Đề thi Học kì 1 Toán 10 năm 2023-2024 Sở Sơn La

$2x^2 + x - 3 < 0$

$x^2 + \dfrac{1}{2x} + 5 \geq 0$

$2x^2 + 3xy - 1 < 0$

$2x - 1 > 0$

$y = x^3 - 4x + 1$

$y = -2x + 3$

$y = x^2 - 2x + 3$

$y = - 4x^4 + 3x^2$

$S = \dfrac{1}{2} ac \sin C$

$S = \dfrac{1}{2} ab \sin A$

$S = \dfrac{1}{2} ab \sin B$

$S = \dfrac{1}{2} ab \sin C$

$x^2 + y^2 < 2$

$x + y \geq 2$

$2x^2 + 3y > 0$

$x + y^2 \geq 0$

5

4

6

3

$\vec{a} . \vec{b} = \left|\vec{a}\right| . \left|\vec{b}\right| . \cos{(\vec{a}, \vec{b})}$

$\vec{a} . \vec{b} = \left|\vec{a}\right| . \left|\vec{b}\right| . \sin{(\vec{a}, \vec{b})}$

$\vec{a} . \vec{b} = a.b . \cos{(\vec{a}, \vec{b})}$

$\vec{a} . \vec{b} = \left|\vec{a}\right| . \left|\vec{b}\right|$

64 chia hết cho 8.

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$\overrightarrow{MA}+\overrightarrow{MB}=2\overrightarrow{MI}$

$\overrightarrow{MA}+\overrightarrow{MB} = \dfrac{1}{2}\overrightarrow{MI}$

$\overrightarrow{MA}+\overrightarrow{MB}=-\dfrac{1}{2}\overrightarrow{MI}$

$\overrightarrow{MA}+\overrightarrow{MB}=-2\overrightarrow{MI}$

$\mathbb{R}$.

$\mathbb{R}\setminus\{-2\}$.

$\mathbb{R}\setminus\{1;2\}$.

$\mathbb{R}\setminus\{2\}$.

$\Delta >0$.

$\Delta=0$.

$\Delta<0$.

$\Delta \geq 0$.

$\sin 150^\circ = -\dfrac{\sqrt{3}}{2}$

$\cos 150^\circ = \dfrac{\sqrt{3}}{2}$

$\tan 150^\circ = - \dfrac{\sqrt{3}}{3}$

$\cot 150^\circ = \sqrt{3}$.

$\overrightarrow{AC}$.

$\overrightarrow{BA}$.

$\overrightarrow{CA}$.

$\overrightarrow{CB}$.

$\overrightarrow{AB} + \overrightarrow{AD} = \overrightarrow{CA}$.

$\overrightarrow{CD} + \overrightarrow{CB} = \overrightarrow{AC}$.

$\overrightarrow{BA} + \overrightarrow{BC} = \overrightarrow{DB}$.

$\overrightarrow{AB} + \overrightarrow{AD} = \overrightarrow{AC}$.

$f(x)=5x+2$.

$f(x)=-2x-6$.

$f(x)=\dfrac{x^2-3x+2}{x+2}$.

$f(x)=x^2+x-6$.

$ax_0^2+bx_0+c>0$.

$ax_0^2+bx_0+c \geq 0$.

$ax_0^2+bx_0+c \leq 0$.

$ax_0^2+bx_0+c <0$.

$[-2;+\infty)$

$(3;7)$

$[3;7]$

$[-2;3]$.

$\left\{-1;\dfrac{7}{4}\right\}$

$\left\{1; -\dfrac{7}{4}\right\}$

$\left\{1; \dfrac{7}{4}\right\}$

$\left\{-1; -\dfrac{7}{4}\right\}$

$(0;0)$

$(1;1)$

$(-1;1)$

$(-1;-1)$

$\sin(180^\circ -\alpha) = -\cos\alpha$

$\sin(180^\circ -\alpha) = -\sin\alpha$

$\sin(180^\circ -\alpha) = \sin\alpha$

$\sin(180^\circ -\alpha) = \cos\alpha$

$x=-1$

$y=-2$

$y=2$

$x=1$

$(-3;-1)$

$(-\infty;-3)\cup(-1;+\infty)$

$[-3;-1]$

$(-\infty;-3]\cup[-1;+\infty)$

$\overrightarrow{MA} = -\dfrac{3}{2}\overrightarrow{MG}$

$\overrightarrow{MA} = -3\overrightarrow{MG}$

$\overrightarrow{MA} = \dfrac{3}{2}\overrightarrow{MG}$

$\overrightarrow{MA} = 3\overrightarrow{MG}$

$T=(-1;+\infty)$

$T=\varnothing$

$T=\mathbb{R}$

$T=(-\infty;-1)$

$f(2)=0$

$f(-4)=0$

$f(3)=0$

$f(1)=0$

-18

36

$18 \sqrt{2}$

-36

$\sqrt{12}$

12

$12 \sqrt{5}$

$\sqrt{5}$

$\overrightarrow{MP}$

$\overrightarrow{NP}$

$\overrightarrow{NQ}$

$\overrightarrow{MQ}$