Đáp án:
\(\dfrac{{{x^2} + 10x - 1}}{{x\left( {x - 1} \right)\left( {x + 2} \right)}}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\left( {\dfrac{1}{{x + 1}} - \dfrac{3}{{{x^3} + 1}} + \dfrac{3}{{{x^2} - x + 1}}} \right).\dfrac{{3{x^2} - 3x + 3}}{{\left( {x + 1} \right)\left( {x - 1} \right)}} - \dfrac{{2x - 2}}{{{x^2} + 2x}}\\
= \left[ {\dfrac{{{x^2} - x + 1 - 3 + 3x + 3}}{{\left( {x + 1} \right)\left( {{x^2} - x + 1} \right)}}} \right].\dfrac{{3\left( {{x^2} - x + 1} \right)}}{{\left( {x + 1} \right)\left( {x - 1} \right)}} - \dfrac{{2x - 2}}{{{x^2} + 2x}}\\
= \dfrac{{{x^2} + 2x + 1}}{{\left( {x + 1} \right)\left( {{x^2} - x + 1} \right)}}.\dfrac{{3\left( {{x^2} - x + 1} \right)}}{{\left( {x + 1} \right)\left( {x - 1} \right)}} - \dfrac{{2x - 2}}{{{x^2} + 2x}}\\
= \dfrac{{{{\left( {x + 1} \right)}^2}}}{{\left( {x + 1} \right)\left( {{x^2} - x + 1} \right)}}.\dfrac{{3\left( {{x^2} - x + 1} \right)}}{{\left( {x + 1} \right)\left( {x - 1} \right)}} - \dfrac{{2x - 2}}{{x\left( {x + 2} \right)}}\\
= \dfrac{3}{{x - 1}} - \dfrac{{2x - 2}}{{x\left( {x + 2} \right)}}\\
= \dfrac{{3\left( {{x^2} + 2x} \right) - \left( {2x - 2} \right)\left( {x - 1} \right)}}{{x\left( {x - 1} \right)\left( {x + 2} \right)}}\\
= \dfrac{{3{x^2} + 6x - 2{x^2} + 2x + 2x - 1}}{{x\left( {x - 1} \right)\left( {x + 2} \right)}}\\
= \dfrac{{{x^2} + 10x - 1}}{{x\left( {x - 1} \right)\left( {x + 2} \right)}}
\end{array}\)
