Đáp án:
$\begin{array}{l}
a)\frac{{6{x^2}{y^2}}}{{8x{y^5}}} = \frac{{2x{y^2}\left( {3x} \right)}}{{2x{y^2}\left( {4{y^3}} \right)}} = \frac{{3x}}{{4{y^3}}}\\
b)\frac{{10x{y^2}\left( {x + y} \right)}}{{15xy{{\left( {x + y} \right)}^3}}} = \frac{{2y}}{{3{{\left( {x + y} \right)}^2}}}\\
c)\frac{{2{x^2} + 2x}}{{x + 1}} = \frac{{2x\left( {x + 1} \right)}}{{x + 1}} = 2x\\
d)\frac{{{x^2} - xy - x + y}}{{{x^2} + xy - x - y}} = \frac{{x\left( {x - y} \right) - \left( {x - y} \right)}}{{x\left( {x + y} \right) - \left( {x + y} \right)}} = \frac{{\left( {x - y} \right)\left( {x - 1} \right)}}{{\left( {x + y} \right)\left( {x - 1} \right)}} = \frac{{x - y}}{{x + y}}
\end{array}$
