Question

Prove that 1/3+root7 + 1/root7+ root5 + 1/root5+root3 + 1/root3 +1 =1

Answer 1

Answer:

Step-by-step explanation:

To prove

$\frac{1}{3+\sqrt{7} }$ + $\frac{1}{\sqrt{7}+\sqrt{5} }$ + $\frac{1}{\sqrt{5}+\sqrt{3} }$ + $\frac{1}{\sqrt{3}+1 }$ = 1

Solution:

$\frac{1}{3+\sqrt{7} }$

Rationalizing factor is $3-\sqrt{7}$

$\frac{1}{3+\sqrt{7} }$ = $\frac{1}{3+\sqrt{7} }$ × $\frac{3-\sqrt{7}}{3-\sqrt{7} }$

= $\frac{3-\sqrt{7}}{9-7 }$

= $\frac{3-\sqrt{7}}{2}$

$\frac{1}{3+\sqrt{7} }$ = $\frac{3-\sqrt{7}}{2}$

$\frac{1}{\sqrt{7}+\sqrt{5} }$

Rationalizing factor is $\sqrt{7}-\sqrt{5}$

$\frac{1}{\sqrt{7}+\sqrt{5} }$ = $\frac{1}{\sqrt{7}+\sqrt{5} }$  ×$\frac{\sqrt{7}- \sqrt{5}}{\sqrt{7}- \sqrt{5} }$

= $\frac{\sqrt{7}- \sqrt{5}}{2}$

$\frac{1}{\sqrt{7}+\sqrt{5} }$ = $\frac{\sqrt{7}- \sqrt{5}}{2}$

$\frac{1}{\sqrt{5}+\sqrt{3} }$

Rationalizing factor is $\sqrt{5}-\sqrt{3}$

$\frac{1}{\sqrt{5}+\sqrt{3} }$  = $\frac{1}{\sqrt{5}+\sqrt{3} }$   ×$\frac{\sqrt{5}- \sqrt{3}}{\sqrt{5}- \sqrt{3} }$

= $\frac{\sqrt{5}- \sqrt{3}}{2}$

$\frac{1}{\sqrt{5}+\sqrt{3} }$ = $\frac{\sqrt{5}- \sqrt{3}}{2}$

$\frac{1}{\sqrt{3}+1 }$

Rationalizing factor is $\sqrt{3}-1$

$\frac{1}{\sqrt{3}+1 }$  =$\frac{1}{\sqrt{3}+1 }$   ×$\frac{\sqrt{3}- 1}{\sqrt{3}- 1 }$

= $\frac{\sqrt{3}- 1}{2}$

$\frac{1}{\sqrt{3}+1 }$ = = $\frac{\sqrt{3}- 1}{2}$

Substituting the values of each term we get,

LHS = $\frac{1}{3+\sqrt{7} }$ + $\frac{1}{\sqrt{7}+\sqrt{5} }$ + $\frac{1}{\sqrt{5}+\sqrt{3} }$ + $\frac{1}{\sqrt{3}+1 }$  =  $\frac{3-\sqrt{7}}{2}$ +  $\frac{\sqrt{7}- \sqrt{5}}{2}$ +  $\frac{\sqrt{5}- \sqrt{3}}{2}$ + $\frac{\sqrt{3}- 1}{2}$

= $\frac{3-\sqrt{7} +\sqrt{7} -\sqrt{5}+\sqrt{5}-\sqrt{3} +\sqrt{3} -1 }{2}$

= $\frac{3-1}{2}$

= 1

=RHS

Hence proved

#SPJ2

Answer 2

Answer:

$\frac{1}{3+\sqrt{7} } +\frac{1}{\sqrt{7} +\sqrt{5} }+ \frac{1}{\sqrt{5} +\sqrt{3} }+\frac{1}{\sqrt{3} +1 }=1$ is proved.

Step-by-step explanation:

Step 1 of 2

To prove:- $\frac{1}{3+\sqrt{7} } +\frac{1}{\sqrt{7} +\sqrt{5} }+ \frac{1}{\sqrt{5} +\sqrt{3} }+\frac{1}{\sqrt{3} +1 }=1$

Consider the left-hand side as follows:

$\frac{1}{3+\sqrt{7} } +\frac{1}{\sqrt{7} +\sqrt{5} }+ \frac{1}{\sqrt{5} +\sqrt{3} }+\frac{1}{\sqrt{3} +1 }$ . . . . . (1)

Rationalize the term $\frac{1}{3+\sqrt{7} }$ as follows:

⇒ $\frac{1}{3+\sqrt{7} }\times\frac{3-\sqrt{7} }{3-\sqrt{7}}$

⇒ $\frac{3-\sqrt{7} }{(3+\sqrt{7} )(3-\sqrt{7} )}$

⇒ $\frac{3-\sqrt{7} }{3^2-(\sqrt{7})^2}$ (Using the identity, $(a+b)(a-b)=a^2-b^2$)

⇒ $\frac{3-\sqrt{7} }{9-7}$

⇒ $\frac{3}{2}-\frac{\sqrt{7} }{2}$

Similarly,

Rationalize the term $\frac{1}{\sqrt{7} +\sqrt{5} }$ as follows:

⇒ $\frac{1}{\sqrt{7}+\sqrt{5} }\times\frac{\sqrt{7}-\sqrt{5} }{\sqrt{7}-\sqrt{5} }$

⇒ $\frac{\sqrt{7}-\sqrt{5} }{(\sqrt{7})^2-(\sqrt{5} )^2}$ (Using the identity, $(a+b)(a-b)=a^2-b^2$)

⇒ $\frac{\sqrt{7}-\sqrt{5} }{7-5}$

⇒ $\frac{\sqrt{7} }{2}-\frac{\sqrt{5} }{2}$

Rationalize the term $\frac{1}{\sqrt{5} +\sqrt{3} }$ as follows:

⇒ $\frac{1}{\sqrt{5}+\sqrt{3} }\times\frac{\sqrt{5}-\sqrt{3} }{\sqrt{5}-\sqrt{3} }$

⇒ $\frac{\sqrt{5}-\sqrt{3} }{(\sqrt{5})^2-(\sqrt{3} )^2}$ (Using the identity, $(a+b)(a-b)=a^2-b^2$)

⇒ $\frac{\sqrt{5}-\sqrt{3} }{5-3}$

⇒ $\frac{\sqrt{5} }{2}-\frac{\sqrt{3} }{2}$

Rationalize the term $\frac{1}{\sqrt{3}+1 }$ as follows:

⇒ $\frac{1}{\sqrt{3} +1 }\times\frac{\sqrt{3} -1 }{\sqrt{3}-1 }$

⇒ $\frac{\sqrt{3}-1 }{((\sqrt{3} )^2-1^2}$ (Using the identity, $(a+b)(a-b)=a^2-b^2$)

⇒ $\frac{\sqrt{3} -1 }{3-1}$

⇒ $\frac{\sqrt{3} }{2}-\frac{1}{2}$

Step 2 of 2

Substitute all the rationalizing values in the equation (1) as follows:

⇒ $\frac{3}{2}-\frac{\sqrt{7} }{2} +\frac{\sqrt{7} }{2}-\frac{\sqrt{5} }{2} +\frac{\sqrt{5} }{2}-\frac{\sqrt{3} }{2} +\frac{\sqrt{3} }{2}-\frac{1 }{2}$

On simplifying, we get

⇒ $\frac{3}{2}-\frac{1}{2}$

⇒ $\frac{2}{2}$

⇒ 1

Thus, LHS = 1

Also, the RHS = 1

Therefore, LHS = RHS

Hence proved.

#SPJ2