\(A=3-4x-x^2=-\left(x^2+4x+4\right)+7=7-\left(x+2\right)^2\ge7\forall x\)
Dấu bằng xảy ra \(\Leftrightarrow x+2=0\Leftrightarrow x=-2\)
Vậy A max là 7 chỉ khi x=-2
b) \(7-x^2-y^2-2\left(x+y\right)\)
\(=7-x^2-y^2-2x-2y\)
\(=-x^2-2x-1-y^2-2y-1+9\)
\(=-\left(x+1\right)^2-\left(y+1\right)^2+9\le9\)
Max = 9 \(\Leftrightarrow\hept{\begin{cases}x+1=0\\y+1=0\end{cases}\Leftrightarrow}x=y=-1\)
Vậy ...................