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

In equation 1, we will rationalize the DENOMINATOR of each of the TERMS to remove root from the denominator and MAKE the calculations easier.

$(\frac{1}{{\sqrt 9- \sqrt 8 }} = \frac{1}{{\sqrt 9- \sqrt 8 }} \times \frac{{\sqrt 9+ \sqrt 8 }}{{\sqrt 9+ \sqrt 8 }} = \frac{{\sqrt 9+ \sqrt 8 }}{{{{\left( {\sqrt 9 } \right)}^2} - {{\left( {\sqrt 8 } \right)}^2}}} = \frac{{\sqrt 9+ \sqrt 8 }}{{9 - 8}} = \sqrt 9+ \sqrt 8 )$

Similarly, $(\frac{1}{{\sqrt 8- \sqrt 7 }} = \sqrt 8+ \sqrt 7)$

$(\BEGIN{array}{l}\frac{1}{{\sqrt 7- \sqrt 6 }} = \sqrt 7+ \sqrt 6 \\\frac{1}{{\sqrt 6- \sqrt 5 }} = \sqrt 6+ \sqrt 5 \\\frac{1}{{\sqrt 5- \sqrt 4 }} = \sqrt 5+ \sqrt 4\end{array})$

Equation 1 will reduce to

$(\begin{array}{l}\left( {\sqrt 9+ \sqrt 8 } \right) - \left( {\sqrt 8+ \sqrt 7 } \right) + \left( {\sqrt 7+ \sqrt 6 } \right) - \left( {\sqrt 6+ \sqrt 5 } \right) + \left( {\sqrt 5+ \sqrt 4 } \right)\\ = \sqrt 9+ \sqrt 4\end{array})$ 

= 3 + 2 = 5