Question

answer this question​

Answer 1

$\Large \textit{Here's the method...!}$

$\frac{1}{\sqrt{9}-\sqrt{8}}-\frac{1}{\sqrt{8}-\sqrt{7}}+\frac{1}{\sqrt{7}-\sqrt{6}}-\frac{1}{\sqrt{6}-\sqrt{5}}+\frac{1}{\sqrt{5}-\sqrt{4}} \\ \\ \\ \frac{1(\sqrt{9}+\sqrt{8})}{(\sqrt{9}-\sqrt{8})(\sqrt{9}+\sqrt{8})}-\frac{1(\sqrt{8}+\sqrt{7})}{(\sqrt{8}-\sqrt{7})(\sqrt{8}+\sqrt{7})}+\frac{1(\sqrt{7}+\sqrt{6})}{(\sqrt{7}-\sqrt{6})(\sqrt{7}+\sqrt{6})}-\frac{1(\sqrt{6}+\sqrt{5})}{(\sqrt{6}-\sqrt{5})(\sqrt{6}+\sqrt{5})}+\frac{1(\sqrt{5}+\sqrt{4})}{(\sqrt{5}-\sqrt{4})(\sqrt{5}+\sqrt{4})}$

$\frac{\sqrt{9}+\sqrt{8}}{(\sqrt{9})^2-(\sqrt{8})^2}-\frac{\sqrt{8}+\sqrt{7}}{(\sqrt{8})^2-(\sqrt{7})^2}+\frac{\sqrt{7}+\sqrt{6}}{(\sqrt{7})^2-(\sqrt{6})^2}-\frac{\sqrt{6}+\sqrt{5}}{(\sqrt{6})^2-(\sqrt{5})^2}+\frac{\sqrt{5}+\sqrt{4}}{(\sqrt{5})^2-(\sqrt{4})^2} \\ \\ \\ \frac{\sqrt{9}+\sqrt{8}}{9-8}-\frac{\sqrt{8}+\sqrt{7}}{8-7}+\frac{\sqrt{7}+\sqrt{6}}{7-6}-\frac{\sqrt{6}+\sqrt{5}}{6-5}+\frac{\sqrt{5}+\sqrt{4}}{5-4}$

$\frac{\sqrt{9}+\sqrt{8}}{1}-\frac{\sqrt{8}+\sqrt{7}}{1}+\frac{\sqrt{7}+\sqrt{6}}{1} -\frac{\sqrt{6}+\sqrt{5}}{1}+\frac{\sqrt{5}+\sqrt{4}}{1} \\ \\ \\ (\sqrt{9}+\sqrt{8})-(\sqrt{8}+\sqrt{7})+(\sqrt{7}+\sqrt{6})-(\sqrt{6}+\sqrt{5})+(\sqrt{5}+\sqrt{4}) \\ \\ \\ \sqrt{9}+\sqrt{8}-\sqrt{8}-\sqrt{7}+\sqrt{7}+\sqrt{6}-\sqrt{6}-\sqrt{5}+\sqrt{5}+\sqrt{4} \\ \\ \\ \sqrt{9}+\sqrt{4} \\ \\ \\ 3+2 = \Large \textbf{\bold{5}}$

$\huge \underline{\underline{\textsc{Thus the answer is}\ \ \textbf{5} \textsc{.}}}$