Question

Show that :
${\displaystyle \frac{1}{3-2\surd 2}-\frac{1}{2\surd 2-\surd 7}+\frac{1}{\surd 7-\surd 6}-\frac{1}{\surd 6-\surd 5}+\frac{1}{\surd 5-2}=5}$

Answer 1

L.H.S = ${\displaystyle \frac{1}{3-2\surd 2}-\frac{1}{2\surd 2-\surd 7}+\frac{1}{\surd 7-\surd 6}-\frac{1}{\surd 6-\surd 5}+\frac{1}{\surd 5-2}}$
= ${\displaystyle \frac{1}{3-\surd 8}-\frac{1}{\surd 8-\surd 7}+\frac{1}{\surd 7-\surd 6}-\frac{1}{\surd 6-\surd 5}+\frac{1}{\surd 5-2}}$

= ${\displaystyle \frac{1}{3-\surd 8}\times \frac{3+\surd 8}{3+\surd 8}-\frac{1}{\surd 8-\surd 7}\times \frac{\surd 8+\surd 7}{\surd 8+\surd 7}+\frac{1}{\surd 7-\surd 6}\times \frac{\surd 7+\surd 6}{\surd 7+\surd 6}-\frac{1}{\surd 6-\surd 5}\times \frac{\surd 6+\surd 5}{\surd 6+\surd 5}+\frac{1}{\surd 5-2}\times \frac{\surd 5+2}{\surd 5+2}}$

= ${\displaystyle \frac{3+\surd 8}{{\left(3\right)}^2-{(\surd 8)}^2} -\frac{\surd 8+\surd 7}{{(\surd 8)}^2-{(\surd 7)}^2}+\frac{\surd 7+\surd 6}{{(\surd 7)}^2-{(\surd 6)}^2}-\frac{\surd 6-\surd 5}{{(\surd 6)}^2-{(\surd 5)}^2}+\frac{\surd 5-2}{{(\surd 5)}^2-{\left(2\right)}^2}}$

= ${\displaystyle \frac{3+\surd 8}{9-8} -\frac{\surd 8+\surd 7}{8-7} +\frac{\surd 7+\surd 6}{7-6}-\frac{\surd 6-\surd 5}{6-5}+\frac{\surd 5-2}{5-4}}$

= 3 + √8 - √8 - √7 + √7 + √6 - √6 - √5 + √5 + 2

= 3 + 2

= 5

= R.H.S.