<s>[INST] The least integer value of $x$, which satisfy $|x| + |\frac{x}{x - 1}| = \frac{x^2}{|x - 1|}$, is... 
Question:  The least integer value of $x$, which satisfy $|x| + |\frac{x}{x - 1}| = \frac{x^2}{|x - 1|}$, is...


MY ATTEMPT:
Case 1: Taking all the modulus positive and solving ...
Case 2: Taking all the modulus negative and solving...
Case 3: Taking different signs of different modulus at a time and solving...
But the Case 3 will arise to many different cases and this will take a lot of time to solve. So, definitely an alternate method is required. Please provide an efficient method to solve this question. Thank you!
P.S.: The answer to this question is x=0

This is a new edit made 8 hours ago. Anyone who posted their answers before that are not wrong and not flawed. Their answer will satisfy this question: The least integer value of $x$, which satisfy $|x| + |\frac{x}{x + 1}| = \frac{x^2}{|x - 1|}$, is.... Sorry for the confusion

 [/INST] Case $3$ is not as many cases as you'd think.
Consider.
Case 1: $x \ge 1$ then $x - 1 \ge 0$ and and the modulus are non-negitive.  However $\frac x{x-1}$ is undefined for $x = 1$ so $x> 1$
$x + \frac x{x-1} = \frac {x(x-1) + x}{x-1} = \frac {x^2}{x-1}$ which is always true so $x > 1$ are solutions.
Case 2: $x \le 0$ so $x -1 < 0$ but $\frac {x}{x-1} \ge  0$.
So $-x + \frac x{x-1} = - \frac {x^2}{x-1}$ which is always true if 
$\frac {-x(x-1) + x}{x-1} = \frac {-x^2 + 2x}{x-1} = \frac {-x^2}{x-1}$ which only happens if so $x = 0$.  So $x = 0$ is   solution.
Case 3: $0 < x < 1$ then $x > 0$ and $x-1 < 0$ and $\frac x{x-1} < 0$.
So we want $x - \frac x{x-1} = -\frac {x^2}{x-1}$ which is true if
$\frac {x(x-1) - x}{x-1} = \frac {-x^2}{x-1}$ which is true if
$\frac {x^2 - 2x}{x-1} = \frac {-x^2}{x-1}$ which is true if and only if
$x^2 - 2x = -x^2$
$2x^2 = 2x$ 
$x^2 = x$ which happens if $x = 0$ or $x =1$ but those are out of range.
So solutions are $x= 0$ or $x > 1$.
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