Answered

Westonci.ca makes finding answers easy, with a community of experts ready to provide you with the information you seek. Get immediate and reliable solutions to your questions from a knowledgeable community of professionals on our platform. Our platform offers a seamless experience for finding reliable answers from a network of knowledgeable professionals.

Question:-
▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬
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.

Your visit means a lot to us. Don't hesitate to return for more reliable answers to any questions you may have. Thank you for visiting. Our goal is to provide the most accurate answers for all your informational needs. Come back soon. Westonci.ca is committed to providing accurate answers. Come back soon for more trustworthy information.