9. As shown in the figure, in $\triangle A B D$, point $C$ is on $A D$, $\angle A B C=\frac{\pi}{2}, \angle D B C=\frac{\pi}{6}, A B=C D=1$, then $A C=$
In $\triangle A B D$,
$$
\frac{A D}{\sin 120^{\circ}}=\frac{1}{\sin D} \Rightarrow \sin D=\frac{\sqrt{3}}{2(x+1)} \text { (where } A D=x \text { ) }
$$

In $\triangle B C D$,
$$
\frac{B C}{\sin D}=\frac{1}{\sin 30^{\circ}} \Rightarrow \frac{\sqrt{x^{2}-1}}{\sin D}=2 \Rightarrow \sin D=\frac{\sqrt{x^{2}-1}}{2}
$$

From (1) and (2) we get
$$
\frac{\sqrt{3}}{2(x+1)}=\frac{\sqrt{x^{2}-1}}{2} \Rightarrow(x+1) \sqrt{x^{2}-1}=\sqrt{3} \Rightarrow x^{4}+2 x^{3}-2 x-4=0 \Rightarrow(x+2)\left(x^{3}-2\right)=0
$$

Since $x+2>0$, we have $x^{3}=2$, i.e., $x=\sqrt[3]{2}$.