Đáp án:
$\begin{array}{l}
A = \left\{ {\left( {x;{x^2}} \right)/x \in \left\{ { - 1;0;1} \right\}} \right\}\\
\Leftrightarrow {x^2} = 0;{x^2} = 1\\
\Leftrightarrow A = \left\{ { - 1;0;1} \right\}\\
B = \left\{ {\left( {x;y} \right)/{x^2} + {y^2} \le 2;x;y \in Z} \right\}\\
\Leftrightarrow \left\{ {{x^2} = {y^2} = 1} \right.\\
\Leftrightarrow B = \left\{ { - 1;1} \right\}
\end{array}$
