Đáp án: $\left( {x;y} \right) = \left\{ {\left( {1;2} \right);\left( {2;1} \right);\left( { - 1; - 2} \right);\left( { - 2; - 1} \right)} \right\}$
Giải thích các bước giải:
$\begin{array}{l}
\left\{ \begin{array}{l}
{x^2} + xy + {y^2} = 7\\
{x^4} + {y^4} + {x^2}{y^2} = 21
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
{x^2} + {y^2} = 7 - xy\\
\left( {{x^4} + 2{x^2}{y^2} + {y^4}} \right) - {x^2}{y^2} = 21
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
{x^2} + 2xy + {y^2} - xy = 7\\
{\left( {{x^2} + {y^2}} \right)^2} - {x^2}{y^2} = 21
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
{\left( {x + y} \right)^2} - xy = 7\\
{\left( {7 - xy} \right)^2} - {x^2}{y^2} = 21
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
{\left( {x + y} \right)^2} = 7 + xy\\
{x^2}{y^2} - 14xy + 49 - {x^2}{y^2} = 21
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
{\left( {x + y} \right)^2} = 7 + xy\\
xy = 2
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
{\left( {x + y} \right)^2} = 9\\
xy = 2
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
xy = 2;x + y = 3\\
xy = 2;x + y = - 3
\end{array} \right.\\
\Rightarrow \left( {x;y} \right) = \left\{ {\left( {1;2} \right);\left( {2;1} \right);\left( { - 1; - 2} \right);\left( { - 2; - 1} \right)} \right\}
\end{array}$
