\(\left(2x+2\right)\left(y-3\right)=11\)
\(\Leftrightarrow2x+2;y-3\inƯ\left(11\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}2x+2=1\\y-3=11\end{matrix}\right.\\\left\{{}\begin{matrix}2x+2=-1\\y-3=-11\end{matrix}\right.\\\left\{{}\begin{matrix}2x+2=11\\y-3=1\end{matrix}\right.\\\left\{{}\begin{matrix}2x+2=-11\\y-3=-1\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=-\dfrac{1}{2}\\y=14\end{matrix}\right.\\\left\{{}\begin{matrix}x=-2\\y=-8\end{matrix}\right.\\\left\{{}\begin{matrix}x=\dfrac{9}{2}\\y=4\end{matrix}\right.\\\left\{{}\begin{matrix}x=-\dfrac{13}{2}\\y=2\end{matrix}\right.\end{matrix}\right.\)
Vậy ...
(2x + 2) . (y - 3) = 11
Ta có : 11 = 11.1= 1.11=(-1).(-11)=(-11).(-1)
