Question

Ребят Помогите Срочно!!!
3/3-y - y/y^2-9​

Answer 1

Ответ:

Объяснение:

$\frac{3}{3-y}-\frac{y}{y^2-9}= \frac{3}{3-y}-\frac{y}{(y-3)(y+3)}= \frac{3}{3-y} + \frac{y}{(3-y)(y+3)} =\\ =\frac{3(3+y)}{(3-y)(3+y)} + \frac{y}{(3-y)(y+3)}=\frac{3(3+y)+y}{9-y^2} = \frac{9+3y+y}{9-y^2} = \frac{9+4y}{9-y^2}$