$\frac{1}{\sqrt{n}+\sqrt{n+1}}=\frac{\sqrt{n}-\sqrt{n+1}}{n-(n+1)}=\sqrt{n+1}-\sqrt{n}$
$\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+...+\frac{1}{\sqrt{n}+\sqrt{n+1}}=(\sqrt{2}-1)+(\sqrt{3}-\sqrt{2})+(\sqrt{4}-\sqrt{3})+...+(\sqrt{n+1}-\sqrt{n})$
$=\sqrt{n+1}-1$
ดังนั้น $\sqrt{n+1}-1=8$
$n+1=81$
$n=80$