3) a) \(\frac{3}{2y+6} + \frac{y^2-y-3}{y^2-9} - 1;\)

Преобразуем выражение:

\(\frac{3}{2y+6} + \frac{y^2-y-3}{y^2-9} - 1\)

\(\frac{3}{2(y+3)} + \frac{y^2-y-3}{(y-3)(y+3)} - 1\)

Приведем дроби к общему знаменателю: \(2(y-3)(y+3)\)

\(\frac{3(y-3)}{2(y-3)(y+3)} + \frac{2(y^2-y-3)}{2(y-3)(y+3)} - \frac{2(y-3)(y+3)}{2(y-3)(y+3)} = \frac{3y - 9 + 2y^2 - 2y - 6 - 2(y^2 - 9)}{2(y-3)(y+3)} = \frac{3y - 9 + 2y^2 - 2y - 6 - 2y^2 + 18}{2(y-3)(y+3)} = \frac{y + 3}{2(y-3)(y+3)} = \frac{1}{2(y-3)}\)

Выражение упрощено.

Ответ: \(\frac{1}{2(y-3)}\)