Đề thi Học kì 1 Toán 10 năm 2024-2025 trường THPT B Bình Lục (Hà Nam)

A. “$\forall x \in \mathbb{R} : x^{2} + 1 = 0$”.

B. “$\exists x \in \mathbb{R} : x^{2} + 1 \leq 0$”.

C. “$\exists x \in \mathbb{R} : x^{2} + 1 > 0$”.

D. “$\forall x \in \mathbb{R} : x^{2} + 1 < 0$”.

A. $\left\{\begin{matrix}a > 0 \\ \Delta \leq 0\end{matrix}\right.$

B. $\left\{\begin{matrix}a > 0 \\ \Delta \geq 0\end{matrix}\right.$

C. $\left\{\begin{matrix}a > 0 \\ \Delta < 0\end{matrix}\right.$

D. $\left\{\begin{matrix}a < 0 \\ \Delta > 0\end{matrix}\right.$

A. $\left(-\infty ; 4\right)$.

B. $\left[4 ; +\infty\right)$.

C. $\left(-\infty ; 4\right].$

D. $\left(4 ; +\infty\right).$

A. $\overrightarrow{AB} - \overrightarrow{BC} = \overrightarrow{CA}$.

B. $\overrightarrow{AB} + \overrightarrow{AC} = \overrightarrow{BC}$.

C. $\overrightarrow{CA} + \overrightarrow{AB} = \overrightarrow{CB}$.

D. $\overrightarrow{CA} - \overrightarrow{BA} = \overrightarrow{BC}$.

A. $\cos A= \dfrac{AB^{2}+CA^{2}-BC^{2}}{2.AB.CA}$.

B. $\cos A= \dfrac{AB^{2}+CA^{2}-BC^{2}}{AB.CA}$.

C. $\cos A= \dfrac{AB^{2}+BC^{2}-CA^{2}}{2.AB.BC}$.

D. $\cos A= \dfrac{AB^{2}+BC^{2}-CA^{2}}{AB.BC}$.

$\mathbf{A}.\; O(0;0)$

$\mathbf{B}.\; M(-2;2)$

$\mathbf{C}.\; P(4;1)$

$\mathbf{D}.\; N(1;-2)$

$\mathbf{A}.\; \cos \left( 180^0 - a \right) = - \cos a$

$\mathbf{B}.\; \cot \left( 180^0 - a \right) = \cot a$

$\mathbf{C}.\; \sin \left( 180^0 - a \right) = - \sin a$

$\mathbf{D}.\; \tan \left( 180^0 - a \right) = \tan a$

$\mathbf{A}.\; 0$

$\mathbf{B}.\; 2$

$\mathbf{C}.\; 1$

$\mathbf{D}.\; -9$

$\mathbf{A}.\; X = (0;1)$

$\mathbf{B}.\; X = (1; + \infty)$

$\mathbf{C}.\; X = (-1;0)$

$\mathbf{D}.\; X = (0; + \infty)$