Đáp án: (x;y)=(2;0) hoặc (0;2)
Giải thích các bước giải:
$\begin{array}{l}
\left\{ \begin{array}{l}
{x^2} + {y^2} + xy = 4\\
x + y + xy = 2
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
{x^2} + {y^2} + xy + x + y + xy = 4 + 2\\
xy = 2 - \left( {x + y} \right)
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
{\left( {x + y} \right)^2} + \left( {x + y} \right) - 6 = 0\\
xy = 2 - \left( {x + y} \right)
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
{\left( {x + y} \right)^2} - 2\left( {x + y} \right) + 3\left( {x + y} \right) - 6 = 0\\
xy = 2 - \left( {x + y} \right)
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
\left( {x + y - 2} \right)\left( {x + y + 3} \right) = 0\\
xy = 2 - \left( {x + y} \right)
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x + y = 2;xy = 0\\
x + y = - 3;xy = 5\left( {vô\,nghiệm} \right)
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = 2;y = 0\\
x = 0;y = 2
\end{array} \right.
\end{array}$
