ĐKXĐ: ...
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{2\left(x^2+y^2\right)}+2\sqrt{xy}=16\\x+y+2\sqrt{xy}=16\end{matrix}\right.\)
\(\Rightarrow\sqrt{2\left(x^2+y^2\right)}=x+y\)
\(\Leftrightarrow2\left(x^2+y^2\right)=x^2+2xy+y^2\)
\(\Leftrightarrow\left(x-y\right)^2=0\Rightarrow x=y\)
Thay vào pt dưới: \(2\sqrt{x}=4\Rightarrow x=4\Rightarrow x=y=4\)