ĐKXĐ:\(\left\{{}\begin{matrix}x\ne1\\x\ne3\end{matrix}\right.\)
\(\dfrac{6}{x-1}-\dfrac{4}{x-3}=\dfrac{8}{\left(x-1\right)\left(3-x\right)}\\ \Leftrightarrow\dfrac{6\left(x-3\right)}{\left(x-1\right)\left(x-3\right)}-\dfrac{4\left(x-1\right)}{\left(x-1\right)\left(x-3\right)}=\dfrac{-8}{\left(x-1\right)\left(x-3\right)}\\ \Leftrightarrow\dfrac{6x-18}{\left(x-1\right)\left(x-3\right)}-\dfrac{4x-4}{\left(x-1\right)\left(x-3\right)}+\dfrac{8}{\left(x-1\right)\left(x-3\right)}=0\\ \Leftrightarrow\dfrac{6x-18-4x+4+8}{\left(x-1\right)\left(x-3\right)}=0\\ \Rightarrow2x-6=0\\ \Leftrightarrow x=3\left(ktm\right)\)
\(ĐK:x\ne1;x\ne3\)
\(\Rightarrow\dfrac{6}{x-1}+\dfrac{4}{3-x}=\dfrac{8}{\left(x-1\right)\left(3-x\right)}\)
\(\Leftrightarrow\dfrac{6\left(3-x\right)+4\left(x-1\right)}{\left(x-1\right)\left(3-x\right)}=\dfrac{8}{\left(x-1\right)\left(3-x\right)}\)
\(\Leftrightarrow6\left(3-x\right)+4\left(x-1\right)=8\)
\(\Leftrightarrow18-6x+4x-4-8=0\)
\(\Leftrightarrow-2x+6=0\)
\(\Leftrightarrow2x=6\)
\(\Leftrightarrow x=3\left(ktm\right)\)
Vậy \(S=\varnothing\)