Question

Question:-
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Prove That Answer is 1
$\sf \dashrightarrow \ \dfrac{1}{3 + \sqrt{7}} \ + \ \dfrac{1}{\sqrt{7} + \sqrt{5}} \ + \ \dfrac{1}{\sqrt{5} + \sqrt{3}} \ + \ \dfrac{1}{\sqrt{3} + 1} = 1$

Spammer Go Away ​

Answer 1

$\large \rm \red{\widetilde{To\ prove:-}}$

$\rm \dfrac{1}{3 + \sqrt{7}} \ + \ \dfrac{1}{\sqrt{7} + \sqrt{5}} \ + \ \dfrac{1}{\sqrt{5} + \sqrt{3}} \ + \ \dfrac{1}{\sqrt{3} + 1} = 1$

$\large \rm \red{\widetilde{Taking\ RHS:-}}$

$\rm :\longmapsto \dfrac{1}{3 + \sqrt{7}} \ + \ \dfrac{1}{\sqrt{7} + \sqrt{5}} \ + \ \dfrac{1}{\sqrt{5} + \sqrt{3}} \ + \ \dfrac{1}{\sqrt{3} + 1}$

Rationalizing, we get

$\rm :\longmapsto \dfrac{1}{3+\sqrt7} \times \dfrac{3-\sqrt7}{3-\sqrt7} + \dfrac{1}{\sqrt7 + \sqrt5} \times \dfrac{\sqrt7 - \sqrt5}{\sqrt7 - \sqrt5}$

$\rm \, \, \, \, \, \; \; \;+ \dfrac{1}{\sqrt5 + \sqrt3} \times \dfrac{\sqrt5 - \sqrt3}{\sqrt5-\sqrt3} + \dfrac{1}{\sqrt3 + 1} \times \dfrac{\sqrt3 - 1}{\sqrt3 -1}$

$\rm :\longmapsto \dfrac{3 - \sqrt7}{(3)^2 - (\sqrt7)^2} + \dfrac{\sqrt7 - \sqrt5}{(\sqrt7)^2 - (\sqrt5)^2} +\dfrac{\sqrt5 - \sqrt3}{(\sqrt5)^2 - (\sqrt3)^2} + \dfrac{\sqrt3-1}{(\sqrt3)^2 - 1^2}$

$\rm :\longmapsto \dfrac{3 - \sqrt7}{9-7} + \dfrac{\sqrt7 - \sqrt5}{7-5} +\dfrac{\sqrt5 - \sqrt3}{5-3} + \dfrac{\sqrt3-1}{3-1}$

$\rm :\longmapsto \dfrac{3 - \sqrt7}{2} + \dfrac{\sqrt7 - \sqrt5}{2} +\dfrac{\sqrt5 - \sqrt3}{2} + \dfrac{\sqrt3-1}{2}$

$\rm :\longmapsto \dfrac{3\cancel{ - \sqrt7 + \sqrt7} \cancel{- \sqrt5 + \sqrt5}\cancel{ - \sqrt3 + \sqrt3} - 1}{2}$

$\rm :\longmapsto \dfrac{2}{2}$

$\rm :\longmapsto 1$

= RHS

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Answer 2

Answer:

Refer To The Attachment

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hope it helps