\(x^2+2y^2+2xy-2y+1=0\)
=> \(x^2+y^2+y^2+2xy-2y+1=0\)
=> \((x^2+2xy+y^2)+(y^2-2y+1)=0\)
=> \(\left(x+y\right)^2+\left(y-1\right)^2=0\)
Ta thấy:
\(\left(x+y\right)^2\ge0\)
\(\left(y-1\right)^2\ge0\)
=> \(\left(x+y\right)^2+\left(y-1\right)^2\ge0\)
Mà \(\left(x+y\right)^2+\left(y-1\right)^2=0\)
=> \(\left\{{}\begin{matrix}\left(x+y\right)^2=0\\\left(y-1\right)^2=0\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x+y=0\\y-1=0\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x=-1\\y=1\end{matrix}\right.\)
Vậy x = -1; y =1
\(x^2+2y^2+2xy-2y+1=0\)
\(\Leftrightarrow\left(x^2+2xy+y^2\right)+\left(y^2-2y+1\right)=0\)
\(\Leftrightarrow\left(x+y\right)^2+\left(y-1\right)^2=0\)
Dễ thấy: \(\left\{{}\begin{matrix}\left(x+y\right)^2\ge0\\\left(y-1\right)^2\ge0\end{matrix}\right.\)
\(\Rightarrow\left(x+y\right)^2+\left(y-1\right)^2\ge0\)
Xảy ra khi \(\left\{{}\begin{matrix}\left(x+y\right)^2=0\\\left(y-1\right)^2=0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=-1\\y=1\end{matrix}\right.\)
