The problem asks us to simplify the expression $\frac{1}{(x+1)(x+2)} - \frac{5}{(x+2)(x-3)}$.

1. Problem Description

The problem asks us to simplify the expression $\frac{1}{(x+1)(x+2)} - \frac{5}{(x+2)(x-3)}$.

2. Solution Steps

We need to find a common denominator to combine the two fractions. The least common denominator (LCD) is $(x+1)(x+2)(x-3)$.

We rewrite the first fraction with the LCD:

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

We rewrite the second fraction with the LCD:

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

Now, we can subtract the two fractions:

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

Simplify the numerator:

$x - 3 - 5(x+1) = x - 3 - 5x - 5 = -4x - 8$

So the expression becomes:

$\frac{-4x - 8}{(x+1)(x+2)(x-3)}$

We can factor out a $-4$ from the numerator:

$-4x - 8 = -4(x+2)$

Thus, the expression becomes:

$\frac{-4(x+2)}{(x+1)(x+2)(x-3)}$

We can cancel the $(x+2)$ term in the numerator and the denominator:

$\frac{-4}{(x+1)(x-3)}$

3. Final Answer

$\frac{-4}{(x+1)(x-3)}$