Question

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Solve Question No. 87

Answer 1

Hey mate !!!


look this attachment :-



hence the correct answer is Option number B


hope it helps Dear!!

thanks

Answer 2

Hey friend, Harish here.

Here is your answer:

To find,

$\frac{1}{1+ \sqrt{2} } + \frac{1}{ \sqrt{2} + \sqrt{3} } ........ + \frac{1}{ \sqrt{8}+ \sqrt{9} }$

Solution,

First we must rationalize the denominators of all terms.

To rationalize them, we must multiply and divide them by their conjugate.

Conjugates of denominators are:

For 1st term it is - (1 - √2)

For 2nd term it is -  (√2 - √3)..

So on for the last term.

Now let us rationalize them,

Then,

$\frac{1}{ 1+ \sqrt{2}} = \frac{1}{1+ \sqrt{2} } \times \frac{1- \sqrt{2} }{1- \sqrt{2} }= \frac{1- \sqrt{2} }{ (1+ \sqrt{2})( 1- \sqrt{2})} = \frac{ 1- \sqrt{2}}{1-2} = -(1- \sqrt{2})$


Here in the denominator to  multiply we use (a+b)(a-b) = a² - b² identity.

⇒ ( 1 + √2) ( 1 - √2) = 1² - (√2)² = 1 - 2 = -1

The same thing happens for all the terms, 

Then,

We get,

$-(1- \sqrt{2})-( \sqrt{2}- \sqrt{3})-( \sqrt{3}- \sqrt{4}) ........... -( \sqrt{8}- \sqrt{9})$

⇒ $-1 + \sqrt{2}- \sqrt{2}+ \sqrt{3}- \sqrt{3}+ \sqrt{4}- \sqrt{4}...... - \sqrt{8}+ \sqrt{9}$

Here all terms except -1 & √9 get canceled..

Then,

$-1 + \sqrt{9} = -1 + 3 = 2$

Therefore the answer is 2. (OPTION-B)
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Hope my answer is helpful to you.