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