Bài giải:

$\sqrt{\frac{x+1}{x-2}}$ có nghĩa $\Leftrightarrow \frac{x+1}{x-2}\ge 0$

TH1: $\left\{ \begin{align}& x+1\ge 0 \\& x-2>0 \\\end{align} \right.$

$\Leftrightarrow \left\{ \begin{align}& x\ge -1 \\& x>2 \\\end{align} \right.\Leftrightarrow x>2$

TH2: $\left\{ \begin{align}& x+1\le 0 \\& x-2 < 0 \\\end{align} \right.$

$\Leftrightarrow \left\{ \begin{align}  & x\le -1 \\& x < 2 \\\end{align} \right.\Leftrightarrow x\le -1$

Vậy $x\le -1$hoặc $x>2$