Question

rationalize the following:1/√2+√5​

Answer 1

${\huge{\gray{\underline{\mathbb{\red{Ans}{\orange{wer}{\blue{:-} }}}}}}}$

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

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

using the identity

x²-y²= (x-y) (x+y)

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

hope it will help you

Answer 2

Answer:

$\frac{1}{ \sqrt{2} + \sqrt{5} } \\ \\ \frac{1}{ \sqrt{2} + \sqrt{5} } \times \frac{ \sqrt{2 } - \sqrt{5} }{ \sqrt{2} - \sqrt{5} } \\ \\ \frac{ \sqrt{2} - \sqrt{5} }{( \sqrt{2} - \sqrt{5} )( \sqrt{2} + \sqrt{5} } \\ \\ using \: identity \\ \\ {x}^{2} - {y}^{2} = (x - y)(x + y) \\ \\ \frac{ \sqrt{2} - \sqrt{5} }{2 - 5} \\ \\ \frac{ \sqrt{2} - \sqrt{5} }{ - 3}$