a) Ta có: \(\left\{\begin{matrix}\left|2x-3\right|\ge0\\\left(y+1\right)^2\ge0\end{matrix}\right.\Rightarrow\left\{\begin{matrix}-\left|2x-3\right|\le0\\-\left(y+1\right)^2\le0\end{matrix}\right.\)
\(\Rightarrow-\left|2x-3\right|-\left(y+1\right)^2\le0\)
\(\Rightarrow A=20-\left|2x-3\right|-\left(y+1\right)^2\le20\)
Dấu " = " khi \(\left\{\begin{matrix}\left|2x-3\right|=0\\\left(y+1\right)^2=0\end{matrix}\right.\Rightarrow\left\{\begin{matrix}2x-3=0\\y+1=0\end{matrix}\right.\Rightarrow\left\{\begin{matrix}x=\frac{3}{2}\\y=-1\end{matrix}\right.\)
Vậy \(MAX_A=20\) khi \(x=\frac{3}{2};y=-1\)
b) Ta có: \(3\left(x+1\right)^2\ge0\)
\(\Rightarrow-3\left(x+1\right)^2\le0\)
\(\Rightarrow B=7-3\left(x+1\right)^2\le7\)
Dấu " = " khi \(3\left(x+1\right)^2=0\Rightarrow x+1=0\Rightarrow x=-1\)
Vậy \(MAX_B=7\) khi x = -1