6.22 Given $x=\frac{1}{2}\left(1991^{\frac{1}{n}}-1991^{-\frac{1}{n}}\right)$ ( $n$ is a natural number). Then the value of $\left(x-\sqrt{1+x^{2}}\right)^{n}$ is
(A) $1991^{-1}$.
(B) $-1991^{-1}$.
(C) $(-1)^{n} 1991$.
(D) $(-1)^{n} 1991^{-1}$.
(China Junior High School Mathematics Competition, 1991)
[Solution] Since $1+x^{2}=1+\frac{1}{4}\left(1991 \frac{2}{n}-2+1991^{-\frac{2}{n}}\right)$
$$
\begin{aligned}
& =\frac{1}{4}\left(1991^{\frac{2}{n}}+2+1991^{-\frac{2}{n}}\right) \\
& =\left[\frac{1}{2}\left(1991^{\frac{1}{n}}+1991^{-\frac{1}{n}}\right)\right]^{2} . \\
\therefore \quad \text { the original expression }= & {\left[\frac{1}{2}\left(1991^{\frac{1}{n}}-1991^{-\frac{1}{n}}\right)-\frac{1}{2}\left(1991^{\frac{1}{n}}+1991^{-\frac{1}{n}}\right)\right]^{n} } \\
= & \left(-1991^{-\frac{1}{n}}\right)^{n}=(-1)^{n} 1991^{-1} .
\end{aligned}
$$

Therefore, the answer is $(D)$.