A)x(x-2)+x-2=0
⇔x(x-2)+(x-2)=0
⇔(x-2)(x+1)=0
⇔\(\left[ \begin{array}{l}x-2=0\\x+1=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=2\\x=-1\end{array} \right.\)
Vậy S={2;-1}
B)5x(x-3)-x+3=0
⇔5x(x-3)-(x-3)=0
⇔(x-3)(5x-1)=0
⇔\(\left[ \begin{array}{l}x-3=0\\5x-1=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=3\\x=$\frac{1}{5}$ \end{array} \right.\)
Vậy S={3;$\frac{1}{5}$