Question

x+9/x-3 - 6/x^2-9 : 6/(x+3)^2 - 3x-3/x-3

Answer 1

$dfrac{x+9}{x-3}-dfrac{6}{x^2-9}:dfrac{6}{(x+3)^2}-dfrac{3x-3}{x-3}=dfrac{x+9}{x-3}-dfrac{6}{(x-3)(x+3)}cdotdfrac{(x+3)^2}{6}-\ \ -dfrac{3x-3}{x-3}=dfrac{x+9}{x-3}-dfrac{x+3}{x-3}-dfrac{3x-3}{x-3}=dfrac{x+9-x-3-3x+3}{x-3}=dfrac{-3x+9}{x-3}=\ \ \ =dfrac{-3(x-3)}{x-3}=-3$