3. `P=A.B` `ĐK: x≥0;x ` `khác ` 1
`⇔P={x-3}/{\sqrt{x}-1}.{\sqrt{x}-1}/{\sqrt{x}+1}`
`⇔P={x-3}/{\sqrt{x}+1}`
`⇔P={x-1-2}/{\sqrt{x}+1}`
`⇔P={(\sqrt{x}+1)(\sqrt{x}-1)-2}/{\sqrt{x}+1}`
`⇔P=\sqrt{x}-1-2/{\sqrt{x}+1}`
Để P nguyên thì `-2/{\sqrt{x}+1}` nguyên và `\sqrt{x}` nguyên
⇒ `\sqrt{x}+1∈ ` Ư(-2)={-1;1;-2,2}
mà ` \sqrt{x}≥0`⇒`\sqrt{x}+1≥1`
`\sqrt{x}+1=1`
⇔`\sqrt{x}=0`
⇔`x=0`
`\sqrt{x}+1=2`
`⇔\sqrt{x}=1`
`⇔x=1(loại)`
Vậy `x=0`