Question

Помогите!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Answer 1

$(\frac{3}{x^2-9}+\frac{1}{3-x})\cdot \frac{x^2-6x+9}{x}=(\frac{3}{(x-3)(x+3)}-\frac{1}{x-3})\cdot \frac{(x-3)^2}{x}=\\\\=\frac{3(x-3)-(x+3)}{(x-3)(x+3)}\cdot \frac{(x-3)^2}{x}=\frac{3x-9-x-3}{x+3}\cdot \frac{x-3}{x}=\frac{2x-12}{x+3}\cdot \frac{x-3}{x}=\\\\= \frac{2(x-6)\cdot (x-3)}{x\cdot (x+3)}$