\(3x+4y-xy=16\)
\(\Leftrightarrow3x+4y-xy-12=16-12\)
\(\Leftrightarrow y\left(4-x\right)-12+3x=4\)
\(\Leftrightarrow y\left(4-x\right)-3\left(4-x\right)=4\)
\(\Leftrightarrow\left(4-x\right)\left(y-3\right)=4\)
\(\Leftrightarrow4-x;y-3\inƯ\left(4\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}4-x=1\\y-3=4\end{matrix}\right.\\\left\{{}\begin{matrix}4-x=2\\y-3=2\end{matrix}\right.\\\left\{{}\begin{matrix}4-x=4\\y-3=1\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=3\\y=7\end{matrix}\right.\\\left\{{}\begin{matrix}x=2\\y=5\end{matrix}\right.\\\left\{{}\begin{matrix}x=0\\y=4\end{matrix}\right.\end{matrix}\right.\)
Vậy ..
\(3x+4y-xy=16\)
\(\Leftrightarrow3x-xy+4y-12=4\)
\(\Leftrightarrow x\left(3-y\right)+4\left(y-3\right)=4\)
\(\Leftrightarrow\left(x-4\right)\left(3-y\right)=4\)
Xảy ra các trường hợp như sau :
TH1 : \(\left\{{}\begin{matrix}x-4=1\\3-y=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=5\\y=1\end{matrix}\right.\)
TH2 : \(\left\{{}\begin{matrix}x-4=-1\\3-y=-4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=7\end{matrix}\right.\)
TH3 : \(\left\{{}\begin{matrix}x-4=4\\3-y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=8\\y=2\end{matrix}\right.\)
TH4 : \(\left\{{}\begin{matrix}x-4=-4\\3-y=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=4\end{matrix}\right.\)
TH5 : \(\left\{{}\begin{matrix}x-4=2\\3-y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=6\\y=1\end{matrix}\right.\)
TH6 : \(\left\{{}\begin{matrix}x-4=-2\\3-y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=5\end{matrix}\right.\)
Vậy ta tìm được các cặp x , y .
