Đề thi Giữa kì 2 Toán 10 năm 2025-2026 trường THPT Duy Tân (Quảng Ngãi)

$M=(1;3)$

$M=(-1;3)$

$M=(3;1)$

$M=(-3;1)$

$D = \mathbb{R}\backslash \left\{ \dfrac{1}{3} \right\}$

$D=\mathbb{R}$

$D=\mathbb{R}\backslash \{3\}$

$D=\mathbb{R}\backslash \{3; -1\}$

$-\dfrac{5}{2}$

$-2$

$-1$

$-5$

$y = 2x - 3$

$y = 5x^{2} - 3x + 4$

$y^{2} = x$

$y = \sqrt{x^{2} - 1}$

$f(x) = 3x^{2} + 4x - 1$

$f(x) = \sqrt{x^{2} + 3x + 4}$

$f(x) = 3x^{2} + 4 + \dfrac{1}{x}$

$f(x) = x + 8$

$S = \{1\}$

$S = \varnothing$

$S = \left\{1; \dfrac{5}{2}\right\}$

$S = \left\{ \dfrac{5}{2}\right\}$

$(-\infty; -2) \cup (3; +\infty)$

$[-2; 3]$

$(-\infty; -2] \cup [3; +\infty)$

$(-2; 3)$

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

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

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

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

$3x - 2y + 4 = 0$

$3x - 2y - 4 = 0$

$2x + 3y + 4 = 0$

$4x + 6y - 8 = 0$

$d : \left\{ \begin{array}{l} x = 1 + 3t \\ y = -2 + 5t \end{array} \right.$

$d : \left\{ \begin{array}{l} x = 1 + 5t \\ y = -2 - 3t \end{array} \right.$

$d : \left\{ \begin{array}{l} x = 3 + t \\ y = 5 - 2t \end{array} \right.$

$d : \left\{ \begin{array}{l} x = 3 + 2t \\ y = 5 + t \end{array} \right.$