Question

Find the value of the expression $=\frac{1}{3-\sqrt{8}}-\frac{1}{\sqrt{8}-\sqrt{7}}+\frac{1}{\sqrt{7}-\sqrt{6}}-\frac{1}{\sqrt{6}-\sqrt{5}}+\frac{1}{\sqrt{5}-\sqrt{4}}$

Answer 1

$\frac{1}{3-\sqrt{8}}=\frac{1}{(3-\sqrt{8})}\times \frac{3+\sqrt{8}}{3+\sqrt{8}}=\frac{3+\sqrt{8}}{9-8}=3+\sqrt{8}$

$\frac{1}{\sqrt{8}-\sqrt{7}}=\frac{1}{\sqrt{8}-\sqrt{7}}\times \frac{\sqrt{8}+\sqrt{7}}{\sqrt{8}+\sqrt{7}}=\frac{\sqrt{8}+\sqrt{7}}{8-7}=\sqrt{8}+\sqrt{7}$

$\frac{1}{\sqrt{7}-\sqrt{6}}=\frac{1}{\sqrt{7}-\sqrt{6}}\times \frac{\sqrt{7}+\sqrt{6}}{\sqrt{7}+\sqrt{6}}=\frac{\sqrt{7}+\sqrt{6}}{7-6}=\sqrt{7}+\sqrt{6}$

$\frac{1}{\sqrt{6}-\sqrt{5}}=\frac{1}{\sqrt{6}-\sqrt{5}}\times \frac{\sqrt{6}+\sqrt{5}}{\sqrt{6}+\sqrt{5}}=\frac{\sqrt{6}+\sqrt{5}}{6-5}=\sqrt{6}+\sqrt{5}$

$\frac{1}{\sqrt{5}-\sqrt{4}}=\frac{1}{\sqrt{5}-\sqrt{4}}\times \frac{\sqrt{5}+\sqrt{4}}{\sqrt{5}+\sqrt{4}}=\frac{\sqrt{5}+\sqrt{4}}{5-4}=\sqrt{5}+\sqrt{4}$

$\therefore$ Given exp. = $3+\sqrt{8}-(\sqrt{8}+\sqrt{7})+\sqrt{7}+\sqrt{6}-(\sqrt{6}+\sqrt{5})+\sqrt{5}+\sqrt{4}$

$=3+\sqrt{8}-\sqrt{8}-\sqrt{7}+\sqrt{7}+\sqrt{6}-\sqrt{6}-\sqrt{5}+\sqrt{5}+\sqrt{4}=3+\sqrt{4}=3+2=5$