\(P=x^2+2y^2+2xy-6x-8y+2027\\ =\left(x^2+y^2+9+2xy-6x-6x\right)+\left(y^2-2y+1\right)+2017\\ =\left(x+y-3\right)^2+\left(y-1\right)^2+2017\)
Do \(\left(x+y-3\right)^2\ge0\forall x;y\)
\(\left(y-1\right)^2\ge0\forall x;y\)
\(\Rightarrow\left(x+y-3\right)^2+\left(y-1\right)^2\ge0\forall x;y\\ \Rightarrow\left(x+y-3\right)^2+\left(y-1\right)^2+2017\ge2017\forall x;y\)
Dấu "=" xảy ra khi: \(\left\{{}\begin{matrix}\left(x+y-3\right)^2=0\\\left(y-1\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+y-3=0\\y-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=3-y\\y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\)
Vậy \(P_{\left(Min\right)}=2017\) khi \(\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\)
P = x2+2y2 +2xy-6x-8y+2027
=x2+2xy+y2+y2-6x-6y-2y+1+9+2017
=(x2+2xy+y2)-(6x+6y)+9+(y2-2y+1)+2017
=(x+y)2-6(x+y)+9+(y-1)2+2017
=[(x+y)2-6(x+y)+9]+(y-1)2 +2017
=(x+y-3)2+(y-1)2+2017
Do (x+y-3)2 \(\ge0\forall x\)
(y-1)2 \(\ge0\forall x\)
=>\(\left(x+y-3\right)^2+\left(y-1\right)^2\ge0\)
=>\(\left(x+y-3\right)^2+\left(y-1\right)^2+2017\ge2017\)=> P\(\ge2017\)
Min P=2017 khi
y-1=0
=> y=1
x+y-3=0
=>x+1-3=0
=> x=2
Vậy GTNN của P=2017 khi y=1 và x=2
