Question

find the value of a and b in √3+√2/√3-√2=a+b√6​

Answer 1

Answer:

thankyou

hope it will be helpful to you

Answer 2

Answer:

this uses comparing and solvin both.

hope this helps

Step-by-step explanation:

$\sqrt{3} +\frac{\sqrt2}{\sqrt3-\sqrt2}=\sqrt3+\frac{\sqrt2(\sqrt3+\sqrt2)}{(\sqrt3)^2-(\sqrt2)^2} \\= \sqrt3+\sqrt2(\sqrt2+\sqrt3) \\= \sqrt3+(\sqrt2)^2+\sqrt6= 2+\sqrt3+\sqrt6$

we have, $a+b\sqrt6$

By comparing we get,

$a=2+\sqrt3\\b=1$