\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+y\right)^3-3xy\left(x+y\right)=19\\8\left(x+y\right)+xy\left(x+y\right)=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+y\right)^3-3xy\left(x+y\right)=19\\24\left(x+y\right)+3xy\left(x+y\right)=6\end{matrix}\right.\)
\(\Rightarrow\left(x+y\right)^3+24\left(x+y\right)-25=0\)
\(\Leftrightarrow\left(x+y-1\right)\left[\left(x+y\right)^2+\left(x+y\right)+25\right]=0\)
\(\Leftrightarrow x+y=1\Rightarrow y=1-x\)
\(\Rightarrow x^3+\left(1-x\right)^3=19\)
\(\Leftrightarrow x^2-x-6=0\)
\(\left\{{}\begin{matrix}x^3+y^3=19\\\left(x+y\right)\left(8+xy\right)=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+y\right)^3-3xy\left(x+y\right)=19\\8\left(x+y\right)+xy\left(x+y\right)=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+y\right)^3-3xy\left(x+y\right)=19\\24\left(x+y\right)+3xy\left(x+y\right)=6\end{matrix}\right.\)
\(\rightarrow\left(x+y\right)^3+24\left(x+y\right)-25=0\)
\(\Leftrightarrow\left(x+y-1\right)[\left(x+y\right)^2+\left(x+y\right)+25]=0\)
\(\Leftrightarrow x+y=1\)
\(\Leftrightarrow y=1-x\)
\(\rightarrow x^3+\left(1-x\right)^3=19\)
\(\Leftrightarrow x^2-x-6=0\)
Từ đây bạn giải Δ hoặc bấm máy
\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
Thay x vào hệ giải ra y
