Ta có:
$2+6y=\dfrac{x}{y}-\sqrt{x-2y}$
$\Leftrightarrow x-y\sqrt{x-2y}-2y-6y^2=0$
$\Leftrightarrow x-2y-y\sqrt{x-2y}+\dfrac{y^2}{4}-\dfrac{25y^2}{4}$
$\Leftrightarrow \left(\sqrt{x-2y}-\dfrac{y}{2}\right)^2-\dfrac{25y^2}{4}=0$
$\Leftrightarrow (\sqrt{x-2y}-3y)(\sqrt{x-2y}+2y)=0$
$\Leftrightarrow \left\{\begin{array}{l}\sqrt{x-2y}=3y\\\sqrt{x-2y}=-2y\end{array}\right.$
Thay vào pt đầu tiên để tìm $x,y$.