`24=(x+1)(x+2)(x+3)(x+4)`
`24=(x^2+5x+4)(x^2+5x+6)`
đẳ `x^2+5x+4=t`
`24=t(t+2)`
`⇔t^2+2t+1=25`
`⇔(t+1)^2=25`
`⇔`\(\left[ \begin{array}{l}t+1=5\\t+1=-5\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}t=4\\t=-6\end{array} \right.\)
với t=4
`⇒x^2+5x+4=4`
`⇒x^2+5x=0`
`⇔x(x+5)=0`
`⇔`\(\left[ \begin{array}{l}x=0\\x+5=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=0\\x=-5\end{array} \right.\)
với t=-6
`⇒x^2+5x+4=-6`
`⇒x^2+5x+10=0`
`⇔4x^2+20x+40=0`
`⇔(2x+5)^2+15=0`
`⇒`vô nghiệm
