Đáp án:
$\begin{array}{l}
a)\dfrac{{10x{y^2}\left( {x + y} \right)}}{{15xy{{\left( {x + y} \right)}^3}}} = \dfrac{{2y}}{{3{{\left( {x + y} \right)}^2}}}\\
b)\dfrac{{{x^2} - xy - x + y}}{{{x^2} + xy - x - y}}\\
= \dfrac{{x\left( {x - y} \right) - \left( {x - y} \right)}}{{x\left( {x + y} \right) - \left( {x + y} \right)}}\\
= \dfrac{{\left( {x - 1} \right)\left( {x - y} \right)}}{{\left( {x - 1} \right)\left( {x + y} \right)}}\\
= \dfrac{{x - y}}{{x + y}}\\
c)\dfrac{{3{x^2} - 12x + 12}}{{{x^4} - 8x}}\\
= \dfrac{{3\left( {{x^2} - 4x + 4} \right)}}{{x\left( {{x^3} - 8} \right)}}\\
= \dfrac{{3{{\left( {x - 2} \right)}^2}}}{{x\left( {x - 2} \right)\left( {{x^2} + 2x + 4} \right)}}\\
= \dfrac{{3\left( {x - 2} \right)}}{{x\left( {{x^2} + 2x + 4} \right)}}\\
d)\dfrac{{7{x^2} + 14x + 7}}{{3{x^2} + 3x}}\\
= \dfrac{{7\left( {{x^2} + 2x + 1} \right)}}{{3x\left( {x + 1} \right)}}\\
= \dfrac{{7{{\left( {x + 1} \right)}^2}}}{{3x\left( {x + 1} \right)}}\\
= \dfrac{{7\left( {x + 1} \right)}}{{3x}}\\
e)\dfrac{{2{a^2} - 2ab}}{{ac + ad - bc - bd}}\\
= \dfrac{{2a\left( {a - b} \right)}}{{a\left( {c + d} \right) - b\left( {c + d} \right)}}\\
= \dfrac{{2a\left( {a - b} \right)}}{{\left( {c + d} \right)\left( {a - b} \right)}}\\
= \dfrac{{2a}}{{c + d}}\\
f)\dfrac{{{x^2} - xy}}{{{y^2} - {x^2}}}\\
= \dfrac{{x\left( {x - y} \right)}}{{ - \left( {x + y} \right)\left( {x - y} \right)}}\\
= \dfrac{{ - x}}{{x + y}}
\end{array}$
