\(P=x^2+10y^2-6xy+4x-14y+2023\)
\(P=x^2-6xy+9y^2+4x-12y+y^2-2y+1+2022\)
\(P=\left(x-3y\right)^2+4\left(x-3y\right)+4+\left(y-1\right)^2+2018\)
\(P=\left(x-3y+2\right)^2+\left(y-1\right)^2+2018\ge2018\forall x;y\)
Dấu "=" xảy ra \(\Leftrightarrow x=y=1\)
\(P=x^2+\left(3y\right)^2+4-6xy+4x-12y+y^2-2y+1+2018\)
\(=\left(x-3y+2\right)^2+\left(y-1\right)^2+2018\)
\(\Rightarrow...\)
\(P=x^2+10y^2-6xy+4x-14y+2023\\ P=\left(x^2+9y^2+4-6xy+4x-12y\right)+\left(y^2-2y+1\right)+2018\\ P=\left(x-3y+2\right)^2+\left(y-1\right)^2+2018\)
Ta có: \(\left(x-3y+2\right)^2+\left(y-1\right)^2\ge0\)
\(\Rightarrow P=\left(x-3y+2\right)^2+\left(y-1\right)^2+2018\ge2018\)
Dấu "=" xảy ra khi:
\(\left\{{}\begin{matrix}\left(x-3y+2\right)^2=0\\\left(y-1\right)^2=0\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x-3y+2=0\\y-1=0\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x-3y=-2\\y=1\end{matrix}\right.\\ \Rightarrow x-3=-2\\ \Rightarrow x=1\)
Vậy MinP = 2018 <=> x = 1, y = 1
