1.104 If $x+\sqrt{x^{2}-1}+\frac{1}{x-\sqrt{x^{2}-1}}=20$, then
$$
x^{2}+\sqrt{x^{4}-1}+\frac{1}{x^{2}+\sqrt{x^{4}-1}}
$$

is equal to
(A) 5.05 .
(B) 20 .
(C) 51.005 .
(D) 61.25 .
(E) 400 .
(42nd American High School Mathematics Examination, 1991)
[Solution] Notice the transformation:
$$
\begin{aligned}
& \frac{1}{x-\sqrt{x^{2}-1}} \\
= & \frac{x+\sqrt{x^{2}-1}}{\left(x-\sqrt{x^{2}-1}\right)\left(x+\sqrt{x^{2}-1}\right)} \\
= & x+\sqrt{x^{2}-1},
\end{aligned}
$$

Thus, from the given condition, we have $x+\sqrt{x^{2}-1}=10$, solving which yields $x=\frac{101}{20}$.
Also,
$$
\begin{aligned}
& x^{2}+\sqrt{x^{4}-1}+\frac{1}{x^{2}+\sqrt{x^{4}-1}} \\
= & 2 x^{2}=2 \cdot \left(\frac{101}{20}\right)^{2}=51.005 .
\end{aligned}
$$

Therefore, the answer is (C).