$(x -2)(x -1)x(x +1) = 24$
$⇔ (x² -x)(x² -x -2) = 24$
$Đặt$ $a = x² -x -1,ta$ $có:$
$(a +1)(a -1) = 24$
$⇔ a² -1 = 24$
$⇔ a² = 25$
$⇔ a = ±5$
$Thay$ $a = x² -x -1,ta$ $được:$
$TH1:$ $x² +x -1 = 5$
$⇔ x² +x -6 = 0$
$⇔ x² +3x -2x -6 = 0$
$⇔ x.(x +3) -2.(x +3) = 0$
$⇔ (x +3).(x -2) = 0$
$⇔ \left[ \begin{array}{l}x +3=0\\x -2=0\end{array} \right. ⇔ \left[ \begin{array}{l}x=-3\\x=2\end{array} \right.$
$TH2:$ $x² +x -1 = -5$
$⇔ x² +x +4 = 0$
$⇔ (x + \frac{1}{2})² + \frac{15}{4} = 0$ $(Vô$ $lí)$
$Vì$ $(x + \frac{1}{2})² + \frac{15}{4} > 0$ $(Vs$ $∀$ $x)$
$Vậy$ $S =$ {$-3; 2$}
