\(P=\left(x^2-4xy+4y^2\right)+2\left(x-2y\right)+x^2-4x+2019\)
\(=\left(x-2y\right)^2+2\left(x-2y\right)+1+\left(x-2\right)^2+2014\)
\(=\left(x-2y+1\right)^2+\left(x-2\right)^2+2014\ge2014\)
Đẳng thức xảy ra khi \(\left\{{}\begin{matrix}x=2\\x=2y-1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=2\\y=\frac{3}{2}\end{matrix}\right.\)
Vậy...
\(P=2x^2+4y^2-4xy-2x-4y+2019\)
\(P=\left(x^2-4xy+4y^2\right)+2\left(x-2y\right)+1+\left(x-2\right)^2+2014\) ( Bước này mình làm hơi tắt , cái này bạn chỉ cần chú ý để tách ra thôi )
\(P=\left(x-2y+1\right)^2+\left(x-2\right)^2+2014\ge2014\)
Dấu '' = '' xảy ra
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-2y+1\right)^2=0\\\left(x-2\right)^2=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-2y+1=0\\x=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3-2y=0\\x=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\frac{3}{2}\\x=2\end{matrix}\right.\)
Vậy Min \(P=2014\Leftrightarrow\left\{{}\begin{matrix}y=\frac{3}{2}\\x=2\end{matrix}\right.\)