Đáp án:
$\begin{array}{l}
x + {y^2} + xy = 1\\
\Leftrightarrow x + {y^2} + xy - 1 = 0\\
\Leftrightarrow x\left( {y + 1} \right) + \left( {y + 1} \right)\left( {y - 1} \right) = 0\\
\Leftrightarrow \left( {y + 1} \right)\left( {x + y - 1} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
y = - 1\\
x = 1 - y
\end{array} \right.\\
Thay\,vào\,pt\,(1)\\
\Rightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
y = - 1\\
{x^2} + 2 = 3
\end{array} \right.\\
\left\{ \begin{array}{l}
x = 1 - y\\
{\left( {1 - y} \right)^2} + 2{y^2} = 3
\end{array} \right.
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
y = - 1\\
x = \pm 1
\end{array} \right.\\
\left\{ \begin{array}{l}
x = 1 - y\\
3{y^2} - 2y - 2 = 0
\end{array} \right.
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
y = - 1\\
x = \pm 1
\end{array} \right.\\
\left\{ \begin{array}{l}
x = 1 - y\\
\left[ {y = \frac{{1 \pm \sqrt 7 }}{3}} \right.
\end{array} \right.
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
y = - 1\\
x = \pm 1
\end{array} \right.\\
\left\{ \begin{array}{l}
x = \frac{{2 \mp \sqrt 7 }}{3}\\
y = \frac{{1 \pm \sqrt 7 }}{3}
\end{array} \right.
\end{array} \right.\\
\Rightarrow \left( {x;y} \right) = \left\{ {\left( {1; - 1} \right);\left( { - 1; - 1} \right);\left( {\frac{{2 \mp \sqrt 7 }}{3};\frac{{1 \pm \sqrt 7 }}{3}} \right)} \right\}
\end{array}$
