`(x+1)(x+2)(x+3)(x+4)=24`
`=>[(x+1)(x+4)][(x+2)(x+3)]=24`
`=>(x^2+5x+4)(x^2+5x+6)=24`
Đặt `x^2+5x+4=a`
`=>a(a+2)=24`
`<=>a^2+2a=24`
`<=>a^2+2a-24=0`
`<=>(a-4)(a+6)=0`
`=>(x^2+5x+4-4)(x^2+5x+6+6)=0`
`<=>(x^2+5x)(x^2+5x+12)=0`
Với `x^2+5x=0=>x(x+5)=0=>` \(\left[ \begin{array}{l}x=0\\x=-5\end{array} \right.\)
Ta có: `x^2+5x+12\ne0`
Vậy `S={-5;0}`