(x - 1)/(2x + 3) - (2x - 1)/(3 - 2x) = 0

To solve the equation

\[ \frac{x - 1}{2x + 3} - \frac{2x - 1}{3 - 2x} = 0, \]

we first need to rearrange the equation:

\[ \frac{x - 1}{2x + 3} = \frac{2x - 1}{3 - 2x}. \]

Next, we can cross-multiply to eliminate the fractions:

\[ (x - 1)(3 - 2x) = (2x - 1)(2x + 3). \]

Expanding both sides:

1. Left side:

\[ (x - 1)(3 - 2x) = 3x - 2x^2 - 3 + 2x = -2x^2 + 5x - 3. \]

2. Right side:

\[ (2x - 1)(2x + 3) = 4x^2 + 6x - 2x - 3 = 4x^2 + 4x - 3. \]

Now we set both expanded sides equal to each other:

\[ -2x^2 + 5x - 3 = 4x^2 + 4x - 3. \]

Next, we move all terms to one side:

\[ -2x^2 + 5x - 3 - 4x^2 - 4x + 3 = 0, \]

which simplifies to:

\[ -6x^2 + x = 0. \]

Factoring out \(x\):

\[ x(-6x + 1) = 0. \]

Setting each factor to zero gives us:

1. \(x = 0\)
2. \(-6x + 1 = 0\) or \(x = \frac{1}{6}\).

So the solutions are:

\[ x = 0 \quad \text{and} \quad x = \frac{1}{6}. \]

To ensure these solutions do not make the original denominators equal to zero, we check:

1. For \(x = 0\):

   • \(2(0) + 3 = 3\) (not zero)
   • \(3 - 2(0) = 3\) (not zero)

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

   • \(2\left(\frac{1}{6}\right) + 3 = \frac{1}{3} + 3 = \frac{10}{3}\) (not zero)
   • \(3 - 2\left(\frac{1}{6}\right) = 3 - \frac{1}{3} = \frac{8}{3}\) (not zero)

Thus, both solutions are valid. Therefore, the final solutions are:

\[ \boxed{0 \quad \text{and} \quad \frac{1}{6}}. \]