(1) (2007 - Fujian) Given that $f(x)$ is a decreasing function on $\mathbf{R}$, then the range of real numbers $x$ that satisfy $f\left(\left|\frac{1}{x}\right|\right)<f(1)$ is ( ).
(A) $(-1,1)$
(B) $(0,1)$
(C) $(-1,0) \cup(0,1)$
(D) $(-\infty,-1) \cup(1,+\infty)$
Given that $\left|\frac{1}{x}\right|>1$, then $|x|<1$ and $|x| \neq 0$, so, $-1<x<0$ or $0<x<1$, the answer is $\mathrm{C}$.