Question

Rationalise the denominator of 1/√7+√6-√13​

Answer 1

Answer:

I'm weak in maths I don't help you

Answer 2

Answer:

$\dfrac{7 \sqrt{6} + 6 \sqrt{7} + \sqrt{546} }{84}$

Step-by-step explanation:

By Rationalizing the denominator , I get the answer .

$\dfrac{1}{ \sqrt{7} + \sqrt{6} - \sqrt{13} } \\ \\ = \frac{1 \times ( \sqrt{7 } + \sqrt{6} + \sqrt{13)} }{( \sqrt{7} + \sqrt{6} - \sqrt{13})( \sqrt{7} + \sqrt{6} - \sqrt{13} )} \\ \\ = \frac{ \sqrt{7} + \sqrt{6} + \sqrt{13} }{ {( \sqrt{7} + \sqrt{6})^{2} - ( \sqrt{13}) }^{2} } \\ \\ = \frac{ \sqrt{7 } + \sqrt{6} + \sqrt{13} }{ { (\sqrt{7} )}^{2} + {( \sqrt{6}) }^{2} + 2 \sqrt{6}. \sqrt{7} - 13} \\ \\ = \frac{ \sqrt{7} + \sqrt{6} + \sqrt{13} }{7 + 6 + 2 \sqrt{42} - 13 } \\ \\ = \frac{ \sqrt{7} + \sqrt{6} + \sqrt{13} }{13 - 13 + 2 \sqrt{42} } \\ \\ = \frac{(\sqrt{7} + \sqrt{6} + \sqrt{13} ) \sqrt{42} }{2 \sqrt{42} \times \sqrt{ 42}} \\ \\ =\frac{ \sqrt{7 \times 42} + \sqrt{6 \times 42} + \sqrt{13 \times 42} }{2 \times 42} \\ \\ = \frac{7 \sqrt{6} + 6 \sqrt{7} + \sqrt{546} }{84}$