x/x-2+ √x-2 x/x+2+ √x-2 x/x^2-4+ √x-2 √1/3-2x √4+2x+3 √-2/x+1 √x^2+1 √4x^2+3 √9x^2-6x-1 √-x^2+2x-1 √- x+5(phép đối x+5) √-2x^2-1

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}$