Đáp án:
\(\dfrac{{{x^2} - 5x}}{{\left( {{x^3} + 1} \right)\left( {x + 2} \right)}}\)
Giải thích các bước giải:
\(\begin{array}{l}
DK:x \ne - 1\\
\left( {\dfrac{1}{{x + 1}} - \dfrac{{6x + 3}}{{{x^3} + 1}} + \dfrac{2}{{{x^2} - x + 1}}} \right):\left( {x + 2} \right)\\
= \left[ {\dfrac{{{x^2} - x + 1 - 6x - 3 + 2\left( {x + 1} \right)}}{{\left( {x + 1} \right)\left( {{x^2} - x + 1} \right)}}} \right].\dfrac{1}{{x + 2}}\\
= \dfrac{{{x^2} - 5x}}{{\left( {x + 1} \right)\left( {{x^2} - x + 1} \right)}}.\dfrac{1}{{x + 2}}\\
= \dfrac{{{x^2} - 5x}}{{\left( {{x^3} + 1} \right)\left( {x + 2} \right)}}
\end{array}\)
