Find the values of 'a' and 'b':

$\dfrac{\sqrt{7} - 2}{\sqrt{7} + 2} = a\sqrt{7} + b$

Given,

Equation : $\dfrac{\sqrt{7} - 2}{\sqrt{7} + 2} = a\sqrt{7} + b$

Rationalizing L.H.S. of the above equation :

$\Rightarrow \dfrac{\sqrt{7} - 2}{\sqrt{7} + 2} \times \dfrac{\sqrt{7} - 2}{\sqrt{7} - 2} \\[1em] \Rightarrow \dfrac{(\sqrt{7} - 2)^2}{(\sqrt{7})^2 - 2^2} \\[1em] \Rightarrow \dfrac{(\sqrt{7})^2 + 2^2 - 2 \times \sqrt{7} \times 2}{7 - 4} \\[1em] \Rightarrow \dfrac{7 + 4 - 4\sqrt{7}}{3} \\[1em] \Rightarrow \dfrac{11 - 4\sqrt{7}}{3} \\[1em] \Rightarrow -\dfrac{4\sqrt{7}}{3} + \dfrac{11}{3}.$

Comparing $-\dfrac{4\sqrt{7}}{3} + \dfrac{11}{3} \text{ with } a\sqrt{7} + b$, we get :

a = $-\dfrac{4}{3}\text{ and } b = \dfrac{11}{3}$.

Hence, a = $-\dfrac{4}{3}\text{ and } b = \dfrac{11}{3}$.