Question

√3+√2/√3-√2=a+b√6 simply ​

Answer 1

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

Rationalising the denominator

$= > \frac{11 + 6 \sqrt{2} }{7} = \frac{11}{7} + \frac{6}{7} \sqrt{2}$

NOW,

$= > \frac{11}{7} + \frac{6}{7} \sqrt{2} = a + b \sqrt{2}$

Comparing, we get

$= > a = \frac{1}{7} \: \: , \: b = \frac{6}{7}$

HOPE THIS HEIPS U......