Đáp án:

$\begin{array}{l}
a)x - 2 - \dfrac{{{x^2} - 10}}{{x + 2}}\\
 = \dfrac{{\left( {x - 2} \right)\left( {x + 2} \right) - {x^2} + 10}}{{x + 2}}\\
 = \dfrac{{{x^2} - 4 - {x^2} + 10}}{{x + 2}}\\
 = \dfrac{6}{{x + 2}}\\
b){x^2} + {y^2} - \dfrac{{2\left( {{x^4} + {y^4}} \right)}}{{{x^2} + {y^2}}}\\
 = \dfrac{{{{\left( {{x^2} + {y^2}} \right)}^2} - 2{x^4} - 2{y^4}}}{{{x^2} + {y^2}}}\\
 = \dfrac{{{x^4} + {y^2} + 2{x^2}{y^2} - 2{x^4} - 2{y^4}}}{{{x^2} + {y^2}}}\\
 =  - \dfrac{{{{\left( {{x^2} - {y^2}} \right)}^2}}}{{{x^2} + {y^2}}}\\
c)\dfrac{{x - 3}}{{4x + 4}} - \dfrac{{x - 1}}{{6x - 30}}\\
 = \dfrac{{\left( {x - 3} \right).2.\left( {x - 5} \right) - 3.\left( {x - 1} \right).\left( {x + 1} \right)}}{{12\left( {x + 1} \right)\left( {x - 5} \right)}}\\
 = \dfrac{{2\left( {{x^2} - 8x + 15} \right) - 3\left( {{x^2} - 1} \right)}}{{12\left( {x + 1} \right)\left( {x - 5} \right)}}\\
 = \dfrac{{ - {x^2} - 16x + 33}}{{12\left( {x + 1} \right)\left( {x - 5} \right)}}\\
d)\dfrac{1}{{x - 5{x^2}}} - \dfrac{{25x - 15}}{{25{x^2} - 1}}\\
 = \dfrac{1}{{x\left( {1 - 5x} \right)}} + \dfrac{{25x - 15}}{{\left( {1 - 5x} \right)\left( {1 + 5x} \right)}}\\
 = \dfrac{{1 + 5x + x\left( {25x - 15} \right)}}{{x\left( {1 - 5x} \right)\left( {1 + 5x} \right)}}\\
 = \dfrac{{25{x^2} - 10x + 1}}{{x\left( {1 - 5x} \right)\left( {1 + 5x} \right)}}\\
 = \dfrac{{{{\left( {5x - 1} \right)}^2}}}{{x\left( {1 - 5x} \right)\left( {1 + 5x} \right)}}\\
 = \dfrac{{1 - 5x}}{{x\left( {1 + 5x} \right)}}\\
e)\dfrac{{x + 9y}}{{{x^2} - 9{y^2}}} - \dfrac{{3y}}{{{x^2} + 3xy}}\\
 = \dfrac{{x + 9y}}{{\left( {x - 3y} \right)\left( {x + 3y} \right)}} - \dfrac{{3y}}{{x\left( {x + 3y} \right)}}\\
 = \dfrac{{x\left( {x + 9y} \right) - 3y\left( {x - 3y} \right)}}{{x\left( {x + 3y} \right)\left( {x - 3y} \right)}}\\
 = \dfrac{{{x^2} + 9xy - 3xy + 9{y^2}}}{{x\left( {x + 3y} \right)\left( {x - 3y} \right)}}\\
 = \dfrac{{{x^2} + 6xy + 9{y^2}}}{{x\left( {x + 3y} \right)\left( {x - 3y} \right)}}\\
 = \dfrac{{{{\left( {x + 3y} \right)}^2}}}{{x\left( {x + 3y} \right)\left( {x - 3y} \right)}}\\
 = \dfrac{{x + 3y}}{{x\left( {x - 3y} \right)}}\\
f)\dfrac{1}{{x + 1}} - \dfrac{1}{{{x^3} + 1}} + \dfrac{1}{{{x^2} - x + 1}}\\
 = \dfrac{{{x^2} - x + 1 - 1 + x + 1}}{{\left( {x + 1} \right)\left( {{x^2} - x + 1} \right)}}\\
 = \dfrac{{{x^2} + 1}}{{{x^3} + 1}}
\end{array}$