若 $\sqrt{5-2 \sqrt{6}}=\sqrt{m}-\sqrt{n}$, 则 $m=$ \$ \qquad \$ , $n=$
解: $\because \sqrt{3}>\sqrt{2}$, 即 $\sqrt{3}-\sqrt{2}>0$,

$\therefore \sqrt{5-2 \sqrt{6}}=\sqrt{2+3-2 \sqrt{6}}$

$=\sqrt{(\sqrt{2})^{2}-2 \cdot \sqrt{2} \cdot \sqrt{3}+(\sqrt{3})^{2}}$

$=\sqrt{(\sqrt{2}-\sqrt{3})^{2}}$

$=|\sqrt{2}-\sqrt{3}|=\sqrt{3}-\sqrt{2}$,

又 $\because \sqrt{5-2 \sqrt{6}}=\sqrt{m}-\sqrt{n}$,

则 $m=3, n=2$.

此题考查了二次根式的化简求值, 其技巧性较强, 灵活变换等式左边的被开方数是解本题的关键.