Đề thi Học kì 2 Toán 10 năm 2024-2025 trường THPT Thị xã Quảng Trị

$y=10x^{2}$

$y^{2}=20x$

$y^{2}=\dfrac{5}{2}x$

$y^{2}=5x$

$I\left(4;-2\right);R=3$.

$I\left(4;-2\right);R=9$.

$I\left(-4;2\right);R=9$.

$I\left(-4;2\right);R=3$.

$\vec{n}=\left(-2;-1\right)$.

$\vec{n} =\left(2;1\right)$.

$\vec{n}=\left(4;-2\right)$.

$\vec{n}=\left(1;2\right)$.

$\dfrac{x^{2}}{9}+\dfrac{y^{2}}{9}=1$.

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

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

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

$y=\left(x^{2}-4x+1\right)\left(x+2\right).$

$y=\dfrac{2x}{x+2}$.

$y=-2x^{2}+3x+1$.

$y=3x+5$

25.

20.

35.

7.

$I(2;-3)$

$I(1;2)$

$I(1;-2)$

$I(-1;-6)$

$\dfrac{x^2}{9}-\dfrac{y^2}{4}=1$

$\dfrac{x^2}{9}+\dfrac{y^2}{4}=1$

$\dfrac{x^2}{9}-\dfrac{y^2}{4}=0$

$y^2=4x$

7

$C_7^3$

$A_7^3$

$\dfrac{7!}{3!}$

$\dfrac{2}{10}$

$\dfrac{5}{10}$

$\dfrac{4}{10}$

$\dfrac{3}{10}$