E(x)=[tex] \frac{x-3}{x+3}- \frac{x+3}{3-x} + \frac{12x}{x^2-9}=
\frac{x-3}{x+3}+ \frac{x+3}{x-3}+ \frac{12x}{(x-3)(x+3)} =
\frac{x^{2}-6x+9+ x^{2} +6x+9+12x }{(x-3)(x+3)}=
\\ \frac{2 x^{2} +18+12x}{(x-3)(x+3)}= \\
\frac{2( x^{2} +6x+9)}{(x-3)(x+3)}= \\
\frac{2(x+3)^{2} }{(x-3)(x+3)}= \frac{2(x+3)}{x-3}
[/tex]