6.232.

$$
\left\{\begin{array}{l}
\sqrt{\frac{20 y}{x}}=\sqrt{x+y}+\sqrt{x-y} \\
\sqrt{\frac{16 x}{5 y}}=\sqrt{x+y}-\sqrt{x-y}
\end{array}\right.
$$
## Solution.

Domain of definition: $\left\{\begin{array}{l}x+y \geq 0, \\ x-y \geq 0, \\ x>0, \\ y>0 .\end{array}\right.$

Dividing the first equation by the second, we get

$$
\sqrt{\frac{20 y}{x} \cdot \frac{5 y}{16 x}}=\frac{\sqrt{x+y}+\sqrt{x-y}}{\sqrt{x+y}-\sqrt{x-y}} \Leftrightarrow \sqrt{\frac{25 y^{2}}{4 x^{2}}}=\frac{\sqrt{x+y}+\sqrt{x-y}}{\sqrt{x+y}-\sqrt{x-y}} \Leftrightarrow
$$

$\Leftrightarrow \frac{5 y}{2 x}=\frac{\sqrt{x+y}+\sqrt{x-y}}{\sqrt{x+y}-\sqrt{x-y}}$, and after multiplying the numerator and denominator

of the right-hand side of the equation by the expression conjugate to the denominator,

$$
\begin{aligned}
& \frac{5 y}{2 x}=\frac{(\sqrt{x+y}+\sqrt{x-y})(\sqrt{x+y}+\sqrt{x-y})}{(\sqrt{x+y}-\sqrt{x-y})(\sqrt{x+y}+\sqrt{x-y})} \Leftrightarrow \\
& \Leftrightarrow \frac{5 y}{2 x}=\frac{x+y+2 \sqrt{x^{2}-y^{2}}+x-y}{x+y-x+y} \Leftrightarrow \frac{5 y^{2}}{2 x}-x=\sqrt{x^{2}-y^{2}} \Rightarrow \\
& \Rightarrow \frac{25 y^{4}}{4 x^{2}}-5 y^{2}+x^{2}=x^{2}-y^{2} \Leftrightarrow \frac{25 y^{4}}{4 x^{2}}-4 y^{2}=0 \Rightarrow 25 y^{2}=16 x^{2}, y=\frac{4}{5} x
\end{aligned}
$$

Thus $\sqrt{\frac{20}{x} \cdot \frac{4}{5} x}=\sqrt{x+\frac{4}{5} x}+\sqrt{x-\frac{4}{5} x}, 4=\sqrt{\frac{9}{5}} x+\sqrt{\frac{x}{5}} \Leftrightarrow$ $\Leftrightarrow \sqrt{\frac{x}{5}}=1$, from which $x=5 ; y=4$.

Answer: $(5 ; 4)$.