6・13 Let $x+\sqrt{x^{2}-1}=a^{\frac{m-n}{2 m n}}$. Then $x-\sqrt{x^{2}-1}$ equals
(A) $a^{\frac{2 m n}{m-n}}$.
(B) $a^{\frac{2 m n}{n-m}}$.
(C) $a^{\frac{n-m}{2 m n}}$.
(D) $a^{\frac{n+m}{2 m n}}$.
(Zhejiang Hangzhou Junior High School Mathematics Competition, 1985;
Jiangsu Junior High School Mathematics Competition, 1990)
[Solution] Notice that $\left(x+\sqrt{x^{2}-1}\right)\left(x-\sqrt{x^{2}-1}\right)=1$, then
$$
\begin{aligned}
x-\sqrt{x^{2}-1} & =\frac{1}{x+\sqrt{x^{2}-1}}=\left(x+\sqrt{x^{2}-1}\right)^{-1} \\
& =\left(a^{\frac{m-n}{2 m n}}\right)^{-1}=a^{\frac{n-m}{2 m n}} .
\end{aligned}
$$

Therefore, the answer is $(C)$.