\(\Leftrightarrow\left\{{}\begin{matrix}x+y+xy=5\\\left(x+y\right)^3-3xy\left(x+y\right)+\left(xy\right)^3=17\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}x+y=a\\xy=b\end{matrix}\right.\) với \(a^2\ge4b\)
\(\Rightarrow\left\{{}\begin{matrix}a+b=5\\a^3-3ab+b^3=17\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}b=5-a\\a^3-3ab+b^3=17\end{matrix}\right.\)
\(\Rightarrow a^3-3a\left(5-a\right)+\left(5-a\right)^3=17\)
\(\Leftrightarrow18a^2-90a+108=0\Rightarrow\left[{}\begin{matrix}a=2\Rightarrow b=3\left(l\right)\\a=3\Rightarrow b=2\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x+y=3\\xy=2\end{matrix}\right.\) theo Viet đảo x;y là nghiệm:
\(t^2-3t+2=0\Rightarrow\left[{}\begin{matrix}t=1\\t=2\end{matrix}\right.\) \(\Rightarrow\left(x;y\right)=\left(1;2\right);\left(2;1\right)\)
