\(A=2x^2+4y^2+4xy+10x+12y+18\)
\(A=x^2+4xy+4y^2+6x+12y+9+x^2+4x+4+5\)
\(A=\left(x+2y\right)^2+2.3\left(x+2y\right)+9+\left(x+2\right)^2+5\)
\(A=\left(x+2y+3\right)^2+\left(x+2\right)^2+5\)
Do : \(\hept{\begin{cases}\left(x+2y+3\right)^2\ge0\forall x\\\left(x+2\right)^2\ge0\forall x\end{cases}}\)
\(\Leftrightarrow\left(x+2y+3\right)^2+\left(x+2\right)^2+5\ge5\)
\("="\Leftrightarrow\hept{\begin{cases}x+2y+3=0\\x+2=0\end{cases}\Leftrightarrow\hept{\begin{cases}y=-\frac{1}{2}\\x=-2\end{cases}}}\)
Vậy \(A_{min}=5\Leftrightarrow\hept{\begin{cases}x=-2\\y=-\frac{1}{2}\end{cases}}\)
Chúc bạn học tốt !!!