The least integer value of x , which satisfies `|x|+|(x)/(x-1)|=(x^(2))/(|x-1|)` is

Answer

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

### Step 1: Identify Cases Based on \( x \)

We need to consider different cases for \( x \) based on the absolute values involved in the equation. The critical points are \( x = 0 \) and \( x = 1 \). Therefore, we will analyze three cases:
1. \( x > 1 \)
2. \( 0 < x < 1 \)
3. \( x < 0 \)

### Step 2: Case 1: \( x > 1 \)

In this case, both \( |x| = x \) and \( |x-1| = x-1 \). Thus, the equation simplifies to:

\[ x + \frac{x}{x-1} = \frac{x^2}{x-1} \]

Multiply through by \( x-1 \) (which is positive for \( x > 1 \)):

\[ (x + \frac{x}{x-1})(x-1) = x^2 \]

This expands to:

\[ x(x-1) + x = x^2 \]

Simplifying gives:

\[ x^2 - x + x = x^2 \]

This simplifies to:

\[ 0 = 0 \]

This is true for all \( x > 1 \). Therefore, any \( x > 1 \) is a solution.

### Step 3: Case 2: \( 0 < x < 1 \)

In this case, \( |x| = x \) and \( |x-1| = 1-x \). The equation becomes:

\[ x + \frac{x}{1-x} = \frac{x^2}{1-x} \]

Multiply through by \( 1-x \):

\[ (x + \frac{x}{1-x})(1-x) = x^2 \]

This expands to:

\[ x(1-x) + x = x^2 \]

Simplifying gives:

\[ x - x^2 + x = x^2 \]

This simplifies to:

\[ 2x - x^2 = x^2 \]

Rearranging gives:

\[ 2x = 2x^2 \]

Dividing both sides by 2 (assuming \( x \neq 0 \)):

\[ x = x^2 \]

This implies:

\[ x^2 - x = 0 \implies x(x-1) = 0 \]

The solutions are \( x = 0 \) or \( x = 1 \). Since we are in the interval \( 0 < x < 1 \), the only valid solution is \( x = 0 \), which is not included in this case.

### Step 4: Case 3: \( x < 0 \)

In this case, \( |x| = -x \) and \( |x-1| = 1-x \). The equation becomes:

\[ -x + \left| \frac{x}{x-1} \right| = \frac{x^2}{1-x} \]

Since \( x < 0 \), \( \frac{x}{x-1} \) is also negative, so \( \left| \frac{x}{x-1} \right| = -\frac{x}{x-1} \). Thus, the equation simplifies to:

\[ -x - \frac{x}{x-1} = \frac{x^2}{1-x} \]

Multiply through by \( 1-x \):

\[ (-x - \frac{x}{x-1})(1-x) = x^2 \]

This expands to:

\[ -x(1-x) - \frac{x(1-x)}{x-1} = x^2 \]

This leads to no valid solutions since \( x < 0 \) does not satisfy the equation.

### Conclusion

From all cases, the only integer solution we found is \( x = 0 \) from the second case. Therefore, the least integer value of \( x \) that satisfies the equation is:

\[ \boxed{0} \]