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Answer 1

Answer:

$\frac{5 \sqrt{3} + \sqrt{2} }{ \sqrt{3} + \sqrt{2} } \\ multiplying \: nr \: and \: dr \: by \: the \: \\ conjugate \: of \: dr \: that \: is \: by \: \\ \sqrt{3} - \sqrt{2} \: we \: find: \\ \frac{5 \sqrt{3} + \sqrt{2} }{ \sqrt{3} + \sqrt{2} } \times \frac{\sqrt{3} - \sqrt{2} }{\sqrt{3} - \sqrt{2} } \\ = \frac{(5 \sqrt{3} + \sqrt{2} ) \times (\sqrt{3} - \sqrt{2}) }{( \sqrt{3} + \sqrt{2} )\times (\sqrt{3} - \sqrt{2})} \\ = \frac{5 \sqrt{3}( \sqrt{3} - \sqrt{2} ) + \sqrt{2} ( \sqrt{3} - \sqrt{2} )}{( { \sqrt{3} })^{2} - {( \sqrt{2} })^{2} } \\ = \frac{5 \times 3 - 5 \sqrt{6} + \sqrt{6} - 2 }{3 - 2} \\ = \frac{15 - 4 \sqrt{6} - 2}{1} \\ = 13 - 4 \sqrt{6}$