Đáp án:
$\begin{array}{l}
a)MSC = 12\\
\left\{ \begin{array}{l}
\dfrac{x}{6} = \dfrac{{2x}}{{12}}\\
\dfrac{{x - 3}}{3} = \dfrac{{4x - 12}}{{12}}\\
\dfrac{{{x^3}}}{4} = \dfrac{{3{x^3}}}{{12}}
\end{array} \right.\\
b)MSC = \left( {x - 1} \right)\left( {x + 1} \right)\\
\left\{ \begin{array}{l}
\dfrac{x}{{x + 1}} = \dfrac{{x\left( {x - 1} \right)}}{{\left( {x - 1} \right)\left( {x + 1} \right)}} = \dfrac{{{x^2} - x}}{{\left( {x - 1} \right)\left( {x + 1} \right)}}\\
\dfrac{{{x^2}}}{{1 - x}} = \dfrac{{ - {x^2}\left( {x + 1} \right)}}{{\left( {x - 1} \right)\left( {x + 1} \right)}} = \dfrac{{ - {x^3} - {x^2}}}{{\left( {x - 1} \right)\left( {x + 1} \right)}}\\
\dfrac{1}{{{x^2} - 1}} = \dfrac{1}{{\left( {x - 1} \right)\left( {x + 1} \right)}}
\end{array} \right.\\
c)MSC = 6\left( {x + 3} \right)\left( {x - 2} \right)\left( {x + 2} \right)\\
\left\{ \begin{array}{l}
\dfrac{2}{{x + 3}} = \dfrac{{2.6\left( {x - 2} \right)\left( {x + 2} \right)}}{{6\left( {x + 3} \right)\left( {x - 2} \right)\left( {x + 2} \right)}} = \dfrac{{12{x^2} - 24}}{{6\left( {x + 3} \right)\left( {x - 2} \right)\left( {x + 2} \right)}}\\
\dfrac{5}{{3{x^2} - 12x}} = \dfrac{{5.2\left( {x + 3} \right)}}{{6\left( {x + 3} \right)\left( {x - 2} \right)\left( {x + 2} \right)}} = \dfrac{{10x + 30}}{{6\left( {x + 3} \right)\left( {x - 2} \right)\left( {x + 2} \right)}}\\
\dfrac{3}{{\left( {2x + 4} \right)\left( {x + 3} \right)}} = \dfrac{{3.3\left( {x - 2} \right)}}{{6\left( {x + 3} \right)\left( {x - 2} \right)\left( {x + 2} \right)}} = \dfrac{{9x - 18}}{{6\left( {x + 3} \right)\left( {x - 2} \right)\left( {x + 2} \right)}}
\end{array} \right.
\end{array}$