=> xy-x+2y-2=5-2(trừ cả 2 vế cho 2 bn nhé)
=> y(x+2)-(x+2)=3
=> (x+2)(y-1)=3
=>\(\hept{\begin{cases}x+2=1\\y-1=3\end{cases}}\)=>\(\orbr{\begin{cases}x=-2\\y=4\end{cases}}\)
\(\hept{\begin{cases}x+2=3\\y-1=1\end{cases}}\)=>\(\hept{\begin{cases}x=1\\y=2\end{cases}}\)
\(\orbr{\begin{cases}x+2=-1\\y-1=-3\end{cases}}\)=> \(\hept{\begin{cases}x=-3\\y=-2\end{cases}}\)
\(\hept{\begin{cases}x+2=-3\\y-1=-1\end{cases}}\)=>\(\hept{\begin{cases}x=-5\\y=0\end{cases}}\)
\(\left(xy-x\right)+\left(2y-2\right)=3\)
\(x.\left(y-1\right)+2.\left(y-1\right)=3\)
\(\left(y-1\right).\left(x+2\right)=3\)
\(\left(y-1\right).\left(x+2\right)=3.1=1.3=-3.-1=-1.-3\)
x+2 | 3 | 1 | -1 | -3 |
| x | 1 | -1 | -3 | -5 |
| y-1 | 1 | 3 | -3 | -1 |
| y | 2 | 4 | -2 | 0 |
vậy các cặp (x;y) phải tìm lag (1;2) ; (-1;4) ; (-3;-2) ; (-5;0)
