2.128 The solution situation of the equation $\sqrt{x-1} \cdot \sqrt{x+1}=-\sqrt{x^{2}-1}$ is
(A) Unique solution.
(B) Two solutions.
(C) Infinitely many solutions.
(D) No real solutions.
(6th "Jinyun Cup" Junior High School Mathematics Invitational, 1989)
［Solution］From the problem, we have $x-1 \geqslant 0$, which means $x \geqslant 1$.
And then only
$$
\begin{array}{l}
\text { and } \quad \sqrt{x-1} \geqslant 0, \quad \sqrt{x+1} \geqslant 0, \\
\text { we have } \\
\therefore \quad \sqrt{x-1} \cdot \sqrt{x+1}=\sqrt{x^{2}-1}=0, \\
\therefore=1 \text { or }-1 \text { (discard as it does not meet the problem's conditions). }
\end{array}
$$

Therefore, the answer is $(A)$.