Đề thi Giữa kì 2 Toán 10 năm 2024-2025 trường THPT Lê Lợi (Kon Tum)

$x=-\dfrac{b}{a}$

$x=\dfrac{b}{2a}$

$x=-\dfrac{b}{2a}$

$y=-\dfrac{b}{2a}$

$x=4$

$x=-2$

$x=0$

$x=3$

3

0

2

1

$\overrightarrow{n}=(2;3)$

$\overrightarrow{n}=(3;2)$

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

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

$\overrightarrow{u}(2;-3)$

$\overrightarrow{u}(2;3)$

$\overrightarrow{u}(-3;2)$

$\overrightarrow{u}(1;2)$

$I\left(-\dfrac{b}{2a};- \dfrac{\Delta}{4a} \right)$.

$I\left(-\dfrac{b}{2a}; \dfrac{\Delta}{4a} \right)$.

$I\left(\dfrac{b}{2a}; \dfrac{\Delta}{4a} \right)$.

$I\left(- \dfrac{b}{a}; - \dfrac{\Delta}{4a} \right)$.

$\left\{\begin{array}{l} x = -2 + 3t \\ y = 1 - 4t \end{array} \ (t \in \mathbb{R}) \right.$

$\left\{\begin{array}{l} x = 3-2t \\ y = -4 + t \end{array} \ (t \in \mathbb{R}) \right.$

$\left\{\begin{array}{l} x = 1 - 2t \\ y = -4 + 3t \end{array} \ (t \in \mathbb{R}) \right.$

$\left\{\begin{array}{l} x = -2 + t \\ y = 3 - 4t \end{array} \ (t \in \mathbb{R}) \right.$

$\mathbb{R} \setminus \{1\}$.

$(1; +\infty)$.

$\mathbb{R} \setminus \{ \pm 1 \}$.

$\mathbb{R} \setminus \{-1\}$.

$y = 2x^4 + x^2 - 1$.

$y = x^2 + 2x -1$.

$y = 2x -1$.

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