设 $0\leqslant x_1\leqslant x_2$，数列 $\{x_n\}$ 满足 $x_{n+2}=x_{n+1}+x_n$（$n\in \mathbb N^{\ast}$）．若 $1\leqslant x_7\leqslant 2$，则 $x_8$ 的取值范围是_______．
答案 $\left[\dfrac{21}{13},\dfrac{13}4\right]$．
解析 设 $x_1=a$，$x_2=a+b$，其中 $a,b\geqslant 0$，则\[\begin{array}{c|c|c|c|c|c|c|c|c}\hline n&1&2&3&4&5&6&7&8\\ \hline x_n&a&a+b&2a+b&3a+2b&5a+3b&8a+5b&13a+8b&21a+13b\\ \hline \end{array}\]于是\[x_8=\dfrac{21}{13}\cdot \left(13a+\dfrac{169}{21}b\right)\geqslant \dfrac{21}{13}x_7\geqslant \dfrac{21}{13},\]等号当 $(a,b)=\left(\dfrac{1}{13},0\right)$ 时取得，且\[x_8=\dfrac{13}8\cdot \left(\dfrac{168}{13}a+8b\right)\leqslant \dfrac{13}8x_7\leqslant \dfrac{13}4,\]等号当 $(a,b)=\left(0,\dfrac 14\right)$ 时取得． 综上所述，$x_8$ 的取值范围是 $\left[\dfrac{21}{13},\dfrac{13}4\right]$．