Đáp án:
`M = |(x - 2020) (x^2 - 16)| + 2x (x - 4) + 8 (4 - x) + 2021`
`⇔ M = |(x - 2020) (x^2 - 16)| + 2x^2 - 8x + 32 - 8x + 2021`
`⇔ M = |(x - 2020) (x^2 - 16)| + 2x^2 - 16x + 32 + 2021`
`⇔ M = |(x - 2020) (x^2 - 16)| + (2x^2 - 16x + 32) + 2021`
`⇔ M = |(x - 2020) (x^2 - 16)| + 2 [x^2 - 8x + 16] + 2021`
`⇔ M = |(x - 2020) (x^2 - 16)| + 2 [x^2 - 2 . 4x + 4^2] + 2021`
`⇔ M = |(x - 2020) (x^2 - 16)| + 2 [x - 4]^2 + 2021`
Vì \(\left\{ \begin{array}{l}|(x-2020)(x^2-16)|≥0∀x\\2 [x-4]^2≥0∀x\end{array} \right.\)
`-> |(x - 2020) (x^2 - 16)| + 2 [x - 4]^2 ≥ 0∀x`
`-> |(x - 2020) (x^2 - 16)| + 2 [x - 4]^2 + 2021 ≥ 0 + 0 + 2021 = 2021`
`-> M ≥ 2021`
`-> M_{min} = 2021`
Dấu "`=`" xảy ra khi :
\(\left\{ \begin{array}{l}(x-2020) (x^2-16)=0\\x-4=0\end{array} \right.\)
$\bullet$ `(x - 2020) (x^2 - 16) = 0`
`->` \(\left[ \begin{array}{l}x-2020=0\\x^2-16=0\end{array} \right.\)
`->` \(\left[ \begin{array}{l}x=2020\\x=4\\x=-4\end{array} \right.\) `(1)`
$\bullet$ `x - 4 = 0`
`-> x = 4` `(2)`
Từ `(1)` và `(2)`
`-> x =4`
Vậy `M_{min} = 2021 ⇔ x = 4`
