e) x² - 2x + 1 = 4
⇔ x² - 2x + 1 - 4 = 0
⇔ x² - 2x - 3 = 0
⇔ x² - ( 3x - x ) - 3 = 0
⇔ x² - 3x + x - 3 = 0
⇔ ( x² - 3x ) + ( x - 3 ) = 0
⇔ x ( x - 3 ) + ( x - 3 ) = 0
⇔ ( x - 3 ) ( x + 1 ) = 0
⇔ \(\left[ \begin{array}{l}x-3=0\\x+1=0\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=3\\x=-1\end{array} \right.\)
Vậy S={3; -1}