Example 5 Given $x=\frac{1}{2}\left(1991^{\frac{1}{n}}-1991^{-\frac{1}{n}}\right)(n$ is an integer). Then, the value of $\left(x-\sqrt{1+x^{2}}\right)^{n}$ is ( ).
(A) $1991^{-1}$
(B) $-1991^{-1}$
(C) $(-1)^{n} 1991$
(D) $(-1)^{n} 1991^{-1}$
Solution: Let $a=1991^{\frac{1}{n}}, b=-1991^{-\frac{1}{n}}$, then $a+b=2 x, a b=-1$.
By the inverse of Vieta's formulas, $a$ and $b$ are the two real roots of the equation
$$
t^{2}-2 x t-1=0
$$

Solving this equation, we get
$$
t=x \pm \sqrt{1+x^{2}} .
$$

Since $b<a$, we have $b=x-\sqrt{1+x^{2}}$.
$$
\begin{array}{l}
\therefore\left(x-\sqrt{1+x^{2}}\right)^{n}=b^{n}=\left(-1991^{-\frac{1}{n}}\right)^{n} \\
\quad=(-1)^{n} 1991^{-1} .
\end{array}
$$

Therefore, the correct choice is (D).