Question

Show that:
$\frac{1}{3 - \sqrt{8} } - \frac{1}{ \sqrt{8 - \sqrt{7} } } + \frac{1}{ \sqrt{7 - \sqrt{6} } } - \frac{1}{ \sqrt{6 - \sqrt{5} } } + \frac{1}{ \sqrt{7 - 6} } = 5$
​

Answer 1

Given : Correct Question is

$\dfrac{1}{3-\sqrt{8} } -\dfrac{1}{\sqrt{8} - \sqrt{7} }+\dfrac{1}{\sqrt{7} - \sqrt{6} }-\dfrac{1}{\sqrt{6} - \sqrt{5} }+\dfrac{1}{\sqrt{5} - 2}=5$

To Find : Prove

Solution:

$\dfrac{1}{3-\sqrt{8} } -\dfrac{1}{\sqrt{8} - \sqrt{7} }+\dfrac{1}{\sqrt{7} - \sqrt{6} }-\dfrac{1}{\sqrt{6} - \sqrt{5} }+\dfrac{1}{\sqrt{5} - 2}=5$

$LHS = \dfrac{1}{3-\sqrt{8} } -\dfrac{1}{\sqrt{8} - \sqrt{7} }+\dfrac{1}{\sqrt{7} - \sqrt{6} }-\dfrac{1}{\sqrt{6} - \sqrt{5} }+\dfrac{1}{\sqrt{5} - 2}$

$\dfrac{1}{3-\sqrt{8} } = \dfrac{1}{3-\sqrt{8} } \times \dfrac{3+\sqrt{8} }{3+\sqrt{8} } = 3+\sqrt{8}$

$\dfrac{1}{\sqrt{8} -\sqrt{7} } = \dfrac{1}{\sqrt{8} -\sqrt{7} } \times \dfrac{\sqrt{8} +\sqrt{7} }{\sqrt{8} + \sqrt{7}} = \sqrt{8}+\sqrt{7}$

$\dfrac{1}{\sqrt{7} -\sqrt{6} } = \dfrac{1}{\sqrt{7} -\sqrt{6} } \times \dfrac{\sqrt{7} +\sqrt{6} }{\sqrt{7} + \sqrt{6}} = \sqrt{7}+\sqrt{6}$

$\dfrac{1}{\sqrt{6} -\sqrt{5} } = \dfrac{1}{\sqrt{6} -\sqrt{5} } \times \dfrac{\sqrt{6} +\sqrt{5} }{\sqrt{6} + \sqrt{5}} = \sqrt{6}+\sqrt{5}$

$\dfrac{1}{\sqrt{5} -2 } = \dfrac{1}{\sqrt{5} -2 } \times \dfrac{\sqrt{5} +2 }{\sqrt{5} + 2} = \sqrt{5}+2$

$LHS = \dfrac{1}{3-\sqrt{8} } -\dfrac{1}{\sqrt{8} - \sqrt{7} }+\dfrac{1}{\sqrt{7} - \sqrt{6} }-\dfrac{1}{\sqrt{6} - \sqrt{5} }+\dfrac{1}{\sqrt{5} - 2}$

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

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

= 3  + 2

= 5

= RHS

QED

Hence Proved

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Answer 2

Answer:

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