1/(3-x)+6/(x+3)<1 решите неравенства

$\frac{1}{3-x}+\frac{6}{x+3}\ \textless \ 1\\\\ -\frac{1}{x-3}+\frac{6}{x+3}-\frac{1}{1}\ \textless \ 0\\\\ -\frac{1*(x+3)}{(x-3)*(x+3)}+\frac{6*(x-3)}{(x+3)*(x-3)}-\frac{1*(x-3)*(x+3)}{(x-3)*(x+3)}\ \textless \ 0\\\\ \frac{-(x+3)+6*(x-3)-(x-3)*(x+3)}{(x-3)*(x+3)}\ \textless \ 0\\\\ \frac{-x-3+6x-18-(x^2-9)}{(x-3)*(x+3)}\ \textless \ 0\\\\ \frac{-x-3+6x-18-x^2+9}{(x-3)*(x+3)}\ \textless \ 0\\\\ \frac{-x^2+5x-12}{(x-3)*(x+3)}\ \textless \ 0\\\\ \frac{x^2-5x+12}{(x-3)*(x+3)}\ \textgreater \ 0\\\\$

$\frac{x^2-2*2.5x+2.5^2-2.5^2+12}{(x-3)*(x+3)}\ \textgreater \ 0\\\\ \frac{(x-2.5)^2-6.25+12}{(x-3)*(x+3)}\ \textgreater \ 0\\\\ \frac{(x-2.5)^2+5.75}{(x-3)*(x+3)}\ \textgreater \ 0\\\\ \frac{1}{(x-3)*(x+3)}\ \textgreater \ 0\\\\ +++++++(-3)--------(3)++++++\ \textgreater \ x\\\\ x\in(-\infty;\ -3)\cup(3;\ +\infty)$

Ответ: $(-\infty;\ -3)\cup(3;\ +\infty)$