Question

If $x=\frac{2ab}{a+b}$, find the value of: $\frac{x+a}{x-a}+\frac{x+b}{x-b}$

Answer 1

$x=\frac{2ab}{a+b}$

$\Rightarrow 1x=\frac{a+b}{2ab}=\frac{1}{2}[\frac{1}{a}+\frac{1}{b}]$

$\Rightarrow \frac{x+a}{x-a}+\frac{x+b}{x-b}$

$\Rightarrow \frac{1+\frac{a}{x}}{1-\frac{a}{x}}+\frac{1+\frac{b}{x}}{1-\frac{b}{x}}$

$\Rightarrow \frac{1+\frac{a+b}{2b}}{1-\frac{a+b}{2b}}+\frac{1+\frac{a+b}{2a}}{1-\frac{a+b}{2a}}$

$\Rightarrow \frac{a+3b}{b-a}+\frac{3a+b}{a-b}$

$\Rightarrow \frac{a+3b}{b-a}-\frac{3a+b}{b-a}$

$\Rightarrow \frac{a+3b-3a-b}{b-a}$

$\Rightarrow \frac{2(b-a)}{b-a}$

$\Rightarrow 2.$

Hence, the answer is $2.$