Đề thi học kỳ 1 Toán 10 năm 2024 — Trường THPT Thanh Bình 1, Đồng ThápMã đề 101

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

$\displaystyle x - y^2 > 3$

$\displaystyle x - y > 3$

$\displaystyle x^2 + y > 3$

$\displaystyle 3^2 + 4^2 = 5^2$

Mấy giờ rồi?

$\displaystyle \exists x \in \mathbb{N}, x+1=0$

$\sqrt{2}$ là số vô tỉ

$\displaystyle 3$

$\displaystyle 4$

$\displaystyle 2$

$\displaystyle 1$

$\displaystyle \tan(180^0 - \alpha) = \tan \alpha$

$\displaystyle \cos(180^0 - \alpha) = \cos \alpha$

$\displaystyle \cot(180^0 - \alpha) = \cot \alpha$

$\displaystyle \sin(180^0 - \alpha) = \sin \alpha$

$\displaystyle 20.250$

$\displaystyle 2030$

$\displaystyle 2020$

$\displaystyle 2025$

$\displaystyle 30$

$\displaystyle 20$

$\displaystyle 10\sqrt{3}$

$\displaystyle 20\sqrt{2}$

$\displaystyle \{1;2;3;4;5\}$

$\displaystyle \{2;4;6;8\}$

$\displaystyle \{1;3;5\}$

$\displaystyle \{1;2;3;4;5;7;9\}$

$\displaystyle \overrightarrow{MP}$

$\displaystyle \overrightarrow{MN}$

$\displaystyle \overrightarrow{MQ}$

$\displaystyle \overrightarrow{QM}$

$\displaystyle 4$

$\displaystyle 6$

$\displaystyle 7$

$\displaystyle 5$

$\displaystyle D = \mathbb{R} \setminus \{-3\}$

$\displaystyle D = \mathbb{R} \setminus \{1; -3\}$

$\displaystyle D = (-3; +\infty)$

$\displaystyle D = (-\infty; -3)$

$\displaystyle \left|\overrightarrow{DE}\right|$

$\displaystyle \overrightarrow{ED}$

$\displaystyle \overrightarrow{DE}$

$\displaystyle DE$

$\displaystyle \cos \alpha < 0$

$\displaystyle \tan \alpha < 0$

$\displaystyle \cot \alpha < 0$

$\displaystyle \sin \alpha > 0$

$\displaystyle x = -3$

$\displaystyle x = 3$

$\displaystyle x = -6$

$\displaystyle x = 6$

$\displaystyle \sqrt{5}$

$\displaystyle \sqrt{7}$

$\displaystyle \sqrt{19}$

$\displaystyle \sqrt{13}$

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

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

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

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

$\displaystyle x + y \le 3$

$\displaystyle x + y < 3$

$\displaystyle x + y \ge 3$

$\displaystyle x + y > 3$

$\displaystyle 2,2$

$\displaystyle 2,3$

$\displaystyle 2,4$

$\displaystyle 2,1$

$\displaystyle \overrightarrow{IA} + \overrightarrow{BI} = \vec{0}$

$\displaystyle \overrightarrow{IA} - \overrightarrow{IB} = \vec{0}$

$\displaystyle \overrightarrow{AI} + \overrightarrow{IB} = \vec{0}$

$\displaystyle \overrightarrow{IA} + \overrightarrow{IB} = \vec{0}$

$\displaystyle y = -x^2 + 2x$

$\displaystyle y = x^2 + 2x$

$\displaystyle y = -x^2 - 2x$

$\displaystyle y = x^2 - 2x$