Solve `abs(x/(x-1))+abs(x)=x^(2)/abs(x-1)`.

Answer

To solve the equation \( \left| \frac{x}{x-1} \right| + |x| = \frac{x^2}{|x-1|} \), we will analyze the equation by considering different cases based on the values of \( x \).

### Step 1: Identify critical points
The critical points for this equation occur where the expressions inside the absolute values change sign. These points are:
- \( x = 0 \) (for \( |x| \))
- \( x = 1 \) (for \( |x-1| \))

### Step 2: Break into cases
We will consider three cases based on the critical points identified above:
1. Case 1: \( x < 0 \)
2. Case 2: \( 0 \leq x < 1 \)
3. Case 3: \( x \geq 1 \)

### Case 1: \( x < 0 \)
In this case:
- \( |x| = -x \)
- \( |x-1| = -(x-1) = -x + 1 \)

Substituting these into the equation gives:
\[ \left| \frac{x}{x-1} \right| - x = \frac{x^2}{-x + 1} \]
This simplifies to:
\[ -\frac{x}{x-1} - x = \frac{x^2}{-x + 1} \]
This case leads to contradictions because the left side will yield a negative result while the right side is non-negative. Thus, there are no solutions in this case.

### Case 2: \( 0 \leq x < 1 \)
In this case:
- \( |x| = x \)
- \( |x-1| = -(x-1) = -x + 1 \)

Substituting these into the equation gives:
\[ \left| \frac{x}{x-1} \right| + x = \frac{x^2}{-x + 1} \]
This simplifies to:
\[ -\frac{x}{x-1} + x = \frac{x^2}{1 - x} \]
Multiplying through by \( (1-x) \) to eliminate the denominator (valid since \( x < 1 \)):
\[ -x + x(1-x) = x^2 \]
This simplifies to:
\[ -x + x - x^2 = x^2 \]
Thus:
\[ 0 = 2x^2 \]
This implies \( x = 0 \), which is valid in this case.

### Case 3: \( x \geq 1 \)
In this case:
- \( |x| = x \)
- \( |x-1| = x - 1 \)

Substituting these into the equation gives:
\[ \frac{x}{x-1} + x = \frac{x^2}{x-1} \]
This simplifies to:
\[ \frac{x + x(x-1)}{x-1} = \frac{x^2}{x-1} \]
This leads to:
\[ \frac{x^2}{x-1} = \frac{x^2}{x-1} \]
This is always true for \( x > 1 \).

### Conclusion
The solutions to the original equation are:
- From Case 2: \( x = 0 \)
- From Case 3: \( x \geq 1 \)

Thus, the final solution set is:
\[ x \in \{0\} \cup [1, \infty) \]