6 Given that $f(x)$ is a function defined on the set of real numbers, $f(3)=-\sqrt{3}$, and $f(x+2)[1-f(x)]=1+f(x)$, then $f(2009)=(\quad)$.
(A) $2+\sqrt{3}$
(B) $2-\sqrt{3}$
(C) $\sqrt{3}$
(D) $-\sqrt{3}$
6) A Hint: When $x=1$,
$$
f(3)[1-f(1)]=1+f(1),
$$

So
$$
-\sqrt{3}[1-f(1)]=1+f(1),
$$

So
$$
f(1)=\frac{1+\sqrt{3}}{\sqrt{3}-1}=2+\sqrt{3} .
$$

Since $f(x+2)=\frac{1+f(x)}{1-f(x)}$,
So
$$
\begin{aligned}
f(x+4) & =\frac{1+f(x+2)}{1-f(x+2)} \\
& =\frac{1+\frac{1+f(x)}{1-f(x)}}{1-\frac{1+f(x)}{1-f(x)}} \\
& =\frac{2}{-2 f(x)} \\
& =-\frac{1}{f(x)} .
\end{aligned}
$$

So
$$
f(x+8)=-\frac{1}{f(x+4)}=-\frac{1}{-\frac{1}{f(x)}}=f(x),
$$

So
$$
f(2009)=f(251 \times 8+1)=f(1)=2+\sqrt{3} .
$$

Therefore, the answer is A.
(Supplied by Zuo Chongjing, Urumqi No.1 High School)