Question

Find value of $a$ and $b$ in $\frac{\sqrt{2}+\sqrt{3}}{3\sqrt{2}+2\sqrt{3}}=a+b\sqrt{6}$.

Answer 1

$\frac{\sqrt{2}+\sqrt{3}}{3\sqrt{2}+2\sqrt{3}}=\frac{(\sqrt{2}+\sqrt{3})(3\sqrt{2}-2\sqrt{3})}{(3\sqrt{2}+2\sqrt{3})(3\sqrt{2}-2\sqrt{3})}=\frac{6-2\sqrt{6}+3\sqrt{6}-6}{18-12}=\frac{\sqrt{6}}{6}$

comparing with $a+b\sqrt{6}$ we get

$a=0$ and $b=\frac{1}{6}$