9. If $(2 x-1)^{8}=a_{8} x^{8}+a_{7} x^{7}+\cdots+a_{1} x$ $+a_{0}$, then $a_{8}+a_{6}+a_{4}+a_{2}=$ $\qquad$
9.3280 .

Let $f(x)=(2 x-1)^{8}$, then
$$
\begin{array}{l}
a_{0}=f(0)=1, \\
a_{8}+a_{7}+\cdots+a_{0}=f(1)=1, \\
a_{8}-a_{7}+a_{6}-a_{5}+a_{4}-a_{3}+a_{2}-a_{1}+a_{0} \\
=f(-1)=(-3)^{8}=6561 .
\end{array}
$$

Therefore, $a_{8}+a_{6}+a_{4}+a_{2}$
$$
=\frac{1}{2}(f(1)+f(-1))-f(0)=3280 \text {. }
$$