\(\left(x+4\right)\left(2-y\right)=-3\)
\(\Rightarrow-3=1.\left(-3\right)=\left(-3\right).1=\left(-1\right).3=3.\left(-1\right)\)
\(\Rightarrow\left[\begin{matrix}x+4=1\Rightarrow x=1-4=-3\\2-y=-3\Rightarrow y=2-\left(-3\right)=5\end{matrix}\right.\)
\(\Rightarrow\left[\begin{matrix}x+4=-3\Rightarrow x=-3-4=-7\\2-y=1\Rightarrow y=2-1=1\end{matrix}\right.\)
\(\Rightarrow\left[\begin{matrix}x+4=-1\Rightarrow x=-1-4=-5\\2-y=3\Rightarrow y=2-3=-1\end{matrix}\right.\)
\(\Rightarrow\left[\begin{matrix}x+4=3\Rightarrow x=3-4=-1\\2-y=-1\Rightarrow y=2-\left(-1\right)=3\end{matrix}\right.\)
\(\Rightarrow x=\left\{-7;-5;-3;-1\right\};y=\left\{\pm1;3;5\right\}\)
b) \(xy-x=19\)
\(\Rightarrow x\left(y-1\right)=19\)
\(\Rightarrow x\inƯ\left(19\right);y-1\inƯ\left(19\right)\)
Ta có: \(Ư\left(19\right)=\left\{\pm1;\pm19\right\}\)
\(\Rightarrow x\in\left\{\pm1;\pm19\right\}\)
\(y-1\in\left\{\pm1;\pm19\right\}\)
Xét các TH sau:
_ Nếu \(x=1\) thì \(y-1=19\)
\(y-1=19\Rightarrow y=20\)
\(x=1\)
_ Nếu \(x=-1\) thì \(y-1=-19\)
\(y-1=-19\Rightarrow y=-18\)
\(x=-1\)
_ Nếu \(x=19\) thì \(y-1=1\)
\(y-1=1\Rightarrow y=2\)
\(x=19\)
_ Nếu \(x=-19\) thì \(y-1=-1\)
\(y-1=-1\) \(\Rightarrow y=0\)
\(x=-19\)
Vậy ta tìm đc các cặp số sau: \(x=1\) và \(y=20\); \(x=-1\) và \(y=-18\); \(x=19\) và \(y=2\);
\(x=-19\) và \(y=0.\)
