Theo đề bài : 2x2 + 2y2 + 2xy - 2x + 2y + 2 = 0
\(\Rightarrow\) ( x2 + 2xy + y2 ) + ( x2 - 2x + 1 ) + ( y2 + 2y + 1 ) = 0
( x + y )2 + ( x - 1 )2 + ( y + 1 )2 = 0
Ta thấy : \(\left(x+y\right)^2\ge0;\forall x,y\in R\)
\(\left(x-1\right)\ge0;\forall x\in R\)
\(\left(y+1\right)^2\ge0;\forall y\in R\)
\(\Rightarrow\left(x+y\right)^2+\left(x-1\right)^2+\left(y+1\right)^2\ge0;\forall x,y\in R\)
Dấu " = " xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}\left(x+y\right)^2=0\\\left(x-1\right)^2=0\\\left(y+1\right)^2=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x+y=0\\x-1=0\\y+1=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=-y\\x=1\\y=-1\end{matrix}\right.\left(\text{Thỏa mãn}\right)\)
Thay \(x=1\) và \(y=-1\) vào \(A=\left(x-2\right)^{2017}+\left(y+1\right)^{2018}\) , ta được :
\(A=\left(x-2\right)^{2017}+\left(y+1\right)^{2018}\)
\(A=\left(1-2\right)^{2017}+\left(-1+1\right)^{2018}\)
\(A=-1+0\)
\(A=-1\)
Vậy \(A=-1\Leftrightarrow\left\{{}\begin{matrix}2x^2+2y^2+2xy-2x+2y+2=0\\x=1\\y=-1\end{matrix}\right.\)
