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