Ta có:
\(\left|x\right|\ge0;y^2\ge0\) mà \(\left|x\right|+y^2=2\)
TH1: \(\left|x\right|=0\Rightarrow x=0\Rightarrow y^2=2\Rightarrow y=\sqrt{2}\), vô lý
TH2:\(\left|x\right|=1\Rightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\Rightarrow y^2=1\Rightarrow\left[{}\begin{matrix}y=1\\y=-1\end{matrix}\right.\)thỏa mãn
TH3: \(\left|x\right|=2\Rightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\Rightarrow y^2=0\Rightarrow y=0\) thỏa mãn
Vậy (x; y) ∈{(1; 1); (1;-1); (-1; 1); (-1; -1); (2;0); (-2;0)} thỏa mãn đề ra
