\(D=-x^2+2xy-4y^2+2x+10y-8\)
\(-D=x^2-2xy+4y^2-2x-10y+8\)
\(-D=\left(x^2-2xy+y^2\right)+3y^2-2x-10y+8\)
\(-D=\left[\left(x-y\right)^2-2\left(x-y\right)+1\right]+3\left(y^2-4y+4\right)-5\)
\(-D=\left(x-y-1\right)^2+3\left(y-2\right)^2-5\)
Mà \(\left(x-y-1\right)^2\ge0\forall x;y\)
\(\left(y-2\right)^2\ge0\forall y\Rightarrow3\left(y-2\right)^2\ge0\forall y\)
\(\Rightarrow-D\ge-5\)
\(\Leftrightarrow D\le5\)
Dấu "=" xảy ra khi : \(\hept{\begin{cases}x-y-1=0\\y-2=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=3\\y=2\end{cases}}\)
Vậy \(D_{Max}=5\Leftrightarrow\left(x;y\right)=\left(3;2\right)\)
\(D=-x^2+2xy-4y^2+2x+10y-8\)
\(=-\left(x^2+y^2+1-2xy+2y-2x\right)-3\left(y^2-4y+4\right)+5\)
\(=-\left(x-y-1\right)^2-3\left(y-2\right)^2+5\le5\)
Vậy MAX \(D=5\)
Dấu "=" xảy ra \(\Leftrightarrow\)\(\hept{\begin{cases}x-y-1=0\\y-2=0\end{cases}}\)\(\Leftrightarrow\)\(\hept{\begin{cases}x=3\\y=2\end{cases}}\)