Đề thi Học kì 2 Toán 10 năm 2025-2026 trường THPT Hướng Hóa (Quảng Trị)

$D=\mathbb{R}\setminus\{2\}$.

$D=(-\infty;2]$.

$D=(2;+\infty)$.

$D=[2;+\infty)$.

$x=\dfrac{1}{4}$.

$x=-\dfrac{1}{4}$.

$x=\dfrac{1}{2}$.

$x=-\dfrac{1}{2}$.

$\Delta_{3}:x+3y+1=0$.

$\Delta_{2}:2x+4y+6=0$.

$\Delta_{1}:2x+4y+1=0$.

$\Delta_{4}:2x-4y+3=0$.

$(2;+\infty)$.

$(-2;1)$.

$(-\infty;-1)$.

$(-1;2)$.

100.

330.

300.

270.

$y=6x^{2}$.

$y=6x$.

$y^{2}=\dfrac{6}{x}$.

$y^{2}=6x$.

$\left(x+3\right)^{2}+\left(y-2\right)^{2}=5$.

$\left(x-3\right)^{2}+\left(y+2\right)^{2}=5$.

$\left(x-3\right)^{2}+\left(y+2\right)^{2}=25$.

$\left(x-3\right)^{2}+\left(y-2\right)^{2}=25$.

$A_{10}^{6}$.

$6^{10}$.

$6!$.

$C_{10}^{6}$.

$\dfrac{x^{2}}{4}-\dfrac{y^{2}}{3}=1$.

$\dfrac{x^{2}}{3}+\dfrac{y^{2}}{4}=1$.

$\dfrac{x}{4}+\dfrac{y}{3}=1$.

$\dfrac{x^{2}}{4}+\dfrac{y^{2}}{3}=1$.