Đáp án:
$\begin{array}{l}
\frac{{x - 1}}{{{x^3}}} - \frac{{x + 1}}{{{x^3} - {x^2}}} + \frac{3}{{{x^3} - 2{x^2} + x}}\\
= \frac{{x - 1}}{{{x^3}}} - \frac{{x + 1}}{{{x^2}\left( {x - 1} \right)}} + \frac{3}{{x{{\left( {x - 1} \right)}^2}}}\\
= \frac{{\left( {x - 1} \right).{{\left( {x - 1} \right)}^2} - x\left( {x + 1} \right).\left( {x - 1} \right) + 3.{x^2}}}{{{x^3}{{\left( {x - 1} \right)}^2}}}\\
= \frac{{{x^3} - 3{x^2} + 3x - 1 - x\left( {{x^2} - 1} \right) + 3{x^2}}}{{{x^3}{{\left( {x - 1} \right)}^2}}}\\
= \frac{{3x - 1 + x}}{{{x^3}{{\left( {x - 1} \right)}^2}}}\\
= \frac{{4x - 1}}{{{x^3}{{\left( {x - 1} \right)}^2}}}\\
\frac{{{x^3} + {x^2} - 2x - 20}}{{{x^2} - 4}} - \frac{5}{{x + 2}} + \frac{3}{{x - 2}}\\
= \frac{{{x^3} + {x^2} - 2x - 20 - 5\left( {x - 2} \right) + 3\left( {x + 2} \right)}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}\\
= \frac{{{x^3} + {x^2} - 4x - 4}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}\\
= \frac{{\left( {x + 1} \right)\left( {{x^2} - 4} \right)}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}\\
= x + 1
\end{array}$
