Đáp án:
$\begin{array}{l}
4)MSC = {\left( {x + 1} \right)^2}{\left( {1 - x} \right)^2}\\
\left\{ \begin{array}{l}
\dfrac{3}{{{x^2} + 2x + 1}} = \dfrac{{3{{\left( {1 - x} \right)}^2}}}{{{{\left( {x + 1} \right)}^2}{{\left( {1 - x} \right)}^2}}}\\
\dfrac{6}{{1 - {x^2}}} = \dfrac{6}{{\left( {x + 1} \right)\left( {1 - x} \right)}} = \dfrac{{6\left( {1 - {x^2}} \right)}}{{{{\left( {x + 1} \right)}^2}{{\left( {1 - x} \right)}^2}}}\\
\dfrac{7}{{{x^2} - 2x + 1}} = \dfrac{{7{{\left( {x + 1} \right)}^2}}}{{{{\left( {x + 1} \right)}^2}{{\left( {1 - x} \right)}^2}}}
\end{array} \right.\\
5)MSC = \left( {a + b} \right)\left( {{a^2} - ab + {b^2}} \right)\\
\left\{ \begin{array}{l}
\dfrac{{2a}}{{{a^3} + {b^3}}} = \dfrac{{2a}}{{\left( {a + b} \right)\left( {{a^2} - ab + {b^2}} \right)}}\\
\dfrac{{ - 3b}}{{a + b}} = \dfrac{{ - 3b\left( {{a^2} - ab + {b^2}} \right)}}{{\left( {a + b} \right)\left( {{a^2} - ab + {b^2}} \right)}}\\
\dfrac{{4c}}{{{a^2} - ab + {b^2}}} = \dfrac{{4c.\left( {a + b} \right)}}{{\left( {a + b} \right)\left( {{a^2} - ab + {b^2}} \right)}}
\end{array} \right.
\end{array}$
