Đáp án:
\[P = \dfrac{{y - x}}{{xy}}\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
P = \dfrac{2}{x} - \left( {\dfrac{{{x^2}}}{{{x^2} - xy}} + \dfrac{{{x^2} - {y^2}}}{{xy}} - \dfrac{{{y^2}}}{{{y^2} - xy}}} \right):\dfrac{{{x^2} - xy + {y^2}}}{{x - y}}\\
= \dfrac{2}{x} - \left( {\dfrac{{{x^2}}}{{x\left( {x - y} \right)}} + \dfrac{{{x^2} - {y^2}}}{{xy}} - \dfrac{{{y^2}}}{{y\left( {y - x} \right)}}} \right):\dfrac{{{x^2} - xy + {y^2}}}{{x - y}}\\
= \dfrac{2}{x} - \left( {\dfrac{x}{{x - y}} + \dfrac{{{x^2} - {y^2}}}{{xy}} - \dfrac{y}{{y - x}}} \right):\dfrac{{{x^2} - xy + {y^2}}}{{x - y}}\\
= \dfrac{2}{x} - \left( {\dfrac{{x + y}}{{x - y}} + \dfrac{{{x^2} - {y^2}}}{{xy}}} \right):\dfrac{{{x^2} - xy + {y^2}}}{{x - y}}\\
= \dfrac{2}{x} - \dfrac{{xy.\left( {x + y} \right) + \left( {{x^2} - {y^2}} \right)\left( {x - y} \right)}}{{\left( {x - y} \right)xy}}.\dfrac{{x - y}}{{{x^2} - xy + {y^2}}}\\
= \dfrac{2}{x} - \dfrac{{{x^2}y + x{y^2} + {x^3} - {x^2}y - x{y^2} + {y^3}}}{{xy}}.\dfrac{1}{{{x^2} - xy + {y^2}}}\\
= \dfrac{2}{x} - \dfrac{{{x^3} + {y^3}}}{{xy\left( {{x^2} - xy + {y^2}} \right)}}\\
= \dfrac{2}{x} - \dfrac{{\left( {x + y} \right)\left( {{x^2} - xy + {y^2}} \right)}}{{xy\left( {{x^2} - xy + {y^2}} \right)}}\\
= \dfrac{2}{x} - \dfrac{{x + y}}{{xy}}\\
= \dfrac{{2y - \left( {x + y} \right)}}{{xy}}\\
= \dfrac{{y - x}}{{xy}}
\end{array}\)
