\(xy+4x+y=3\)
\(\Leftrightarrow x\left(y+4\right)+\left(y+4\right)=7\)
\(\Leftrightarrow\left(x+1\right)\left(y+4\right)=7\)
Vì x ; y nguyên nên x + 1 nguyên , y + 4 nguyên
Ta có bảng
| x + 1 | -7 | -1 | 1 | 7 |
| y + 4 | -1 | -7 | 7 | 1 |
| x | -8 | -2 | 0 | 6 |
| y | -5 | -11 | 3 | -3 |
Vậy ,.............
\(xy+4x+y=3\)
\(\Rightarrow x\left(y+4\right)+\left(y+4\right)=3+4\)
\(\Rightarrow\left(x+1\right)\left(y+4\right)=7\)
\(\Rightarrow\left(x+1\right);\left(y+4\right)\inƯ\left(7\right)=\left\{\pm1;\pm7\right\}\)
Ta có các trường hợp sau
\(TH1:\hept{\begin{cases}x+1=1\\y+4=7\end{cases}\Leftrightarrow\hept{\begin{cases}x=0\\y=3\end{cases}}}\) \(TH2:\hept{\begin{cases}x+1=-1\\y+4=-7\end{cases}\Leftrightarrow\hept{\begin{cases}x=-2\\y=-11\end{cases}}}\)
\(TH3:\hept{\begin{cases}x+1=7\\y+4=1\end{cases}\Leftrightarrow\hept{\begin{cases}x=6\\y=-3\end{cases}}}\) \(TH4:\hept{\begin{cases}x+1=-7\\y+4=-1\end{cases}\Leftrightarrow\hept{\begin{cases}x=-8\\y=-5\end{cases}}}\)
Vậy\(\left(x;y\right)\in\left\{\left(0;3\right);\left(-2;-11\right);\left(6;-3\right);\left(-8;-5\right)\right\}\)