Turning $\frac{2x-y}{2x+y}\frac1{\left(1-\frac{y}{2x}\right)^2}$ into $\frac1{\frac{y}{x}(1-\frac{y}{2x})+(1-\frac{y}{2x})^2}$

Question

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 ?

Answer 1

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

Answer 2

Hint:

Let $t= {y\over 2x}$ and express $y= t\cdot 2x$. Now put this $y$ in to your equation...

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Now $$LHS = {2x-2xt\over 2x+2xt}\cdot {1\over (1-t)^2} = {1\over 1-t^2} $$

and $$RHS = {1\over 2t(1-t)+(1-t)^2} = {1\over (1-t)(2t+1-t)}=...$$