Question

1/(3 - sqrt(8)) - 1/(sqrt(8) - sqrt(7)) + 1/(sqrt(7) - sqrt(6)) - 1/(sqrt(6) - sqrt(5)) + 1/(sqrt(7) - 6)

please tell me correct answer​

Answer 1

Answer:

0.46580266694

Step-by-step explanation:

correct ans correct ans correct ans correct ans

Answer 2

Concept: This question is about 'Simplification'. Here some fractions have been given, we have to solve them and find the value. In this question first, we have to rationalize the denominator.

Given: $\frac{1}{3-\sqrt{8} } -\frac{1}{\sqrt{8}-\sqrt{7} }+\frac{1}{\sqrt{7}- \sqrt{6} }-\frac{1}{\sqrt{6}-\sqrt{5} } +\frac{1}{\sqrt{7}-6 }$

To find: Solution to the question

Solution:

rationalize all terms separately

$\frac{1}{3-\sqrt{8} }$ × $\frac{3+\sqrt{8} }{3+\sqrt{8}}$ $= \frac{3\sqrt{8} }{9-8} =3+\sqrt{8}$

$\frac{1}{\sqrt{8}-\sqrt{7} }$ × $\frac{\sqrt{8}+\sqrt{7} }{\sqrt{8}+\sqrt{7}}= \frac{\sqrt{8}+ \sqrt{7} }{8-7}= \sqrt{8}+ \sqrt{7}$

$\frac{1}{\sqrt{7}-\sqrt{6} }$ × $\frac{\sqrt{7}+\sqrt{6} }{\sqrt{7}+\sqrt{6}} = \frac{\sqrt{7}+\sqrt{6}}{7-6}=\sqrt{7}+\sqrt{6}$

$\frac{1}{\sqrt{6}-\sqrt{5} }$ × $\frac{\sqrt{6}+\sqrt{5}}{\sqrt{6}+\sqrt{5}} =\frac{\sqrt{6}+\sqrt{5}}{6-5}=\sqrt{6}+\sqrt{5}$

$\frac{1}{\sqrt{7}-6 }$ × $\frac{\sqrt{7}+6 }{\sqrt{7} +6}=\frac{\sqrt{7}+6 }{7-36}= \frac{\sqrt{7}+6 }{-29}$

Now put all terms in the given question

  $3+\sqrt{8}-(\sqrt{8}+\sqrt{7})+(\sqrt{7}+\sqrt{6})-(\sqrt{6}+\sqrt{5})-\frac{\sqrt{7}+6 }{29}$

= $3+\sqrt{8}-\sqrt{8}-\sqrt{7}+\sqrt{7}+\sqrt{6}-\sqrt{6}-\sqrt{5}-\frac{\sqrt{7}+6 }{29}$

= $3-\sqrt{5}-\frac{\sqrt{7}+6 }{29}$

= $\frac{87+29\sqrt{5}-29\sqrt{7}+174 }{29}$

= $\frac{261+29\sqrt{5}-29\sqrt{7}}{29}$

= $29(\frac{9+\sqrt{5}-\sqrt{7} }{29})$

= $9+\sqrt{5}-\sqrt{7}$

Hence the right answer to this question is ($9+\sqrt{5}-\sqrt{7}$)