Đáp án:
`min{(x-1).(x+2).(x+3).(x+6)}=-36` khi `x∈{0;-5}`
Giải thích các bước giải:
`(x-1).(x+2).(x+3).(x+6)`
`=[(x-1).(x+6)].[(x+2).(x+3)]`
`=(x^2+6x-x-6).(x^2+3x+2x+6)`
`=(x^2+5x-6).(x^2+5x+6)`
`=(x^2+5x)^2-6^2`
`=(x^2+5x)^2-36`
Vì `(x^2+5x)^2≥0∀x`
`->(x^2+5x)^2-36≥-36∀x`
`->(x-1).(x+2).(x+3).(x+6)≥-36∀x`
Dấu `'='` xảy ra: `<=>x^2+5x=0<=>x.(x+5)=0`
`<=>`\(\left[ \begin{array}{l}x=0\\x+5=0\end{array} \right.\)`<=>` \(\left[ \begin{array}{l}x=0\\x=-5\end{array} \right.\)
Vậy `min{(x-1).(x+2).(x+3).(x+6)}=-36` khi `x∈{0;-5}`
