2. Given $x=\frac{1}{\sqrt{3}-\sqrt{2}}$. Then $x^{6}-2 \sqrt{2} x^{5}-x^{4}$ $+x^{3}-2 \sqrt{3} x^{2}+2 x-\sqrt{2}$ is $\qquad$.
2. $\sqrt{3}$.
$$
\begin{array}{l}
\text { Given } x=\frac{1}{\sqrt{3}-\sqrt{2}}=\sqrt{3}+\sqrt{2} \text {, we get } x^{2}-2 \sqrt{2} x-1=0 \text {, } \\
x^{2}-2 \sqrt{3} x+1=0 \text {. } \\
\therefore \text { the original expression }=x^{4}\left(x^{2}-2 \sqrt{2} x-1\right)+x\left(x^{2}-2 \sqrt{3} x\right. \\
+1)+x-\sqrt{2}=x-\sqrt{2}=\sqrt{3} \text {. } \\
\end{array}
$$