Đáp án + giải thích bước giải :
Ta có :
`⇔` \(\left\{ \begin{array}{l}|x-2019|^{20}≥0∀x\\(x-y)^2≥0∀y\end{array} \right.\)
`⇔ |x - 2019|^{20} + (x- y)^2 ≥ 0 ∀x,y`
Dấu "`=`" xảy ra khi :
`⇔` \(\left\{ \begin{array}{l}x-2019=0\\x-y=0\end{array} \right.\)
`⇔` \(\left\{ \begin{array}{l}x=0+2019\\y=x-0\end{array} \right.\)
`⇔` \(\left\{ \begin{array}{l}x=2019\\y=2019-0\end{array} \right.\)
`⇔` \(\left\{ \begin{array}{l}x=2019\\y=2019\end{array} \right.\)
Vậy `(x;y) ∈ {2019; 2019}`
