12. Calculate:
\[
\begin{array}{l}
\sqrt{2005 \times 2006 \times 2007 \times 2008+1}-2006^{2} \\
=
\end{array}
\]
12.2005 .

Let $n=2005$. Then
$$
\begin{array}{l}
\text { Original expression }=\sqrt{n(n+1)(n+2)(n+3)+1}-(n+1)^{2} \\
=\sqrt{\left(n^{2}+3 n\right)\left(n^{2}+3 n+2\right)+1}-(n+1)^{2} \\
=\sqrt{\left(n^{2}+3 n+1\right)^{2}}-(n+1)^{2}=n=2005 .
\end{array}
$$