Turning $\frac{2x-y}{2x+y}\frac1{\left(1-\frac{y}{2x}\right)^2}$ into $\frac1{\frac{y}{x}(1-\frac{y}{2x})+(1-\frac{y}{2x})^2}$ I stopped at this equality in the solution of a question in my book
$$\left(\frac{2x-y}{2x+y}\right)\left(\frac{1}{\left(1-\frac{y}{2x}\right)^2}\right) = \frac{1}{\big( \frac{y}{x}\big) \Big(1-\frac{y}{2x} \Big) +\Big( 1-\frac{y}{2x}\Big)^2}$$
I already verified that starting in R.H.S will give the L.H.S but I am wondering how can we simplify the "L.H.S" to arrive to the R.H.S ?

Divide top and bottom by $2x-y$. The numerator becomes 1 and the denominator becomes 
$$\frac{2x+y}{2x-y}\left(1-\frac{y}{2x}\right)^2$$
$$=\frac{1-\frac{y}{2x}+\frac{y}{x}}{1-\frac{y}{2x}}\left(1-\frac{y}{2x}\right)^2$$
$$=\frac{y}{x}\left(1-\frac{y}{2x}\right)+\left(1-\frac{y}{2x}\right)^2$$
