Question

$\frac{1}{ \sqrt{2} + \sqrt{3} + \sqrt{5} }$
Rationalise the denominator.

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Answer 1

Here is your answer! Hope it helps! :)

I'll give uh full ans personally! due to some connection problem I'm unable to attach full answer! Hope uh understand! :)

Answer 2

${\huge{\boxed{\bf{ANSWER}}}}$

☆ To Rationalize the denominator of -

$\longrightarrow \frac{1}{ \sqrt{2} + \sqrt{3} + \sqrt{5} }$

☆ Solution -

Consider (√2 + √3 ) as a single term ..

$\implies \: \frac{1}{( \sqrt{2} + \sqrt{3}) + \sqrt{5} }$

Now , Multiply and divide by its conjugate..

$\implies \: \frac{1}{( \sqrt{2} + \sqrt{3} ) + \sqrt{5} } \times \frac{( \sqrt{2} + \sqrt{3}) - \sqrt{5} }{ (\sqrt{2} + \sqrt{3}) - \sqrt{5} }$

$\implies \: \frac{ \sqrt{2} + \sqrt{3} - \sqrt{5} }{ {( \sqrt{2} + \sqrt{3} )}^{2} - {( \sqrt{5} )}^{2} }$

$\implies \: \frac{ \sqrt{2} + \sqrt{3} - \sqrt{5} }{2 + 3 + 2 \sqrt{6} - 5}$

$\implies \frac{ \sqrt{2} + \sqrt{3} - \sqrt{5} }{2 \sqrt{6} }$

Still , it's not rationalized as , we have square root term in denominator...

so, multiply & divide by √6

$\implies \: \frac{ (\sqrt{2} + \sqrt{3} - \sqrt{5} ) \times \sqrt{6} }{2 \sqrt{6} \times \sqrt{6} }$

$\implies \: \frac{ \sqrt{12} + \sqrt{18} - \sqrt{30} }{12}$

$\implies \: \frac{2 \sqrt{3} + 3 \sqrt{2} - \sqrt{30} }{12}$