5+ Đề thi Giữa kì 1 Toán 10 Bắc Ninh năm 2026 (các năm)

$-2\in \left(-2;5\right]$

$-2\subset \left[-2;5\right]$

$\{1\}\not\subset \left[-2;5\right]$

$\{-2\}\subset \left[-2;5\right]$

$\overline{P} : \text{\textquotedbl} \exists x \in \mathbb{R}, 2x^2 \neq 1 \text{\textquotedbl}$

$\overline{P} : \text{\textquotedbl} \forall x \in \mathbb{R}, 2x^2 = 1 \text{\textquotedbl}$

$\overline{P} : \text{\textquotedbl} \forall x \in \mathbb{R}, 2x^2 \neq 1 \text{\textquotedbl}$

$\overline{P} : \text{\textquotedbl} \exists x \in \mathbb{R}, 2x^2 \ge 1 \text{\textquotedbl}$

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

$\left\{0;1;2;3;4\right\}$

$\left\{1;2;3;4\right\}$

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

$(A \cap B) \subset A$

$(A \cap B) \cup (A \setminus B) = A$

$(A \setminus B) \subset A$

$(A \cap B) \cap (A \setminus B) = A$

$\left\{1;2;3;4;5;6;8\right\}$

$\left\{1;6;8\right\}$

$\left\{1;3;5\right\}$

$\left\{3;5\right\}$

$\left(3;1\right)$

$\left(0;3\right)$

$\left(1;3\right)$

$\left(4;1\right)$

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

$b^2 = a^2 + c^2 - 2ac \cos B$

$\frac{a}{\cos A} = \frac{b}{\cos B} = \frac{c}{\cos C}$

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

$\dfrac{\sqrt{6}}{3}$ cm.

$\dfrac{\sqrt{6}}{2}$ cm.

$\sqrt{3}$ cm.

$\sqrt{2}$ cm.

$\overrightarrow{AB} = \overrightarrow{CD}$.

$\left|\overline{AB}\right| = \left|\overline{CD}\right|.$

$\overrightarrow{BC} = \overrightarrow{AD}.$

$\left|\overline{AB}\right| = \left|\overline{BC}\right|.$

$\overrightarrow{AB} - \overrightarrow{BC} = \overrightarrow{AC}$.

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

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

$\overrightarrow{AB} = \overrightarrow{CB} - \overrightarrow{CA}$.