Đề kiểm tra giữa kỳ 2 Toán 7 năm 2025 — Trường THCS Uy Nỗ, Hà Nội

$\displaystyle x = \frac{y}{2} = \frac{z}{3}$

$\displaystyle \frac{x}{2} = \frac{y}{1} = \frac{z}{3}$

$\displaystyle \frac{x}{3} = \frac{y}{2} = \frac{z}{1}$

$\displaystyle \frac{x}{3} = \frac{y}{1} = \frac{z}{2}$

$\displaystyle y = \frac{1}{3}x$

$\displaystyle y = 3x$

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

$\displaystyle y = -3x$

$\displaystyle \frac{x}{2} = \frac{y}{3}$

$\displaystyle \frac{2}{x} = \frac{y}{3}$

$\displaystyle \frac{3}{y} = \frac{2}{x}$

$\displaystyle \frac{x}{3} = \frac{y}{2}$

$\displaystyle AB - BC > AC$

$\displaystyle AB + BC > AC$

$\displaystyle AB + BC = AC$

$\displaystyle AB + BC < AC$

$\displaystyle \widehat{A} < \widehat{B} < \widehat{C}$

$\displaystyle \widehat{B} < \widehat{C} < \widehat{A}$

$\displaystyle \widehat{C} < \widehat{A} < \widehat{B}$

$\displaystyle \widehat{C} < \widehat{B} < \widehat{A}$

$\displaystyle x + y$

$\displaystyle \frac{x}{2} + y$

$\displaystyle \frac{x + y}{2}$

$\displaystyle x + \frac{y}{2}$

$\displaystyle x = 2023$

$\displaystyle x = 2024$

$\displaystyle x = 2025$

$\displaystyle x = 2026$

$\displaystyle x^{2} + 2xy - 2$

$\displaystyle 2x^{2} + xy + 2$

$\displaystyle 2x^{2} - 5x + 1$

$\displaystyle -x^{2} + 3x + 2$

$\displaystyle a = 4$

$\displaystyle a = 8$

$\displaystyle a = -16$

$\displaystyle a = -4$

$\displaystyle x^{2} + 2x + 1$

$\displaystyle x^{3} + 2x^{2} + x$

$\displaystyle x^{2} - 1$

$\displaystyle x^{4} + 2x - 3$

$\displaystyle AB$

$\displaystyle BC$

$\displaystyle AC$

$\displaystyle BA$