Đặt \(A=-x^2-3y^2-2xy+10x+14y-18\)
Ta có : \(-A=x^2+3y^2+2xy-10x-14y+18\)
\(-A=\left(x^2+2xy+y^2\right)+2y^2-10x-14y+18\)
\(-A=\left[\left(x+y\right)^2-2\left(x+y\right)\times5+25\right]+2y^2-4y+7\)
\(-A=\left(x+y-5\right)^2+2\left(y^2-2y+1\right)+5\)
\(-A=\left(x+y-5\right)^2+2\left(y-1\right)^2+5\)
Mà \(\left(x+y-5\right)^2\ge0\forall x;y\in R\)
\(\left(y-1\right)^2\ge0\forall y\in R\Rightarrow2\left(y-1\right)^2\ge0\forall y\in R\)
\(\Rightarrow-A\ge5\)
\(\Leftrightarrow A\le-5\)
Dấu " = " xảy ra khi:
\(\hept{\begin{cases}x+y-5=0\\y-1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=4\\y=1\end{cases}}\)
Vậy Max A = - 5 khi ( x ; y ) = ( 4 ; 1 )
