\(\dfrac{x}{8}-\dfrac{1}{y}=\dfrac{1}{4}\)
\(\Leftrightarrow\dfrac{x}{8}-\dfrac{1}{4}=\dfrac{1}{y}\)
\(\Leftrightarrow\dfrac{x}{8}-\dfrac{2}{8}=\dfrac{1}{y}\)
\(\Leftrightarrow\dfrac{x-2}{8}=\dfrac{1}{y}\)
\(\Leftrightarrow\left(x-2\right)y=8\)
\(\Leftrightarrow x-2;y\inƯ\left(8\right)\)
Ta có bảng :
| \(x-2\) | \(1\) | \(2\) | \(4\) | \(8\) | \(-1\) | \(-2\) | \(-4\) | \(-8\) |
| \(x\) | \(3\) | \(4\) | \(6\) | \(10\) | \(1\) | \(0\) | \(-2\) | \(-6\) |
| \(y\) | \(8\) | \(4\) | \(2\) | \(1\) | \(-8\) | \(-4\) | \(-2\) | \(-1\) |
| Đk \(x,y\in Z\) | tm | tm | tm | tm | tm | tm | tm | tm |
Vậy ...
\(\dfrac{x}{8}-\dfrac{1}{y}=\dfrac{1}{4}\)
\(\Rightarrow\dfrac{xy-8}{8y}=\dfrac{1}{4}\)
\(\Rightarrow\dfrac{xy-8}{2y}=1\)
\(\Rightarrow xy-8=2y\)
\(\Rightarrow y\left(x-2\right)=8\)
Do đó y là ước của 8 \(\Rightarrow y\in\left\{\pm1;\pm2;\pm4;\pm8\right\}\). Ta có bảng sau:
| y | 1 | 2 | 4 | 8 | -1 | -2 | -4 | -8 |
| x - 2 | 8 | 4 | 2 | 1 | -8 | -4 | -2 | -1 |
| x | 10 | 6 | 4 | 3 | -6 | -2 | 0 | 1 |
Vậy ta có các cặp x, y là: (10, 1); (6, 2); (4, 4); (3, 8); (-6, -1); (-2, -2); (0, -4); (1; -8)
