Question 72

# $$\frac{1}{3-\sqrt8}-\frac{1}{\sqrt8-\sqrt7}+\frac{1}{\sqrt7-\sqrt6}-\frac{1}{\sqrt6-\sqrt5}+\frac{1}{\sqrt5-\sqrt4}$$ is equal to

Solution

Expression : $$\frac{1}{3-\sqrt8}-\frac{1}{\sqrt8-\sqrt7}+\frac{1}{\sqrt7-\sqrt6}-\frac{1}{\sqrt6-\sqrt5}+\frac{1}{\sqrt5-\sqrt4}$$

Rationalizing the denominator, we get :

= $$(\frac{1}{3-\sqrt8}\times\frac{3+\sqrt8}{3+\sqrt8})-(\frac{1}{\sqrt8-\sqrt7}\times\frac{\sqrt8+\sqrt7}{\sqrt8+\sqrt7})+(\frac{1}{\sqrt7-\sqrt6}\times\frac{\sqrt7+\sqrt6}{\sqrt7+\sqrt6})-(\frac{1}{\sqrt6-\sqrt5}\times\frac{\sqrt6+\sqrt5}{\sqrt6+\sqrt5})+(\frac{1}{\sqrt5-\sqrt4}\times\frac{\sqrt5+\sqrt4}{\sqrt5+\sqrt4})$$

Using, $$(a-b)(a+b)=a^2-b^2$$

= $$\frac{3+\sqrt8}{9-8}-\frac{\sqrt8+\sqrt7}{8-7}+\frac{\sqrt7+\sqrt6}{7-6}-\frac{\sqrt6+\sqrt5}{6-5}+\frac{\sqrt5+\sqrt4}{5-4}$$

= $$(3+\sqrt8)+(-\sqrt8-\sqrt7)+(\sqrt7+\sqrt6)+(-\sqrt6-\sqrt5)+(\sqrt5+\sqrt4)$$

= $$3+\sqrt4=3+2=5$$

=> Ans - (A)