Đáp án:
`x = -1; y = 1`
Giải thích các bước giải:
\(x^2+2y^2+2xy-2y+1=0\)
\(\Leftrightarrow\left(x+y\right)^2+\left(y-1\right)^2=0\)
\(\left\{{}\begin{matrix}\left(x+y\right)^2\ge0\\\left(y-1\right)^2\ge0\end{matrix}\right.\Leftrightarrow\left(x+y\right)^2+\left(y-1\right)^2\ge0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+y\right)^2=0\\\left(y-1\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=1\end{matrix}\right.\)
Vậy `x = -1; y = 1`