Solve: (3x-15)/(x+4) : (x^2-25)/(3x+12)

We have to solve the expression: $$\frac{3x-15}{x+4} : \frac{x^2-25}{3x+12}$$

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

$$\frac{3x-15}{x+4} \cdot \frac{3x+12}{x^2-25}$$

Next, we can factor the polynomials:

$$3x - 15 = 3(x-5)$$ $$x+4 = x+4$$ $$3x+12 = 3(x+4)$$ $$x^2 - 25 = (x-5)(x+5)$$

So, the expression becomes:

$$\frac{3(x-5)}{x+4} \cdot \frac{3(x+4)}{(x-5)(x+5)}$$

Now, we can cancel common factors:

$$\frac{3\cancel{(x-5)}}{\cancel{x+4}} \cdot \frac{3\cancel{(x+4)}}{\cancel{(x-5)}(x+5)} = \frac{3 \cdot 3}{x+5}$$

$$\frac{9}{x+5}$$

Ответ: $$\frac{9}{x+5}$$