Ta có: \(A=x^2+6xy+10y^2-4x-18y+2018\)
\(=x^2+6xy+9y^2-4x-12y+y^2-6y+2018\)
\(=\left(x+3y\right)^2-4\left(x+3y\right)+4+y^2-6y+9+2005\)
\(=\left(x+3y-2\right)^2+\left(y-3\right)^2+2005\ge2005\forall x,y\)
Dấu '=' xảy ra khi \(\begin{cases}y-3=0\\ x+3y-2=0\end{cases}\Rightarrow\begin{cases}y=3\\ x=-3y+2=-3\cdot3+2=-9+2=-7\end{cases}\)

