Ta có : \(K=\frac{\sqrt{x}+1}{\sqrt{x}-3}=\frac{\sqrt{x}-3+4}{\sqrt{x}-3}=1+\frac{4}{\sqrt{x}-3}\)
4 \(⋮\)(\(\sqrt{x}-3\))
=> \(\sqrt{x}-3\inƯ\left(4\right)\)
=> \(\sqrt{x}-3\in\left\{\pm1;\pm2;\pm4\right\}\)
Lập bảng :
\(\sqrt{x}-3\) | 1 | -1 | 2 | -2 | 4 | -4 |
\(\sqrt{x}\) | 4 | 2 | 5 | 1 | 7 | -1 |
\(x\) | 16 | 4 | 25 | 1 | 49 | \(\varnothing\) |
Vậy : ...