1)
a) \(x^2-4x+y^2+2y+5=0\)
\(\left(x^2-4x+4\right)+\left(y^2+2x+1\right)=0\)
\(\left(x-2\right)^2+\left(y+1\right)^2=0\)
\(\Rightarrow\left\{{}\begin{matrix}x=2\\y=-1\end{matrix}\right.\)
b) \(x^2+2y^2+2xy-2y+1=0\)
\(\left(x^2+2xy+y^2\right)+\left(y^2-2y+1\right)=0\)
\(\left(x+y\right)^2+\left(y-1\right)^2=0\)
\(\Rightarrow\left\{{}\begin{matrix}x+y=0\\y-1=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=-y\\y=1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-1\\y=1\end{matrix}\right.\)
c) \(x^2+2y^2+2xy=2y-2\)
\(x^2+2y^2+2xy-2y+2=0\)
\(\left(x^2+2xy+y^2\right)+\left(y^2-2y+1\right)+1=0\)
\(\left(x+y\right)^2+\left(y-1\right)^2+1=0\)
Ta thấy \(\left\{{}\begin{matrix}\left(x+y\right)^2\ge0\\\left(y-1\right)^2\ge0\\1>0\end{matrix}\right.\)
\(\Rightarrow\) \(\left(x+y\right)^2+\left(y-1\right)^2+1>0\)
Vậy ko có x và y thoả mãn bài toán
