Ta có : \(2x+\frac{1}{7}=\frac{1}{y}\) \(\Rightarrow\frac{14x}{7}+\frac{1}{7}=\frac{1}{y}\) \(\Rightarrow\frac{14x+1}{7}=\frac{1}{y}\)
\(\Rightarrow\left(14x+1\right)\cdot y=7\)
\(\Rightarrow14x+1;y\in\text{Ư}\left(7\right)\)
Ta co bang:
| 14x+1 | 1 | -1 | 7 | -7 |
| x | 0 | \(\frac{-1}{7}\) | \(\frac{3}{7}\) | \(\frac{-4}{7}\) |
| y | 7 | -7 | 1 | -1 |
Do x, y \(\in\)Z\(\Rightarrow\)x=0; y=7
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