2. If the positive real numbers $x, y$ satisfy $x-2 \sqrt{y}=\sqrt{2 x-y}$, then the maximum value of $x$ is $\qquad$ .
2. 10 .

Let $a=\sqrt{2 x-y}, b=\sqrt{y}$. Then $a^{2}+b^{2}=2 x, x=a+2 b$.
Also, $(a-1)^{2}+(b-2)^{2}=5(a, b \geqslant 0)$, so $(2 x)_{\max }=(\sqrt{5}+\sqrt{5})^{2}=20$.
Therefore, the maximum value of $x$ is 10.