\(\Leftrightarrow2xy+3x-5y-11=0\)
\(\Leftrightarrow x\left(2y+3\right)=5y+11\)
\(\Rightarrow x=\frac{5y+11}{2y+3}=2+\frac{y+5}{2y+3}\)
Để x nguyên thì \(y+5⋮2y+3\)
\(\Rightarrow2y+10⋮2y+3\)
\(\Rightarrow7⋮2y+3\)\(\Rightarrow2y+3\inƯ\left(7\right)=\left\{\pm1;\pm7\right\}\)
\(\Rightarrow y\in\left\{-5;-2;-1;2\right\}\)
\(\Rightarrow x\in\left\{2;-1;6;3\right\}\)
Vậy (x;y)=\(\left(2;-5\right);\left(-1;-2\right);\left(6;-1\right);\left(3;2\right)\)
2xy -11 = 5y - 3x
=> 4xy - 22 - 10y+6x=0
\(\Rightarrow2x\left(2x+3\right)-5\left(2y+3\right)-7=0\)
\(\Rightarrow\left(2x-5\right)\left(2y+3\right)=7\)
Bảng
| 2x-5 | 1 | -1 | 7 | -7 |
| 2y+3 | 7 | -7 | 1 | -1 |
| x | 3 | 2 | 6 | -1 |
| y | 2 | -5 | -1 | -2 |
Vây.....
