1. Given $x=\frac{1}{2-\sqrt{5}}$, then $x^{3}+3 x^{2}-5 x+1=$
1. 0 .

Solution: $\because x=-2-\sqrt{5}$, we get $(x+2)^{2}=5$
$$
\therefore x^{2}+4 x-1=0
$$

And $x^{3}+3 x^{2}-5 x+1=(x-1) \cdot\left(x^{2}+4 x-1\right)=0$.