Đề thi Học kì 1 Toán 10 năm 2025-2026 trường chuyên Lê Quý Đôn (Đà Nẵng)

$10 \sqrt{2}$

$5 \sqrt{6}$

$\dfrac{5\sqrt{6}}{3}$

$\dfrac{10\sqrt{6}}{3}$

23748000

23748590

23748600

23749000

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

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

$\overrightarrow{AB} - \overrightarrow{AC} = \overrightarrow{CB}$

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

$(-2;1)$

$(1;2)$

$(1;-2)$

$(2;-1)$

$\vec{u}.\vec{v} = (xx'; yy')$

$\vec{u}.\vec{v} = xx' + yy'$

$\vec{u}.\vec{v} = (x + x'; y + y')$

$\vec{u}.\vec{v} = xy + x'y'$

$\overrightarrow{IA} = \overrightarrow{IB}$

$\overrightarrow{IA} = \dfrac{1}{2} \overrightarrow{AB}$

$\overrightarrow{IA} + \overrightarrow{IB} = \overrightarrow{AB}$

$\overrightarrow{IB} = \dfrac{1}{2} \overrightarrow{AB}$

$A \bigcup B=[2;4].$

$A \bigcup B=(-3;5]$

$A \bigcup B=\{-3;5\}.$

$A \bigcup B=(-3;2).$

$\forall n \in \mathbb{N}, n^{2} \leq 2025.$

$\exists n \in \mathbb{N}, n^{2} \leq 2025.$

$\forall n \in \mathbb{N}, n^{2} < 2025.$

$\forall n \in \mathbb{N}, n^{2} > 2025.$

$\sqrt{\dfrac{(x_1-\overline{x})^2+(x_2-\overline{x})^2+...+(x_n-\overline{x})^2}{n}}.$

$\dfrac{(x_1-\overline{x})^2+(x_2-\overline{x})^2+...+(x_n-\overline{x})^2}{n}$.

$\sqrt{(x_1-\overline{x})^2+(x_2-\overline{x})^2+...+(x_n-\overline{x})^2}$.

$\sqrt{\dfrac{(x_1-\overline{x})^2+(x_2-\overline{x})^2+...+(x_n-\overline{x})^2}{n-1}}.$

$2x+0.y \geq 3.$

$x^{2}+y < 7.$

$0.x+0.y \leq 1.$

$3x-\sqrt{y} \geq 1.$