`x(x-1)(x+1)(x+2)=24`
`⇔[x(x+1)][(x-1)(x+2)]=24`
`⇔(x^2+x)(x^2+x-2)=24`
$\text{Đặt $x^2$ + x = y}$
`⇒y(y-2)=24`
`⇔y^2-2y-24=0`
`⇔y^2-6y+4y-24=0`
`⇔y(y-6)+4(y-6)=0`
`⇔(y+4)(y-6)=0`
$\text{Quay trở lại ta có:}$
`⇒(x^2+x+4)(x^2+x-6)=0`
`⇔` `⇔`\(\left[ \begin{array}{l}x^2+x+4=0\\x^2+x-6=0\end{array} \right.\) \(\left[ \begin{array}{l}x∈∅\\x=2; x = -3\end{array} \right.\)