How to simplify $\frac{1-\frac{1-x}{1-2x}}{1-2\frac{1-x}{1-2x}}$? $$\frac{1-\frac{1-x}{1-2x}}{1-2\frac{1-x}{1-2x}}$$
I have been staring at it for ages and know that it simplifies to $x$, but have been unable to make any significant progress.
I have tried doing $(\frac{1-x}{1-2x})(\frac{1+2x}{1+2x})$ but that doesn't help as it leaves $\frac{1+x-2x^2}{1-4x^2}$ any ideas? 

Hint:$f^{-1}(f(x))=x$ where $f(x)=\frac{1-x}{1-2x}$,$(x\ne 1/2)$
Solution:Here $f^{-1}(x)=\frac{1-x}{1-2x}$.We know that $f^{-1}(f(x))=x$ and as  $\frac{1-\frac{1-x}{1-2x}}{1-2\frac{1-x}{1-2x}}=f^{-1}(f(x))$ so it follows that $\frac{1-\frac{1-x}{1-2x}}{1-2\frac{1-x}{1-2x}}=f^{-1}(f(x))=x$ So we dont need to simplify.
(Here $f'(x)=\frac{1}{(1-2x)^2}>0$ so $f^{-1} $ exists.)
