4) a) 9y+1 y-8 +; y²-4 4-y2 y²-4

4) a) $$\frac{9y+1}{y^2-4} + \frac{y-8}{4-y^2} + \frac{1-7y}{y^2-4} = \frac{9y+1}{y^2-4} - \frac{y-8}{y^2-4} + \frac{1-7y}{y^2-4} = \frac{9y+1 -y+8 +1 -7y}{y^2-4} = \frac{y+10 -y -7y}{y^2-4} = \frac{y -7y + 10}{y^2-4} = \frac{y - 7y + 10}{y^2 - 4} = \frac{10 + y}{y^2-4}$$

Ответ: $$\frac{10+y}{y^2-4}$$