Сколько ответ [latex] \frac{3}{x+3}+ \frac{3}{x^2-3x}+ \frac{2x}{9-x^2}=[/latex]

[latex] \frac{3}{x+3}+ \frac{3}{x^2-3x}+ \frac{2x}{9-x^2}=\\\\= \frac{3}{x+3}+ \frac{3}{x(x-3)}- \frac{2x}{(x-3)(x+3)}=\\\\= \frac{3x(x-3)+3(x+3)-2x^2}{x(x+3)(x-3)}=\\\\= \frac{3x^2-9x+3x+9-2x^2}{x(x+3)(x-3)}=\\\\= \frac{x^2-6x+9}{x(x+3)(x-3)}=\\\\= \frac{(x-3)^2}{x(x+3)(x-3)}=\\\\= \frac{x-3}{x(x+3)} [/latex]