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Prove That:
[tex]\begin{gathered}\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\\ \end{gathered}[/tex]


Spammer Go Away ​​

Sagot :

YourHelperAdi

Step-by-step explanation:

[tex]\large \rm \red{\widetilde{Taking\ RHS:-}}[/tex]

[tex]\rm :\longmapsto \dfrac{1}{3 + \sqrt{7}} \ + \ \dfrac{1}{\sqrt{7} + \sqrt{5}} \ + \ \dfrac{1}{\sqrt{5} + \sqrt{3}} \ + \ \dfrac{1}{\sqrt{3} + 1}[/tex]

Rationalizing, we get

[tex]\rm :\longmapsto \dfrac{1}{3+\sqrt7} \times \dfrac{3-\sqrt7}{3-\sqrt7} + \dfrac{1}{\sqrt7 + \sqrt5} \times \dfrac{\sqrt7 - \sqrt5}{\sqrt7 - \sqrt5}[/tex]

[tex]\rm + \dfrac{1}{\sqrt5 + \sqrt3} \times \dfrac{\sqrt5 - \sqrt3}{\sqrt5-\sqrt3} + \dfrac{1}{\sqrt3 + 1} \times \dfrac{\sqrt3 - 1}{\sqrt3 -1}[/tex]

[tex]\rm :\longmapsto \dfrac{3 - \sqrt7}{(3)^2 - (\sqrt7)^2}[/tex]

[tex] + \dfrac{\sqrt7 - \sqrt5}{(\sqrt7)^2 - (\sqrt5)^2} [/tex]

[tex]+\dfrac{\sqrt5 - \sqrt3}{(\sqrt5)^2 - (\sqrt3)^2} +[/tex]

[tex]\dfrac{\sqrt3-1}{(\sqrt3)^2 - 1^2}[/tex]

SalmonWorldTopper

Given info:- Rationalisie the denominator of the following 1/(3+√7) + 1/(√7+√5) + 1/(√5+√3) + 1/(√3+1) = 1, also show that Left hand side is equal to Right hand side that means LHS = RHS.

Explanation:-

On rationalising the denominator, we get

First term: 1/(3+√7)

⇛{1/(3+√7)}×{(3-√7)/(3-√7)}

⇛{1(3-√7)}/{(3+√7)(3-√7)}

⇛(3-√7)/{(3)²-(√7)²}

⇛(3-√7)/{(3*3)-(√7*7)}

⇛(3-√7)/(9 - 7)

⇛(3-√7)/2

Second term: 1/(√7+√5)

⇛{1/√7+√5)}×{(√7-√5)/(√7-√5)}

⇛{1(√7-√5)}/{(√7+√5)(√7-√5)}

⇛(√7-√5)/{(√7)²-(√5)²}

⇛(√7-√5)/{(√7*7)-(√5*5)}

⇛(√7-√5)/(7-5)

⇛(√7-√5)/2

Third term: 1/(√5+√3)

⇛{1/(√5+√3)}×{(√5-√3)/(√5-√3)}

⇛{1(√5-√3)}/{(√5+√3)(√5-√3)}

⇛(√5-√3}/{(5)²-(√3)²}

⇛(√5-√3)/{(√5*5)-(√3*3)}

⇛(√5-√3)/(5-3)

⇛(√5-√3)/2

Fourth term: 1/(√3+1)

⇛{1/(√3+1)}×{(√3-1)/(√3-1)}

⇛{1(√3-1)}/{(√3+1)(√3-1)}

⇛(√3-1)/{(√3)²-(1)²}

⇛(√3-1)/{(√3*3)-(1*1)}

⇛(√3-1)/3-1

⇛(√3-1)/2

Hence, LHS = 1/(3+√7) + 1/(√7+√5) + 1/(√5+√3) + 1/(√3+1)

Now, Substitute their rationalised value in expression, we get

= (3-√7)/2 + (√7-√5)/2 + (√5-√3)/2 + (√3-1)/2

Take the LCM of all the denominator 2,2,2 and 2 is "2".

= (3-√7+√7-√5+√5-√3+√3-1)/2

= (3-1)/2

= 2/2

= 1

= RHS proved.

Hope this may help you.

If you have any doubt, then you can ask me in the comments.

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