Rationalise the denominator and simplify :

$\dfrac{3}{4 - \sqrt3}$

Since, the denominator = $4 - \sqrt3$, its rationalizing factor = $4 + \sqrt3$.

$\dfrac{3}{4 - \sqrt3} = \dfrac{3}{4 - \sqrt3} \times \dfrac{4 + \sqrt3}{4 + \sqrt3}\\[1em] = \dfrac{3(4 + \sqrt3)}{4^2 - (\sqrt3)^2} \\[1em] = \dfrac{3(4 + \sqrt3)}{16 - 3} \\[1em] = \dfrac{3(4 + \sqrt3)}{13}$

Hence, $\dfrac{3}{4 - \sqrt3} = \dfrac{3(4 + \sqrt3)}{13}$.