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