Đề kiểm tra giữa kỳ 2 Toán 8 năm 2024 — Trường THCS Quỳnh Xuân, Nghệ An

$\displaystyle B \neq 0$

$\displaystyle B \geq 0$

$\displaystyle B \leq 0$

$\displaystyle A = 0$

(g.g)

(c.g.c)

(c.c.g)

(c.c.c)

$\displaystyle \frac{3}{x - 3}$

$\displaystyle 2$

$\displaystyle \frac{3x + y}{x}$

$\displaystyle \frac{x + y}{0}$

$\displaystyle 3x$

$\displaystyle 3x + y$

$\displaystyle y$

$\displaystyle 2y$

$\displaystyle A.D = B.C$

$\displaystyle A.B = D.C$

$\displaystyle A.C = B.D$

$\displaystyle A = D$

$\displaystyle \frac{X}{Y} = \frac{-X}{Y}$

$\displaystyle \frac{X}{Y} = \frac{-X}{-Y}$

$\displaystyle \frac{X}{Y} = \frac{X}{-Y}$

$\displaystyle \frac{X}{Y} = \frac{Y}{-X}$

$\displaystyle 2y$

$\displaystyle 2xy$

$\displaystyle \frac{1}{2x}$

$\displaystyle \frac{1}{2y}$

$\displaystyle x \neq 2$

$\displaystyle x > 2$

$\displaystyle x \neq -2$

$\displaystyle x \neq 0$

$\displaystyle \frac{x}{y}$

$\displaystyle \frac{-x}{y}$

$\displaystyle \frac{-x}{-y}$

$\displaystyle \frac{y}{x}$

Hình 4

Hình 3

Hình 2

Hình 1

$\displaystyle \frac{3y^2}{9xy^2}$

$\displaystyle \frac{y^2}{9xy^2}$

$\displaystyle \frac{3y^2}{9xy}$

$\displaystyle \frac{3y}{9xy^2}$

$\displaystyle 15x$

$\displaystyle x(x-1)$

$\displaystyle x^2 -1$

$\displaystyle x^2 +1$