3х/х-1 - 4/х = 3/х^2-х

To solve the equation:

\[ \frac{3x}{x - 1} - \frac{4}{x} = \frac{3}{x^2 - x} \]

First, we can rewrite the right-hand side of the equation:

\[ \frac{3}{x^2 - x} = \frac{3}{x(x - 1)} \]

Now the equation becomes:

\[ \frac{3x}{x - 1} - \frac{4}{x} = \frac{3}{x(x - 1)} \]

Next, we will find a common denominator for the left side of the equation. The common denominator for \(\frac{3x}{x - 1}\) and \(\frac{4}{x}\) is \(x(x - 1)\). We can rewrite the left side as:

\[ \frac{3x \cdot x - 4(x - 1)}{x(x - 1)} = \frac{3x^2 - 4(x - 1)}{x(x - 1)} \]

This simplifies to:

\[ \frac{3x^2 - 4x + 4}{x(x - 1)} \]

Now the equation looks like:

\[ \frac{3x^2 - 4x + 4}{x(x - 1)} = \frac{3}{x(x - 1)} \]

Since the denominators are the same, we can equate the numerators:

\[ 3x^2 - 4x + 4 = 3 \]

Now let's move everything to one side:

\[ 3x^2 - 4x + 4 - 3 = 0 \] \[ 3x^2 - 4x + 1 = 0 \]

Next, we can use the quadratic formula to solve for \(x\):

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

where \(a = 3\), \(b = -4\), and \(c = 1\):

\[ b^2 - 4ac = (-4)^2 - 4(3)(1) = 16 - 12 = 4 \]

Putting it into the formula gives:

\[ x = \frac{4 \pm \sqrt{4}}{2 \cdot 3} \] \[ x = \frac{4 \pm 2}{6} \]

So, we have two solutions:

1. \(x = \frac{6}{6} = 1\)
2. \(x = \frac{2}{6} = \frac{1}{3}\)

Let's check if either of these values makes the original equation undefined. The original equation has denominators \(x(x - 1)\) and \(x - 1\), which are undefined for \(x = 0\) and \(x = 1\). Thus, \(x = 1\) cannot be accepted as a solution.

Therefore, the final solution is:

\[ \boxed{\frac{1}{3}} \]