Đáp án:
$\begin{array}{l}
a)\dfrac{{2x + 1}}{2} + 3 \ge \dfrac{{3 - 5x}}{3} - \dfrac{{4x + 1}}{4}\\
\Rightarrow \dfrac{{6\left( {2x + 1} \right) + 3.12}}{{12}} \ge \dfrac{{4\left( {3 - 5x} \right) - 3.\left( {4x + 1} \right)}}{{12}}\\
\Rightarrow 12x + 6 + 36 \ge 12 - 20x - 12x - 3\\
\Rightarrow 44x \ge 33\\
\Rightarrow x \ge \dfrac{3}{4}\\
b)\dfrac{{3 - 5x}}{{ - 4}} \le 0\\
\Rightarrow 3 - 5x \ge 0\\
\Rightarrow 5x \le 3\\
\Rightarrow x \le \dfrac{3}{5}\\
c)\dfrac{{2x\left( {3x - 5} \right)}}{{{x^2} + 1}} < 0\\
\Rightarrow 2x\left( {3x - 5} \right) < 0\left( {do:{x^2} + 1 > 0} \right)\\
\Rightarrow 0 < x < \dfrac{5}{3}
\end{array}$
