The problem is to simplify the expression: $\frac{a^2 - 3a}{a^2 - 25} \div \frac{a^2 - 9}{a^2 + 5a}$.

1. Problem Description

The problem is to simplify the expression:

$\frac{a^2 - 3a}{a^2 - 25} \div \frac{a^2 - 9}{a^2 + 5a}$.

2. Solution Steps

First, we rewrite the division as multiplication by the reciprocal:

$\frac{a^2 - 3a}{a^2 - 25} \div \frac{a^2 - 9}{a^2 + 5a} = \frac{a^2 - 3a}{a^2 - 25} \cdot \frac{a^2 + 5a}{a^2 - 9}$.

Next, we factor each of the expressions:

$a^2 - 3a = a(a - 3)$

$a^2 - 25 = (a - 5)(a + 5)$

$a^2 + 5a = a(a + 5)$

$a^2 - 9 = (a - 3)(a + 3)$

Substituting these factorizations into the expression, we get:

$\frac{a(a - 3)}{(a - 5)(a + 5)} \cdot \frac{a(a + 5)}{(a - 3)(a + 3)}$.

Now we can cancel common factors in the numerator and denominator:

$\frac{a(a - 3)}{(a - 5)(a + 5)} \cdot \frac{a(a + 5)}{(a - 3)(a + 3)} = \frac{a}{a - 5} \cdot \frac{a}{a + 3} = \frac{a^2}{(a - 5)(a + 3)}$.

Expanding the denominator gives:

$(a - 5)(a + 3) = a^2 + 3a - 5a - 15 = a^2 - 2a - 15$.

Therefore, the simplified expression is $\frac{a^2}{a^2 - 2a - 15}$.

3. Final Answer

$\frac{a^2}{(a-5)(a+3)} = \frac{a^2}{a^2 - 2a - 15}$