(2x+4/x^2-x)-(x-4/x^2+x)=0 решите уравнение

[latex] \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+4x+2x+4-(x^2-4x-x+4)}{x(x-1)(x+1)}=0\\\\ \frac{2x^2+6x+4-x^2+5x-4}{x(x-1)(x+1)}=0\\\\ \frac{x^2+11x}{x(x-1)(x+1)}=0\\\\x \neq 0, x \neq б1\\\\x^2+11x=0\\x(x+11)=0\\x \neq 0\\x+11=0\\x=-11 [/latex] Ответ: -11

2х+4/х^2-x-x+4x/x^2-x=0 2(4/x^2)-x=0 2(4/x^2)=x 4/x^2=x/2 x*x^2=4*2 x^3=8 x=