\(4x^2-4xy+y^2+x^2+2x+1-1< 0\)
\(\Leftrightarrow\left(2x-y\right)^2+\left(x+1\right)^2< 1\)
\(\Leftrightarrow\left(x+1\right)^2< 1\Rightarrow x+1=0\Rightarrow x=-1\)
\(\Rightarrow\left(y+2\right)^2< 1\Rightarrow y+2=0\Rightarrow y=-2\)
Vậy \(\left(x;y\right)=\left(-1;-2\right)\)