Đáp án:

\(\begin{array}{l}
Bai\,\,3:\\
a)\,\,S = \left\{ 1 \right\}.\\
b)\,\,S = \left\{ 2 \right\}.\\
c)\,\,S = \left\{ 3 \right\}.\\
d)\,\,S = \emptyset .\\
Bai\,\,4:\\
a)\,\,\,S = \left\{ 0 \right\}.\\
b)\,\,\,S = \left\{ {\frac{3}{2}} \right\}.\\
c)\,\,\,S = \left\{ 5 \right\}.\\
d)\,\,S = \emptyset .
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
Bai\,\,3:\\
a)\,\,\sqrt {3 - x}  + x = \sqrt {3 - x}  + 1.\\
DKXD:\,\,3 - x \ge 0 \Leftrightarrow x \le 3\\
Pt \Leftrightarrow x = 1\,\,\left( {tm} \right)\\
Vay\,\,S = \left\{ 1 \right\}.\\
b)\,\,x + \sqrt {x - 2}  = \sqrt {2 - x}  + 2\\
DKXD:\,\,\left\{ \begin{array}{l}
x - 2 \ge 0\\
2 - x \ge 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ge 2\\
x \le 2
\end{array} \right. \Leftrightarrow x = 2\\
Thu\,\,lai:\,\,2 + \sqrt {2 - 2}  = \sqrt {2 - 2}  + 2\,\,\left( {Dung} \right).\\
Vay\,\,S = \left\{ 2 \right\}.\\
c)\,\,\frac{{{x^2}}}{{\sqrt {x - 1} }} = \frac{9}{{\sqrt {x - 1} }}\\
DKXD:\,\,x - 1 > 0 \Leftrightarrow x > 1\\
Pt \Leftrightarrow {x^2} = 9 \Leftrightarrow \left[ \begin{array}{l}
x = 3\,\,\left( {tm} \right)\\
x =  - 3\,\,\left( {ktm} \right)
\end{array} \right.\\
Vay\,\,S = \left\{ 3 \right\}.\\
d)\,\,{x^2} - \sqrt {1 - x}  = \sqrt {x - 2}  + 3\\
DKXD:\,\,\left\{ \begin{array}{l}
1 - x \ge 0\\
x - 2 \ge 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \le 1\\
x \ge 2
\end{array} \right. \Rightarrow x \in \emptyset \\
Vay\,\,S = \emptyset .\\
Bai\,\,4:\\
a)\,\,x + 1 + \frac{2}{{x + 3}} = \frac{{x + 5}}{{x + 3}}\,\,\left( {x \ne  - 3} \right)\\
 \Leftrightarrow x + 1 = \frac{{x + 5}}{{x + 3}} - \frac{2}{{x + 3}}\\
 \Leftrightarrow x + 1 = \frac{{x + 5 - 2}}{{x + 3}}\\
 \Leftrightarrow x + 1 = 1 \Leftrightarrow x = 0\,\,\left( {tm} \right).\\
Vay\,\,S = \left\{ 0 \right\}.\\
b)\,\,2x + \frac{3}{{x - 1}} = \frac{{3x}}{{x - 1}}\,\,\left( {x \ne 1} \right)\\
 \Leftrightarrow 2x = \frac{{3x - 3}}{{x - 1}} = \frac{{3\left( {x - 1} \right)}}{{x - 1}} = 3\\
 \Leftrightarrow x = \frac{3}{2}\,\,\left( {tm} \right).\\
Vay\,\,S = \left\{ {\frac{3}{2}} \right\}.\\
c)\,\,\frac{{{x^2} - 4x - 2}}{{\sqrt {x - 2} }} = \sqrt {x - 2} \,\,\left( {x > 2} \right)\\
 \Leftrightarrow {x^2} - 4x - 2 = {\left( {\sqrt {x - 2} } \right)^2} = x - 2\\
 \Leftrightarrow {x^2} - 5x = 0\\
 \Leftrightarrow x\left( {x - 5} \right) = 0 \Leftrightarrow \left[ \begin{array}{l}
x = 0\,\,\left( {ktm} \right)\\
x = 5\,\,\left( {tm} \right)
\end{array} \right.\\
Vay\,S = \left\{ 5 \right\}.\\
d)\,\,\frac{{2{x^2} - x - 3}}{{\sqrt {2x - 3} }} = \sqrt {2x - 3} \,\,\left( {x > \frac{3}{2}} \right)\\
 \Leftrightarrow 2{x^2} - x - 3 = 2x - 3\\
 \Leftrightarrow 2{x^2} - 3x = 0\\
 \Leftrightarrow x\left( {2x - 3} \right) = 0 \Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x = \frac{3}{2}
\end{array} \right.\,\,\left( {ktm} \right)\\
Vay\,\,S = \emptyset .
\end{array}\)