

A. $3^{\\log{x} + \\log{y}} = 3^{\\log{x}} + 3^{\\log{y}}$ \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 B. $3^{\\log{(x+y)}} = 3^{\\log{x}} \\cdot 3^{\\log{y}}$<\/p>\n

C. $3^{\\log{(xy)}} = 3^{\\log{x}} \\cdot 3^{\\log{y}}$\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 D. $3^{\\log{x} \\cdot \\log{y}} = 3^{\\log{x}} + 3^{\\log{y}}$<\/p>\n

Ph\u1ea7n 2 (2 \u0111i\u1ec3m): C\u00e2u tr\u1eafc nghi\u1ec7m \u0111\u00fang sai<\/strong><\/p>\n

C\u00e2u 1.<\/strong> Cho h\u00e0m s\u1ed1 $y = 5^x$. C\u00e1c m\u1ec7nh \u0111\u1ec1 sau \u0111\u00fang hay sai?<\/p>\n

a)] T\u1eadp x\u00e1c \u0111\u1ecbnh c\u1ee7a h\u00e0m s\u1ed1 l\u00e0 kho\u1ea3ng $(0; +\\infty)$.
\nb) T\u1eadp gi\u00e1 tr\u1ecb c\u1ee7a h\u00e0m s\u1ed1 l\u00e0 $\\mathbb{R}$.
\nc) H\u00e0m s\u1ed1 \u0111\u00e3 cho \u0111\u1ed3ng bi\u1ebfn tr\u00ean kho\u1ea3ng $(0;1)$.
\nd) \u0110\u1ed3 th\u1ecb h\u00e0m s\u1ed1 $y = 5^{-x}$ c\u1eaft \u0111\u01b0\u1eddng th\u1eb3ng $y = -x + 2$ t\u1ea1i \u0111i\u1ec3m c\u00f3 ho\u00e0nh \u0111\u1ed9 d\u01b0\u01a1ng.<\/p>\n

C\u00e2u 2.<\/strong> Cho h\u00e0m s\u1ed1 $y = f(x) = \\log_3(2x + 3)$. C\u00e1c m\u1ec7nh \u0111\u1ec1 sau \u0111\u00fang hay sai?<\/p>\n

a) T\u1eadp x\u00e1c \u0111\u1ecbnh c\u1ee7a h\u00e0m s\u1ed1: $D = \\left[ -\\frac{3}{2}, +\\infty \\right)$.
\nb) Nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh $f(x) = 1$ l\u00e0 $x = 0$.
\nc) T\u1eadp nghi\u1ec7m c\u1ee7a b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh $f(x) < 2$ c\u00f3 \u0111\u00fang 3 s\u1ed1 nguy\u00ean.
\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 $[0;3]$ l\u00e0 3.<\/p>\n

Ph\u1ea7n 3 (2 \u0111i\u1ec3m): Tr\u1eafc nghi\u1ec7m tr\u1ea3 l\u1eddi ng\u1eafn<\/strong><\/p>\n

C\u00e2u 1.<\/strong> C\u00f4ng th\u1ee9c $M = M_0 \\left( \\frac{1}{2} \\right)^{\\frac{t}{T}}$ cho bi\u1ebft kh\u1ed1i l\u01b0\u1ee3ng c\u1ee7a m\u1ed9t ch\u1ea5t ph\u00f3ng x\u1ea1 sau th\u1eddi gian $t$ k\u1ec3 t\u1eeb th\u1eddi \u0111i\u1ec3m n\u00e0o \u0111\u00f3 (g\u1ecdi l\u00e0 th\u1eddi \u0111i\u1ec3m ban \u0111\u1ea7u), $M_0$ l\u00e0 kh\u1ed1i l\u01b0\u1ee3ng ban \u0111\u1ea7u, $T$ l\u00e0 chu k\u00ec b\u00e1n r\u00e3 c\u1ee7a ch\u1ea5t ph\u00f3ng x\u1ea1 \u0111\u00f3 (c\u1ee9 sau m\u1ed7i chu k\u00ec, kh\u1ed1i l\u01b0\u1ee3ng ch\u1ea5t ph\u00f3ng x\u1ea1 gi\u1ea3m \u0111i m\u1ed9t n\u1eeda). Trong m\u1ed9t ph\u00f2ng th\u00ed nghi\u1ec7m, v\u1edbi kh\u1ed1i l\u01b0\u1ee3ng $200$ g radon ban \u0111\u1ea7u, sau $16$ ng\u00e0y ch\u1ec9 c\u00f2n l\u1ea1i $11$ g. Chu k\u00ec b\u00e1n r\u00e3 c\u1ee7a radon b\u1eb1ng bao nhi\u00eau ng\u00e0y? (l\u00e0m tr\u00f2n k\u1ebft qu\u1ea3 \u0111\u1ebfn h\u00e0ng ph\u1ea7n ch\u1ee5c).<\/p>\n

C\u00e2u 2.<\/strong> Cho h\u00ecnh ch\u00f3p $S.ABCD$ c\u00f3 \u0111\u00e1y l\u00e0 h\u00ecnh vu\u00f4ng c\u1ea1nh $a$, $SA \\perp (ABCD)$, $SA = a\\sqrt{2}$. G\u1ecdi $H$ l\u00e0 h\u00ecnh chi\u1ebfu vu\u00f4ng g\u00f3c c\u1ee7a $A$ l\u00ean c\u1ea1nh $SB$. T\u00ednh $\\tan$ g\u00f3c gi\u1eefa hai \u0111\u01b0\u1eddng th\u1eb3ng $DH$ v\u00e0 $BC$? (l\u00e0m tr\u00f2n k\u1ebft qu\u1ea3 qua \u0111\u1ebfn h\u00e0ng ph\u1ea7n tr\u0103m).<\/p>\n

Ph\u1ea7n 4 (3 \u0111i\u1ec3m): T\u1ef1 lu\u1eadn<\/strong><\/p>\n

C\u00e2u 1. (1 \u0111i\u1ec3m)<\/strong> Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh: $2\\log_3 (x+1) = 1 \u2013 \\log_{\\frac{1}{3}} (x+7)$<\/p>\n

C\u00e2u 2. (2 \u0111i\u1ec3m)<\/strong> Cho h\u00ecnh ch\u00f3p $S.ABC$ c\u00f3 \u0111\u00e1y $ABC$ l\u00e0 tam gi\u00e1c vu\u00f4ng t\u1ea1i $B$, $SA = AB$, $SA \\perp (ABC)$. G\u1ecdi $H$ l\u00e0 h\u00ecnh chi\u1ebfu vu\u00f4ng g\u00f3c c\u1ee7a $A$ l\u00ean $SB$, $K$ l\u00e0 trung \u0111i\u1ec3m c\u1ea1nh $AC$.<\/p>\n

a) Ch\u1ee9ng minh $\\triangle SBC$ vu\u00f4ng.<\/p>\n

b) Ch\u1ee9ng minh $AB \\perp HK$.<\/p>\n

T\u1ea3i \u0110\u1ec1 ki\u1ec3m tra gi\u1eefa k\u1ef3 2 To\u00e1n 11 n\u0103m 2024 \u2013 2025 tr\u01b0\u1eddng THPT Nguy\u1ec5n T\u1ea5t Th\u00e0nh TP HCM<\/span><\/strong><\/h2>\n

\u0110\u1ec1 ki\u1ec3m tra gi\u1eefa h\u1ecdc k\u00ec 2 To\u00e1n 11 n\u0103m 2024 \u2013 2025 tr\u01b0\u1eddng THPT H\u00f9ng V\u01b0\u01a1ng TP HCM <\/strong>b\u00e1m s\u00e1t ch\u01b0\u01a1ng tr\u00ecnh To\u00e1n 11, gi\u00fap b\u1ea1n l\u00e0m quen v\u1edbi c\u1ea5u tr\u00fac \u0111\u1ec1, n\u00e2ng cao k\u1ef9 n\u0103ng gi\u1ea3i b\u00e0i t\u1eadp v\u00e0 c\u1ea3i thi\u1ec7n \u0111i\u1ec3m s\u1ed1 m\u1ed9t c\u00e1ch hi\u1ec7u qu\u1ea3. H\u00e3y t\u1ea3i ngay b\u1ed9 \u0111\u1ec1 thi th\u1eed c\u1ef1c k\u1ef3 h\u1eefu \u00edch n\u00e0y v\u00e0 r\u00e8n luy\u1ec7n m\u1ed7i ng\u00e0y \u0111\u1ec3 chu\u1ea9n b\u1ecb th\u1eadt t\u1ed1t cho k\u1ef3 ki\u1ec3m tra gi\u1eefa k\u1ef3 2 s\u1eafp t\u1edbi.<\/p>\n