6. Given that $a$ is a constant, $x \in \mathbf{R}, f(x)=\frac{f(x-a)-1}{f(x-a)+1},$ the period of $f(x)$ is $\qquad$ .
6. $4 a$.
$f(x-a)=\frac{f(x-2 a)-1}{f(x-2 a)+1}, f(x)=-\frac{1}{f(x-2 a)}, f(x-2 a)=-\frac{1}{f(x-4 a)}$, so $f(x)=f(x-4 a), f(x)$ has a period of $4 a$.