`P = ( sqrt {x} - 1 )/( sqrt {x} - 3 ) ( Đk : x ≥ 0 ; x \ne 9 )`
`a) \text{Để} P < 1`
`<=> ( sqrt {x} - 1 )/( sqrt {x} - 3 ) < 1`
`<=> ( sqrt {x} - 1 )/( sqrt {x} - 3 ) - 1 < 0`
`<=> ( sqrt {x} - 1 - sqrt {x} + 3 )/( sqrt {x} - 3 ) < 0`
`<=> 2/( sqrt {x} - 3 ) < 0`
`<=> sqrt {x} - 3 < 0 \text{( vì 2 > 0 )}`
`<=> sqrt {x} < 3`
`<=> x < 9`
`\text{Vậy} 0 ≤ x < 9 \text{thì} P<1`
`b) P ∈ ZZ`
`=> ( sqrt {x} - 1 )/( sqrt {x} - 3 ) ∈ ZZ`
`<=> ( sqrt {x} - 3 + 2 )/( sqrt {x} - 3 ) ∈ ZZ`
`<=> 1 + 2/( sqrt {x} - 3 ) ∈ ZZ`
`<=> sqrt {x} - 3 ∈ Ư(2) = { ±1 ; ±2 }`
\begin{array}{|c|c|c|}\hline \text{$\sqrt{x}$-3}&\text{-2}&\text{-1}&\text{1}&\text{2}\\\hline \text{$\sqrt{x}$}&\text{1}&\text{2}&\text{4}&\text{5}\\\hline \text{x}&\text{1}&\text{4}&\text{16}&\text{25}\\\hline\text{}&\text{tmđk}&\text{tmđk}&\text{tmđk}&\text{tmđk}\\\hline\end{array}
`\text{Vậy} x ∈ { 1 ; 4 ; 16 ; 25 }`
