\(\Leftrightarrow4x^2-12xy+12y^2=12y\)
\(\Leftrightarrow\left(2x-3y\right)^2=12y-3y^2\)
Do \(\left(2x-3y\right)^2\ge0;\forall x;y\Rightarrow12y-3y^2\ge0\)
\(\Rightarrow y^2-4y+4\le4\)
\(\Rightarrow\left(y-2\right)^2\le4\)
\(\Rightarrow\left[{}\begin{matrix}\left(y-2\right)^2=0\\\left(y-2\right)^2=1\\\left(y-2\right)^2=4\end{matrix}\right.\) \(\Rightarrow y=\left\{0;1;2;3;4\right\}\)
Lần lượt thế vào pt ban đầu ta được các cặp nghiệm:
\(\left(x;y\right)=\left(0;0\right);\left(0;1\right);\left(3;1\right);\left(3;3\right);\left(6;3\right);\left(6;4\right)\)