Question

Tính GTNN của N=\(2x^2+2xy+5y^2-8x-22y+2019\)

Answer 1

\(2N=4x^2+4xy+10y^2-16x-44y+4038\)

\(=4x^2+4x\left(y-4\right)+\left(y-4\right)^2-\left(y-4\right)^2+10y^2-44y+4038\)

\(=\left(2x+y-4\right)^2+9y^2-36y^2+36+3986\)

\(=\left(2x+y-4\right)^2+\left(3y-6\right)^2+3986\ge3986\forall x,y\)

\(\Rightarrow N\ge1993\forall x,y\)

Dấu "=" \(\Leftrightarrow\left\{{}\begin{matrix}\left(2x+y-4\right)^2=0\\\left(3y-6\right)^2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=2\end{matrix}\right.\)