Đáp án:
$\begin{array}{l}
a)\dfrac{{2x + 8}}{{{x^2} - 4x + 4}} - \dfrac{7}{{x - 2}}\\
= \dfrac{{2x + 8}}{{{{\left( {x - 2} \right)}^2}}} - \dfrac{7}{{x - 2}}\\
= \dfrac{{2x + 8 - 7\left( {x - 2} \right)}}{{{{\left( {x - 2} \right)}^2}}}\\
= \dfrac{{22 - 5x}}{{{{\left( {x - 2} \right)}^2}}}\\
c)\dfrac{{a + b}}{{{a^2} - ab + {b^2}}} - \dfrac{1}{{a + b}}\\
= \dfrac{{{{\left( {a + b} \right)}^2} - {a^2} + ab - {b^2}}}{{\left( {a + b} \right)\left( {{a^2} - ab + {b^2}} \right)}}\\
= \dfrac{{3ab}}{{{a^3} + {b^3}}}\\
d)a - 2 + \dfrac{{4a}}{{a - 2}} - \dfrac{{{a^3} + b}}{{{a^2} - 2a}}\\
= \dfrac{{\left( {a - 2} \right).a\left( {a - 2} \right) + 4a.a - {a^3} - b}}{{a\left( {a - 2} \right)}}\\
= \dfrac{{{a^3} - 4{a^2} + 4a + 4{a^2} - {a^3} - b}}{{a\left( {a - 2} \right)}}\\
= \dfrac{{4a - b}}{{a\left( {a - 2} \right)}}\\
2)\dfrac{{3{a^2} + 3}}{{{a^3} - 1}} - \dfrac{{a - 1}}{{{a^2} + a + 1}} + \dfrac{2}{{1 - a}}\\
= \dfrac{{3{a^2} + 3}}{{\left( {a + 1} \right)\left( {{a^2} + a + 1} \right)}} - \dfrac{{a - 1}}{{{a^2} + a + 1}} - \dfrac{2}{{a - 1}}\\
= \dfrac{{3{a^2} + 3 - \left( {a - 1} \right)\left( {a - 1} \right) - 2\left( {{a^2} + a + 1} \right)}}{{\left( {a + 1} \right)\left( {{a^2} + a + 1} \right)}}\\
= \dfrac{{3{a^2} + 3 - {a^2} + 2a - 1 - 2{a^2} - 2a - 2}}{{\left( {a + 1} \right)\left( {{a^2} + a + 1} \right)}}\\
= \dfrac{0}{{\left( {a + 1} \right)\left( {{a^2} + a + 1} \right)}}\\
= 0
\end{array}$
1)
$\begin{array}{l}
b)x - \dfrac{{xy}}{{x + y}} - \dfrac{{{x^3}}}{{\left( {x + y} \right)\left( {x - y} \right)}}\\
= \dfrac{{x\left( {x - y} \right)\left( {x + y} \right) - xy\left( {x - y} \right) - {x^3}}}{{\left( {x + y} \right)\left( {x - y} \right)}}\\
= \dfrac{{x\left( {{x^2} - {y^2}} \right) - {x^2}y + x{y^2} - {x^3}}}{{\left( {x + y} \right)\left( {x - y} \right)}}\\
= \dfrac{{{x^3} - x{y^2} - {x^2}y + x{y^2} - {x^3}}}{{\left( {x + y} \right)\left( {x - y} \right)}}\\
= \dfrac{{ - {x^2}y}}{{{x^2} - {y^2}}}
\end{array}$
