{"id":410,"date":"2026-05-05T04:44:19","date_gmt":"2026-05-05T04:44:19","guid":{"rendered":"https:\/\/79.buffdemo.com\/index.php\/2026\/05\/05\/10-de-on-tap-kiem-tra-giua-hoc-ki-2-toan-11-ctm-theo-form-bgd-2025\/"},"modified":"2026-05-05T07:17:04","modified_gmt":"2026-05-05T07:17:04","slug":"10-de-on-tap-kiem-tra-giua-hoc-ki-2-toan-11-ctm-theo-form-bgd-2025","status":"publish","type":"post","link":"https:\/\/79.buffdemo.com\/index.php\/2026\/05\/05\/10-de-on-tap-kiem-tra-giua-hoc-ki-2-toan-11-ctm-theo-form-bgd-2025\/","title":{"rendered":"10 \u0111\u1ec1 \u00f4n t\u1eadp ki\u1ec3m tra gi\u1eefa h\u1ecdc k\u00ec 2 To\u00e1n 11 CTM theo Form BGD 2025"},"content":{"rendered":"

B\u1ed9 t\u00e0i li\u1ec7u 10 \u0111\u1ec1 \u00f4n t\u1eadp ki\u1ec3m tra gi\u1eefa h\u1ecdc k\u00ec 2 To\u00e1n 11 CTM theo Form BGD 2025<\/strong> g\u1ed3m 10 \u0111\u1ec1 \u00f4n t\u1eadp b\u00e1m s\u00e1t c\u1ea5u tr\u00fac chu\u1ea9n c\u1ee7a B\u1ed9 Gi\u00e1o d\u1ee5c & \u0110\u00e0o t\u1ea1o t\u1eeb c\u01a1 b\u1ea3n \u0111\u1ebfn n\u00e2ng cao. C\u00e1c \u0111\u1ec1 thi bao g\u1ed3m c\u00e1c chuy\u00ean \u0111\u1ec1 tr\u1ecdng t\u00e2m nh\u01b0 H\u00e0m s\u1ed1 logarit, Gi\u1ea3i b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh m\u0169, H\u00ecnh h\u1ecdc kh\u00f4ng gian v\u00e0 \u1ee8ng d\u1ee5ng th\u1ef1c t\u1ebf. H\u00e3y c\u00f9ng T\u00e0i Li\u1ec7u \u00d4n Thi<\/strong><\/a> kh\u00e1m ph\u00e1 ngay b\u1ed9 \u0111\u1ec1 n\u00e0y nh\u00e9!<\/p>\n

N\u1ed9i dung 10 \u0111\u1ec1 \u00f4n t\u1eadp ki\u1ec3m tra gi\u1eefa h\u1ecdc k\u00ec 2 To\u00e1n 11 CTM theo Form BGD 2025<\/span><\/strong><\/h2>\n

10 \u0111\u1ec1 \u00f4n t\u1eadp ki\u1ec3m tra gi\u1eefa h\u1ecdc k\u00ec 2 To\u00e1n 11 CTM theo Form BGD 2025 <\/strong>g\u1ed3m <\/span>136 trang <\/strong>v\u1edbi <\/span>10 \u0111\u1ec1 ki\u1ec3m tra \u0111a d\u1ea1ng. <\/strong>C\u00e1c \u0111\u1ec1 thi \u0111\u01b0\u1ee3c bi\u00ean so\u1ea1n k\u1ef9 l\u01b0\u1ee1ng, b\u00e1m s\u00e1t n\u1ed9i dung s\u00e1ch gi\u00e1o khoa v\u00e0 cung c\u1ea5p nhi\u1ec1u d\u1ea1ng c\u00e2u h\u1ecfi gi\u00fap h\u1ecdc sinh c\u1ee7ng c\u1ed1 ki\u1ebfn th\u1ee9c v\u00e0 ph\u00e1t tri\u1ec3n t\u01b0 duy to\u00e1n h\u1ecdc. B\u00ean d\u01b0\u1edbi m\u1ed7i \u0111\u1ec1 \u0111\u1ec1u \u0111i k\u00e8m \u0111\u00e1p \u00e1n chi ti\u1ebft \u0111\u1ec3 c\u00e1c b\u1ea1n h\u1ecdc sinh tham kh\u1ea3o.\u00a0<\/span>M\u1ed7i \u0111\u1ec1 thi bao g\u1ed3m 4 ph\u1ea7n ch\u00ednh (Tr\u1eafc nghi\u1ec7m nhi\u1ec1u \u0111\u00e1p \u00e1n, \u0110\u00fang\/sai, tr\u1ea3 l\u1eddi ng\u1eafn v\u00e0 t\u1ef1 lu\u1eadn) <\/strong>\u0111\u1ea3m b\u1ea3o \u0111\u00e1nh gi\u00e1 to\u00e0n di\u1ec7n ki\u1ebfn th\u1ee9c v\u00e0 kh\u1ea3 n\u0103ng v\u1eadn d\u1ee5ng to\u00e1n h\u1ecdc c\u1ee7a h\u1ecdc sinh:<\/p>\n

PH\u1ea6N I: C\u00c2U TR\u1eaeC NGHI\u1ec6M NHI\u1ec0U PH\u01af\u01a0NG \u00c1N L\u1ef0A CH\u1eccN \u2013 12 c\u00e2u<\/strong><\/p>\n

C\u00e2u 1:<\/strong> Cho $a > 0$ v\u00e0 $a \\neq 1$. T\u00ednh gi\u00e1 tr\u1ecb c\u1ee7a bi\u1ec3u th\u1ee9c $P = \\log_a \\left( a \\cdot \\sqrt[3]{a} \\right)$<\/p>\n

A. $\\dfrac{1}{3}$ \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0B. $3$ \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0C. $4$ \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0D. $\\dfrac{4}{3}$<\/p>\n

C\u00e2u 2:<\/strong> T\u1eadp nghi\u1ec7m $S$ c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh $\\log_3 (x \u2013 1) = 2$.<\/p>\n

A. $S = \\{10\\}$ \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0B. $S = \\emptyset$ \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0C. $S = \\{7\\}$ \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0D. $S = \\{6\\}$<\/p>\n

PH\u1ea6N II: C\u00c2U TR\u1eaeC NGHI\u1ec6M \u0110\u00daNG SAI \u2013 2 c\u00e2u, m\u1ed7i c\u00e2u c\u00f3 4 \u00fd\u00a0<\/strong><\/p>\n

C\u00e2u 1:<\/strong> Cho h\u00e0m s\u1ed1 $y = \\log_3(5x \u2013 3)$.<\/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 $\\left(\\dfrac{3}{5}; +\\infty\\right)$.
\nc) \u0110\u1ed3 th\u1ecb h\u00e0m s\u1ed1 \u0111i qua \u0111i\u1ec3m $M(6;3)$.
\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 $\\left[ \\dfrac{4}{5}, \\dfrac{12}{5} \\right]$ l\u00e0 $2$.<\/p>\n

C\u00e2u 2:<\/strong> Cho h\u00ecnh ch\u00f3p $S.ABCD$ c\u00f3 \u0111\u00e1y l\u00e0 h\u00ecnh ch\u1eef nh\u1eadt v\u00e0 $SA$ vu\u00f4ng g\u00f3c v\u1edbi m\u1eb7t ph\u1eb3ng \u0111\u00e1y. G\u1ecdi $H,K$ theo th\u1ee9 t\u1ef1 l\u00e0 h\u00ecnh chi\u1ebfu c\u1ee7a $A$ tr\u00ean c\u00e1c c\u1ea1nh $SB, SD$.<\/p>\n\n\n\n\n\n\n\n
<\/td>\nM\u1ec7nh \u0111\u1ec1<\/strong><\/td>\n\u0110\u00fang<\/strong><\/td>\nSai<\/strong><\/td>\n<\/tr>\n
a)<\/td>\nTam gi\u00e1c $SBC$ vu\u00f4ng<\/td>\n<\/td>\n<\/td>\n<\/tr>\n
b)<\/td>\nTam gi\u00e1c $SCD$ vu\u00f4ng<\/td>\n<\/td>\n<\/td>\n<\/tr>\n
c)<\/td>\n$SC \\perp (AHK)$<\/td>\n<\/td>\n<\/td>\n<\/tr>\n
d)<\/td>\n$HK \\perp SC$<\/td>\n<\/td>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

PH\u1ea6N III: C\u00c2U TR\u1eaeC NGHI\u1ec6M TR\u1ea2 L\u1edcI NG\u1eaeN \u2013 4 c\u00e2u<\/strong><\/p>\n

C\u00e2u 1:<\/strong> Ph\u01b0\u01a1ng tr\u00ecnh $\\log_{\\sqrt{3}}(x \u2013 2) + \\log_3 (x \u2013 4)^2 = 0$ c\u00f3 hai nghi\u1ec7m $x_1, x_2$. T\u00ednh gi\u00e1 tr\u1ecb c\u1ee7a bi\u1ec3u th\u1ee9c S = (x_1 \u2013 x_2)^2<\/p>\n

C\u00e2u 2:<\/strong> Kim t\u1ef1 th\u00e1p Kheops \u1edf Ai C\u1eadp c\u00f3 d\u1ea1ng l\u00e0 h\u00ecnh ch\u00f3p t\u1ee9 gi\u00e1c \u0111\u1ec1u c\u00f3 c\u1ea1nh \u0111\u00e1y d\u00e0i 262 m\u00e9t, c\u1ea1nh b\u00ean d\u00e0i 230 m\u00e9t. H\u00e3y t\u00ednh g\u00f3c t\u1ea1o b\u1edfi m\u1eb7t b\u00ean v\u00e0 m\u1eb7t \u0111\u00e1y c\u1ee7a kim t\u1ef1 th\u00e1p. (\u0111\u01a1n v\u1ecb \u0111\u1ed9, k\u1ebft qu\u1ea3 l\u00e0m tr\u00f2n \u0111\u1ebfn h\u00e0ng \u0111\u01a1n v\u1ecb)<\/em><\/p>\n

PH\u1ea6N IV: T\u1ef0 LU\u1eacN \u2013 4 c\u00e2u<\/strong><\/p>\n

C\u00e2u 1:<\/strong> Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh: $\\log_2 \\left( x^2 + 3x \\right) = 2$.<\/p>\n

C\u00e2u 2:<\/strong> Cho h\u00ecnh ch\u00f3p $S.ABCD$ c\u00f3 $SA \\perp (ABCD)$ v\u00e0 \u0111\u00e1y $ABCD$ l\u00e0 h\u00ecnh ch\u1eef nh\u1eadt. G\u1ecdi $H,K$ l\u1ea7n l\u01b0\u1ee3t l\u00e0 h\u00ecnh chi\u1ebfu c\u1ee7a \u0111i\u1ec3m $A$ tr\u00ean c\u00e1c c\u1ea1nh $SB,SD$. Ch\u1ee9ng minh r\u1eb1ng $(AHK) \\perp (SAC)$.<\/p>\n

T\u1ea3i 10 \u0111\u1ec1 \u00f4n t\u1eadp ki\u1ec3m tra gi\u1eefa h\u1ecdc k\u00ec 2 To\u00e1n 11 CTM theo Form BGD 2025<\/span><\/strong><\/h2>\n

B\u1ed9 t\u00e0i li\u1ec7u 10 \u0111\u1ec1 \u00f4n t\u1eadp ki\u1ec3m tra gi\u1eefa h\u1ecdc k\u00ec 2 To\u00e1n 11 CTM theo Form BGD 2025 <\/strong>b\u00e1m s\u00e1t c\u1ea5u tr\u00fac \u0111\u1ec1 thi m\u1edbi nh\u1ea5t c\u1ee7a B\u1ed9 Gi\u00e1o d\u1ee5c v\u00e0 \u0110\u00e0o t\u1ea1o bao g\u1ed3m \u0111\u1ea7y \u0111\u1ee7 c\u00e1c d\u1ea1ng c\u00e2u h\u1ecfi t\u1eeb c\u01a1 b\u1ea3n \u0111\u1ebfn n\u00e2ng cao. \u0110\u1eb7c bi\u1ec7t, m\u1ed7i \u0111\u1ec1 thi \u0111\u1ec1u c\u00f3 h\u01b0\u1edbng d\u1eabn gi\u1ea3i chi ti\u1ebft, gi\u00fap h\u1ecdc sinh hi\u1ec3u s\u00e2u h\u01a1n v\u1ec1 ph\u01b0\u01a1ng ph\u00e1p l\u00e0m b\u00e0i v\u00e0 c\u1ea3i thi\u1ec7n k\u1ebft qu\u1ea3 h\u1ecdc t\u1eadp. H\u00e3y t\u1ea3i ngay b\u1ed9 \u0111\u1ec1 thi to\u00e1n gi\u1eefa h\u1ecdc k\u00ec 2 theo \u0111\u01b0\u1eddng link d\u01b0\u1edbi \u0111\u00e2y \u0111\u1ec3 luy\u1ec7n t\u1eadp m\u1ed7i ng\u00e0y v\u00e0 chinh ph\u1ee5c \u0111i\u1ec3m cao trong k\u1ef3 thi s\u1eafp t\u1edbi!<\/p>\n