2.294. $\sqrt{6 m+2 \sqrt{9 m^{2}-n^{2}}}-\sqrt{6 m-2 \sqrt{9 m^{2}-n^{2}}}=2 \sqrt{3 m-n}$.
Solution.

Domain of definition: $\left\{\begin{array}{l}6 m+2 \sqrt{9 m^{2}-n^{2}} \geq 0, \\ 6 m-2 \sqrt{9 m^{2}-n^{2}} \geq 0, \\ 3 m-n \geq 0, \\ 3 m+n \geq 0,\end{array},\left\{\begin{array}{l}m>0, \\ n \leq 3 m .\end{array}\right.\right.$

$$
\begin{aligned}
& \sqrt{6 m+2 \sqrt{9 m^{2}-n^{2}}}-\sqrt{6 m-2 \sqrt{9 m^{2}-n^{2}}}= \\
& =\sqrt{3 m+n+2 \sqrt{(3 m+n)(3 m-n)}+3 m-n}- \\
& -\sqrt{3 m+n-2 \sqrt{(3 m+n)(3 m-n)}+3 m-n}= \\
& =\sqrt{(\sqrt{3 m+n}+\sqrt{3 m-n})^{2}}-\sqrt{(\sqrt{3 m+n}-\sqrt{3 m-n})^{2}}=
\end{aligned}
$$

$$
\begin{aligned}
& =\sqrt{3 m+n}+\sqrt{3 m-n}-|\sqrt{3 m+n}-\sqrt{3 m-n}|= \\
& =\left\{\begin{array}{l}
\sqrt{3 m+n}+\sqrt{3 m-n}+\sqrt{3 m+n}-\sqrt{3 m-n}=2 \sqrt{3 m+n} \text { for } n<0, \\
\sqrt{3 m+n}+\sqrt{3 m-n}-\sqrt{3 m+n}+\sqrt{3 m-n}=2 \sqrt{3 m-n} \text { for } n \geq 0.
\end{array}\right.
\end{aligned}
$$