Solve $\frac{|x|}{|x-1|}+|x|=\frac{x^2}{|x-1|}$ Solve $\frac{|x|}{|x-1|}+|x|=\frac{x^2}{|x-1|}$.What will be the easiest techique to solve this sum ?
Just wanted to share a special type of equation and the fastest way to solve it.I am not asking for an answer and i have solved it in my answer given below.Thank You for viewing.

To me the easiest and most systematic way to solve it is to explicitly write out the absolute value which means breaking the equation into regions:
$$
\frac{|x|}{|x - 1|} + |x| = \frac{x^2}{|x - 1|} \rightarrow \begin{cases}
\left(\frac{1}{x - 1} + 1\right)x = \frac{x^2}{x - 1} & 1 < x < \infty \\ 
\left(\frac{1}{1 - x} + 1\right)x = \frac{x^2}{1 - x} & 0 \leq x < 1 \\
-\left(\frac{1}{1 - x} + 1\right)x = \frac{x^2}{1 - x} & -\infty < x < 0
\end{cases}
$$
$\frac{1}{x - 1} + 1 = \frac{1 + x - 1}{x - 1} = \frac{x}{x - 1}$ and $\frac{1}{1 - x} + 1 = \frac{1 + 1 - x}{1 - x} = \frac{2 - x}{1 - x}$ which gives:
$$
\frac{|x|}{|x - 1|} + |x| = \frac{x^2}{|x - 1|} \rightarrow \begin{cases}
\frac{x^2 - x^2}{x - 1} = 0 & 1 < x < \infty \\ 
\frac{2x - x^2 - x^2}{1 - x} = 0 & 0 \leq x < 1 \\
\frac{-2x + x^2 - x^2}{1 - x} = 0 & -\infty < x < 0
\end{cases}
$$
Which gives:
$$
\frac{|x|}{|x - 1|} + |x| = \frac{x^2}{|x - 1|} \rightarrow \begin{cases}
0 = 0 & 1 < x < \infty \\ 
2x\frac{1 - x}{1 - x} = 0 & 0 \leq x < 1 \\
-\frac{2x}{1 - x} = 0 & -\infty < x < 0
\end{cases}
$$
Which means that this equation is true for all (real) values $x > 1$ and $x = 0$--there are no negative values of $x$ which satisfy this equation since the third case only has a solution at $x = 0$.
