5) 5y - 15; y2-9 y²-9'

5) $$\frac{5y}{y^2-9} - \frac{15}{y^2-9} = \frac{5y - 15}{y^2 - 9} = \frac{5(y - 3)}{(y - 3)(y + 3)} = \frac{5}{y + 3}$$

Ответ: $$\frac{5}{y+3}$$