Question

find a and b if 3+√6/√3+√2= a+b√3​

Answer 1

$= > \frac{3 + \sqrt{6} }{ \sqrt{3} + \sqrt{2} } = a + b \sqrt{3}$

$= > \frac{3 + \sqrt{6} }{ \sqrt{3} + \sqrt{2} } \times \frac{ \sqrt{3} - \sqrt{2} }{ \sqrt{3} - \sqrt{2} }$

$= > \frac{3 \times \sqrt{3} - \sqrt{2} + \sqrt{6} \times \sqrt{3} - \sqrt{2} }{ { \sqrt{3} }^{2} - { \sqrt{2} }^{2} }$

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

$= > \frac{ \sqrt{3} }{1}$

$= > \sqrt{3}$

Therefore a=0 , b=1

Answer 2

Answer:

there for a =0,b=1.........

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