Đáp án:
$\begin{array}{l}
\frac{3}{{x - 2}} \le \frac{5}{{2x - 1}}\\
\Rightarrow \frac{3}{{x - 2}} - \frac{5}{{2x - 1}} \le 0\\
\Rightarrow \frac{{3\left( {2x - 1} \right) - 5\left( {x - 2} \right)}}{{\left( {x - 2} \right)\left( {2x - 1} \right)}} \le 0\\
\Rightarrow \frac{{x + 7}}{{\left( {x - 2} \right)\left( {2x - 1} \right)}} \le 0\\
\Rightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x + 7 \le 0\\
\left( {x - 2} \right)\left( {2x - 1} \right) > 0
\end{array} \right.\\
\left\{ \begin{array}{l}
x + 7 \ge 0\\
\left( {x - 2} \right)\left( {2x - 1} \right) < 0
\end{array} \right.
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x \le - 7\\
\left[ \begin{array}{l}
x > 2\\
x < \frac{1}{2}
\end{array} \right.
\end{array} \right.\\
\left\{ \begin{array}{l}
x \ge - 7\\
\frac{1}{2} < x < 2
\end{array} \right.
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
x \le - 7\\
\frac{1}{2} < x < 2
\end{array} \right.\\
Vậy\,x \le - 7\,hoặc\,\frac{1}{2} < x < 2\\
\left| {2x - 1} \right| < 3x + 5\\
\Rightarrow \left\{ \begin{array}{l}
3x + 5 > 0\\
- 3x - 5 < 2x - 1 < 3x + 5
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
x > \frac{{ - 5}}{3}\\
x > - \frac{4}{5}
\end{array} \right. \Rightarrow x > - \frac{4}{5}\\
Vậy\,x > - \frac{4}{5}
\end{array}$
