Đề thi Học kì 1 Toán 10 năm 2025-2026 trường THPT Đô Lương 1 (Nghệ An)

21

$\sqrt{21}$

11

121

$\vec{c} = (12; -2)$

$\vec{c} = (-10; -3)$

$\vec{c} = (-1; 3)$

$\vec{c} = (1; -3)$

$a = 2R \cdot \tan A$

$a = 2R \cdot \sin 4$

$a = R \cdot \sin A$

$a = 2R \cdot \cos A$

$A \cup B = (-1;0]$

$A \cup B = [-1;3]$

$A \cup B = (0;1)$

$A \cup B = [-1;1)$

$x - y < 0$

$2x - 3y - 1 > 0$

$4x < y$

$x - 3y - 2 \ge 0$

“$\forall x \in \mathbb{R} | x^2 + x + 1 \le 0$”.

“$\exists x \in \mathbb{R} | x^2 + x + 1 \le 0$”.

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

“$\exists x \in \mathbb{R} | x^2 + x + 1 < 0$”.

$\overrightarrow{BD}$

$\overrightarrow{DB}$

$\overrightarrow{AC}$

$\overrightarrow{CA}$

$45^\circ$

$30^\circ$

$90^\circ$

$60^\circ$

$\vec{a} = 3\vec{i} + 2\vec{j}$

$\vec{a} = 3\vec{j}$

$\vec{a} = 2\vec{i}$

$\vec{a} = 2\vec{i} + 3\vec{j}$

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

B. $\vec{a} \cdot \vec{b} = \overrightarrow{a} \cdot \overrightarrow{b} \cdot \cos\left(\overrightarrow{a,b}\right)$

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

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