a) `x(x+1)+2(x+1)=0`
⇔`(x+1)(x+2)=0`
⇔\(\left[ \begin{array}{l}x+1=0\\x+2=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=-1\\x=-2\end{array} \right.\)
Vậy `S={-1,-2}`
b) `x(x+1)-x-1=0`
⇔`x(x+1)-(x+1)=0`
⇔`(x+1)(x-1)=0`
⇔\(\left[ \begin{array}{l}x+1=0\\x-1=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=-1\\x=1\end{array} \right.\)
Vậy `S={1,-1}`
