Giải giúp tớ hệ pt này với x+y+x ² +y ²=8 xy(x+1)(y+1)=12

$$\eqalign{ & \left\{ \matrix{ x + y + {x^2} + {y^2} = 8 \hfill \cr xy\left( {x + 1} \right)\left( {y + 1} \right) = 12 \hfill \cr} \right. \cr & \Leftrightarrow \left\{ \matrix{ x\left( {x + 1} \right) + y\left( {y + 1} \right) = 8 \hfill \cr x\left( {x + 1} \right)y\left( {y + 1} \right) = 12 \hfill \cr} \right. \cr & \Rightarrow x\left( {x + 1} \right)\,\,va\,\,y\left( {y + 1} \right)\,\,la\,\,nghiem\,\,cua\,\,pt \cr & {X^2} - 8X + 12 = 0 \Leftrightarrow \left[ \matrix{ X = 6 \hfill \cr X = 2 \hfill \cr} \right. \cr & TH1:\,\,\left\{ \matrix{ x\left( {x + 1} \right) = 6 \hfill \cr y\left( {y + 1} \right) = 2 \hfill \cr} \right. \Leftrightarrow \left\{ \matrix{ \left[ \matrix{ x = 2 \hfill \cr x = - 3 \hfill \cr} \right. \hfill \cr \left[ \matrix{ y = 1 \hfill \cr y = - 2 \hfill \cr} \right. \hfill \cr} \right. \cr & TH2:\,\,\left\{ \matrix{ x\left( {x + 1} \right) = 2 \hfill \cr y\left( {y + 1} \right) = 6 \hfill \cr} \right. \Leftrightarrow \left\{ \matrix{ \left[ \matrix{ x = 1 \hfill \cr x = - 2 \hfill \cr} \right. \hfill \cr \left[ \matrix{ y = 2 \hfill \cr y = - 3 \hfill \cr} \right. \hfill \cr} \right. \cr & \Rightarrow S = \left\{ {\left( {2;1} \right);\left( {2; - 2} \right);\left( { - 3;1} \right);\left( { - 3; - 2} \right);\left( {1;2} \right);\left( {1; - 3} \right);\left( { - 2;2} \right);\left( { - 2; - 3} \right)} \right\} \cr} $$

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