1. Let $a_{0}=0, a_{1}=1, a_{n+2}=a_{n+1}+a_{n}(n=0,1,2, \cdots)$. Also let $x_{1} \in \mathbf{R}, x_{n}=\frac{a_{n-1}+a_{n-2} \cdot x_{1}}{a_{n}+a_{n-1} \cdot x_{1}}(n=2,3, \cdots)$. If $x_{2004}=\frac{1}{x_{1}}-1$, find $x_{1}$.
Let $x_{n}=\frac{1}{x_{1}}-1(n=2004)$, then $\frac{a_{n-1}+a_{n-2} \cdot x}{a_{n}+a_{n-1} \cdot x_{1}}=\frac{1}{x_{1}}-1$, i.e., $\frac{1}{x_{1}}=\frac{a_{n+1}+a_{n} \cdot x_{1}}{a_{n}+a_{n-1} \cdot x_{1}}$, which means $a_{n+1} \cdot x_{1}+a_{n} \cdot x_{1}^{2}=a_{n}+a_{n-1} \cdot x_{1}$, thus $a_{n} \cdot x_{1}^{2}+a_{n} \cdot x_{1}-a_{n}=0$, i.e., $x_{1}^{2}+x_{1}-1=0$, hence $x_{1}=\frac{1}{2}(-1 \pm \sqrt{5})$.