6. 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) $F\left(x^{-1}\right)=F(x)$;
(D) $F(F(x))=-x$.
6. (A)

Let $u=\frac{1-x}{1+x}$, solving for $x$ gives $x=\frac{1-u}{1+u}$.
Therefore, $F(u)=\frac{1-u}{1+u}$, which means $F(x)=\frac{1-x}{1+x}$.
Then, verify one by one, we find
$$
\begin{aligned}
F(-2-x) & =\frac{1-(-2-x)}{1+(-2-x)}=-\frac{3+x}{1+x} \\
& =-2-\frac{1-x}{1+x}=-2-F(x) .
\end{aligned}
$$

Choose (A).
Alternatively, this problem can also be solved using the method of special value verification and elimination.
From $F\left(\frac{1-x}{1+x}\right)=x$,
let $x=1$, we get
$$
F(0)=1,
$$

let $x=0$, we get
$$
F(1)=0 \text {. }
$$

In the equation of (B), let $x=0$, we have $F(0)=F(1)$. Therefore, (B) should be excluded. In the equation of (C), $x=0$ is undefined, so (C) is excluded. In the equation (D), the left side $F(F(1))=F(0)=1 \neq-1$, so (D) is also excluded.