Simplify : $$\frac{1}{{\sqrt 3 + \sqrt 4 }} \,+ $$   $$\frac{1}{{\sqrt 4 + \sqrt 5 }} \,+ $$   $$\frac{1}{{\sqrt 5 + \sqrt 6 }} \,+ $$   $$\frac{1}{{\sqrt 6 + \sqrt 7 }} \,+ $$   $$\frac{1}{{\sqrt 7 + \sqrt 8 }}\, + $$   $$\frac{1}{{\sqrt 8 + \sqrt 9 }} = ?$$

Solution (By Examveda Team)

$$\eqalign{ & \frac{1}{{\sqrt 3 + \sqrt 4 }} \cr & = \frac{1}{{\sqrt 4 + \sqrt 3 }} \times \frac{{\sqrt 4 - \sqrt 3 }}{{\sqrt 4 - \sqrt 3 }} \cr & = \frac{{\sqrt 4 - \sqrt 3 }}{1} \cr & = \sqrt 4 - \sqrt 3 \cr & {\text{Similarly}} \cr & \frac{1}{{\sqrt 4 + \sqrt 5 }} = \sqrt 5 - \sqrt 4 \cr & \Rightarrow \frac{1}{{\sqrt 5 + \sqrt 6 }} = \sqrt 6 - \sqrt 5 \cr & \Rightarrow \frac{1}{{\sqrt 6 + \sqrt 7 }} = \sqrt 7 - \sqrt 6 \cr & \Rightarrow \frac{1}{{\sqrt 7 + \sqrt 8 }} = \sqrt 8 - \sqrt 7 \cr & \Rightarrow \frac{1}{{\sqrt 8 + \sqrt 9 }} = \sqrt 9 - \sqrt 8 \cr & {\text{Now put values}} \cr & \Rightarrow \sqrt 4 - \sqrt 3 + \sqrt 5 - \sqrt 4 + \sqrt 6 - \sqrt 5 + \sqrt 7 - \sqrt 6 + \sqrt 8 - \sqrt 7 + \sqrt 9 - \sqrt 8 \cr & \Rightarrow \sqrt 9 - \sqrt 3 \cr & \Rightarrow 3 - \sqrt 3 \cr} $$