Em tham khảo:
$x^{2}+y^2+4x-4y=0$
⇔$x^{2}+x+4+y^2-4x+4-8=0$
⇔$(x+2)^{2}+(y-2)^2=2^2+2^2 (x;y∈Z)$
⇔$\left \{ {{(x+2)^2=2^2} \atop {(y-2)^=2^2}} \right.$
⇔$\left \{ {{\left[ \begin{array}{l}x=-4\\x=0\end{array} \right.} \atop {\left[ \begin{array}{l}y=4\\x=0\end{array} \right.}} \right.$
Học tốt