Đề thi Giữa kì 2 Toán 10 năm 2024-2025 trường THPT Mường Than (Lai Châu)

$x^{2}-y^{2}=4$

$y=2x-1$

$x^{2}+y^{2}=9$

$y^{2}=x$

$(1; + \infty)$

$(-\infty;1]$

$[1; + \infty)$

$R \setminus \{1\}$

$y=x^{2}+x$

$y=\dfrac{1}{x^{2}}$

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

$y=x^{3}+x+1$

$\vec{n}_1= (3; -2)$

$\vec{n}_2= (2; 3)$

$\vec{n}_3= (2; -3)$

$\vec{n}_4= (3; 2)$

$2x+3y+1=0$

$2x-y+1=0$

$2x-4y-2=0$

$x-2y+1=0$

$\dfrac{x^2}{16} + \dfrac{y^2}{7} = -1$

$\dfrac{x^2}{25} - \dfrac{y^2}{7} = 1$

$\dfrac{x^2}{25} + \dfrac{y^2}{16} = 1$

$\dfrac{x^2}{7} - \dfrac{y^2}{16} = -1$

$\dfrac{x^2}{16} - \dfrac{y^2}{7} = -1$

$\dfrac{x^2}{1} + \dfrac{y^2}{6} = -1$

$\dfrac{x^2}{25} + \dfrac{y^2}{16} = 1$

$\dfrac{x^2}{7} - \dfrac{y^2}{16} = 1$

$x^2 = -6y$

$x^2 = 6y$

$y^2 = x$

$y^2 = -6x$