\(\left\{{}\begin{matrix}\left(x^2+1\right)\left(y^2+1\right)=10\\\left(x+y\right)\left(xy-1\right)=3\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x^2y^2+x^2+y^2+1=10\\\left(x+y\right)\left(xy-1\right)=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2y^2+\left(x+y\right)^2-2xy+1=10\\\left(x+y\right)\left(xy-1\right)=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(xy-1\right)^2+\left(x+y\right)^2=10\\\left(x+y\right)\left(xy-1\right)=3\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}x+y=a\\xy-1=b\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a^2+b^2=10\\ab=3\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a^2+b^2=10\\2ab=6\end{matrix}\right.\)
\(\Rightarrow\left(a+b\right)^2=16\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}a+b=4\\ab=3\end{matrix}\right.\\\left\{{}\begin{matrix}a+b=-4\\ab=3\end{matrix}\right.\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\left(a;b\right)=\left(1;3\right);\left(3;1\right)\\\left(a;b\right)=\left(-1;-3\right);\left(-3;-1\right)\end{matrix}\right.\)
\(\Rightarrow...\)
Đến đây bạn tự thế vào và giải theo Viet đảo, dạng cơ bản rồi, 4 trường hợp nên làm biếng
