Đáp án:
$\begin{array}{l}
a)\dfrac{{6x + 3}}{{2x - 1}} > 0\\
\Rightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
6x + 3 > 0\\
2x - 1 > 0
\end{array} \right.\\
\left\{ \begin{array}{l}
6x + 3 < 0\\
2x - 1 < 0
\end{array} \right.
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x > - \dfrac{1}{2}\\
x > \dfrac{1}{2}
\end{array} \right.\\
\left\{ \begin{array}{l}
x < - \dfrac{1}{2}\\
x < \dfrac{1}{2}
\end{array} \right.
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
x > \dfrac{1}{2}\\
x < - \dfrac{1}{2}
\end{array} \right.\\
b)\dfrac{{5 - 2x}}{{3 + x}} < 0\\
\Rightarrow \dfrac{{2x - 5}}{{x + 3}} > 0\\
\Rightarrow \left[ \begin{array}{l}
x > \dfrac{5}{2}\\
x < - 3
\end{array} \right.\\
c)\dfrac{{9 - 5x}}{{2x + 1}} > 0\\
\Rightarrow \dfrac{{5x - 9}}{{2x + 1}} < 0\\
\Rightarrow - \dfrac{1}{2} < x < \dfrac{9}{5}\\
d)\dfrac{{10 - x}}{{5 + x}} < 0\\
\Rightarrow \dfrac{{x - 10}}{{x + 5}} > 0\\
\Rightarrow \left[ \begin{array}{l}
x > 10\\
x < - 5
\end{array} \right.\\
e)\dfrac{{9 - 3x}}{{x + 1}} > 0\\
\Rightarrow \dfrac{{3x - 9}}{{x + 1}} < 0\\
\Rightarrow - 1 < x < 3
\end{array}$
