Đáp án:

$\begin{array}{l}
4a)\\
 + \dfrac{{{x^2}}}{{1 - {x^2}}} = \dfrac{{{x^2}}}{{\left( {1 - x} \right)\left( {1 + x} \right)}} = \dfrac{{ - {x^2}}}{{\left( {x - 1} \right)\left( {x + 1} \right)}}\\
 + \dfrac{1}{{x - 1}} = \dfrac{{x + 1}}{{\left( {x - 1} \right)\left( {x + 1} \right)}}\\
b)\\
 + \dfrac{1}{{x - 1}} = \dfrac{{x + 1}}{{\left( {x - 1} \right)\left( {x + 1} \right)}}\\
 + \dfrac{1}{{x + 1}} = \dfrac{{x - 1}}{{\left( {x - 1} \right)\left( {x + 1} \right)}}\\
c)\\
 + \dfrac{{2x}}{{x - 2}}\\
 + \dfrac{{3x + 2}}{{2 - x}} = \dfrac{{ - 3x - 2}}{{x - 2}}\\
d)\\
 + \dfrac{x}{{x - 2}} = \dfrac{{x\left( {x + 2} \right)}}{{\left( {x - 2} \right)\left( {x + 2} \right)}} = \dfrac{{{x^2} + 2x}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}\\
 + \dfrac{1}{{x + 2}} = \dfrac{{x - 2}}{{\left( {x + 2} \right)\left( {x - 2} \right)}}\\
e)\\
 + \dfrac{1}{{{x^2} - 4}} = \dfrac{1}{{\left( {x - 2} \right)\left( {x + 2} \right)}}\\
 = \dfrac{{{x^2} + 2x + 4}}{{\left( {x - 2} \right)\left( {x + 2} \right)\left( {{x^2} + 2x + 4} \right)}}\\
 + \dfrac{1}{{{x^3} - 8}} = \dfrac{1}{{\left( {x - 2} \right)\left( {{x^2} + 2x + 4} \right)}}\\
 = \dfrac{{x + 2}}{{\left( {x - 2} \right)\left( {x + 2} \right)\left( {{x^2} + 2x + 4} \right)}}\\
 + \dfrac{1}{{x + 2}} = \dfrac{{\left( {x - 2} \right)\left( {{x^2} + 2x + 4} \right)}}{{\left( {x + 2} \right)\left( {x - 2} \right)\left( {{x^2} + 2x + 4} \right)}}\\
 = \dfrac{{{x^3} - 8}}{{\left( {x + 2} \right)\left( {x - 2} \right)\left( {{x^2} + 2x + 4} \right)}}\\
5)\\
a)\\
 + \dfrac{6}{{a - 1}}\\
 + \dfrac{2}{a} = \dfrac{{2.3}}{{3a}} = \dfrac{6}{{3a}}\\
b)\\
 + \dfrac{1}{a} = \dfrac{{a - 1}}{{a\left( {a - 1} \right)}}\\
 + \dfrac{{a - 1}}{{a - 2}}\\
c)\dfrac{{x + y}}{x} = \dfrac{{\left( {x + y} \right)\left( {x - y} \right)}}{{x\left( {x - y} \right)}}\\
\dfrac{{{x^2} - {y^2}}}{{{x^2} + xy}} = \dfrac{{\left( {x - y} \right)\left( {x + y} \right)}}{{{x^2} + xy}}\\
d)\dfrac{{{x^3} - 1}}{{{x^2} + 1}} = \dfrac{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}}{{{x^2} + 1}}\\
\dfrac{{x - 1}}{{x + 1}} = \dfrac{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}}{{\left( {x + 1} \right)\left( {{x^2} + x + 1} \right)}}
\end{array}$