$6) \dfrac{x}{2(x-3)}+\dfrac{x}{2(x+1)}=\dfrac{2x}{(x+1)(x-3)} (x\ne 3; -1)$
$\to x(x+1)+x(x-3)-2x.2=0$
$\to x^2+x+x^2-3x-4x=0$
$\to 2x^2-6x=0$
$\to 2x(x-3)=0$
$ \to \begin{cases}2x=0 \\x-3=0\\\end{cases} \to \begin{cases} x=0\\x=3 \text{ (loại)}\\\end{cases}$
Vậy $S={0}$