Đáp án:
$\begin{array}{l}
8)\dfrac{{3x - 2}}{{{x^3} + 1}}.\dfrac{{x - 1 - {x^2}}}{{4 - 9{x^2}}}\\
= \dfrac{{3x - 2}}{{\left( {x + 1} \right)\left( {{x^2} - x + 1} \right)}}.\dfrac{{ - \left( {{x^2} - x + 1} \right)}}{{\left( {2 - 3x} \right)\left( {2 + 3x} \right)}}\\
= \dfrac{{ - 1}}{{ - \left( {2x + 3} \right)\left( {x + 1} \right)}}\\
= \dfrac{1}{{\left( {2x + 3} \right)\left( {x + 1} \right)}}\\
9)\dfrac{{4x + 8}}{{{{\left( {x - 10} \right)}^2}}}.\dfrac{{2x - 20}}{{\left( {x + 2} \right)}}\\
= \dfrac{{4\left( {x + 2} \right)}}{{{{\left( {x - 10} \right)}^2}}}.\dfrac{{2\left( {x - 10} \right)}}{{x + 2}}\\
= \dfrac{8}{{x - 10}}
\end{array}$
