Question

Prove that 1/3+√7 + 1/√7+√5 + 1/√5+√3 + 1/√3+1 = 1

Answer 1

1/(3+√7)+1/(√7+√5)+1/(√5+√3)+1/(√3+1)
=1/(3+√7)×(3-√7)/(3-√7)+1/(√7+√5)×(√7-√5)/(√7-√5)+1/(√5+√3)×(√5-√3)/(√5-√3)+1/(√3+1)× (√3-1)/(√3-1)
=(3-√7)/(3^2-√7^2)+(√7-√5)/(√7^2-√5^2)+(√5-√3)/(√5^2-√3^2)+(√3-1)/(√3^2-1^2)
=3-√7)/(9-7)+(√7-√5)/(7-5)+(√5-√3)/(5-3)+(√3-1)/(3-1)
=3-√7+√7-√5+√5-√3+√3-1/2
=3-1/2
=2/2
=1
therefore L.H.S=R.H S

Answer 2

$\bf \frac{1}{3 + \sqrt{7} } + \frac{1}{ \sqrt{7} + \sqrt{5} } + \frac{1}{ \sqrt{5} + \sqrt{3} } + \frac{1}{ \sqrt{3} + 1} = 1 \\ \\ \tt \large \: RATIONALISING \: \: THE \: \: ABOVE \\ \\ \bf \frac{3 - \sqrt{7} }{9 - 7 } + \frac{ \sqrt{7} - \sqrt{5} }{ 7 - 5 } + \frac{ \sqrt{5} - \sqrt{3} }{5 - 3 } + \frac{ \sqrt{3} - 1}{ 3 - 1} = 1 \\ \\ \bf \frac{3 - \cancel{\sqrt{7} + \sqrt{7} - \sqrt{5} + \sqrt{5} - \sqrt{3} + \sqrt{3} }- 1}{ 2} = 1 \\ \\ \huge \bf \fcolorbox{red}{grey}{1 = 1}$