6. Given real numbers $x, y$ satisfy $\left(x-\sqrt{x^{2}-2008}\right)\left(y-\sqrt{y^{2}-2008}\right)=2008$. Then the value of $3 x^{2}-2 y^{2}+3 x-3 y-2007$ is ( ).
(A) -2008
(B) 2008
(C) -1
(D) 1
6.D.

From $\left(x-\sqrt{x^{2}-2008}\right)\left(y-\sqrt{y^{2}-2008}\right)=2008$,
we get
$$
\begin{array}{l}
x-\sqrt{x^{2}-2008} \\
=\frac{2008}{y-\sqrt{y^{2}-2008}}=y+\sqrt{y^{2}-2008}, \\
y-\sqrt{y^{2}-2008}=\frac{2008}{x-\sqrt{x^{2}-2008}} \\
=x+\sqrt{x^{2}-2008} .
\end{array}
$$

From the above two equations, we can get $x=y$.
Therefore, $\left(x-\sqrt{x^{2}-2008}\right)^{2}=2008$.
Solving this, we get $x^{2}=2008$.
$$
\begin{array}{l}
\text { Hence } 3 x^{2}-2 y^{2}+3 x-3 y-2007 \\
=3 x^{2}-2 x^{2}+3 x-3 x-2007 \\
=x^{2}-2007=1 \text {. }
\end{array}
$$