1. If $f(x)=x^{6}-2 \sqrt{2006} x^{5}-x^{4}+x^{3}-$ $2 \sqrt{2007} x^{2}+2 x-\sqrt{2006}$, then $f(\sqrt{2006}+\sqrt{2007})=$ $\qquad$
$$
\begin{array}{l}
\text { 2. } \sqrt{2007}. \\
\text { Consider the quadratic equations } \\
x^{2}-2 \sqrt{2006} x-1=0, \\
x^{2}-2 \sqrt{2007} x+1=0 .
\end{array}
$$

From equation (1), we get $x^{2}-2 \sqrt{2006} x+2006-2007=0$, which is
$$
(x-\sqrt{2006}+\sqrt{2007})(x-\sqrt{2006}-\sqrt{2007})=0.
$$

From equation (2), we get $x^{2}-2 \sqrt{2007} x+2007-2006=0$, which is
$$
(x-\sqrt{2007}+\sqrt{2006})(x-\sqrt{2007}-\sqrt{2006})=0.
$$

Therefore, $x_{0}=\sqrt{2006}+\sqrt{2007}$ is a common root of equations (1) and (2).
$$
\begin{array}{r}
\text { Also, } f(x)=x^{4}\left(x^{2}-2 \sqrt{2006} x-1\right)+x\left(x^{2}-\right. \\
2 \sqrt{2007} x+1)+(x-\sqrt{2006}-\sqrt{2007})+\sqrt{2007},
\end{array}
$$

then $f\left(x_{0}\right)=x_{0}^{4} \cdot 0+x_{0} \cdot 0+0+\sqrt{2007}=\sqrt{2007}$.