Đáp án:
\(\dfrac{{{x^4} - 2{x^2}y + x{y^3} - {x^2}{y^2}}}{{ - {{\left( {x - y} \right)}^2}\left( {{x^2} - 2y} \right)}}\)
Giải thích các bước giải:
\(\begin{array}{l}
A = \left( {\dfrac{x}{{{y^2} - xy}} + \dfrac{y}{{{x^2} - 2y}}} \right):\dfrac{{{x^2} - {y^2}}}{{{x^2}y + x{y^2}}}\\
= \left[ {\dfrac{{x\left( {{x^2} - 2y} \right) + y\left( {{y^2} - xy} \right)}}{{y\left( {y - x} \right)\left( {{x^2} - 2y} \right)}}} \right]:\dfrac{{\left( {x - y} \right)\left( {x + y} \right)}}{{xy\left( {x + y} \right)}}\\
= \dfrac{{{x^3} - 2xy + {y^3} - x{y^2}}}{{y\left( {y - x} \right)\left( {{x^2} - 2y} \right)}}.\dfrac{{xy}}{{x - y}}\\
= \dfrac{{{x^3} - 2xy + {y^3} - x{y^2}}}{{\left( {y - x} \right)\left( {{x^2} - 2y} \right)}}.\dfrac{x}{{x - y}}\\
= \dfrac{{{x^4} - 2{x^2}y + x{y^3} - {x^2}{y^2}}}{{ - {{\left( {x - y} \right)}^2}\left( {{x^2} - 2y} \right)}}
\end{array}\)
