Giải thích các bước giải:
\(\begin{array}{l}
\dfrac{{3x + 1}}{{{{\left( {x - 1} \right)}^2}}} - \dfrac{1}{{x + 1}} + \dfrac{{x + 3}}{{1 - {x^2}}}\\
= \dfrac{{3x + 1}}{{{{\left( {x - 1} \right)}^2}}} - \dfrac{1}{{\left( {x + 1} \right)}} - \dfrac{{\left( {x + 3} \right)}}{{\left( {x - 1} \right)\left( {x + 1} \right)}}\\
= \dfrac{{3x\left( {x + 1} \right) - {{\left( {x - 1} \right)}^2} - \left( {x + 3} \right)\left( {x - 1} \right)}}{{{{\left( {x - 1} \right)}^2}\left( {x + 1} \right)}}\\
= \dfrac{{3{x^2} + 3x - {x^2} + 2x - 1 - {x^2} - 2x + 3}}{{{{\left( {x - 1} \right)}^2}\left( {x + 1} \right)}}\\
= \dfrac{{{x^2} + 3x + 2}}{{{{\left( {x - 1} \right)}^2}\left( {x + 1} \right)}} = \dfrac{{{x^2} + 2x + x + 2}}{{{{\left( {x - 1} \right)}^2}\left( {x + 1} \right)}}\\
= \dfrac{{x\left( {x + 2} \right) + \left( {x + 2} \right)}}{{{{\left( {x - 1} \right)}^2}\left( {x + 1} \right)}} = \dfrac{{\left( {x + 2} \right)\left( {x + 1} \right)}}{{{{\left( {x - 1} \right)}^2}\left( {x + 1} \right)}}\\
= \dfrac{{x + 2}}{{{{\left( {x - 1} \right)}^2}}}
\end{array}\)
