Đề kiểm tra giữa kỳ 1 Toán 10 năm 2025 — Trường THPT Edison, Thái NguyênMã đề 125

$\displaystyle B = \left(-\dfrac{1}{2}; 3\right]$

$\displaystyle B = \left[-\dfrac{1}{2}; 3\right)$

$\displaystyle B = \left(-\dfrac{1}{2}; 3\right)$

$\displaystyle B = \left[-\dfrac{1}{2}; 3\right]$

"$\exists x \in \mathbb{R}: x^2 + 1 \ne 0$"

"$\forall x \in \mathbb{R}: x^2 + 1 \le 0$"

"$\exists x \in \mathbb{R}: x^2 + 1 \le 0$"

"$\exists x \in \mathbb{R}: x^2 + 1 < 0$"

$\displaystyle A \cap B = \{5;7;9\}$

$\displaystyle A \cap B = \{5;9\}$

$\displaystyle A \cap B = \{2;3;8\}$

$\displaystyle A \cap B = \{7\}$

$\displaystyle 2x^2 + 5y > 3$

$\displaystyle 3x^2 + 2x - 4 > 0$

$\displaystyle 2x \le 5y + 3z$

$\displaystyle 2x - 3y > 5$

$\displaystyle 76$

$\displaystyle 2\sqrt{19}$

$\displaystyle 14$

$\displaystyle 6\sqrt{2}$

$\displaystyle (3;2)$

$\displaystyle (5;-1)$

$\displaystyle (-2;5)$

$\displaystyle (4;0)$

$\displaystyle \overrightarrow{DC}$

$\displaystyle \overrightarrow{BA}$

$\displaystyle \overrightarrow{BC}$

$\displaystyle \overrightarrow{DA}$

$\displaystyle \tan \alpha < 0$

$\displaystyle \cot \alpha > 0$

$\displaystyle \sin \alpha < 0$

$\displaystyle \cos \alpha > 0$

$\displaystyle \cot(90^{\circ} - \alpha) = -\tan \alpha$

$\displaystyle \cos(90^{\circ} - \alpha) = \sin \alpha$

$\displaystyle \sin(90^{\circ} - \alpha) = -\cos \alpha$

$\displaystyle \tan(90^{\circ} - \alpha) = -\cot \alpha$

$\overrightarrow{MP}$ và $\overrightarrow{PN}$

$\overrightarrow{NM}$ và $\overrightarrow{NP}$

$\overrightarrow{MN}$ và $\overrightarrow{PN}$

$\overrightarrow{MN}$ và $\overrightarrow{MP}$

$\displaystyle 3\text{cm}$

$\displaystyle 4\text{cm}$

$\displaystyle 5\text{cm}$

$\displaystyle 6\text{cm}$