Bài giải:

$\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+\sqrt{16}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=\frac{(\sqrt{2}+\sqrt{3}+2)+(2+\sqrt{6}+\sqrt{8})}{\sqrt{2}+\sqrt{3}+\sqrt{4}}$

$=\frac{(\sqrt{2}+\sqrt{3}+\sqrt{4})+(\sqrt{4}+\sqrt{6}+\sqrt{8})}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=\frac{(\sqrt{2}+\sqrt{3}+\sqrt{4})+\sqrt{2}.(\sqrt{2}+\sqrt{3}+\sqrt{4})}{\sqrt{2}+\sqrt{3}+\sqrt{4}}$

$=\frac{(\sqrt{2}+\sqrt{3}+\sqrt{4})(\sqrt{2}+1)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=\sqrt{2}+1$