\(\left\{{}\begin{matrix}\left(x^2-x\right)\left(2y-y^2\right)=20\\x^2-x+y^2-2y=19\end{matrix}\right.\).
Đặt \(a=x^2-x,b=y^2-2y\), ta có hệ:
\(\left\{{}\begin{matrix}-ab=20\\a+b=19\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}ab=-20\\b=19-a\end{matrix}\right.\)\(\Rightarrow a\left(19-a\right)=-20\)\(\Leftrightarrow-a^2+19a+20=0\)\(\Leftrightarrow\left[{}\begin{matrix}a=20\\a=-1\end{matrix}\right.\).
Với a = 20 suy ra b = 19 - 20 = -1.
Ta có \(\left\{{}\begin{matrix}x^2-x=20\\y^2-2y=-1\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x^2-x-20=0\\y^2-2y+1=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+4\right)\left(x-5\right)=0\\\left(y-1\right)^2=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x=-4\\x=5\end{matrix}\right.\\y=1\end{matrix}\right.\).
Ta có hai cặp nghiệm \(\left(x,y\right)=\left(-4,1\right);\left(x,y\right)=\left(5,1\right)\).
Với a = -1 suy ra \(x^2-x=-1\Leftrightarrow x^2-x+1=0\) (vô nghiệm).
Vậy hệ phương trình có hai cặp nghiệm \(\left(x,y\right)=\left(-4,1\right);\left(x,y\right)=\left(5,1\right)\).