If $\frac{1}{\sqrt{5}-\sqrt{5-\sqrt{24}}}+\frac{1}{\sqrt{5}+\sqrt{5+\sqrt{24}}}=\frac{k}{\sqrt{2}}$ , then $k^2+4=$

$=\frac{1}{\sqrt{5}-\left(\sqrt{3}-\sqrt{2}\right)}+\frac{1}{\sqrt{5}+\left(\sqrt{3}+\sqrt{2}\right)}$

$=\frac{1}{\left(\sqrt{5}+\sqrt{2}\right)-\sqrt{3}}+\frac{1}{\left(\sqrt{5}+\sqrt{2}\right)+\sqrt{3}}$

$=\frac{1\left(\sqrt{5}+\sqrt{2}+\sqrt{3}\right)+1\left(\sqrt{5}+\sqrt{2}-\sqrt{3}\right)}{{\left(\sqrt{5}+\sqrt{2}\right)}^2-{\left(\sqrt{3}\right)}^2}$

$=\frac{2\left(\sqrt{5}+\sqrt{2}\right)}{4+2\sqrt{10}}$

$=\frac{2\left(\sqrt{5}+\sqrt{2}\right)}{4+2\sqrt{10}}$

$=\frac{2\left(\sqrt{5}+\sqrt{2}\right)}{2\sqrt{2}\left(\sqrt{2}+\sqrt{5}\right)}$

$=\frac{1}{\sqrt{2}}$

$\Rightarrow \frac{k}{\sqrt{2}}=\frac{1}{\sqrt{2}}$

$\therefore k^2+4=1^2+4=5$