$\begin{cases}x^2y-5xy^2+xy=0 \, (1)\\x-2y+xy=6 \, (2) \end{cases}$
$(1) \Leftrightarrow xy(x-5y+1)=0$
$\Leftrightarrow$ \(\left[ \begin{array}{l}x=0\\y=0\\x-5y+1=0\end{array} \right.\)
$\Rightarrow$ \(\left[ \begin{array}{l}y=-3\\x=6\\x=5y-1 \,(*)\end{array} \right.\)
Thay $(*)$ vào $(2)$ ta được:
$(5y - 1) -2y + (5y - 1)y = 6\\\Leftrightarrow5y^2+2y-7=0$
$\Leftrightarrow$ \(\left[ \begin{array}{l}y=1\\y=-\dfrac{7}{5}\end{array} \right.\)
$\Rightarrow$ \(\left[ \begin{array}{l}x=4\\x=-8\end{array} \right.\)
Vậy hệ có các cặp nghiệm $(x;y) = \left\{(0;-3),(6;0),(4;1),(-8;-\dfrac{7}{5})\right\}$
