\(Q=x^2+2y^2+2xy-2x-6y+2017\\ Q=x^2+y^2+y^2+xy+xy-x-x-y-y-4y+1+4+2012\\ Q=\left(x^2+xy-x\right)+\left(y^2+xy-y\right)-\left(x+y-1\right)+\left(y^2-4y+4\right)+2012\\ Q=x\left(x+y-1\right)+y\left(x+y-1\right)-\left(x+y-1\right)+\left(y^2-2\cdot y\cdot2+2^2\right)+2012\\ Q=\left(x+y-1\right)\left(x+y-1\right)+\left(y-2\right)^2+2012\\ Q=\left(x+y-1\right)^2+\left(y-2\right)^2+2012\\ Do\left(x+y-1\right)^2\ge0\forall x;y\\ \left(y-2\right)^2\ge0\forall y\\ \Rightarrow\left(x+y-1\right)^2+\left(y-2\right)^2\ge0\forall x;y\\ \Rightarrow Q=\left(x+y-1\right)^2+\left(y-2\right)^2+2012\ge2012\forall x;y\\ \text{Dấu “=” xảy ra khi :}\left\{{}\begin{matrix}\left(x+y-1\right)^2=0\\\left(y-2\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+y-1=0\\y-2=0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x+y=1\\y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+2=1\\y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=2\end{matrix}\right.\\ \text{Vậy }Q_{\left(Min\right)}=2012\text{ }khi\text{ }x=-1;y=2\)
\(Q=\left(x^2+y^2+1^2+2xy-2y-2x\right)+\left(y^2-4y+4\right)+2012\)
\(Q=\left(x+y-1\right)^2+\left(y-2\right)^2+2012\ge2012\)
Dấu \("="\) xảy ra khi y-2=0 và x+y-1=0
=>y=2
Ta có y=2=>x=-1
Vậy GTNN của Q là 2012 khi \(x=-1;y=2\)
Q = x2 + 2y2 + 2xy - 2x - 6y + 2017
=x2+xy-x+y2+xy-y-x-y+1+y2-4y+4+2012
=(x2+xy-x)+(y2+xy-y)-(y+x-1)+(y2-4y+4)+2012
=x(x+y-1)+y(y+x-1)-(y+x-1)+(y-2)2+2012
=(x+y-1)(x+y-1)+(y-2)2+2012
Q=(x+y-1)2+(y-2)2+2012
do (x+y-1)2\(\ge0\forall x;y\)
(y-2)2\(\ge0\forall y\)
=>(x+y-1)2+(y-2)2\(\ge0\)
=>(x+y-1)2+(y-2)2+2012\(\ge2012\)
=>MinQ =2012 khi :
x+y-1=0
=>x+y=1(1)
y-2=0
=>y=2 (2)
thay (2)vào(1) ta có
x+2=1
=>x=-3
