Question

rationalize denominator root 6/root3- root 2

Answer 1

Hey friend !!!!!

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

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