Đáp án:
$\begin{array}{l}
\sqrt {\dfrac{{x - 2}}{{x + 1}}} \\
Dkxd:\dfrac{{x - 2}}{{x + 1}} \ge 0\\
\Rightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x - 2 \ge 0\\
x + 1 > 0
\end{array} \right.\\
\left\{ \begin{array}{l}
x - 2 \le 0\\
x + 1 < 0
\end{array} \right.
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x \ge 2\\
x > - 1
\end{array} \right.\\
\left\{ \begin{array}{l}
x \le 2\\
x < - 1
\end{array} \right.
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
x \ge 2\\
x < - 1
\end{array} \right.\\
\Rightarrow TXD:D = \left( { - \infty ; - 1} \right) \cup \left[ {2; + \infty } \right)
\end{array}$
