\(A=3-\left(x+2\right)^2\)
\(\left(x+2\right)^2\ge0\forall x\)
\(\Rightarrow3-\left(x+2\right)^2\le3\)
Dấu "=" xảy ra khi:
\(\left(x+2\right)^2=0\Rightarrow x+2=0\Rightarrow x=-2\)
\(\Rightarrow MAX_A=3\) khi \(x=-2\)
\(B=-\left(x-2\right)^2-\left(y+1\right)^2-9\) (đã sửa đề)
\(\left\{{}\begin{matrix}\left(x-2\right)^2\ge0\forall x\\\left(y+1\right)^2\ge0\forall y\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}-\left(x-2\right)^2\le0\\-\left(y+1\right)^2\le0\end{matrix}\right.\)
\(\Rightarrow-\left(x-2\right)^2-\left(y+1\right)^2\le0\)
\(\Rightarrow-\left(x-2\right)^2-\left(y+1\right)^2-9\le9\)
Dấu "=" xảy ra khi:
\(\left\{{}\begin{matrix}-\left(x-2\right)^2=0\Rightarrow x-2=0\Rightarrow x=2\\-\left(y+1\right)^2=0\Rightarrow y+1=0\Rightarrow y=-1\end{matrix}\right.\)
\(\Rightarrow MAX_B=9\) khi \(x=2;y=-1\)
\(\)
a. Vì \(\left(x+2\right)^2\ge0\forall x\Rightarrow-\left(x+2\right)^2\le0\)
=> \(3- \left(x+2\right)^2\le3\)
Dấu ''='' xảy ra khi \(\left(x+2\right)^2=0\Rightarrow x=-2\)
Vậy \(A_{MAX}=3\) khi x = -2
b. Đề sai