Đề thi Học kì 1 Toán 9 năm 2025-2026 trường THCS Vĩnh Yên (Phú Thọ)

3

-3

9

$3\sqrt{3}$

$\left\{\begin{matrix}x+3y=7\\2x-y=2\end{matrix}\right.$

$\left\{\begin{matrix}x+\dfrac{1}{y}=12\\2x-y=4\end{matrix}\right.$

$\left\{\begin{matrix}x^2 - 3y = 0\\x-2y=26\end{matrix}\right.$

$\left\{\begin{matrix}\sqrt{x}+3y=0\\0,2x-y=21\end{matrix}\right.$

$\cos F=\dfrac{DE}{EF}$

$\cot F=\dfrac{DF}{EF}$

$\tan E=\dfrac{DF}{EF}$

$\cos E=\dfrac{DE}{EF}$

$x<3$

$x > 3$

$x \le 3$

$x \ge 3$

$a + 2 > b + 2$

$3.a < 3.b$

$-5a < 5b$

$a - 3 > b - 2$

$2x + 3y^2 = 0$

$x^3 + y = 5$

$xy - x = 1$

$2x - 3y = 4$

$\left\{\begin{matrix}x+y=0\\2y - x = 3;\end{matrix}\right.$

$\left\{\begin{matrix}x - 2y = 3\\2x + y = -1;\end{matrix}\right.$

$\left\{\begin{matrix}-x + 3y = -4\\3x - 2y = 1;\end{matrix}\right.$

$\left\{\begin{matrix}2x + y = 1\\x - 3y = 4.\end{matrix}\right.$

$x = -1$

$x = \dfrac{8}{3}$

$x = 1$

$x = \dfrac{3}{8}$

$\sqrt{7}a$

$-\sqrt{7}a^2$

$\sqrt{49a^2}$

$\sqrt{7}a^2$

$\sqrt{3}$

$\dfrac{\sqrt{3}}{2}$

$\dfrac{1}{\sqrt{3}}$

1

$576\pi cm^2$

$144\pi cm^2$

$24\pi cm^2$

$48\pi cm^2$