Đề thi Học kì 2 Toán 11 năm 2025-2026 trường THPT số 1 Ngô Gia Tự (Đắk Lắk)

log₂(ab)=log₂ a.log₂ b

log₂(ab)=log₂ a + log₂ b

$log_2(\dfrac{a}{b})=log_2 a.log_2 b$

log₂(a + b)=log₂ a + log₂ b

$f'(6)=\lim\limits_{x\to 6} \dfrac{f(x) - f(6)}{x - 6}$

$f'(6)=\lim\limits_{x\to 6} \dfrac{f(x) + f(6)}{x - 6}$

$f'(6)=\lim\limits_{x\to 6} \dfrac{f(x) + f(6)}{x + 6}$

$f'(6)=\lim\limits_{x\to 6} \dfrac{f(x) - f(6)}{x + 6}$

$V=Sh$

$V=\dfrac{1}{3}Sh$

$V=3Sh$

$V=\dfrac{4}{3}Sh$

$SA \perp (ABC)$

$AC \perp (SBC)$

$SO \perp (ABCD)$

$AC \perp (SCD)$

$y' = \dfrac{1}{\cos^2 x} - 2$

$y' = \dfrac{1}{\sin^2 x} - 2$

$y' = \dfrac{-1}{\cos^2 x} - 2x$

$y' = \dfrac{-1}{\sin^2 x} - 2$

$\dfrac{2}{5}$

$\dfrac{1}{6}$

$\dfrac{3}{4}$

$\dfrac{1}{4}$

$x=7^3$

$x=\dfrac{7}{3}$

$x=\log_3 7$

$x=\log_7 3$

$P\left(A \cup B\right) = P\left(A\right) - P\left(B\right) + P\left(A\right).P\left(B\right)$

$P\left(A \cup B\right) = P\left(A\right) + P\left(B\right)$

$P\left(A \cup B\right) = P\left(A\right).P\left(B\right)$

$P\left(A \cup B\right) = P\left(A\right) - P\left(B\right)$

$\left(x^n\right)' = n.x^{n-1}$

$\left(c\right)' = 0$

$\left(\sqrt{x}\right)' = \dfrac{1}{\sqrt{x}}$ với $x > 0$

$\left(x\right)' = 1$

0,2

0,6

0,7

0,8

$y' = - e^x.\cos x$

$y' = e^x.\sin x + e^x.\cos x$

$y' = e^x.\sin x - e^x.\cos x$

$y' = e^x.\cos x$