2017/026) if $x=\frac{4ab}{a+b}$ find the value of $\frac{x+2a}{x-2a}+ \frac{x+2b}{x-2b}$

we have
$x=\dfrac{4ab}{a+b}$
hence $\dfrac{x}{2a} = \dfrac{2b}{a+b}$
using componendo dividendo we get
 $\dfrac{x+ 2a}{x-  2a} = \dfrac{2b + (a +b)  }{2b - (a+b) } =  \dfrac{3b + a }{b- a}\cdots(1)$
similarly   $\dfrac{x+ 2b}{x-  2b} = \dfrac{2a + (a +b)  }{2a - (a+b) } =  \dfrac{3a + b}{a-b}\cdots(2)$
Adding (1) and (2) we get
 $\dfrac{x+2a}{x-2a}+ \dfrac{x+2b}{x-2b} =  \dfrac{3b + a}{b-  a}  +  \dfrac{3a + b}{a- b} $
$= \dfrac{3b + a}{b-  a}  -  \dfrac{3a + b}{b-a}$
$= \dfrac{3b + a-3a -b}{b-  a} =  \dfrac{2b-2a}{b-a} = \dfrac{2(b-a)}{b-a} = 2$