8 y² + 8y 4y-4 4-y² 4-y²

8. $$ \frac{y^2+8y}{4-y^2} - \frac{4y-4}{4-y^2} = \frac{y^2+8y - (4y-4)}{4-y^2} = \frac{y^2+8y - 4y + 4}{4-y^2} = \frac{y^2+4y+4}{4-y^2} = \frac{(y+2)^2}{(2-y)(2+y)} = \frac{y+2}{2-y} $$

Ответ: $$ \frac{y+2}{2-y} $$