Đề thi Giữa kì 2 Toán 12 năm 2024-2025 trường THPT Hướng Hóa (Quảng Trị)

$F'(x) = f(x), \forall x \in K$ .

$F''(x) = f(x), \forall x \in K$ .

$F(x) = f'(x), \forall x \in K$ .

$F(x) = f''(x), \forall x \in K$ .

$\displaystyle \int_a^c f(x)\mathrm{d}x + \int_c^b f(x)\mathrm{d}x = \int_a^b f(x)\mathrm{d}x$ .

$\displaystyle \int_a^b f(x)\mathrm{d}x + \int_a^c f(x)\mathrm{d}x = \int_c^b f(x)\mathrm{d}x$ .

$\displaystyle \int_a^b f(x)\mathrm{d}x - \int_a^c f(x)\mathrm{d}x = \int_c^b f(x)\mathrm{d}x$ .

$\displaystyle \int_a^c f(x)\mathrm{d}x + \int_c^b f(x)\mathrm{d}x = \int_b^a f(x)\mathrm{d}x$ .

$\displaystyle \int_c^d [f(x)+g(x)]\mathrm{d}x=\int_c^d f(x)\mathrm{d}x + \int_c^d g(x)\mathrm{d}x$ .

$\displaystyle \int_c^d kf(x)\mathrm{d}x = k \int_c^d f(x)\mathrm{d}x$ .

$\displaystyle \int_c^d [f(x) - g(x)]\mathrm{d}x = \int_c^d f(x)\mathrm{d}x - \int_c^d g(x)\mathrm{d}x$ .

$\displaystyle \int_c^d [f(x) \cdot g(x)]\mathrm{d}x = \int_c^d f(x)\mathrm{d}x \cdot \int_c^d g(x)\mathrm{d}x$ .

2 .

-2 .

12 .

-12 .

$-xy - z - 3 = 0$ .

$-x + y - z^2 + 1 = 0$ .

$-2x + 2y - z - 3 = 0$ .

$\dfrac{1}{x} + \dfrac{1}{y} - \dfrac{3}{z} + 2 = 0$ .

$(-2;1;0)$

$(-2;1;-5)$

$(1;7;5)$

$(-2;2;-5)$

$e^x + x^2 + C$

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

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

$e^x + 1 + C$

6

-3

9

5

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

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

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

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

$x - 2y - 2z - 1 = 0$

$-x + y - 2z - 1 = 0$

$x - 2y - 2z + 7 = 0$

$-x + y - 2z + 1 = 0$

-3

1

2

7

7

8

9

10

12

24

20

10