Bài 4: `ĐKXĐ:x\geq0`
`A={\sqrt{x}-2}/{\sqrt{x}+2}={\sqrt{x}+2-4}/{\sqrt{x}+2}={\sqrt{x}+2}/{\sqrt{x}+2}-{4}/{\sqrt{x}+2}=1-{4}/{\sqrt{x}+2}`
`A\inZ`
`⇔{4}/{\sqrt{x}+2}\inZ`
`⇔\sqrt{x}+2\inƯ(4)={-4,-2,-1,1,2,4}`
`⇔\sqrt{x}\in{-6,-4,-3,-1,0,2}`
Mà `\sqrt{x}\geq0`
`⇔\sqrt{x}\in{0,2}`
`⇔x\in{0,4}` (Thỏa mãn ĐKXĐ)
Vậy với `x\in{0,4}` thì `A\inZ`
Bài 5: `ĐKXĐ:x\geq0`
`A={3\sqrt{x}}/{\sqrt{x}+1}={3\sqrt{x}+3-3}/{\sqrt{x}+1}={3\sqrt{x}+3}/{\sqrt{x}+1}-{3}/{\sqrt{x}+1}={3(\sqrt{x}+1)}/{\sqrt{x}+1}-{3}/{\sqrt{x}+1}=3-{3}/{\sqrt{x}+1}`
`A\inZ`
`⇔{3}/{\sqrt{x}+1}\inZ`
`⇔\sqrt{x}+1\inƯ(3)={-3,-1,1,3}`
`⇔\sqrt{x}\in{-4,-2,0,2}`
Mà `\sqrt{x}\geq0`
`⇔\sqrt{x}\in{0,2}`
`⇔x\in{0,4}` (Thỏa mãn ĐKXĐ)
Vậy với `x\in{0,4}` thì `A\inZ`
