Question

1/(3-x)+6/(x+3)<1 решите неравенства

Answer 1

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

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

Ответ: $(-infty; -3)cup(3; +infty)$