Ta có:\(\left\{{}\begin{matrix}\left(x^2-2x\right)\left(y^2-2y\right)=45\\\left(y-1\right)\left(x-1\right)=8\end{matrix}\right.\)
<=>\(\left\{{}\begin{matrix}x^2y^2-2xy^2-2x^2y+4xy=45\\xy-x-y+1=8\end{matrix}\right.\)
* xy = 7 + x + y
Thế vào phương trình ở trên ta được:
(x+y+7)2 - 2(x+y+7)y - 2(x+y+7)x + 4(x+y+7) = 45
(x+y+7)2 - 2(x+y+7)(x+y) + 4(x+y+7) = 45
(x+y+7)[(x+y+7)-2(x+y)+4] = 45
(x+y+7)(x+y+7-2x-2y+4) = 45
(x+y+7)(-x-y+11)=45
Ta có: 45=1*45=3*15=5*9
\(\left[{}\begin{matrix}\left\{{}\begin{matrix}x+y+7=1\\11-x-y=45\end{matrix}\right.\\\left\{{}\begin{matrix}x+y+7=45\\11-x-y=1\end{matrix}\right.\end{matrix}\right.\)
\(\left[{}\begin{matrix}\left\{{}\begin{matrix}x+y+7=3\\11-x-y=15\end{matrix}\right.\\\left\{{}\begin{matrix}x+y+7=15\\11-x-y=3\end{matrix}\right.\end{matrix}\right.\)
\(\left[{}\begin{matrix}\left\{{}\begin{matrix}x+y+7=5\\11-x-y=9\end{matrix}\right.\\\left\{{}\begin{matrix}x+y+7=9\\11-x-y=5\end{matrix}\right.\end{matrix}\right.\)
