Đề thi thử vào lớp 10 môn Toán năm 2025–2026 — Trường THCS Quảng Long, Quảng BìnhMã đề 1

$\displaystyle x^2 + 3y = 4$

$\displaystyle x - 3y^2 = 5$

$\displaystyle x + \frac{1}{y} = 2$

$\displaystyle 2x - y = 3$

$\displaystyle x + 2y > 0$

$\displaystyle \frac{1}{x} - 3 > 0$

$\displaystyle x^2 + 1 > 0$

$\displaystyle \frac{x}{2} + 1 > 0$

$\displaystyle x > -100$

$\displaystyle x \ge -100$

$\displaystyle x < -100$

$\displaystyle x \le -100$

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

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

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

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

$\displaystyle (-\sqrt{5})^2$

$\displaystyle \sqrt{5^2}$

$\displaystyle \sqrt{(-5)^2}$

$\displaystyle -(\sqrt{5})^2$

$\displaystyle (-1; 2)$

$\displaystyle (1; 2)$

$\displaystyle (1; 1)$

$\displaystyle (-2; 2)$

$\displaystyle 3x^2 - 5x + 2 = 0$

$\displaystyle 4x - 10 = 0$

$\displaystyle x^2 - 5x = x^2 + 1$

$\displaystyle \frac{x^2 - 6}{x} = 0$

$\displaystyle \sin C = 0,75$

$\displaystyle \sin C = 0,8$

$\displaystyle \sin C = 0,6$

$\displaystyle \sin C = 1,3$

$\displaystyle 36^{\circ}52'$

$\displaystyle 60^{\circ}52'$

$\displaystyle 35^{\circ}$

$\displaystyle 60^{\circ}$

$\displaystyle 100\pi (cm^2)$

$\displaystyle 10\pi (cm^2)$

$\displaystyle 20\pi (cm^2)$

$\displaystyle 100\pi^2 (cm^2)$

$\displaystyle 45^{\circ}$

$\displaystyle 90^{\circ}$

$\displaystyle 60^{\circ}$

$\displaystyle 120^{\circ}$

$\displaystyle 100^{\circ}$

$\displaystyle 110^{\circ}$

$\displaystyle 180^{\circ}$

$\displaystyle 90^{\circ}$

27 cm

27,25 cm

25 cm

25,27 cm

$5\pi$ $(cm^{2})$

$25\pi$ $(cm^{2})$

$50\pi$ $(cm^{2})$

$\frac{25}{2}\pi$ $(cm^{2})$