Giải thích các bước giải:

 Ta có:

\(\begin{array}{l}
1,\\
\left( {\sqrt 2  - \sqrt 3 } \right)x \le \sqrt 3  + \sqrt 2 \\
\sqrt 2  - \sqrt 3  < 0 \Rightarrow x \ge \frac{{\sqrt 3  + \sqrt 2 }}{{\sqrt 2  - \sqrt 3 }} \Leftrightarrow x \ge  - 5 - 2\sqrt 6 \\
1,\\
\left\{ \begin{array}{l}
\left( {x - 3} \right)\left( {\sqrt 2  - x} \right) > 0\\
\frac{{4x - 3}}{2} < x + 3
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
\left( {x - 3} \right)\left( {x - \sqrt 2 } \right) < 0\\
2x - \frac{3}{2} < x + 3
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
\sqrt 2  < x < 3\\
2x - x < 3 + \frac{3}{2}
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
\sqrt 2  < x < 3\\
x < \frac{9}{2}
\end{array} \right.\\
 \Leftrightarrow \sqrt 2  < x < 3\\
 \Rightarrow S = \left( {\sqrt 2 ;3} \right)\\
2,\\
\left\{ \begin{array}{l}
\frac{2}{{2x - 1}} \le \frac{1}{{3 - x}}\\
\left| x \right| < 1
\end{array} \right.\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( \begin{array}{l}
x \ne \frac{1}{2}\\
x \ne 3
\end{array} \right)\\
 \Leftrightarrow \left\{ \begin{array}{l}
\frac{2}{{2x - 1}} - \frac{1}{{3 - x}} \le 0\\
\left| x \right| < 1
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
\frac{{2\left( {3 - x} \right) - \left( {2x - 1} \right)}}{{\left( {2x - 1} \right)\left( {3 - x} \right)}} \le 0\\
 - 1 < x < 1
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
\frac{{7 - 4x}}{{\left( {2x - 1} \right)\left( {3 - x} \right)}} \le 0\\
 - 1 < x < 1
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
\frac{{4x - 7}}{{\left( {2x - 1} \right)\left( {x - 3} \right)}} \le 0\\
 - 1 < x < 1
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
\left[ \begin{array}{l}
x < \frac{1}{2}\\
\frac{7}{4} \le x < 3
\end{array} \right.\\
 - 1 < x < 1
\end{array} \right.\\
 \Leftrightarrow  - 1 < x < \frac{1}{2}\\
 \Rightarrow S = \left( { - 1;\frac{1}{2}} \right)
\end{array}\)