Đề giữa học kì 2 Toán 12 năm 2024 – 2025 trường THPT Sầm Sơn – Thanh Hóa

8.

2.

$\dfrac{1}{8}.$

$\dfrac{1}{2}.$

$S = \int\limits_a^b {\left| {f\left( x \right)} \right|dx}$.

$S = \int\limits_a^b {f\left( x \right)dx}$.

$S = - \int\limits_a^b {f\left( x \right)dx}$.

$S = \left| {\int\limits_b^a {f\left( x \right)dx} } \right|$.

$S = - {\int\limits_{- 1}^{1}{f(x)\text{d}x -}}{\int\limits_{1}^{5}{f(x)\text{d}x}}$.

$S = {\int\limits_{- 1}^{1}{f(x)\text{d}x +}}{\int\limits_{1}^{5}{f(x)\text{d}x}}$.

$S = {\int\limits_{- 1}^{1}{f(x)\text{d}x -}}{\int\limits_{1}^{5}{f(x)\text{d}x}}$.

$S = - {\int\limits_{- 1}^{1}{f(x)\text{d}x +}}{\int\limits_{1}^{5}{f(x)\text{d}x}}$.

$S = \pi{\int\limits_{0}^{1}{3x^{2}\text{d}x}}$.

$S = {\int\limits_{0}^{1}{x^{2}\text{d}x}}$.

$S = \pi{\int\limits_{0}^{1}{\left( {3x^{2}} \right)^{2}\text{d}x}}$.

$S = 3{\int\limits_{0}^{1}{x^{2}\text{d}x}}$.

$\overset{\rightarrow}{n_{3}} = \left( {1;,, - 3;,, 0} \right)$.

$\overset{\rightarrow}{n_{2}} = \left( {0;,, 1;,, - 3} \right)$.

$\overset{\rightarrow}{n_{1}} = \left( {1;, - , 3;,, 12} \right)$.

$\overset{\rightarrow}{n_{4}} = \left( {1;,, 0;,, - 3} \right)$.

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

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

$x + xy + 2z + 3 = 0$.

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

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

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

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

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