9. Let $0 \leqslant x_{1} \leqslant x_{2}$, and the sequence $\left\{x_{n}\right\}$ satisfies $x_{n+2}=x_{n+1}+x_{n}, n \geqslant 1$. If $1 \leqslant x_{7} \leqslant 2$, then the range of $x_{8}$ is
$\qquad$ .
$$
\begin{array}{l}
x_{3}=x_{2}+x_{1}, x_{4}=x_{2}+x_{3}=2 x_{2}+x_{1}, \\
x_{5}=x_{4}+x_{3}=3 x_{2}+2 x_{1}, x_{6}=x_{5}+x_{4}=5 x_{2}+3 x_{1}, \\
x_{7}=x_{6}+x_{5}=8 x_{2}+5 x_{1}, x_{8}=x_{7}+x_{6}=13 x_{2}+8 x_{1} .
\end{array}
$$

We can find $A\left(0, \frac{1}{4}\right), B\left(\frac{1}{13}, \frac{1}{13}\right)$, noting that $\frac{5}{8}>\frac{8}{13}$, then $x_{8}=13 x_{2}+8 x_{1}$ achieves its maximum and minimum values at $A, B$. Therefore, the range of $x_{8}$ is $\left[\frac{21}{13}, \frac{13}{4}\right]$.