We need to solve the equation $\frac{3-x}{x-2} = 1 - \frac{2x-5}{x+2}$ for $x$.

1. Problem Description

We need to solve the equation $\frac{3-x}{x-2} = 1 - \frac{2x-5}{x+2}$ for $x$.

2. Solution Steps

First, multiply both sides of the equation by $(x-2)(x+2)$ to eliminate the fractions. We must have $x \ne 2$ and $x \ne -2$.

$(x-2)(x+2) \cdot \frac{3-x}{x-2} = (x-2)(x+2) \cdot \left(1 - \frac{2x-5}{x+2}\right)$

$(x+2)(3-x) = (x-2)(x+2) - (x-2)(2x-5)$

$3x - x^2 + 6 - 2x = x^2 - 4 - (2x^2 - 5x - 4x + 10)$

$x - x^2 + 6 = x^2 - 4 - (2x^2 - 9x + 10)$

$x - x^2 + 6 = x^2 - 4 - 2x^2 + 9x - 10$

$x - x^2 + 6 = -x^2 + 9x - 14$

Add $x^2$ to both sides:

$x + 6 = 9x - 14$

Subtract $x$ from both sides:

$6 = 8x - 14$

Add 14 to both sides:

$20 = 8x$

Divide both sides by 8:

$x = \frac{20}{8} = \frac{5}{2}$

Since $x = \frac{5}{2}$ is not equal to 2 or -2, it is a valid solution.

3. Final Answer

$\frac{5}{2}$