5) 2b/2b+c - 4b²/4b²+4bc+c²;

5) $$\frac{2b}{2b+c} - \frac{4b^2}{4b^2+4bc+c^2} = \frac{2b}{2b+c} - \frac{4b^2}{(2b+c)^2} = \frac{2b(2b+c) - 4b^2}{(2b+c)^2} = \frac{4b^2 + 2bc - 4b^2}{(2b+c)^2} = \frac{2bc}{(2b+c)^2}$$

Ответ: $$\frac{2bc}{(2b+c)^2}$$