$\text{Bạn xem hình:}$
$\text{Bổ sung câu d):}$
$\dfrac{2}{-x^2+6x-8}-\dfrac{x-1}{x-2}=\dfrac{x+3}{x-4}$
$\to \dfrac{-2}{(x-2)(x-4)}-\dfrac{x-1}{x-2}-\dfrac{x-3}{x-4}=0$
$\to -2-(x-1)(x-4)-(x+3)(x-2)=0$
$\to -2-x^2+5x-4-x^2-x+6=0$
$\to -2x^2+4x=0$
$\to -2x(x-2)=0$
$\to -2x=0 \to x=0$
$\to x-2=0 \to x=2 \text{(loại)}$
$\text{Vậy S={0}}$