35. Given $a=\frac{\sqrt{5}-1}{\sqrt{5}+1}$, then the value of $a^{4}-5 a^{3}+10 a^{2}-11 a+4$ is
Answer: 1
Solution: Given $a=\frac{\sqrt{5}-1}{\sqrt{5}+1}=\frac{3-\sqrt{5}}{2}$, we get $\quad 3-2 a=\sqrt{5}$,

Therefore,
$$
(2 a-3)^{2}=5 \text {, }
$$

which means
$$
4 a^{2}-12 a+4=0 \text {, }
$$

or equivalently
$$
a^{2}-3 a+1=0,
$$

Thus, $a^{4}-5 a^{3}+10 a^{2}-11 a+4=a^{2}\left(a^{2}-3 a+1\right)-2 a\left(a^{2}-3 a+1\right)+3\left(a^{2}-3 a+1\right)+1=1$