\(\left\{{}\begin{matrix}x+y+xy=5\\\left(x+y\right)^2-xy=7\end{matrix}\right.\) \(\Rightarrow\) đặt \(\left\{{}\begin{matrix}x+y=a\\x.y=b\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}a+b=5\\a^2-b=7\end{matrix}\right.\) \(\Rightarrow a^2+a-12=0\Rightarrow\left[{}\begin{matrix}a=-4\Rightarrow b=9\\a=3\Rightarrow b=2\end{matrix}\right.\)
TH1: \(\left\{{}\begin{matrix}a=-4\\b=9\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x+y=-4\Rightarrow y=-x-4\\xy=9\end{matrix}\right.\) \(\Rightarrow x\left(-x-4\right)-9=0\)
\(\Rightarrow x^2+4x+9=0\) \(\Rightarrow\) vô nghiệm
TH2: \(\left\{{}\begin{matrix}a=3\\b=2\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x+y=3\Rightarrow y=3-x\\x.y=2\end{matrix}\right.\) \(\Rightarrow x\left(3-x\right)-2=0\Rightarrow-x^2+3x-2=0\)
\(\Rightarrow\left[{}\begin{matrix}x=1\Rightarrow y=2\\x=2\Rightarrow y=1\end{matrix}\right.\)
Vậy pt có 2 cặp nghiệm \(\left(x;y\right)=\left(1;2\right);\left(2;1\right)\)