`@` `\text {Ans}`
`\downarrow`
Mình nghĩ \(2x-5y+5xy=14\) chứ c nhỉ ;-;?
`=> 2x - 5y + 5xy = 12 + 2`
`=> 2x - 2 - 5y + 5xy = 12`
`=> (2x - 2) - (5y - 5xy) = 12`
`=> 2(x - 1) - 5y(1 - x) = 12`
`=> 2(x - 1) + 5y(x - 1) = 12`
`=> (2+5y)(x - 1) = 12`
`=> (2+5y)(x-1) \in \text {Ư(12)} =`\(\left\{\pm1;\pm2;\pm3;\pm4;\pm6;\pm12\right\}\)
Để `2+5y \in \text {Ư(12)}` thì `(2+5y) \in {2; -3; 12}`
Ta có bảng sau:
| \(2+5y\) | `2` | `-3` | `12` |
| `x-1` | `6` | `-4` | `1` |
| \(x\) | `0` | `-1` | `2` |
| \(y\) | `7` | `-3` | `2` |
Vậy, ta có các cặp số nguyên thỏa mãn \(\left\{0;7\right\};\left\{-1;-3\right\};\left\{2;2\right\}.\)
\(\text{2x-5y+5xy=-14 }\)
\(\Rightarrow\text{5y(x-1)+2x-2+2=-14 }\)
\(\Rightarrow\text{5y(x-1)+2(x-1)=-16}\)
\(\Rightarrow\left(x-1\right)\left(5y+2\right)=-16\)
\(\Rightarrow\left(x-1\right)\&\left(5y+2\right)\in\left\{-1;1;-2;2;-4;4;-8;8;-16;16\right\}\)
\(\Rightarrow\left(x;y\right)\in\left\{\left(0;\dfrac{12}{5}\right);\left(2;-\dfrac{16}{5}\right);\left(-1;\dfrac{6}{5}\right);\left(3-2;\right);\left(-3;\dfrac{2}{5}\right);\left(5;-\dfrac{6}{5}\right);\left(-7;0\right);\left(9;-\dfrac{4}{5}\right);\left(-15;-\dfrac{1}{5}\right);\left(17;-\dfrac{3}{5}\right)\right\}\)
\(\Rightarrow\left(x;y\right)\in\left\{\left(3-2;\right);\left(-7;0\right)\right\}\left(x;y\in Z\right)\)