$\frac{\sqrt{5}+\sqrt{3}}{\sqrt{8-2 \sqrt{15}}}+\frac{\sqrt{11+2 \sqrt{30}}}{\sqrt{6}-\sqrt{5}}$

Look at the first term first:

$\frac{\sqrt{5}+\sqrt{3}}{\sqrt{8-2 \sqrt{15}}}$

If you look closely, you'll see that:

$8-2\sqrt{15}=(\sqrt{5})^2+(\sqrt{3})^2-2(\sqrt5)(\sqrt3)$

We know that $(a-b)^2=a^2+b^2-2ab$

So we can write the above expression as $(\sqrt5-\sqrt3)^2$

So, the first term:

$\frac{\sqrt{5}+\sqrt{3}}{\sqrt5-\sqrt3}$

Multiply numerator and denominator by $(\sqrt{5}+\sqrt3)$

$\frac{(\sqrt{5}+\sqrt{3})^2}{(\sqrt5-\sqrt3)(\sqrt5+\sqrt3)}$

We know that $(a+b)(a-b)=a^2-b^2$ and that $(a+b)^2=a^2+b^2+2ab$

So the above expression can be written as: $\frac{5+3+2\sqrt15}{5-3}=\frac{5+3+2\sqrt{15}}{2}=\frac{8+2\sqrt{15}}{2}=4+\sqrt{15}$

Now for the second term, it's the same approach.

$11+2\sqrt{30}=(\sqrt6)^2+(\sqrt5)^2+2(\sqrt6)(\sqrt5)=(\sqrt6+\sqrt5)^2$

So, second term:

$\frac{\sqrt6+\sqrt5}{\sqrt{6}-\sqrt{5}}$

Multiply numerator and denominator by $(\sqrt6+\sqrt5)$

$\frac{(\sqrt6+\sqrt5)^2}{(\sqrt6-\sqrt5)(\sqrt6+\sqrt5)}$

Open the brackets using the same identity

$\frac{6+5+2\sqrt{30}}{6-5}=\frac{6+5+2\sqrt{30}}{1}=11+2\sqrt{30}$

Just add the two terms now, and we'll get

$4+\sqrt{15}+11+2\sqrt{30}=15+\sqrt{15}+2\sqrt{30}$