\(\left(2x+5\right)\left(y-3\right)=22\\ \Rightarrow\left(2x+5\right);\left(y-3\right)\inƯ\left(22\right)=\left\{1;2;11;22\right\}\\ TH1:2x+5=1\Rightarrow x=-2\left(loại\right);\left(y-3\right)=22\Rightarrow y=25\\ TH2:2x+5=2\Rightarrow x=-\dfrac{3}{2}\left(loại\right);\left(y-3\right)=11\Rightarrow y=14\\ TH3:2x+5=11\Rightarrow x=3;\left(y-3\right)=2\Rightarrow y=5\\ TH4:2x+5=22\Rightarrow x=\dfrac{17}{2}\left(loại\right);\left(y-3\right)=1\Rightarrow y=4\\Vậy:\left(x;y\right)=\left(3;5\right)\)
(2x + 5)(y - 3) = 22
Ư(22) = {-1,-2,-11,-22,1,2,11,22}
=> Ta có bảng:
| 2x+5 | -1 | -2 | -11 | -22 | 1 | 2 | 11 | 22 |
| y-3 | -22 | -11 | -2 | -1 | 22 | 11 | 2 | 1 |
| x | -3 | -7/2 | -8 | -27/2 | -2 | -3/2 | 3 | 17/2 |
| y | -19 | -8 | 1 | 2 | 25 | 14 | 5 | 4 |
Vậy ta có cặp x,y nguyên dương thỏa mãn là: (3;5)
