\(\Leftrightarrow\left(x^2-2xy+y^2\right)+\left(4x^2-4x+1\right)+\left(y^2-2y+1\right)^2\le2\)
\(\Leftrightarrow\left(x-y\right)^2+\left(2x-1\right)^2+\left(y-1\right)^2\le2\)
\(\Rightarrow\left(2x-1\right)^2\le2\)
Mà 2x-1 luôn lẻ \(\Rightarrow\left(2x-1\right)^2=1\Rightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
- Với \(x=0\Rightarrow2y^2\le2y\Rightarrow y\left(y-1\right)\le0\)
\(\Rightarrow0\le y\le1\Rightarrow y=\left\{0;1\right\}\)
- Với \(x=1\Rightarrow2y^2+5\le4y+4\Rightarrow2y^2-4y+1\le0\)
\(\Rightarrow2\left(y-1\right)^2\le1\)
\(\Rightarrow\left(y-1\right)^2\le\dfrac{1}{2}\)
\(\Rightarrow\left(y-1\right)^2=0\)
\(\Rightarrow y=1\)
Vậy \(\left(x;y\right)=\left(0;0\right);\left(0;1\right);\left(1;1\right)\)
