1. (2008 National College Entrance Examination, Science Question 3) Let the odd function $f(x)$ be increasing on $(0,+\infty)$, and $f(1)=0$. Then the solution set of the inequality $\frac{f(x)-f(-x)}{x}<0$ is
A. $(-1,0) \cup(1,+\infty)$
B. $(-\infty,-1) \cup(0,1)$
C. $(-\infty,-1) \cup(1,+\infty)$
D. $(-1,0) \cup(0,1)$
1. $\mathrm{D}$

When $x>0$, we have $f(x)0=f(1)$. That is, $-f(x)<f(1), f(-x)<f(1)$, because $f(x)$ is an increasing function on $(0,+\infty)$. Therefore, $-x<1$ and $x<0$, which means $-1<x<0$.
In summary, we get $-1<x<0$ or $0<x<1$, so the answer is D.
(2) As shown in the graph on the right, construct a function graph that satisfies the problem,
$$
\frac{f(x)-f(-x)}{x}<0 \Leftrightarrow \frac{2 f(x)}{x}<0 \Leftrightarrow x f(x)<0 .
$$

According to the graph, we can observe that $-1<x<0$ or $0<x<1$. Therefore, the answer is D.
(3) Construct a function that satisfies the problem, $f(x)=x(x-1)(x+1)(x \neq 0)$,
From the given, we have $\frac{f(x)-f(-x)}{x}=\frac{2 f(x)}{x}=\frac{2 x(x-1)(x+1)}{x}<0$,
Solving this, we get $-1<x<0$ or $0<x<1$, so the answer is $\mathrm{D}$.