Does the expansion of

Suppose $x^{9}$ occurs in the given expression at the $(r+1)$ th term.

Then, we have:

$T_{r+1}={ }^{20} C_{r}\left(2 x^{2}\right)^{20-r}\left(\frac{-1}{x}\right)^{r}$

$=(-1)^{r}{ }^{20} C_{r}(2)^{20-r}(x)^{40-2 r-r}$

For this term to contain $x^{9}$, we must have

$40-3 r=9$

$\Rightarrow 3 r=31$

$\Rightarrow r=\frac{31}{3}$

It is not possible, as $r$ is not an integer.

Hence, there is no term with $x^{9}$ in the given expression.