Đáp án:
$\begin{array}{l}
a)\dfrac{{x - 1}}{{\sqrt {2x - 4} }}\\
Dkxd:2x - 4 > 0 \Leftrightarrow x > 2\\
Vậy\,x > 2\\
b)\dfrac{{3x}}{{\sqrt {{x^2}} }}\\
Dkxd:{x^2} > 0\\
\Leftrightarrow x\# 0\\
Vậy\,x\# 0\\
c)\dfrac{4}{{\sqrt {x + 5} }}\\
Dkxd:x + 5 > 0\\
\Leftrightarrow x > - 5\\
Vậy\,x > - 5\\
5)a)\\
\sqrt {2x + 2} .\sqrt {2x - 4} \\
Dkxd:\left\{ \begin{array}{l}
2x + 2 \ge 0\\
2x - 4 \ge 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x \ge - 1\\
x \ge 2
\end{array} \right.\\
Vậy\,x \ge 2\\
b)\sqrt {\left( {2x + 2} \right)\left( {2x - 4} \right)} \\
Dkxd:\left( {2x + 2} \right)\left( {2x - 4} \right) \ge 0\\
\Leftrightarrow 2.\left( {x + 1} \right).2\left( {x - 2} \right) \ge 0\\
\Leftrightarrow 4\left( {x + 1} \right)\left( {x - 2} \right) \ge 0\\
\Leftrightarrow \left[ \begin{array}{l}
x \ge 2\\
x \le - 1
\end{array} \right.\\
Vậy\,x \le - 1\,hoac\,x \ge 2\\
c)\dfrac{{\sqrt {3x + 1} }}{{\sqrt {x - 1} }}\\
Dkxd:\left\{ \begin{array}{l}
3x + 1 \ge 0\\
x - 1 > 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ge - \dfrac{1}{3}\\
x > 1
\end{array} \right. \Leftrightarrow x > 1\\
Vậy\,x > 1\\
d)\sqrt {\dfrac{{3x + 1}}{{x - 1}}} \\
Dkxd:\dfrac{{3x + 1}}{{x - 1}} \ge 0\\
\Leftrightarrow \left[ \begin{array}{l}
x > 1\\
x \le - \dfrac{1}{3}
\end{array} \right.\\
Vậy\,x \le - \dfrac{1}{3}\,hoac\,x > 1
\end{array}$
