Đề thi Giữa kì 1 Toán 9 năm 2024-2025 trường THCS Trần Quang Khải (TP. HCM)

$\dfrac{x}{3} + \dfrac{y}{2} - 1 = 0$

$3y - 2 = -4(y-2)$

$x^{2} + 2y - 1 = 0$

$3\sqrt{x} + y^{2} = 0$

$(0;2)$

$(2;1)$

$(0;-2)$

$(-3;0)$

$(-1;2)$

$(1;2)$

$(2;-1)$

$(2;1)$

$(-4;2)$

$(2;5)$

$(5;2)$

$(2;-4)$

$x > -\dfrac{5}{3}$

$x < -\dfrac{5}{3}$

$x > -\dfrac{3}{5}$

$x < -\dfrac{3}{5}$

$x - 5 \ge y - 5$

$5 - 4x \le 5 - 4y$

$3x - \dfrac{1}{2} \le 3y - \dfrac{1}{2}$

$5 - x \le 5 - y$

$-4x^{2} + 5 > 6$

$\dfrac{3}{x} + 1 \le 0$

$0x + 5 \ge 0$

$- \dfrac{x}{2} + 6 < 0$

$x < 3$

$x \ge \dfrac{1}{3}$

$x < \dfrac{1}{3}$

$x \ge 3$

$\dfrac{MN}{NP}$

$\dfrac{MP}{NP}$

$\dfrac{MN}{MP}$

$\dfrac{MP}{MN}$

$\cot B = \sin C$

$\tan B = \cos C$

$\tan B = \cot C$

$\tan B = \tan C$

A. $DE = EF . \text{cos}F$

B. $DE = EF . \text{tan}E$

C. $DE = EF . \text{sin}F$

D. $DE = EF . \text{cot}F$

A. $\sin B = \dfrac{2}{3}$

B. $\cos B = \dfrac{2}{3}$

C. $\tan C = \dfrac{2}{3}$

D. $\cot C = \dfrac{2}{3}$