Question

rationalise the denominators of :-

(i) 3/2+√5

(ii) 6/√7

(iii) 9/√7-√2

Answer 1

тнєяє ιѕ уσυяѕ ѕσℓυтισи


нσρє ιт нєℓρѕ

Answer 2

hiii!!!

here's ur answer...

$(i) \: \frac{3}{2 + \sqrt{5} } = \frac{3}{2 + \sqrt{5} } \times \frac{2 - \sqrt{5} }{2 - \sqrt{5} } \\ \\ = \frac{3(2 - \sqrt{5} )}{(2 + \sqrt{5} )(2 - \sqrt{5} )} \\ \\ = \frac{3(2 - \sqrt{5}) }{ {2}^{2} - ( { \sqrt{5} })^{2} } \\ \\ = \frac{3(2 - \sqrt{5} ) }{4 - 5} \\ \\ = \frac{3(2 - \sqrt{3} )}{-1} \\ \\ = \frac{6 - 3 \sqrt{5}}{-1} \\ \\ \\ (ii) \: \frac{6}{ \sqrt{7} } = \frac{6}{ \sqrt{7} } \times \frac{ \sqrt{7} }{ \sqrt{7} } \\ \\ = \frac{6 \sqrt{7} }{7} \\ \\ \\ (iii) \: \frac{9}{ \sqrt{7} - \sqrt{2} } = \frac{9}{ \sqrt{7} - \sqrt{2} } \times \frac{ \sqrt{7} + \sqrt{2} }{ \sqrt{7} + \sqrt{2} } \\ \\ = \frac{9( \sqrt{7} + \sqrt{2)} }{( \sqrt{7 } - \sqrt{2} )( \sqrt{7} + \sqrt{2}) } \\ \\ = \frac{9( \sqrt{7} + \sqrt{2} ) }{ ({ \sqrt{7}) }^{2} - { (\sqrt{2} )}^{2} } \\ \\ = \frac{9( \sqrt{7} + \sqrt{2} )}{7 - 2} \\ \\ \frac{9( \sqrt{7} + \sqrt{2}) }{5} \\ \\ = \frac{9 \sqrt{7} + 9 \sqrt{2} }{5} \\ \\ hope \: this \: helps...$

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