Solution:

Here,

\(\frac{\sqrt3+\sqrt2}{\sqrt3-\sqrt2}\)

Multiplying both numerator and denominator by \(\sqrt3+\sqrt2\) , we get

\(\frac{\sqrt3+\sqrt2}{\sqrt3-\sqrt2}\times\frac{\sqrt3+\sqrt2}{\sqrt3+\sqrt2}\)

= \(\frac{{(\sqrt3+\sqrt2)}^2}{{(\sqrt3)}^2-{(\sqrt2)}^2}\)

= \(\frac{{(\sqrt3)}^2+2\times\sqrt3\times\sqrt2+{(\sqrt2)}^2}{3-2}\)

= \(\frac{3+2\times\sqrt6+2}{3-2}\)

= \(\frac{5+2\times\sqrt6}{1}\)

= \(5+2\sqrt6\)

Is the required answer