$$
\begin{array}{l}
\text { 9. For } \frac{1}{2} \leqslant x \leqslant 1 \text {, when } \\
(1+x)^{5}(1-x)(1-2 x)^{2}
\end{array}
$$

reaches its maximum value, $x=$ $\qquad$ .
9. 8

The maximum value of the expression $[\alpha(1+x)]^{5}[\beta(1-x)][\gamma(2 x-1)]^{2}$, where $\alpha, \beta, \gamma$ are positive integers satisfying $5 \alpha-\beta+4 \gamma=0$, $\alpha(1+x)=\beta(1-x)=\gamma(2 x-1)$. It is given that
$$
\begin{array}{l}
\beta-\alpha \\
\beta+\alpha
\end{array}=\frac{\beta+\gamma}{2 \gamma+\beta} .
$$

Substituting $\beta=5 \alpha+4 \gamma$, we get
$$
0=2(5 \alpha-2 \gamma)(\alpha+\gamma) \text {. }
$$

Taking $(\alpha, \beta, \gamma)=(2,30,5)$, by the AM-GM inequality, we have
$[2(1+x)]^{5}[30(1-x)][5(2 x-1)]^{2} \leqslant\binom{ 15}{4}^{8}$.
Equality holds if and only if $x=\frac{7}{8}$.
Therefore, when $x=\frac{7}{8}$, the maximum value of $(1+x)^{5}(1-x)(1-2 x)^{2}$ is $\frac{3^{7} \times 5^{5}}{2^{2}}$.