9. Given $x>0, y>0$, and $\frac{1}{2 x+y}+\frac{1}{y+1}=1$. Then the minimum value of $x+2 y$ is $\qquad$ .
9. $\sqrt{3}+\frac{1}{2}$.
Let $x+2 y=\lambda_{1}(2 x+y)+\lambda_{2}(y+1)+t$
$\Rightarrow \lambda_{1}=\frac{1}{2}, \lambda_{2}=\frac{3}{2}, t=-\frac{3}{2}$.
Then $x+2 y$
$=\frac{1}{2}(2 x+y)+\frac{3}{2}(y+1)-\frac{3}{2}$
$=\left(\frac{1}{2}(2 x+y)+\frac{3}{2}(y+1)\right)$.
$\quad\left(\frac{1}{2 x+y}+\right.$ $\geqslant \sqrt{3}+\frac{1}{2}$,

Equality holds if and only if $x=\frac{1}{2}+\frac{\sqrt{3}}{3}, y=\frac{\sqrt{3}}{3}$.