Simplify : $\dfrac{1}{\sqrt{4} + \sqrt{5}} + \dfrac{1}{\sqrt{5} + \sqrt{6}} + \dfrac{1}{\sqrt{6} + \sqrt{7}} + \dfrac{1}{\sqrt{7} + \sqrt{8}} +\dfrac{1}{\sqrt{8} + \sqrt{9}}$.

$\dfrac{1}{\sqrt{4} + \sqrt{5}} + \dfrac{1}{\sqrt{5} + \sqrt{6}} + \dfrac{1}{\sqrt{6} + \sqrt{7}} + \dfrac{1}{\sqrt{7} + \sqrt{8}} +\dfrac{1}{\sqrt{8} + \sqrt{9}} \qquad \text{….(i)}$

Simplifying each term individually,

$\dfrac{1}{\sqrt{4} + \sqrt{5}}$

Let us rationalise its denominator,

Then,

$\dfrac{1}{\sqrt{4} + \sqrt{5}} = \dfrac{1}{\sqrt{4} + \sqrt{5}} × \dfrac{\sqrt{4} - \sqrt{5}}{\sqrt{4} - \sqrt{5}} \\[1.5em] \Rightarrow\dfrac{{\sqrt{4} - \sqrt{5}}}{(\sqrt{4})^2 - (\sqrt{5})^2} \\[1.5em] \Rightarrow\dfrac{{\sqrt{4} - \sqrt{5}}}{4 - 5} \\[1.5em] \Rightarrow{-(\sqrt{4} - \sqrt{5})} \\[1.5em] \Rightarrow{(\sqrt{5} - \sqrt{4})} \qquad \text{….(ii)} \\[1.5em]$

$\dfrac{1}{\sqrt{5} + \sqrt{6}}$

Let us rationalise its denominator,

Then,

$\dfrac{1}{\sqrt{5} + \sqrt{6}} = \dfrac{1}{\sqrt{5}+\sqrt{6}}×\dfrac{\sqrt{5} - \sqrt{6}}{\sqrt{5} - \sqrt{6}} \\[1.5em] \Rightarrow\dfrac{{\sqrt{5} - \sqrt{6}}}{(\sqrt{5})^2 - (\sqrt{6})^2} \\[1.5em] \Rightarrow\dfrac{{\sqrt{5} - \sqrt{6}}}{5 - 6} \\[1.5em] \Rightarrow{-(\sqrt{5} - \sqrt{6})} \\[1.5em] \Rightarrow{(\sqrt{6} - \sqrt{5})} \qquad \text{….(iii)} \\[1.5em]$

$\dfrac{1}{\sqrt{6} + \sqrt{7}}$

Let us rationalise its denominator,

Then,

$\dfrac{1}{\sqrt{6} + \sqrt{7}} = \dfrac{1}{\sqrt{6}+ \sqrt{7}} × \dfrac{\sqrt{6} - \sqrt{7}}{\sqrt{6} - \sqrt{7}} \\[1.5em] \Rightarrow\dfrac{{\sqrt{6} - \sqrt{7}}}{(\sqrt{6})^2 - (\sqrt{7})^2} \\[1.5em] \Rightarrow\dfrac{{\sqrt{6} - \sqrt{7}}}{6-7} \\[1.5em] \Rightarrow{-(\sqrt{6} - \sqrt{7})} \\[1.5em] \Rightarrow{(\sqrt{7} - \sqrt{6})} \qquad \text{….(iv)} \\[1.5em]$

$\dfrac{1}{\sqrt7+\sqrt8}$

Let us rationalise its denominator,

Then,

$\dfrac{1}{\sqrt{7} + \sqrt{8}} = \dfrac{1}{\sqrt{7}+ \sqrt{8}} × \dfrac{\sqrt{7} - \sqrt{8}}{\sqrt{7} - \sqrt{8}} \\[1.5em] \Rightarrow\dfrac{{\sqrt{7} - \sqrt{8}}}{(\sqrt{7})^2 - (\sqrt{8})^2} \\[1.5em] \Rightarrow\dfrac{{\sqrt{7} - \sqrt{8}}}{7-8} \\[1.5em] \Rightarrow{-(\sqrt{7} - \sqrt{8})} \\[1.5em] \Rightarrow{(\sqrt{8} - \sqrt{7})} \qquad \text{….(v)} \\[1.5em]$

$\dfrac{1}{\sqrt{4} + \sqrt{5}}$

Let us rationalise its denominator,

Then,

$\dfrac{1}{\sqrt{8} + \sqrt{9}} = \dfrac{1}{\sqrt{8}+ \sqrt{9}} × \dfrac{\sqrt{8} - \sqrt{9}}{\sqrt{8} - \sqrt{9}} \\[1.5em] \Rightarrow\dfrac{{\sqrt {8} - \sqrt{9}}}{(\sqrt{8})^2 - (\sqrt{9})^2} \\[1.5em] \Rightarrow\dfrac{{\sqrt{8} - \sqrt{9}}}{8 - 9} \\[1.5em] \Rightarrow{-(\sqrt{8} - \sqrt{9})} \\[1.5em] \Rightarrow{(\sqrt{9}-\sqrt{8})} \qquad \text{….(vi)} \\[1.5em]$

Using (ii) , (iii) , (iv) , (v) , (vi) in equation (i):

$\dfrac{1}{\sqrt4+\sqrt5} + \dfrac{1}{\sqrt5+\sqrt6} + \dfrac{1}{\sqrt6+\sqrt7} + \dfrac{1}{\sqrt7+\sqrt8} +\dfrac{1}{\sqrt8+\sqrt9} \\[1.5em] = \sqrt{5} - \sqrt{4} + \sqrt{6} - \sqrt{5} + \sqrt{7} - \sqrt{6} + \sqrt{8} - \sqrt{7} + \sqrt{9} - \sqrt{8} \\[1.5em] = \sqrt{9} - \sqrt{4} \\[1.5em] = \bold{3 - 2 = 1 } \\[1.5em]$