2. Given $a=\sqrt{2006}-\sqrt{2005}$,
$$
\begin{array}{l}
b=\sqrt{2007}-\sqrt{2006}, \\
c=\sqrt{2008}-\sqrt{2007} .
\end{array}
$$

Then the correct conclusion is ( ).
(A) $a>b>c$
(B) $c>b>a$
(C) $b>a>c$
(D) $b>c>a$
2.A.

Since $a=\frac{1}{\sqrt{2006}+\sqrt{2005}}$,
$$
b=\frac{1}{\sqrt{2007}+\sqrt{2006}}, c=\frac{1}{\sqrt{2008}+\sqrt{2007}},
$$

Therefore, $a>b>c$.