`(x² -4x)² - (x² -4x) = 20`
`⇔ (x² -4x).(x² -4x -1) = 20`
Đặt `a = x² -4x`, ta có:
`⇔ a.(a -1) = 20`
`⇔ a² -a -20 = 0`
`⇔ a.(a -5) +4.(a -5) = 0`
`⇔ (a -5).(a +4) = 0`
$⇔ \left[ \begin{array}{l}a -5=0\\a +4=0\end{array} \right. ⇔ \left[ \begin{array}{l}x=5\\x=-4\end{array} \right.$
Thay `a = x² -4x`, ta được:
`TH1:` `x² -4x = 5`
`⇔ x² -4x -5 = 0`
`⇔ x.(x -5) +(x -5) = 0`
`⇔ (x -5).(x +1) = 0`
$⇔ \left[ \begin{array}{l}x -5=0\\x +1=0\end{array} \right. ⇔ \left[ \begin{array}{l}x=5\\x=-1\end{array} \right.$
`TH2:` `x² -4x = -4`
`⇔ x² -4x +4 = 0`
`⇔ (x -2)² = 0`
`⇔ x -2 = 0`
`⇔ x = 2`
`Vậy` `S = {5; -1; 2}`
