2. Simplify $\frac{\sqrt{3+2 \sqrt{2}}}{\sqrt{17+12 \sqrt{2}}}-\frac{\sqrt{3-2 \sqrt{2}}}{\sqrt{17-12 \sqrt{2}}}$, the result is ( ).
(A) $\sqrt{2}$
(B) $-\sqrt{2}$
(C) 2
(D) -2
2. D.

Notice that
$$
\begin{array}{l}
3+2 \sqrt{2}=(\sqrt{2}+1)^{2}, \\
3-2 \sqrt{2}=(\sqrt{2}-1)^{2}, \\
17+12 \sqrt{2}=(3+2 \sqrt{2})^{2}, \\
17-12 \sqrt{2}=(3-2 \sqrt{2})^{2} .
\end{array}
$$

Therefore, the original expression
$$
\begin{array}{l}
=\frac{1}{\sqrt{3+2 \sqrt{2}}}-\frac{1}{\sqrt{3-2 \sqrt{2}}} \\
=\frac{1}{\sqrt{2}+1}-\frac{1}{\sqrt{2}-1}=-2 .
\end{array}
$$