Question

Помогите решить срочно

Answer 1

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