\(\left\{{}\begin{matrix}x+y=4\\\left(x^2+y^2\right)\left(x^3+y^3\right)=280\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+y=4\\\left[\left(x+y\right)^2-2xy\right]\left[\left(x+y\right)\left[\left(x+y\right)^2-3xy\right]\right]=280\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+y=4\\\left(16-2xy\right)4\cdot\left(16-3xy\right)=280\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x+y=4\\\left(16-2xy\right)\left(16-3xy\right)=70\left(1\right)\end{matrix}\right.\)
đặt xy = a (4a ≤16)
pt (1) trở thành: \(\left(16-2a\right)\left(16-3a\right)=70\)
\(\Leftrightarrow256-48a-32a+6a^2=70\)
\(\Leftrightarrow6a^2-80a+186=0\Leftrightarrow\left[{}\begin{matrix}a=3\left(tm\right)\\a=\frac{31}{3}\left(loai\right)\end{matrix}\right.\)
với a = 3 ta có: \(\left\{{}\begin{matrix}x+y=4\\xy=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=4-y\\y^2-4y+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=3\\y=1\end{matrix}\right.\\\left\{{}\begin{matrix}x=1\\y=3\end{matrix}\right.\end{matrix}\right.\)
vậy........
