Bài 2:
a) \(A=\frac{10n}{5n-3}=\frac{2\left(5n-3\right)+6}{5n-3}=2+\frac{6}{5n-3}\)
Vậy để A nguyên thì \(5n-3\inƯ\left(6\right)\)
Mà Ư(6)={1;-1;2;-2;3;-3;6;-6}
=>5n-3={1;-1;2;-2;3;-3;6;-6}
Ta có bảng sau:
5n-3 | 1 | -1 | 2 | -2 | 3 | -3 | 6 | -6 |
n | \(\frac{4}{5}\) | \(\frac{2}{5}\) | 1 | \(\frac{1}{5}\) | \(\frac{6}{5}\) | 0 | \(\frac{9}{5}\) | -\(\frac{3}{5}\) |
Vậy \(x=\left\{\frac{4}{5};\frac{2}{5};1;\frac{1}{5};\frac{6}{5};0;\frac{9}{5};-\frac{3}{5}\right\}\) thì A nguyên