Question

rationalize denominator root6 / root 3 -root 2

Answer 1

hence it is rationalised ....

Answer 2

Hey friend !!!!

•°• Here's your answer •°•

$que = \frac{ \sqrt{6} }{ \sqrt{3} - \sqrt{2} } \\ \\ = \frac{ \sqrt{6} }{ \sqrt{3} - \sqrt{2} } \times \frac{ \sqrt{3 } + \sqrt{2} }{ \sqrt{3} + \sqrt{2} } \\ \\ = \frac{ \sqrt{6}( \sqrt{3} + \sqrt{2} )}{ { (\sqrt{3}) }^{2} - ({ \sqrt{2} )}^{2} } \\ \\ = \frac{ \sqrt{18} + \sqrt{12} }{3 - 2} \\ \\ = \frac{ \sqrt{2 \times 3 \times 3} + \sqrt{2 \times 2 \times 3} }{1} \\ \\ = 3 \sqrt{2} + 2 \sqrt{3}$

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