9. $$\frac{2B}{2B+C} - \frac{4B^2}{4B^2+4BC+C^2} =$$

$$\frac{2B}{2B+C} - \frac{4B^2}{(2B+C)^2} = \frac{2B(2B+C) - 4B^2}{(2B+C)^2} = \frac{4B^2 + 2BC - 4B^2}{(2B+C)^2} = \frac{2BC}{(2B+C)^2}$$ $$\textbf{Ответ: } \frac{2BC}{(2B+C)^2}$$