$x.(x -1).(x² -x +1) = 6$
$⇔ (x² -x).(x² -x +1) = 6$
$\text {Đặt t = x² -x + $\dfrac{1}{2}$, ta có:}$
$(t - \dfrac{1}{2}).(t + \dfrac{1}{2}) = 6$
$⇔ t² - \dfrac{1}{4} = 6$
$⇔ t² = \dfrac{25}{4}$
$⇔ t = ±\dfrac{5}{2}$
$\text {Thay t = x² -x + $\dfrac{1}{2}$, ta được:}$
$\text {TH1: x² -x + $\dfrac{1}{2}$ = $\dfrac{5}{2}$ }$
$⇔ x² -x -2 = 0$
$⇔ x.(x -2) +x -2 = 0$
$⇔ (x -2).(x +1) = 0$
$⇔ \left[ \begin{array}{l}x -2=0\\x +1=0\end{array} \right. ⇔ \left[ \begin{array}{l}x=2\\x=-1\end{array} \right.$
$\text {TH2: x² -x + $\dfrac{1}{2}$ = $\dfrac{-5}{2}$ }$
$⇔ x² -x +3 = 0$
$⇔ (x - \dfrac{1}{2})² + \dfrac{11}{4} = 0$ $\text {(Vô lý)}$
$\text {Vì (x - $\dfrac{1}{2}$)² + $\dfrac{11}{4}$ > 0 (vs ∀ x) }$
$\text {Vậy S = {2; -1}}$
