Example 1 Given real numbers $x, y$ satisfy
$$
\begin{array}{l}
\left(x-\sqrt{x^{2}-2008}\right)\left(y-\sqrt{y^{2}-2008}\right) \\
=2008 .
\end{array}
$$

Then the value of $3 x^{2}-2 y^{2}+3 x-3 y-2007$ is ). [1]
(A) -2008
(B) 2008
(C) -1
(D) 1
(2008, National Junior High School Mathematics Competition)
Solve: From the given, we have
$$
\begin{array}{l}
x-\sqrt{x^{2}-2008}=\frac{2008}{y-\sqrt{y^{2}-2008}} \\
=y+\sqrt{y^{2}-2008}, \\
y-\sqrt{y^{2}-2008}=\frac{2008}{x-\sqrt{x^{2}-2008}} \\
=x+\sqrt{x^{2}-2008} .
\end{array}
$$

From equations (1) and (2), we know
$$
x=y \text {. }
$$

Substituting equation (3) into equation (1), we get
$$
\begin{array}{l}
2 \sqrt{x^{2}-2008}=0 \\
\Rightarrow x= \pm \sqrt{2008} .
\end{array}
$$

Therefore, $3 x^{2}-2 y^{2}+3 x-3 y-2007$
$$
=x^{2}-2007=1 \text {. }
$$

Thus, the answer is (D).