Đáp án + Các bước giải:....
`(x+1)(x+2)(x+3)(x+4)=840`
`⇔[(x+1)(x+4)][(x+2)(x+3)]=840`
`⇔(x²+x+4x+4)(x²+2x+3x+6)=840`
`⇔(x²+5x+4)(x^2+5x+6)=840`
`⇔(x²+5x+4)(x^2+5x+4+2)=840`
Đặt `x^2+5x+4=t`
`⇔t(t+2)=840`
`⇔t^2+2t-840=0`
`⇔t²-28t+30t-840=0`
`⇔t(t-28)+30(t-28)=0`
`⇔(t+30)(t-28)=0`
`⇔`\(\left[ \begin{array}{l}t-28=0\\t+30=0\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}t=28\\t=-30\end{array} \right.\)
Với `t=28`
`x²+5x+4=28`
`⇔x²+5x-24=0`
`⇔x²-3x+8x-24=0`
`⇔x(x-3)+8(x-3)=0`
`⇔(x-3)(x+8)=0`
`⇔`\(\left[ \begin{array}{l}x-3=0\\x+8=0\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}x=3\\x=-8\end{array} \right.\)
Với `t=-30`
`x²+5x+4=-30`
`⇔x^2+5x+34=0`
`⇔x^2+5x+25/4+111/4=0`
`⇔[x^2+2*x*5/2+(5/2)^2]+111/4=0`
`⇔(x+5/2)^2=-111/4`
Mà `(x+5/2)^2 >= 0 ∀ x`
Vây `S={3;-8}`
