hpt <=>\(\left\{{}\begin{matrix}x^2+y^2+xy=7\\\left(x^2+y^2\right)^2-x^2y^2=21\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}x^2+y^2=u\\xy=v\end{matrix}\right.\)
Có hệ \(\left\{{}\begin{matrix}u+v=7\\u^2-v^2=21\end{matrix}\right.\)
<=>\(\left\{{}\begin{matrix}u+v=7\\\left(u+v\right)\left(u+v\right)=21\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}u+v=7\\u-v=3\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}u=5\\v=2\end{matrix}\right.\) <=>\(\left\{{}\begin{matrix}x^2+y^2=5\\xy=2\end{matrix}\right.\)
<=> \(\left\{{}\begin{matrix}\left(x+y\right)^2-2xy=5\\xy=2\end{matrix}\right.\) <=>\(\left\{{}\begin{matrix}\left(x+y\right)^2=5+2.2=9\\xy=2\end{matrix}\right.\)
<=>\(\left\{{}\begin{matrix}\left[{}\begin{matrix}x+y=3\\x+y=-3\end{matrix}\right.\\xy=2\end{matrix}\right.\)
TH1: \(\left\{{}\begin{matrix}x+y=3\\xy=2\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}x=3-y\\\left(3-y\right)y=2\end{matrix}\right.\) <=>\(\left\{{}\begin{matrix}x=3-y\\y^2-3y+2=0\end{matrix}\right.\)
<=>\(\left\{{}\begin{matrix}x=3-y\\\left(y-2\right)\left(y-1\right)=0\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}x=3-y\\\left[{}\begin{matrix}y=2\\y=1\end{matrix}\right.\end{matrix}\right.\)<=> \(\left[{}\begin{matrix}\left\{{}\begin{matrix}x=1\\y=2\end{matrix}\right.\\\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\end{matrix}\right.\)
TH2: \(\left\{{}\begin{matrix}x+y=-3\\xy=2\end{matrix}\right.\) <=>\(\left\{{}\begin{matrix}x=-2-y\\\left(-2-y\right)y=2\end{matrix}\right.\) <=>\(\left\{{}\begin{matrix}x=-2-y\\y^2+2y+2=0\end{matrix}\right.\)
<=> \(\left\{{}\begin{matrix}x=-2-y\\\left(y+1\right)^2+1=0\end{matrix}\right.\)(vô nghiệm)
Vậy hpt có hai nghiệm duy nhất (1,2),(2,1)
