Question

Answer this!

○ Solve this ○
$\frac {1}{\sqrt7-\sqrt6}$
​ ​

Answer 1

Question :

Solve :

$\frac{1}{ \sqrt{7} - \sqrt{6} }$

Solution :

$\frac{1}{ \sqrt{7} - \sqrt{6} }$

Now rationalise the denominator

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

$= \frac{ \sqrt{7} + \sqrt{6} }{( \sqrt{7 } - \sqrt{6} )( \sqrt{7} + \sqrt{6} )}$

we know that $a{}^{2}-b{}^{2}=(a+b)(a-b)$

$= \frac{ \sqrt{7} + \sqrt{6} }{( \sqrt{7}) {}^{2} - ( \sqrt{6}) {}^{2} }$

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

$= \sqrt{7} + \sqrt{6}$

it is the required solution!

Answer 2

${\huge {\underline {\underline {\mathfrak {\green {Answer }}}}}}$

_________________________

${\bigstar {\underline {\bf {\pink {Given\:in\:the\:question:-}}}}}$

● $\frac {1}{\sqrt7-\sqrt6}$

_____________________________________

${\bigstar {\underline {\bf {\pink {Solution\:of\:the\:question:-}}}}}$

= $\displaystyle\frac {1}{\sqrt7-\sqrt6}$

= $\displaystyle\frac {1}{\sqrt7-\sqrt6}× \displaystyle\frac {\sqrt7+\sqrt6}{\sqrt7+\sqrt6}$

=$\displaystyle\frac {\sqrt7+\sqrt6}{({\sqrt7})^{2}+({\sqrt6})^{2}}$

= $\displaystyle\frac {\sqrt7+\sqrt6}{7-6}$

= $\sqrt {7}+\sqrt{6}$

_________________________