Đặt $\begin{cases}a=\sqrt{x}\geq 0\\ b=\sqrt{y}\geq 0\end{cases}$Hệ $\Leftrightarrow \begin{cases}\frac{1}{4}+\frac{2a+b}{a^2+b^2}=\frac{2}{\sqrt{a}} \\ \frac{1}{4}-\frac{2a+b}{a^2+b^2}=\frac{1}{\sqrt{b}} \end{cases}$
$\xrightarrow[-]{+}\begin{cases}\frac{1}{2}=\frac{2}{\sqrt{a}}+\frac{1}{\sqrt{b}} \\ \frac{4a+2b}{a^2+b^2}=\frac{2}{\sqrt{a}}-\frac{1}{\sqrt{b}} \end{cases} $
$\overset{\times }{\rightarrow}\frac{4}{a}-\frac{1}{b}=\frac{2a+b}{a^2+b^2}\Leftrightarrow a=2b$
$\Rightarrow (x;y)=(1088+768\sqrt{2};272+192\sqrt{2})$