\(x^2+2y^2+2xy-6x-8y+2018\)
\(=x^2+y^2+9+2xy-6x-6y+y^2-2y+1+2008\)
\(=\left(3-x-y\right)^2+\left(y-1\right)^2+2008\) \(\ge2008\)
Dấu '=' xảy ra \(\Leftrightarrow\)\(\hept{\begin{cases}3-x-y=0\\y-1=0\end{cases}}\) \(\Leftrightarrow\)\(\hept{\begin{cases}x=2\\y=1\end{cases}}\)
Vậy Min P = 2008 <=> x=2; y=1
\(p=\left(x^2+2xy+y^2\right)-\left(6x+6y\right)+9+\left(y^2-2y+1\right)+2008\)
\(=\left(x+y\right)^2-6\left(x+y\right)+9+\left(y-1\right)^2+2008\)
\(=\left(x+y-3\right)^2+\left(y-1\right)^2+2008\)\(\ge2008\)với \(\forall x,y\)
Dấu "=" xảy ra khi y = 1; x = 2
\(P=x^2+2y^2+2xy-6x-8y+2018\)
\(P=\left(x^2+2xy+y^2\right)+y^2-6x-8y+2018\)
\(P=\left[\left(x+y\right)^2-2\left(x+y\right)\times3+9\right]+\left(y^2-2y+1\right)+2008\)
\(P=\left(x+y-3\right)^2+\left(y-1\right)^2+2008\)
Mà \(\left(x+y-3\right)^2\ge0\)
\(\left(y-1\right)^2\ge0\)
\(\Rightarrow P\ge2008\)
Dấu " = " xảy ra khi :
\(\hept{\begin{cases}x+y-3=0\\y-1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=2\\y=1\end{cases}}\)
Vậy Min P = 2008 khi ( x;y ) = ( 2;1 )
