Đáp án:
$\begin{array}{l}
a)MSC:5 - 2x\\
\Rightarrow \left\{ \begin{array}{l}
\dfrac{{5{x^2} + 2}}{{2x - 5}} = \dfrac{{ - 5{x^2} - 2}}{{5 - 2x}}\\
\dfrac{{8x + 7}}{{5 - 2x}}
\end{array} \right.\\
b)MSC:\left( {x - 3} \right)\left( {x + 3} \right) = {x^2} - 9\\
\left\{ \begin{array}{l}
\dfrac{{5x}}{{x - 3}} = \dfrac{{5x\left( {x + 3} \right)}}{{\left( {x - 3} \right)\left( {x + 3} \right)}} = \dfrac{{5{x^2} + 15x}}{{{x^2} - 9}}\\
\dfrac{{7x}}{{x + 3}} = \dfrac{{7x\left( {x - 3} \right)}}{{{x^2} - 9}} = \dfrac{{21{x^2} - 21x}}{{{x^2} - 9}}
\end{array} \right.\\
c)MSC:2{\left( {x + 5} \right)^2}\\
\left\{ \begin{array}{l}
\dfrac{5}{{{x^2} + 10x + 25}} = \dfrac{{5.2}}{{2.{{\left( {x + 5} \right)}^2}}} = \dfrac{{10}}{{2{{\left( {x + 5} \right)}^2}}}\\
\dfrac{{x - 5}}{{2x + 10}} = \dfrac{{\left( {x - 5} \right)\left( {x + 5} \right)}}{{2{{\left( {x + 5} \right)}^2}}} = \dfrac{{{x^2} - 25}}{{2{{\left( {x + 5} \right)}^2}}}
\end{array} \right.\\
d)MSC:\left( {x - 1} \right)\left( {x + 2} \right)\left( {x - 2} \right) = \left( {x - 1} \right)\left( {{x^2} - 4} \right)\\
\left\{ \begin{array}{l}
\dfrac{{x + 1}}{{\left( {x - 1} \right)\left( {x - 2} \right)}} = \dfrac{{\left( {x + 1} \right)\left( {x + 2} \right)}}{{\left( {x - 1} \right)\left( {{x^2} - 4} \right)}} = \dfrac{{{x^2} + 3x + 2}}{{\left( {x - 1} \right)\left( {{x^2} - 4} \right)}}\\
\dfrac{{x + 3}}{{\left( {x - 1} \right)\left( {x + 2} \right)}} = \dfrac{{\left( {x + 3} \right)\left( {x - 2} \right)}}{{\left( {x - 1} \right)\left( {{x^2} - 4} \right)}} \Rightarrow \dfrac{{{x^2} + x - 6}}{{\left( {x - 1} \right)\left( {{x^2} - 4} \right)}}
\end{array} \right.
\end{array}$
