b: \(\frac{\sqrt2+\sqrt3+2}{\sqrt2+\sqrt3+\sqrt6+\sqrt8+4}\)
\(=\frac{\sqrt2+\sqrt3+\sqrt4}{\sqrt2+\sqrt3+\sqrt4+\sqrt4+\sqrt6+\sqrt8}\)
\(=\frac{\sqrt2+\sqrt3+\sqrt4}{\left(\sqrt2+\sqrt3+\sqrt4\right)\left(1+\sqrt2\right)}\)
\(=\frac{1}{\sqrt2+1}=\sqrt2-1\)

