Vì \(x+y=4\Rightarrow x-2+y=2\Rightarrow x-2=2-y\left(1\right)\)
Thay (1) vào A ta được:
\(A=\left(2-y\right)y+2017\)
\(=-y^2+2y+2017\)
\(=-\left(y^2-2y-2017\right)\)
\(=-\left(y^2-y-y-2017\right)\)
\(=-\left[\left(y^2-y\right)-y+1-2018\right]\)
\(=-\left[y\left(y-1\right)-\left(y-1\right)-2018\right]\)
\(=-\left[\left(y-1\right)^2-2018\right]\)
\(=-\left(y-1\right)^2+2018\)
Vì \(-\left(y-1\right)^2\le0\forall y\)
\(\Rightarrow-\left(y-1\right)^2+2018\le2018\forall y\)
Dấu \("="\) xảy ra khi \(\left(y-1\right)^2=0\)
\(\Rightarrow y=1\)
Khi đó: \(x=4-y=4-1=3\)
Vậy \(Min_{bt}=2018\) khi \(\left\{{}\begin{matrix}x=3\\y=1\end{matrix}\right.\).
Bài này trường mk cũng thi nè!
