\(P=2x^2+y^2+2xy-6x-2y+10\)
\(P=\left(x^2+y^2+1^2-2y-2x\right)+\left(x^2-4x+4\right)+5\)
\(P=\left(x+y-1\right)^2+\left(x-2\right)^2+5\)
\(\left\{{}\begin{matrix}\left(x+y-1\right)^2\ge0\\\left(x-2\right)^2\ge0\end{matrix}\right.\) \(\Rightarrow P\ge5\) đẳng thức khi \(\left\{{}\begin{matrix}x-2=0\\x+y-1=0\end{matrix}\right.\) => x=2 và y=-1
2x2 + y2 + 2xy - 6x - 2y + 10
= x2 + y2 + 12 + 2xy - 2x - 2y + x2 - 4x + 4 + 5
= (x + y - 1)2 + (x - 2)2 + 5 \(\ge\) 5
Dấu ''='' xảy ra khi \(\left\{{}\begin{matrix}x+y-1=0\\x-2=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}y=-1\\x=2\end{matrix}\right.\)
Vậy Min = 5 khi x = 2 và y = - 1
