Question

$\frac{ \sqrt{3} - \sqrt{2} }{ \sqrt{3} + \sqrt{2} }$
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Answer 1

Answer:

$\frac{ \sqrt{3} - \sqrt{2} }{ \sqrt{3} + \sqrt{2} } \\ \\ by \: rationalising \: \\ \\ \frac{ \sqrt{3} - \sqrt{2} }{ \sqrt{3} + \sqrt{2} } \times \frac{ \sqrt{3} - \sqrt{2} }{ \sqrt{3} - \sqrt{2} } \\ \\ \frac{({\sqrt{3} - \sqrt{2})}^{2} }{3 - 2} \\ \\ { (\sqrt{3} - \sqrt{2}) }^{2}$

Answer 2

$\frac{ \sqrt{3} - \sqrt{2} }{ \sqrt{3} + \sqrt{2} } \\ = \frac{( \sqrt{3} - \sqrt{2} )( \sqrt{3} - \sqrt{2} )}{( \sqrt{3} + \sqrt{2} )( \sqrt{3} - \sqrt{2} )} \\ = \frac{ { (\sqrt{3}) }^{2} + {(\sqrt{2}) }^{2} - 2( \sqrt{3})( \sqrt{2}) }{ {( \sqrt{ 3}) }^{2} - {( \sqrt{2} )}^{2} } \\ = \frac{3 + 2 - 2 \sqrt{6} }{3 - 2} \\ = \frac{5 - 2 \sqrt{6} }{1} \\ = 5 - 2 \sqrt{6}$