\(xy-x+2y=5\)
\(\Rightarrow xy-x+2y-2=3\)
\(\Rightarrow x\left(y-1\right)+2\left(y-1\right)=3\)
\(\Rightarrow\left(x+2\right)\left(y-1\right)=3\)
Xét ước nha
\(x\left(y+2\right)+y=1\)
\(\Rightarrow xy+2x+y=1\)
\(\Rightarrow xy+2x+y+2=3\)
\(\Rightarrow x\left(y+2\right)+1\left(y+2\right)=3\)
\(\Rightarrow\left(x+1\right)\left(y+2\right)=3\)
Xét ước
\(xy=x-y\)
\(\Rightarrow x-y-xy=0\)
\(\Rightarrow x-y-xy+1=1\)
\(\Rightarrow x\left(1-y\right)+1\left(1-y\right)=1\)
\(\Rightarrow\left(x+1\right)\left(1-y\right)=1\)
Xét ước
đề làm tìm số hữu tỉ x và y?
\(xy-x+2y=5\)
\(\Rightarrow x\left(y-1\right)+2\left(y-1\right)=3\)
\(\Rightarrow\left(x+2\right)\left(y-1\right)=3\)
\(\Rightarrow\left[\left(x+2\right);\left(y-1\right)\right]\inƯ\left(3\right)\)
Xét các trường hợp
\(x\left(y+2\right)+y=1\)
\(\Rightarrow xy+2x+y=1\)
\(\Rightarrow y\left(x+1\right)+2\left(x+1\right)=3\)
\(\Rightarrow\left(y+2\right)\left(x+1\right)=3\)
...
\(xy=x-y\)
\(\Rightarrow2xy=2x-2y\)
\(\Rightarrow2x=2xy+2y\)
\(\Rightarrow2x=2y\left(x+1\right)\)
\(\Rightarrow2\left(x+1\right)-2=2y\left(x +1\right)-2\)
\(\Rightarrow\left(2-2y\right)\left(x+1\right)=0\)
...
Theo cách nghĩ của mk, sai thì thôi, ko người nào đó lại...
Ta có : xy-x+2y=5
=> x.(y-1)+2y=5
=> x.(y-1)+2.(y-1)+2=5
=> (y-1).x+2)=3
=> \(\left(y-1\right);\left(x+2\right)\inƯ_{\left(3\right)}=\left\{\pm1;\pm3\right\}\)
Ta có bảng :
| y-1 | 1 | -1 | 3 | -3 |
| y | 2 | 0 | 4 | -2 |
| x+2 | 3 | -3 | 1 | -1 |
| x | 1 | -5 | -1 | -3 |
Vậy ..............................
