a,\(xy-7y+y=-22\)
\(=xy-7x+y-7+7=-22\)
\(=\left(xy-7x\right)+\left(y-7\right)=-29\)
\(=x\left(y-7\right)+\left(y-7\right)=-29\)
\(=\left(y-7\right)\left(x+1\right)=-29\)
Vì \(x,y\varepsilon Z\)nên\(\left(y-7\right),\left(x+1\right)\varepsilon Z\)
\(\Rightarrow\left(y-7\right);\left(x+1\right)\varepsilon B\left(-29\right)\)
Mà \(-29=-1.29=1.\left(-29\right)\)
Ta có 4TH :\(1,\hept{\begin{cases}y-7=-1\\x+1=29\end{cases}}\Rightarrow\hept{\begin{cases}y=6\\x=28\end{cases}}\left(TM\right)\)
\(2,\hept{\begin{cases}y-7=1\\x+1=-29\end{cases}\Rightarrow\hept{\begin{cases}y=8\\x=-30\end{cases}}}\)
\(3,\hept{\begin{cases}y-7=29\\x+1=-1\end{cases}\Rightarrow\hept{\begin{cases}y=36\\x=-2\end{cases}}}\)
\(4,\hept{\begin{cases}y-7=-29\\x+1=1\end{cases}\Rightarrow\hept{\begin{cases}y=-22\\x=0\end{cases}}}\)
Vậy có 4 cặp (x, y): \(\left(6;28\right);\left(8;-30\right);\left(36;-2\right);\left(-22;0\right)\)
Vì dài quá nên mk chỉ làm từng này thôi nhé, nếu mk đúng nha!
b, xy - 3x + y= -20
=> x(y - 3) + (y - 3) = -23
=> (x + 1)(y - 3) = -23
ta có bảng :
| x+1 | -1 | 1 | -23 | 23 |
| y-3 | 23 | -23 | 1 | - |
| x | -2 | 0 | -24 | 22 |
| y | 26 | -20 | 4 | 2 |
c, xy - 5y - 2x = -41
=> y(x - 5) - 2x + 10 = -31
=> y(x - 5) - 2(x - 5) = -31
=> (y - 2)(x - 5) = -31
| y-2 | -1 | 1 | -31 | 31 |
| x-5 | 31 | -31 | 1 | -1 |
| y | 1 | 3 | -29 | 33 |
| x | 36 | -26 | 6 | 4 |
