1. The function defined on $(-\infty,-2) \cup(2,+\infty)$
$$
f(x)=\frac{x+2+\sqrt{x^{2}-4}}{x+2-\sqrt{x^{2}-4}}+\frac{x+2-\sqrt{x^{2}-4}}{x+2+\sqrt{x^{2}-4}}
$$
( $|x|>2)$ has the following property regarding its odd and even nature ( ).
(A) It is an odd function but not an even function
(B) It is an even function but not an odd function
(C) It is both an odd and an even function
(D) It is neither an odd nor an even function
$-、 1 .(A)$.
For $|x|>2$, we have
$$
\begin{array}{r}
4=x^{2}-\left(x^{2}-4\right)=\left(x+\sqrt{x^{2}-4}\right)\left(x-\sqrt{x^{2}-4}\right) \\
\frac{1}{1} \frac{x+\sqrt{x^{2}-4}}{2}=\frac{2}{x-\sqrt{x^{2}}-4}=\frac{x+2+\sqrt{x^{2}-4}}{x+2-\sqrt{x^{2}-4}} .
\end{array}
$$

Thus, $f(x)=\frac{x+\sqrt{x^{2}-4}}{2}+\frac{x-\sqrt{x^{2}-4}}{2}$,
which means $f(x)=x(|x|>2)$ is an odd function, not an even function.