Đáp án:
$\begin{array}{l}
a)MSC:2\left( {x + 1} \right)\left( {x - 1} \right)\\
\Rightarrow \left\{ \begin{array}{l}
\dfrac{5}{{2x + 2}} = \dfrac{5}{{2\left( {x + 1} \right)}} = \dfrac{{5.\left( {x - 1} \right)}}{{2\left( {x + 1} \right)\left( {x - 1} \right)}} = \dfrac{{5x - 5}}{{2\left( {x + 1} \right)\left( {x - 1} \right)}}\\
\dfrac{9}{{{x^2} - 1}} = \dfrac{9}{{\left( {x + 1} \right)\left( {x - 1} \right)}} = \dfrac{{9.2}}{{2\left( {x + 1} \right)\left( {x - 1} \right)}} = \dfrac{{18}}{{2\left( {x + 1} \right)\left( {x - 1} \right)}}
\end{array} \right.\\
b)\dfrac{1}{{4 - 2x}};\dfrac{3}{{{x^2} - 4}}\\
MSC:2\left( {x - 2} \right)\left( {x + 2} \right)\\
\Rightarrow \left\{ \begin{array}{l}
\dfrac{1}{{4 - 2x}} = \dfrac{{ - 1}}{{2\left( {x - 2} \right)}} = \dfrac{{ - x - 2}}{{2\left( {x - 2} \right)\left( {x + 2} \right)}}\\
\dfrac{3}{{{x^2} - 4}} = \dfrac{3}{{\left( {x - 2} \right)\left( {x + 2} \right)}} = \dfrac{6}{{2\left( {x - 2} \right)\left( {x + 2} \right)}}
\end{array} \right.\\
c)\dfrac{{x - 1}}{{{x^2} - 9}};\dfrac{{2x + 1}}{{2x + 6}}\\
MSC:2\left( {x + 3} \right)\left( {x - 3} \right)\\
\Rightarrow \left\{ \begin{array}{l}
\dfrac{{x - 1}}{{{x^2} - 9}} = \dfrac{{\left( {x - 1} \right).2}}{{2\left( {x + 3} \right)\left( {x - 3} \right)}} = \dfrac{{2x - 2}}{{2\left( {x + 3} \right)\left( {x - 3} \right)}}\\
\dfrac{{2x + 1}}{{2x + 6}} = \dfrac{{2x + 1}}{{2\left( {x + 3} \right)}} = \dfrac{{\left( {2x + 1} \right)\left( {x - 3} \right)}}{{2\left( {x + 3} \right)\left( {x - 3} \right)}}
\end{array} \right.
\end{array}$
