Đáp án:
$1) \left\{\begin{array}{l} x<-1\\ 0\le x<1\end{array} \right. \\ 2) x>2$
Giải thích các bước giải:
$1)\sqrt{\dfrac{-3x}{x^2-1}}\\ \text{ĐKXĐ:}\left\{\begin{array}{l} \dfrac{-3x}{x^2-1} \ge 0\\ x^2-1 \ne 0\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} \left[\begin{array}{l} \left\{\begin{array}{l} -3x\ge0\\ x^2-1\ge 0\end{array} \right. \\ \left\{\begin{array}{l} -3x \le 0\\ x^2-1 \le 0\end{array} \right. \end{array} \right.\\ (x-1)(x+1) \ne 0\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} \left[\begin{array}{l} \left\{\begin{array}{l} x \le 0\\ (x-1)(x+1) \ge 0\end{array} \right. \\ \left\{\begin{array}{l} x \ge 0\\ (x-1)(x+1)\le 0\end{array} \right. \end{array} \right.\\ (x-1)(x+1) \ne 0\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} \left[\begin{array}{l} \left\{\begin{array}{l} x \le 0\\ \left[\begin{array}{l}x>1 \\ x<-1\end{array} \right.\end{array} \right. \\ \left\{\begin{array}{l} x\ge0\\ -1<x<1\end{array} \right. \end{array} \right.\\ x \ne \pm 1\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} x<-1\\ 0\le x<1\end{array} \right. \\ 2)\dfrac{1}{\sqrt{x-2}}\\ ĐKXĐ: \left\{\begin{array}{l} x-2 \ge 0 \\ \sqrt{x-2} \ne 0\end{array} \right. \\ \Leftrightarrow \left\{\begin{array}{l} x \ge 2 \\ x \ne 2\end{array} \right. \\ \Leftrightarrow x>2$
