Question

rationalize the denominator

√2-√3/√2+√3​

Answer 1

Answer:

hope this helps you dear

Answer 2

Hii sanju

$= \frac{ \sqrt{2} - \sqrt{3} }{ \sqrt{2} + \sqrt{3} }$

rationalize the denominator

$= \frac{ \sqrt{2} - \sqrt{3} }{ \sqrt{2} + \sqrt{3} } \times \frac{ \sqrt{2} - \sqrt{3} }{ \sqrt{2} - \sqrt{3} }$

$= \frac{ {( \sqrt{2} - \sqrt{3} })^{2} }{2 - 3}$

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

$= \frac{2 - 2 \sqrt{6} + 6 }{ - 1}$

$= \frac{8 - 2 \sqrt{6} }{ - 1}$

$= 2 \sqrt{6} - 8$