Đáp án:
$\begin{array}{l}
a)\dfrac{x}{{x - 2}} + \sqrt {x - 2} \\
Dkxd:\left\{ \begin{array}{l}
x - 2\# 0\\
x - 2 \ge 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x\# 2\\
x \ge 2
\end{array} \right. \Leftrightarrow x > 2\\
Vậy\,x > 2\\
b)\dfrac{x}{{x + 2}} + \sqrt {x - 2} \\
Dkxd:\left\{ \begin{array}{l}
x + 2\# 0\\
x - 2 \ge 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x\# - 2\\
x \ge 2
\end{array} \right. \Leftrightarrow x \ge 2\\
Vậy\,x \ge 2\\
c)\dfrac{x}{{{x^2} - 4}} + \sqrt {x - 2} \\
Dkxd:\left\{ \begin{array}{l}
{x^2} - 4\# 0\\
x - 2 \ge 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
{x^2}\# 4\\
x \ge 2
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x\# 2;x\# - 2\\
x \ge 2
\end{array} \right.\\
\Leftrightarrow x > 2\\
Vậy\,x > 2\\
d)\sqrt {\dfrac{1}{{3 - 2x}}} \\
Dkxd:\dfrac{1}{{3 - 2x}} \ge 0\\
\Leftrightarrow 3 - 2x > 0\\
\Leftrightarrow x < \dfrac{3}{2}\\
Vậy\,x < \dfrac{3}{2}\\
e)\sqrt {4 + 2x + 3} \\
Dkxd:4 + 2x + 3 \ge 0\\
\Leftrightarrow x \ge - \dfrac{7}{2}\\
Vậy\,x \ge \dfrac{{ - 7}}{2}\\
f)\sqrt {\dfrac{{ - 2}}{{x + 1}}} \\
Dkxd:\dfrac{{ - 2}}{{x + 1}} \ge 0\\
\Leftrightarrow x + 1 < 0\\
\Leftrightarrow x < - 1\\
Vậy\,x < - 1\\
g)\sqrt {{x^2} + 1} \\
DKxd:{x^2} + 1 \ge 0\left( {tm} \right)\\
Vậy\,x \in R\\
h)\sqrt {4{x^2} + 3} \\
DKxd:4{x^2} + 3 \ge 0\left( {tm} \right)\\
Vậy\,x \in R\\
k)\sqrt {9{x^2} - 6x - 1} \\
Dkxd:9{x^2} - 6x - 1 \ge 0\\
\Leftrightarrow 9{x^2} - 6x + 1 \ge 2\\
\Leftrightarrow {\left( {3x - 1} \right)^2} \ge 2\\
\Leftrightarrow \left[ \begin{array}{l}
3x - 1 \ge \sqrt 2 \\
3x - 1 \le - \sqrt 2
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x \ge \dfrac{{\sqrt 2 + 1}}{3}\\
x \le \dfrac{{1 - \sqrt 2 }}{3}
\end{array} \right.\\
Vậy\,x \ge \dfrac{{\sqrt 2 + 1}}{3}\,hoac\,x \le \dfrac{{1 - \sqrt 2 }}{3}\\
l)\sqrt { - {x^2} + 2x - 1} \\
Dkxd: - {x^2} + 2x - 1 \ge 0\\
\Leftrightarrow {x^2} - 2x + 1 \le 0\\
\Leftrightarrow {\left( {x - 1} \right)^2} \le 0\\
\Leftrightarrow x - 1 = 0\\
\Leftrightarrow x = 1\\
Vậy\,x = 1\\
m)\sqrt { - x + 5} \\
Dkxd: - x + 5 \ge 0\\
\Leftrightarrow x \le 5\\
Vậy\,x \le 5\\
n)\sqrt { - 2{x^2} - 1} \\
Dkxd: - 2{x^2} - 1 \ge 0\\
\Leftrightarrow 2{x^2} \le - 1\left( {ktm} \right)\\
Vậy\,x \in \emptyset
\end{array}$
