Đề thi Giữa kì 2 Toán 12 năm 2024-2025 trường THPT Nguyễn Tất Thành (Tp.HCM)

$\vec{n}=(-6;2;4)$

$\vec{n}=(3;-1;-1)$

$\vec{n}=(6;-2;2)$

$\vec{n}=(3;-1;1)$

$F(x) = 3e^x + 2 + C$

$F(x) = 3e^x + 2x^2 + C$

$F(x) = 3e^x + x^2 + C$

$F(x) = 3 \dfrac{e^{x+1}}{x+1} + 2 + C$

$\dfrac{\pi^2 + 16 \pi + 8 \sqrt{2} - 16}{16}$

$\dfrac{\pi^2 + 16 \pi + 2 \sqrt{2} - 4}{16}$

$\dfrac{\pi^2 + 16 \pi + 8 \sqrt{2}}{16}$

$\dfrac{\pi^2 + 16 \pi - 16}{16}$

$I = \int_{-1}^2 (2x - 4) \, dx - \int_2^3 (2x - 4) \, dx$

$I = - \int_{-1}^2 (2x - 4) \, dx + \int_2^3 (2x - 4) \, dx$

$I = - \int_{-1}^2 (2x - 4) \, dx - \int_2^3 (2x - 4) \, dx$

$I = \int_{-1}^2 (2x - 4) \, dx + \int_2^3 (2x - 4) \, dx$

$m=1$

$m=-1$

$m=2$

$m=-2$

$F(x) = x^3 - 2x + 5$

$F(x) = x^3 - 2x - 2$

$F(x) = x^3 - 2x + 3$

$F(x) = x^3 - 2x + 4$

$\vec{c} = (a_3b_1 - a_1b_3, a_1b_2 - a_2b_1, a_2b_3 - a_3b_1)$

$\vec{c} = (a_2b_3 - a_3b_2, a_3b_1 - a_1b_3, a_1b_2 - a_2b_1)$

$\vec{c} = (a_1b_3 - a_3b_1, a_2b_3 - a_3b_2, a_3b_1 - a_1b_3)$

$\vec{c} = (a_1b_3 - a_3b_1, a_2b_2 - a_1b_2, a_3b_2 - a_2b_3)$

6

2

8

4

$\dfrac{1}{2} \ln x + C$

$\dfrac{1}{2} \ln^2 x + C$

$-\dfrac{1}{x^3} + C$

$\ln^2 x + C$