Question

simplify 2underrootsix over underroot2 plus underroot3​

Answer 1

Answer:

$\frac{2 \sqrt{6} }{ \sqrt{2} + \sqrt{3} } \\ multiplying \: numerator \: and \: \\ denominatr \: by \: the \: conjugate \: \\ of \: \sqrt{2} + \sqrt{3} \: we \: find : \\ \frac{2 \sqrt{6} }{ \sqrt{2} + \sqrt{3} } = \frac{2 \sqrt{6} }{ \sqrt{2} + \sqrt{3} } \times \frac{ \sqrt{2} - \sqrt{3}}{ \sqrt{2} - \sqrt{3}} \\ = \frac{2 \sqrt{6} ( \sqrt{2} - \sqrt{3})}{ { (\sqrt{2} })^{2} - (\sqrt{3}) ^{2} } \\ = \frac{2 \sqrt{6} ( \sqrt{2} - \sqrt{3})}{ { (\sqrt{2} })^{2} - (\sqrt{3}) ^{2} } \\ = \frac{2( \sqrt{6 \times 2} - \sqrt{6 \times 3}) }{2 - 3} \\ = \frac{2( \sqrt{2 \times 2 \times 3) } - \sqrt{2 \times 3 \times 3}) }{2 - 3} \\ = \frac{2( 2\sqrt{3} - 3\sqrt{2}) }{ - 1} \\ = - (4 \sqrt{3} - 6 \sqrt{2} ) \\ = 6 \sqrt{2} - 4 \sqrt{3}$