ĐK x ≤ 0 , x ≠ 36
Ta có A=\(\frac{\sqrt{x}+6}{\sqrt{x}-6}\)
=\(\frac{\sqrt{x}-6+12}{\sqrt{x}-6}\)
=\(1+\frac{12}{\sqrt{x}-6}\)
Để x ∈ Z thì \(\frac{12}{\sqrt{x}-6}\in Z\)
⇒ \(\sqrt{x}-6\in\text{Ư}_{12}=1,-1,2,-2,3,-3,4,-4,6,-6,12,-12\)
Ta có bảng
| \(\sqrt{x}-6\) | 1 | -1 | 2 | -2 | 3 | -3 | 4 | -4 | 6 | -6 | 12 | -12 |
| \(\sqrt{x}\) | 7 | 5 | 8 | 4 | 9 | 3 | 10 | 2 | 12 | 0 | 18 | -6(ktm) |
| \(x\) | 49(tm) | 25(tm) | 64(tm) | 16(tm) | 81(tm) | 9(tm) | 100(tm) | 4(tm) | 114(tm) | 0(tm) | 324(tm) |