If $$x = \frac{{4ab}}{{a + b}}{\text{ }}a \ne b,$$     the value of $$\frac{{x + 2a}}{{x - 2a}}$$   + $$\frac{{x + 2b}}{{x - 2b}}$$   is?

$$\eqalign{ & {\text{ }}x = \frac{{4ab}}{{a + b}}{\text{ }} \cr & \Rightarrow \frac{x}{{2a}} = {\text{ }}\frac{{2b}}{{a + b}}{\text{ }} \cr & \frac{{x + 2a}}{{x - 2a}} \cr & = \frac{{2b + a + b}}{{2b - a - b}} \cr & = \frac{{3b + a}}{{b - a}} \cr} $$
( By Componendo and Dividendo rule )
$$\eqalign{ & \Rightarrow {\text{again}}\frac{x}{{2b}} = \frac{{2a}}{{a + b}} \cr & \frac{{x + 2b}}{{x - 2b}} \cr & = \frac{{2a + a + b}}{{2a - a - b}} \cr & = \frac{{3a + b}}{{a - b}} \cr & \Rightarrow \frac{{x + 2a}}{{x - 2a}}{\text{ + }}\frac{{x + 2b}}{{x - 2b}} \cr & \Rightarrow \frac{{3b + a}}{{b - a}} - \frac{{3a + b}}{{a - b}} \cr & \Rightarrow \frac{{3b + a - 3a - b}}{{b - a}} \cr & \Rightarrow \frac{{2b - 2a}}{{b - a}} \cr & \Rightarrow \frac{{2\left( {b - a} \right)}}{{\left( {b - a} \right)}} \cr & \Rightarrow 2 \cr} $$