Resposta :
HOLA.
[tex] \frac{1}{x+1} + \frac{1}{x+6} + \frac{1}{2x} = \frac{1}{x} \\m.m.c.((x+1),(x+6),(2x),(x))=2x*(x+1)*(x+6)\\2x*(x+6)+2x*(x+1)+(x+1)*(x+6) = 2*(x+1)*(x+6)\\2x^2 +12x +2x^2+2x+7x+6 = 2x^2+14x+12\\ 2x^2 +12x +2x^2+2x+7x+6 -2x^2-14x-12=0, reduza~ os~ termos~ semelhantes\\3x^2+7x-6=0,por~Baskara,~temos:\\x'=-3,descarte~e\\\boxed{x''= \frac{2}{3}}[/tex]
[tex] \frac{1}{x+1} + \frac{1}{x+6} + \frac{1}{2x} = \frac{1}{x} \\m.m.c.((x+1),(x+6),(2x),(x))=2x*(x+1)*(x+6)\\2x*(x+6)+2x*(x+1)+(x+1)*(x+6) = 2*(x+1)*(x+6)\\2x^2 +12x +2x^2+2x+7x+6 = 2x^2+14x+12\\ 2x^2 +12x +2x^2+2x+7x+6 -2x^2-14x-12=0, reduza~ os~ termos~ semelhantes\\3x^2+7x-6=0,por~Baskara,~temos:\\x'=-3,descarte~e\\\boxed{x''= \frac{2}{3}}[/tex]