Question

√2/√2+√3-√5 rationalize the denominator​

Answer 1

Answer:

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

rationalize the denominator

$\frac{ \sqrt{2} }{ \sqrt{2} + \sqrt{3} - \sqrt{5} } \times \frac{ \sqrt{2} + \sqrt{3} + \sqrt{5} }{ \sqrt{2} + \sqrt{3} + \sqrt{5} } </p><p>\\ = \frac{ \sqrt{2} ( \sqrt{2} + \sqrt{3} + \sqrt{5}) }{ {( \sqrt{2} + \sqrt{3} )}^{2} - 5} \\ = \frac{2 + \sqrt{6} + \sqrt{10} }{2 + 2 \sqrt{6} + 3 - 5} \\ = \frac{ \sqrt{2}( \sqrt{2} + \sqrt{3} + \sqrt{5} )}{2 \sqrt{6} } \\ = \frac{ \sqrt{2} + \sqrt{3} + \sqrt{5} }{2 \sqrt{3} } \\ = \frac{ \sqrt{2} + \sqrt{3} + \sqrt{5} }{2 \sqrt{3} } \times \frac{ \sqrt{3} }{ \sqrt{3} } \\ = \frac{ \sqrt{3}( \sqrt{2} + \sqrt{3} + \sqrt{5} )}{2 \times 3} \\ = \frac{ \sqrt{3} ( \sqrt{2} + \sqrt{3} + \sqrt{5} )}{2 \sqrt{3} \sqrt{3} } \\ = \frac{ \sqrt{6} + 3+ \sqrt{15} }{2 \times 3 } \\ = \frac{ \sqrt{6} + 3 + \sqrt{15} }{6} \\ hence \: rationalised$