a) x(x+1)(x-1)(x+2)=24
⇔(x²+x)(x²+x-2)=24
Đặt y=x²+x-1
⇔(y+1)(y-1)=24
⇔y²-1-24=0
⇔y²-25=0
⇔(y-5)(y+5)=0
⇔(x²+x-6)(x²+x+4)=0
⇔[(x²+3x)-(2x-6)].($x^2$+x+$\frac{1}{4}+$ $\frac{15}{4}$ )
⇔x(x+3)-2(x+3)=0 (vì ($x^2$+x+$\frac{1}{4}+$ $\frac{15}{4}$ ) ∦0)
⇔(x+3)(x-2)=0
⇔\(\left[ \begin{array}{l}x+3=0\\x-2=0\end{array} \right.\) =>\(\left[ \begin{array}{l}x=-3\\x=2\end{array} \right.\)
Vậy S={-3;2}
b, Ảnh dưới nha
