\(A=-\left(4x^2-4xy+y^2\right)-\left(y^2-2y+1\right)+4\)
\(A=4-\left(2x-y\right)^2-\left(y-1\right)^2\le4\)
\(A_{max}=4\) khi \(\hept{\begin{cases}x=\frac{1}{2}\\y=1\end{cases}}\)
Chúc bạn học tốt !!!
\(-4x^2+4xy-2y^2+2y+3\)
\(=-\left(4x^2+4xy+y^2\right)-\left(y^2-2y+1\right)+4\)
\(=-\left(2x+y\right)^2-\left(y-1\right)^2+4\)
Ta có \(\left(2x+y\right)^2\ge0\) \(\forall x,y\) \(;\left(y-1\right)^2\ge0\) \(\forall y\)
=> \(\left(2x+y\right)^2+\left(y-1\right)^2\ge0\) \(\forall x,y\)
=> \(-\left(2x+y\right)^2-\left(y-1\right)^2\le0\) \(\forall x,y\)
=> \(-\left(2x+y\right)-\left(y-1\right)^2+4\le4\) \(\forall x,y\)
\(MaxA=4\Leftrightarrow\hept{\begin{cases}\left(y-1\right)^2=0\\\left(2x+y\right)^2=0\end{cases}}\) \(\Leftrightarrow\hept{\begin{cases}y-1=0\\2x+y=0\end{cases}\Leftrightarrow\hept{\begin{cases}y=1\\x=-\frac{1}{2}\end{cases}}}\)
