$2 \cdot 91$ The solution to the equation $x^{2}-1 x-1=0$ is
(A) $\frac{1 \pm \sqrt{5}}{2}$.
(B) $\frac{-1 \pm \sqrt{5}}{2}$.
(C) $\frac{1 \pm \sqrt{5}}{2}$ or $-\frac{1 \pm \sqrt{5}}{2}$.
(D) $\pm \frac{1+\sqrt{5}}{2}$.
(China Junior High School Mathematics League, 1991)
[Solution] The original equation is $|x|^{2}-|x|-1=0$.
So $|x|=\frac{1+\sqrt{5}}{2}$ (where $\frac{1-\sqrt{5}}{2}<0$, which has been discarded).
Thus,
$$
x= \pm \frac{1+\sqrt{5}}{2} \text {. }
$$

Therefore, the answer is $(D)$.