$$\frac{1-\frac{1-x}{1-2x}}{1-2\frac{1-x}{1-2x}}$$
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.)
$$\frac{1-\frac{1-x}{1-2x}}{1-2\frac{1-x}{1-2x}}$$ (take L.C.M. in numerator and denominator) $$\dfrac{\dfrac{{(1-2x)}-{(1-x)}}{1-2x}}{\dfrac{(1-2x)-2{(1-x)}}{1-2x}}$$ $$\dfrac{\dfrac{1-2x-1+x}{1-2x}}{\dfrac{1-2x-2+2x}{1-2x}}$$ $$\dfrac{\dfrac{-x}{1-2x}}{\dfrac{-1}{1-2x}}$$ since $$\dfrac{\dfrac ab}{\dfrac cd}=\dfrac ab\times\dfrac {d}{c}$$ so
$${\dfrac{-x}{1-2x}}\times{\dfrac{1-2x}{-1}}$$ $$x$$
Notice that $$\frac{1-\frac{1-x}{1-2x}}{1-2\frac{1-x}{1-2x}} = \frac{\frac{-x}{1-2x}}{\frac{-1}{1-2x}}.$$
When four storey fractions are involved, it's better to do computations separately. Write $$ N=1-\frac{1-x}{1-2x},\qquad D=1-2\frac{1-x}{1-2x} $$ and work on $N$ and $D$: \begin{align} N&=1-\frac{1-x}{1-2x}\\ &=\frac{(1-2x)-(1-x)}{1-2x}\\ &=\frac{1-2x-1+x}{1-2x}\\ &=\frac{-x}{1-2x} \end{align} and \begin{align} D&=1-2\frac{1-x}{1-2x}\\ &=\frac{(1-2x)-2(1-x)}{1-2x}\\ &=\frac{1-2x-2+2x}{1-2x}\\ &=\frac{-1}{1-2x} \end{align} Now you know that your fraction is $$ \frac{N}{D}=N\cdot\frac{1}{D}=\frac{-x}{1-2x}\cdot\frac{1-2x}{-1} =\frac{-x}{-1}=x $$
$$\frac{1 - \frac{1-x}{1 - 2x}}{1 - 2\frac{1-x}{1-2x}}$$
Multiplying through by $1-2x$ gives
$$\frac{1 - 2x - (1 - x)}{1 - 2x - 2(1 - x)}.$$
Expanding the brackets and then simplifying gives
$$\frac{1 - 2x - 1 + x}{1 - 2x - 2 + 2x}$$ $$= \frac{-x}{-1} = x.$$
NOTE: This method works well because you have the same "denominator" of $1 - 2x$ every where. If you had different ones, I would recommend the finding the lowest common denominator method that is your accepted answer.
$$\frac{1-\frac{1-x}{1-2x}}{1-2\frac{1-x}{1-2x}} = \frac{\frac{-x}{1-2x}}{\frac{-1}{1-2x}} = \frac{-x}{1-2x} \frac{1-2x}{-1} = \frac{-x(1-2x)}{-1(1-2x)} = \frac{-x}{-1} = x.$$