Đáp án:
$\begin{array}{l}
3)\dfrac{{3{x^2} + 5x + 1}}{{{x^3} - 1}} - \dfrac{{1 - x}}{{{x^2} + x + 1}} - \dfrac{3}{{x - 1}}\\
= \dfrac{{3{x^2} + 5x + 1 - \left( {1 - x} \right)\left( {x - 1} \right) - 3\left( {{x^2} + x + 1} \right)}}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}}\\
= \dfrac{{3{x^2} + 5x + 1 + {x^2} - 2x + 1 - 3{x^2} - 3x - 3}}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}}\\
= \dfrac{{{x^2} - 1}}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}}\\
= \dfrac{{x + 1}}{{{x^2} + x + 1}}\\
4)\dfrac{1}{{x - 2}} - \dfrac{6}{{{x^3} - 8}} + \dfrac{{x - 2}}{{{x^2} + 2x + 4}}\\
= \dfrac{{{x^2} + 2x + 4 - 6 + \left( {x - 2} \right)\left( {x - 2} \right)}}{{\left( {x - 2} \right)\left( {{x^2} + 2x + 4} \right)}}\\
= \dfrac{{{x^2} + 2x - 2 + {x^2} - 4x + 4}}{{\left( {x - 2} \right)\left( {{x^2} + 2x + 4} \right)}}\\
= \dfrac{{2{x^2} - 2x + 2}}{{\left( {x - 2} \right)\left( {{x^2} + 2x + 4} \right)}}\\
5)\dfrac{{3x + 1}}{{{{\left( {x - 1} \right)}^2}}} - \dfrac{1}{{x + 1}} + \dfrac{{x + 3}}{{1 - {x^2}}}\\
= \dfrac{{\left( {3x + 1} \right)\left( {x + 1} \right) - {{\left( {x - 1} \right)}^2} - \left( {x + 3} \right)\left( {x - 1} \right)}}{{{{\left( {x - 1} \right)}^2}\left( {x + 1} \right)}}\\
= \dfrac{{3{x^2} + 4x + 1 - {x^2} + 2x - 1 - {x^2} - 2x + 3}}{{{{\left( {x - 1} \right)}^2}\left( {x + 1} \right)}}\\
= \dfrac{{{x^2} + 4x + 3}}{{{{\left( {x - 1} \right)}^2}\left( {x + 1} \right)}}\\
= \dfrac{{\left( {x + 1} \right)\left( {x + 3} \right)}}{{{{\left( {x - 1} \right)}^2}\left( {x + 1} \right)}}\\
= \dfrac{{x + 3}}{{{{\left( {x - 1} \right)}^2}}}
\end{array}$
