4. 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) $(-1)^{n} 1991^{-1}$.
4. (D)
$$
\text{Solution} \begin{aligned}
1+x^{2} & =1+\frac{1}{4}\left(1991^{\frac{2}{n}}-2+1991^{-\frac{2}{n}}\right) \\
& =\left[\frac{1}{2}\left(1991^{\frac{1}{n}}+1991^{-\frac{1}{n}}\right)\right]^{2},
\end{aligned}
$$

Therefore, the original expression $=\left(-1991^{-\frac{1}{n}}\right)^{n}$
$$
=(-1)^{n} 1991^{-1} \text{. }
$$