解:由题意得$\sqrt{x-2y+9}+(x-y-3)^2=0$
$\because \sqrt{x-2y+9}≥0,$$(x-y-3)^2≥0$
$\therefore x-2y+9=0,$$x-y-3=0$
联立方程组$\begin{cases}x-2y+9=0\\x-y-3=0\end{cases}$
解得$\begin{cases}x=15\\y=12\end{cases}$
$\therefore 2x-\frac{1}{3}y=2×15-\frac{1}{3}×12=26$
$\therefore 2x-\frac{1}{3}y$的算术平方根为$\sqrt{26}$