Đáp án:
$\begin{array}{l}
1)\dfrac{1}{{\sqrt {x + 2} }}\\
Dkxd:x + 2 > 0\\
\Leftrightarrow x > - 2\\
Vậy\,x > - 2\\
2)\sqrt {\dfrac{3}{{ - 2 + x}}} \\
Dkxd:\dfrac{3}{{ - 2 + x}} \ge 0\\
\Leftrightarrow - 2 + x > 0\\
\Leftrightarrow x > 2\\
Vậy\,x > 2\\
3)\sqrt {\dfrac{{x + 1}}{x}} \\
Dkxd:\dfrac{{x + 1}}{x} \ge 0\\
\Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x + 1 \ge 0\\
x > 0
\end{array} \right.\\
\left\{ \begin{array}{l}
x + 1 \le 0\\
x < 0
\end{array} \right.
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x > 0\\
x \le - 1
\end{array} \right.\\
Vậy\,x \le - 1\,hoặc\,x > 0\\
4)\dfrac{2}{{\sqrt {1 - x} }}\\
Dkxd:1 - x > 0\\
\Leftrightarrow x < 1\\
Vậy\,x < 1\\
5)\dfrac{1}{{\sqrt x }} - \dfrac{1}{{\sqrt x - 2}}\\
Dkxd:\left\{ \begin{array}{l}
x > 0\\
\sqrt x \# 2
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x > 0\\
x\# 4
\end{array} \right.\\
Vậy\,x > 0;x\# 4
\end{array}$
