Đáp án:
$\begin{array}{l}
a) + )\frac{{8xy}}{{{x^2} + 4xy + {y^2}}} = \frac{{8xy\left( {{y^2} - xy} \right)}}{{\left( {{x^2} + 4xy + {y^2}} \right)\left( {{y^2} - xy} \right)}} = \frac{{8x{y^3} - 8{x^2}{y^2}}}{{\left( {{x^2} + 4xy + {y^2}} \right)\left( {{y^2} - xy} \right)}}\\
+ )\frac{{18xy}}{{{y^2} - xy}} = \frac{{18xy\left( {{x^2} + 4xy + {y^2}} \right)}}{{\left( {{x^2} + 4xy + {y^2}} \right)\left( {{y^2} - xy} \right)}}\\
b)\frac{{4xy}}{{{y^2} + 2xy + {x^2}}} = \frac{{4xy\left( {y - x} \right)}}{{\left( {{y^2} + 2xy + {x^2}} \right)\left( {y - x} \right)}}\\
\frac{{8xy}}{{y - x}} = \frac{{8xy\left( {{y^2} + 2xy + {x^2}} \right)}}{{\left( {{y^2} + 2xy + {x^2}} \right)\left( {y - x} \right)}}
\end{array}$
