Đáp án:
$\begin{array}{l}
\left( {Dkxd:x,y \ge 0} \right)\\
x - 2y = \sqrt {xy} \\
\Leftrightarrow x - \sqrt {xy} - 2y = 0\\
\Leftrightarrow {\left( {\sqrt x } \right)^2} - 2\sqrt {xy} + \sqrt {xy} - 2{\left( {\sqrt y } \right)^2} = 0\\
\Leftrightarrow \sqrt x \left( {\sqrt x - 2\sqrt y } \right) + \sqrt y \left( {\sqrt x - 2\sqrt y } \right) = 0\\
\Leftrightarrow \left( {\sqrt x - 2\sqrt y } \right)\left( {\sqrt x + \sqrt y } \right) = 0\\
\Leftrightarrow \sqrt x - 2\sqrt y = 0\left( {do:\sqrt x + \sqrt y > 0} \right)\\
\Leftrightarrow \sqrt x = 2\sqrt y \\
\Leftrightarrow x = 4y\\
Thay\,x = 4y\\
\Rightarrow P = \dfrac{{x - y}}{{x + y}} = \dfrac{{4y - y}}{{4y + y}} = \dfrac{{3y}}{{5y}} = \dfrac{3}{5}\\
Vậy\,P = \dfrac{3}{5}
\end{array}$
