Question

If $x=\frac{4ab}{a+b}$, let us show that $\frac{x+2a}{x-2a}+\frac{x+2b}{x-2b}=2$ [given $a\ne 0,b\ne 0$ and $a\ne b$].

Answer 1

Given,

$x=\frac{4ab}{a+b}$

or, $\frac{x}{2a}=\frac{2b}{a+b}$

or, $\frac{x+2a}{x-2a}=\frac{a+3b}{b-a}$ [ By dividendo and componendo ].......(1).

Again,

$x=\frac{4ab}{a+b}$

or, $\frac{x}{2b}=\frac{2a}{a+b}$

or, $\frac{x+2b}{x-2b}=\frac{3a+b}{a-b}$ [ By dividendo and componendo ].......(2).

Now adding (1) and (2) we get,

$\frac{x+2a}{x-2a}+\frac{x+2b}{x-2b}=\frac{a+3b}{b-a}+\frac{3a+b}{a-b}$

or, $\frac{x+2a}{x-2a}+\frac{x+2b}{x-2b}=\frac{2(b-a)}{b-a}=2$.