Question

Rationalize the denominator:

√2+√3

√2−√3

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Answer 1

Step-by-step explanation:

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Answer 2

$\dfrac{ \sqrt{2} + \sqrt{3} }{ \sqrt{2} - \sqrt{3} } \\ \\ \to \tt \dfrac{ \sqrt{2} + \sqrt{3} }{ \sqrt{2} - \sqrt{3} } \times \dfrac{ \sqrt{2} + \sqrt{3} }{ \sqrt{2} + \sqrt{3} } \\ \\ \to \tt \frac{ {( \sqrt{2} + \sqrt{3} )}^{2} }{ {( \sqrt{2} )}^{2} - {( \sqrt{3} )}^{2} } \\ \\ \to \tt \frac{ {( \sqrt{2} )}^{2} + {( \sqrt{3} )}^{2} + 2( \sqrt{2} )( \sqrt{3} ) }{2 - 3} \\ \\ \to \tt \frac{2 + 3 + 2 \sqrt{6} }{ - 1} \\ \\ \to \tt \frac{5 + 2 \sqrt{6} }{ - 1} \\ \\ \to \tt \frac{5 + 2 \sqrt{6} }{ - 1} \times \frac{ - 1}{ - 1} \\ \\ \to \tt \frac{ - (5 + 2 \sqrt{6} )}{ {( - 1)}^{2} } \\ \\ \to \tt \frac{ - 5 - 2 \sqrt{6} }{1} \: \: = \red{ - (5 + 2 \sqrt{6} )}$