\[\begin{array}{l}
\left\{ \begin{array}{l}
x + y + {x^2} + {y^2} = 8\\
xy\left( {x + 1} \right)\left( {y + 1} \right) = 12
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x\left( {x + 1} \right) + y\left( {y + 1} \right) = 8\\
x\left( {x + 1} \right).y\left( {y + 1} \right) = 12
\end{array} \right.\,\,\,\left( I \right)\\
Dat\,\,\left\{ \begin{array}{l}
x\left( {x + 1} \right) = u\\
y\left( {y + 1} \right) = v
\end{array} \right.\,\,\,\left( {{u^2} \ge 4v} \right)\\
\Rightarrow \left( I \right) \Leftrightarrow \left\{ \begin{array}{l}
u + v = 8\\
uv = 12
\end{array} \right.\\
\Rightarrow u,\,\,v\,\,\,la\,\,\,2\,\,nghiem\,\,\,cua\,\,pt:\,\,\,{t^2} - 8t + 12 = 0\\
\Leftrightarrow \left[ \begin{array}{l}
t = 6\\
t = 2
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
u = 6\\
v = 2
\end{array} \right.\\
\left\{ \begin{array}{l}
u = 2\\
v = 6
\end{array} \right.
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x\left( {x + 1} \right) = 6\\
y\left( {y + 1} \right) = 2
\end{array} \right.\\
\left\{ \begin{array}{l}
x\left( {x + 1} \right) = 2\\
y\left( {y + 1} \right) = 6
\end{array} \right.
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
{x^2} + x - 6 = 0\\
{y^2} + y - 2 = 0
\end{array} \right.\\
\left\{ \begin{array}{l}
{x^2} + x - 2 = 0\\
{y^2} + y - 6 = 0
\end{array} \right.
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
\left[ \begin{array}{l}
x = 2\\
x = - 3
\end{array} \right.\\
\left[ \begin{array}{l}
y = 1\\
y = - 2
\end{array} \right.
\end{array} \right.\\
\left\{ \begin{array}{l}
\left[ \begin{array}{l}
x = 1\\
x = - 2
\end{array} \right.\\
\left[ \begin{array}{l}
y = 2\\
y = - 3
\end{array} \right.
\end{array} \right.
\end{array} \right.
\end{array}\]
Em kết luận nghiệm nhé.