Question

Помогите реши второе задание.

Answer 1

$\frac{2x+4}{x^{2}-x }-\frac{x-4}{x^{2}+x}=0\\\\\frac{2x+4}{x(x-1)}-\frac{x-4}{x(x+1)}=0\\\\\frac{(2x+4)(x+1)-(x-4)(x-1)}{x(x-1)(x+1)}=0\\\\\frac{2x^{2}+2x+4x+4-x^{2}+x+4x-4}{x(x-1)(x+1)}=0\\\\\frac{x^{2}+11x }{x(x-1)(x+1)}=0\\\\\left \{ {{x(x+11)=0} \atop {x\neq0;x\neq-1;x\neq1}} \right.\\\\x+11=0\\\\x=-11\\\\Otvet:\boxed{-11}$