\(A=3\left[6-\left|y-1\right|-\left(x-2\right)^2\right]\)
\(=18-3\left|y-1\right|-3\left(x-2\right)^2\)
Ta thấy: \(\left\{\begin{matrix}\left|y-1\right|\ge0\\\left(x-2\right)^2\ge0\end{matrix}\right.\)\(\Rightarrow\left\{\begin{matrix}3\left|y-1\right|\ge0\\3\left(x-2\right)^2\ge0\end{matrix}\right.\)\(\Rightarrow\left\{\begin{matrix}-3\left|y-1\right|\le0\\-3\left(x-2\right)^2\le0\end{matrix}\right.\)
\(\Rightarrow-3\left|y-1\right|-3\left(x-2\right)^2\le0\)
\(\Rightarrow18-3\left|y-1\right|-3\left(x-2\right)^2\le18\)
\(\Rightarrow A\le18\)
Dấu "=" xảy ra khi \(\Rightarrow\left\{\begin{matrix}-3\left|y-1\right|=0\\-3\left(x-2\right)^2=0\end{matrix}\right.\)\(\Rightarrow\left\{\begin{matrix}\left|y-1\right|=0\\\left(x-2\right)^2=0\end{matrix}\right.\)\(\Rightarrow\left\{\begin{matrix}y=1\\x=2\end{matrix}\right.\)
Vậy \(Max_A=18\) khi \(\left\{\begin{matrix}y=1\\x=2\end{matrix}\right.\)
