Bộ 10 Đề kiểm tra giữa kì 2 Toán 10 có đáp án (Mới nhất) (Đề 10)

$\left[\mathrm{2021,}+\infty \right)$.

$\left(-\infty \mathrm{,2021}\right)$.

$\left\{2021\right\}$.

$\varnothing$.

$\left(\frac{3}{4}-\frac{\sqrt{23}}{4};\frac{3}{4}+\frac{\sqrt{23}}{4}\right)$.

$\left(-\infty ;\frac{3}{4}-\frac{\sqrt{23}}{4}\right)\cup \left(\frac{3}{4}+\frac{\sqrt{23}}{4};+\infty \right)$.

$\left(-\frac{2}{3};+\infty \right)$.

$\left(-\infty ;-\frac{2}{3}\right)$.

$\left|x\right|\ge a\iff -a\le x\le a$.

$\left|x\right|\le a\iff x\le a$.

$\left|x\right|>a\iff x>a$.

$\left|x\right|\ge a\iff [\begin{array}{c}x\le -a\\ x\ge a\end{array}$.

1.

4.

2.

3.

$2x-5y+3z\le 0$.

$3x^2+2x-4>0$.

$2x^2+5y>3$.

$2x+3y<5$.

$x\le 2$.

$\{\begin{array}{c}x\ne 2\\ x\ne -4\end{array}$.

$\{\begin{array}{c}x<2\\ x\ne -4\end{array}$.

x < 2.

a – c > b – d

a + c > b + d

ac > bd

$\frac{a}{c}>\frac{b}{d}$

$S=\left(-\infty ;-2\right]\cup \left[4;+\infty \right)$.

$S=\left[-2;4\right]$.

$S=\left[2;4\right]$.

$S=\left(-\infty ;-2\right)\cup \left(4;+\infty \right)$.

$f\left(x\right)=x-2$.

$f\left(x\right)=2-4x$.

$f\left(x\right)=16-8x$.

$f\left(x\right)=-x-2$.

$m_a^2=\frac{b^2+c^2}{2}-\frac{a^2}{4}$.

$a^2=b^2+c^2+2bc\cos A$.

$S=\frac{abc}{4R}$.

$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R$.

(-2;1).

(3;-7).

(0;1).

(0;0).

0 < m < 4.

m < 0 hoặc m > 4.

m > 2.

m < 2.

$\left(-\infty ;-4\right]\cup \left[3;+\infty \right)$.

$\left(-\infty ;-3\right]\cup \left[4;+\infty \right)$.

$\left[-3;4\right]$.

$\varnothing$.

$\hat{A}={30}^0.$

$\hat{A}={45}^0.$

$\hat{A}={60}^0.$

$\hat{A}={75}^0$ .

$S=\frac{3}{4}(a^2+b^2+c^2)$.

$S=(a^2+b^2+c^2)$.

$S=\frac{3}{2}(a^2+b^2+c^2)$.

$S=3(a^2+b^2+c^2)$.

$\overrightarrow{n}\left(-2;-1\right)$.

$\overrightarrow{n}\left(2;-1\right)$.

$\overrightarrow{n}\left(-1;2\right)$.

$\overrightarrow{n}\left(1;2\right)$.

$\overrightarrow{u}=\left(4;-2\right)$.

$\overrightarrow{v}=\left(-2;-1\right)$.

$\overrightarrow{m}=\left(2;1\right)$.

$\overrightarrow{q}=\left(4;2\right)$.

$d:\frac{x}{a}-\frac{y}{b}=1$.

$d:\frac{x}{b}+\frac{y}{a}=1$.

$d:\frac{x}{a}+\frac{y}{b}=1$.

$d:\frac{x}{a}+\frac{y}{b}=0$.

$\Delta :2x-y-5=0.$

$\Delta :\frac{x-1}{1}=\frac{y+3}{2}\cdot$

$\Delta :\{\begin{array}{c}x=1+t\\ y=-3+2t\end{array}\cdot$

$\Delta :\frac{x+1}{1}=\frac{y-3}{2}\cdot$

$\{\begin{array}{c}x+y\le 2\\ x-2y\ge -4\end{array}.$

$\{\begin{array}{c}x+y\ge 2\\ x-2y\ge -4\end{array}.$

$\{\begin{array}{c}x+y\ge 2\\ x-2y\le -4\end{array}.$

$\{\begin{array}{c}x+y\le 2\\ x-2y\le -4\end{array}.$

$S\le \sqrt{2}$

$S\ge \sqrt{2}$

$-\sqrt{2}\le S\le \sqrt{2}$

$-1\le S\le 1$

$S=(1;2]$

$S=[1;2]$

$S=[1;2)$

$S=\left(-\infty ;1\right]\cup \left(2;+\infty \right)$

$S=\left[-\frac{8}{3};+\infty \right)$

$S=\left(-\frac{8}{3};16\right)$

$S=\left(-\infty ;16\right]$

$S=\left[-\frac{8}{3};16\right]$

3.

1.

2.

0.

$\left(\frac{2}{3};+\infty \right)$.

$\left[\frac{2}{3};+\infty \right)$.

$\left(-\infty ;\frac{2}{3}\right)$.

$ℝ$.

$\frac{3}{a}=\frac{2}{b}-\frac{1}{c}$.

3a = 2b + c.

3a = 2b – c.

$\frac{3}{a}=\frac{2}{b}+\frac{1}{c}$.

a² = b² + c² – 3bc.

a² = b² + c² + bc.

a² = b² + c² + 3bc.

a² = b² + c² – bc.

$D=(-2;+\infty ).$

$D=(-2;3).$

$D=\left(3;+\infty \right).$

$D=(-\infty ;-2].$

x > 1.

x < 1.

x > 4.

$x\in ℝ.$

x = 9.

x = 8.

x = 7.

x = 6.

24x + 7y – 52 = 0.

x = 4, y = 4.

y = 4, 24x – 7y + 4 = 0.

x = 4, 24x + 7y – 52 = 0.

m = 0.

$m\in \varnothing$.

m = 2.

$m=\frac{3}{8}\cdot$

2.

$\frac{5}{2}$.

$2\sqrt{2}$.

3.

3.

15.

7,5.

5.

$(\sqrt{2}-1)x+y=0$.

$x+(\sqrt{2}+1)y-2\sqrt{2}=0$.

$(\sqrt{2}-1)x-y+2\sqrt{2}-1=0$.

$(\sqrt{2}-1)x+y-\sqrt{2}=0$.