Giải thích các bước giải:
`E=(x+1)(x^2−4)(x+5)+2014`
`=(x+1)(x-2)(x+2)(x+5)+2014`
`=[(x+1)(x+2)].[(x-2)(x+5)]+2014`
`=(x^2+3x+2)(x^2+3x-10)+2014`
`=(x^2+3x-4+6)(x^2+3x-4-6)+2014`
`=(x^2+3x-4)^2-36+2014`
`=[(x-1)(x+4)]^2+1978>=1978`
Dấu `=` xảy ra `<=>`\(\left[ \begin{array}{l}x-1=0\\x+4=0\end{array} \right.\) `=>`\(\left[ \begin{array}{l}x=1\\x=-4\end{array} \right.\)
Vậy `Emin=1978<=>x in{1;-4}.`
