Cho x>y>0 thỏa mãn 2(x²+y²) = 5xy. Tính giá trị P=$\frac{x-y}{x+y}$$và$$Q=\frac{x+2y}{x+y}$

$\displaystyle  \begin{array}{{>{\displaystyle}l}}
\ \ \ \ \ \ 2.\left( x^{2} +y^{2}\right) =5xy\\
\Leftrightarrow 2x^{2} +4xy+2y^{2} =9xy\\
\Leftrightarrow 2.( x+y)^{2} =9xy\\
\Leftrightarrow x+y=\frac{3\sqrt{2}}{2} .\sqrt{xy}
\end{array}$
$\displaystyle  \begin{array}{{>{\displaystyle}l}}
\ \ \ \ \ \ 2.\left( x^{2} +y^{2}\right) =5xy\\
\Leftrightarrow 2x^{2} -4xy+2y^{2} =xy\\
\Leftrightarrow 2.( x-y)^{2} =xy\\
\Leftrightarrow x-y=\frac{\sqrt{2}}{2} .\sqrt{xy}
\end{array}$
$\displaystyle  \begin{array}{{>{\displaystyle}l}}
( x+y) -( x-y) =\frac{3\sqrt{2}}{2} .\sqrt{xy} -\frac{\sqrt{2}}{2} .\sqrt{xy} =\sqrt{2} .\sqrt{xy}\\
\Rightarrow y=\frac{\sqrt{2}}{2} .\sqrt{xy}
\end{array}$
$\displaystyle  \begin{array}{{>{\displaystyle}l}}
P=\frac{x-y}{x+y}\\
\ \ \ =\frac{\frac{\sqrt{2}}{2} .\sqrt{xy}}{\frac{3\sqrt{2}}{2} .\sqrt{xy}} =\frac{1}{3}
\end{array}$
$\displaystyle  \begin{array}{{>{\displaystyle}l}}
Q=\frac{x+2y}{x+y}\\
\ \ \ \ =1+\frac{y}{x+y}\\
\ \ \ \ \ =1+\frac{\frac{\sqrt{2}}{2} .\sqrt{xy}}{\frac{3\sqrt{2}}{2} .\sqrt{xy}}\\
\ \ \ \ \ =1+\frac{1}{3} =\frac{4}{3}
\end{array}$