Giúp e với ạ e xin đa tạ ạ

Đáp án:

$36) D=[1;+\infty;) \setminus \{4\}\\ 37) A \cup B=(-\infty;-3] \cup (-2;2) \cup  [3;+\infty).$

Giải thích các bước giải:

$36)\\y=\sqrt{3x-3}+\dfrac{x}{x-4}\\\text{ĐKXĐ: } \left\{\begin{array}{l} 3x-3 \ge 0 \\ x-4 \ne 0\end{array} \right.\\\Leftrightarrow \left\{\begin{array}{l} 3x \ge 3 \\ x \ne 4\end{array} \right.\\\Leftrightarrow \left\{\begin{array}{l} x \ge 1 \\ x \ne 4\end{array} \right.\\\Rightarrow \text{TXĐ: } D=[1;+\infty;) \setminus \{4\}\\37)\\A=\{x \in \mathbb{R} | -1 < x^2 <4\}\\-1 < x^2 <4\\\Leftrightarrow x^2 <4 (\text{Do } -1 < x^2 \ \forall \ x \in \mathbb{R})\\\Leftrightarrow |x| <2\\\Leftrightarrow -2<x <2\\\Rightarrow A=(-2;2)\\B=\{x \in \mathbb{R} ||x| \ge 3\}\\|x| \ge 3 \Leftrightarrow  \left[\begin{array}{l} x \ge 3 \\ x \le -3\end{array} \right.\\\Rightarrow B=(-\infty;-3] \cup [3;+\infty)\\A \cup B=(-\infty;-3] \cup (-2;2) \cup  [3;+\infty).$