\(x^2-4xy+5y^2+2y+1=0\)
\(\Rightarrow\left(x^2-4xy+4y^2\right)+\left(y^2+2y+1\right)=0\)
\(\Rightarrow\left(x+2y\right)^2+\left(y+1\right)^2=0\)
\(\Rightarrow\left[{}\begin{matrix}x+2y=0\\y+1=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=2\\y=-1\end{matrix}\right.\)
Vậy \(\left(x;y\right)=\left(2;-1\right)\)