Question

1 + 2(17-6x)/х2-6х+8 = 1-2х/х-4 - 11/х-2

Answer 1

$1+\frac{2(17-6x)}{x^{2}-6x+8 } =\frac{1-2x}{x-4}-\frac{11}{x-2}\\1+\frac{34-12x}{(x-4)(x-2)}-\frac{1-2x}{x-4}+\frac{11}{x-2}=0\\\frac{x^{2}-6x+8+34-12x-x+2+2x^{2}-4x+11x-44}{(x-4)(x-2)}=0\\\frac{3x^{2}-12x }{(x-4)(x-2)}=0\\\frac{3(x^{2}-4x) }{(x-4)(x-2)}=0\\x^{2}-4x=0,x-2\neq0,x-4\neq 0\\x(x-4)=0,x\neq2,x\neq 4\\x_{1}=0\\x_{2}=4$- не уд.

Ответ : 0