$$ \frac{x^3 - 4x}{x+3} \cdot \frac{3x+9}{x^2 + 4x + 4} : \frac{x^2 - 2x}{2+x} = \frac{x(x^2 - 4)}{x+3} \cdot \frac{3(x+3)}{(x+2)^2} \cdot \frac{x+2}{x(x-2)} = \frac{x(x-2)(x+2)}{x+3} \cdot \frac{3(x+3)}{(x+2)^2} \cdot \frac{x+2}{x(x-2)} = \frac{3(x-2)x(x+2)(x+3)}{(x+3)x(x-2)(x+2)^2} = \frac{3}{x+2}$$
Ответ: $$\frac{3}{x+2}$$