The problem is to solve the equation $\frac{2m-5}{(m-1)(x+2)} - \frac{3}{x+1} = \frac{3x+4}{x^2+3x+2}$ for $m$.

1. Problem Description

The problem is to solve the equation $\frac{2m-5}{(m-1)(x+2)} - \frac{3}{x+1} = \frac{3x+4}{x^2+3x+2}$ for $m$.

2. Solution Steps

First, we factor the denominator on the right-hand side:

$x^2+3x+2 = (x+1)(x+2)$.

So the equation becomes:

$\frac{2m-5}{(m-1)(x+2)} - \frac{3}{x+1} = \frac{3x+4}{(x+1)(x+2)}$.

We multiply both sides of the equation by $(m-1)(x+1)(x+2)$ to eliminate the denominators.

$(x+1)(2m-5) - 3(m-1)(x+2) = (m-1)(3x+4)$

$(2m-5)(x+1) - 3(m-1)(x+2) = (m-1)(3x+4)$

$2mx+2m-5x-5 - 3(mx+2m-x-2) = 3mx+4m-3x-4$

$2mx+2m-5x-5 - 3mx-6m+3x+6 = 3mx+4m-3x-4$

$-mx-4m-2x+1 = 3mx+4m-3x-4$

$-mx-3mx-4m-4m = -3x+2x-4-1$

$-4mx-8m = -x-5$

$m(-4x-8) = -x-5$

$m = \frac{-x-5}{-4x-8} = \frac{x+5}{4x+8} = \frac{x+5}{4(x+2)}$

3. Final Answer

$m = \frac{x+5}{4(x+2)}$