Ta nhận thấy : \(\left(x-2y\right)^2\ge0\)
\(\left(x-3\right)^2\ge0\)
\(\left(y-1\right)^2\ge0\)
\(\Rightarrow A=\left(x-2y\right)^2+\left(x-3\right)^2+\left(y-1\right)^2+3\ge3\)
Min A = 3 \(\Leftrightarrow\begin{cases}x-2y=0\\x-3=0\\y-1=0\end{cases}\Leftrightarrow\begin{cases}x-2y=0\\x=3\\y=1\end{cases}\Leftrightarrow}\begin{cases}x=3\\y=1\\\end{cases}}\)