Example 5 (1984 National High School League Question) If $F\left(\frac{1-x}{1+x}\right)=x$, then which of the following equations is correct? ( ).
A. $F(-2-x)=-2-F(x)$
B. $F(-x)=F\left(\frac{1-x}{1+x}\right)$
C. $F\left(\frac{1}{x}\right)=F(x)$
D. $F[F(x)]=-x$
Solution: Choose A. Reason: From the given $F\left(\frac{1-x}{1+x}\right)=x$, we know $F(x)=\frac{1-x}{1+x}$, whose graph is centrally symmetric about the point $(-1,-1)$. According to Conclusion 9, we have $F(x)+F(-2-x)=-2$.

Or from $F(-2-x)=\frac{1-(-2-x)}{1+(-2-x)}=-\frac{3+x}{1+x}=-2-\frac{1-x}{1+x}=-2-F(x)$, we can also get it. Or verify and eliminate using special values: From $F\left(\frac{1-x}{1+x}\right)=x$, let $x=1$ to get $F(0)=1$, let $x=0$ to get $F(1)=0$. In the equation of B, let $x=0$, we have $F(0)=F(1)$, which can exclude B; in the equation of C, $x=0$ is meaningless, exclude C; in the equation of D, the left side $F[F(1)]=F(0)=1 \neq-1$, exclude D.