Đề thi Giữa kì 1 Toán 10 năm 2023-2024 Sở Bắc Ninh

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

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

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

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

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

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

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

$\overline{P} : "\exists x \in \mathbb{R}, 2x^2 \geq 1"$

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

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

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

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

$(A \cap B) \subset A$

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

$(A \setminus B) \subset A$

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

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

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

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

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

$A \cup B = \left(-2;2\right)$

$A \cup B = \left(-2;2\right]$

$A \cup B = \mathbb{R}$

$A \cup B = \varnothing$

$(3;1)$

$(0;3)$

$(1;3)$

$(4;1)$

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

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

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

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

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

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

$\sqrt{3}$ cm.

$\sqrt{2}$ cm.

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

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

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

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

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

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

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

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