Sorry for the delay, I had to ask my girlfriend how to do this...
We are given:
$\displaystyle \frac{\sum_{k=1}^{99}(\sqrt{10+\sqrt{k}})}{\sum_{k =1}^{99}(\sqrt{10-\sqrt{k}})}$
Let:
$\displaystyle r=\sqrt{10+\sqrt{k}}-\sqrt{10-\sqrt{k}}$
$\displaystyle r^2=10+\sqrt{k}-2\sqrt{100-k}+10-\sqrt{k}$
$\displaystyle r^2=2(10-\sqrt{100-k})$
Since $\displaystyle 0<r$ we may write:
$\displaystyle r=\sqrt{2}\sqrt{10-\sqrt{100-k}}$
Hence, we may write:
$\displaystyle \frac{\sum_{k=1}^{99}(\sqrt{10+\sqrt{k}})}{\sum_{k =1}^{99}(\sqrt{10-\sqrt{k}})}=$
$\displaystyle \frac{\sum_{k=1}^{99}(r+\sqrt{10-\sqrt{k}})}{\sum_{k=1}^{99}(\sqrt{10-\sqrt{k}})}=$
$\displaystyle \frac{\sum_{k=1}^{99}(r)+\sum_{k=1}^{99}(\sqrt{10-\sqrt{k}})}{\sum_{k=1}^{99}(\sqrt{10-\sqrt{k}})}=$
$\displaystyle \frac{\sum_{k=1}^{99}(r)}{\sum_{k=1}^{99}(\sqrt{10-\sqrt{k}})}+\frac{\sum_{k=1}^{99}(\sqrt{10-\sqrt{k}})}{\sum_{k=1}^{99}(\sqrt{10-\sqrt{k}})}=$
$\displaystyle \frac{\sum_{k=1}^{99}(\sqrt{2}\sqrt{10-\sqrt{100-k}})}{\sum_{k=1}^{99}(\sqrt{10-\sqrt{k}})}+1=$
$\displaystyle \sqrt{2}\frac{\sum_{k=1}^{99}(\sqrt{10-\sqrt{k}})}{\sum_{k=1}^{99}(\sqrt{10-\sqrt{k}})}+1=$
$\displaystyle \sqrt{2}+1$