Question

rationalise the denominator of


1/√2+√3+√5

Answer 1

Answer :


$\frac{1}{ \sqrt{2} + \sqrt{3} + \sqrt{5} } \\ \\ \frac{1}{ \sqrt{2} + \sqrt{3} + \sqrt{5} } \times \frac{( \sqrt{2 } + \sqrt{3} - \sqrt{5}) }{( \sqrt{2} + \sqrt{3} - \sqrt{5} )} \\ \\ \frac{ \sqrt{2} + \sqrt{3} - \sqrt{5} }{( \sqrt{2} + \sqrt{3} + \sqrt{5} )( \sqrt{2} + \sqrt{3} - \sqrt{5} )} \\ \\ using \: (a - b)(a + b) = a {}^{2} - b {}^{2}$
$\frac{ \sqrt{2} + \sqrt{3} - \sqrt{5} }{ (\sqrt{2} + \sqrt{3} ) {}^{2} - 5} \\ \\ using \: \: (a + b)(a - b) = {a}^{2} - b {}^{2} \\ \\ \frac{ \sqrt{2} + \sqrt{3} - \sqrt{5} }{2 + 2 \sqrt{6} + 3- 5} \\ \\ \frac{ \sqrt{2} + \sqrt{3} - \sqrt{5} }{0 + 2 \sqrt{6} } \\ \\ \frac{ \sqrt{2} + \sqrt{3} - \sqrt{5} }{2 \sqrt{6} } \\ \\ rationalize \: the \: denominator \\ \\ \frac{ \sqrt{2} + \sqrt{3} - \sqrt{5} }{2 \sqrt{6} } \times \frac{ \sqrt{6} }{ \sqrt{6} } \\ \\ \frac{( \sqrt{2} + \sqrt{3} - \sqrt{5} )( \sqrt{6}) }{2 \sqrt{6} \sqrt{6} } \\ \\ \frac{( \sqrt{2} + \sqrt{3} - \sqrt{5} )( \sqrt{6}) }{2( { \sqrt{6} )}^{2} } \\ \\ \frac{ \sqrt{12} + \sqrt{18} - \sqrt{30} }{2 \times 6} \\ \\ \frac{ 2\sqrt{3} + \sqrt{18} - \sqrt{30} }{12}$



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