$\dfrac{1}{P}=\dfrac{1}{x-\sqrt{x}+1}$
Để $\dfrac{1}{P}$ có giá trị nguyên `<=>`$x-\sqrt{x}+1∈Ư(1)=\{±1\}$
Vì $P=x-\sqrt{x}+1=x-2.\sqrt{x}.\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{3}{4}=(\sqrt{x}-\dfrac{1}{2})^{2}+\dfrac{3}{4}\ge\dfrac{3}{4}>0$
`=>`$x-\sqrt{x}+1=1$
`<=>`$x-\sqrt{x}=0$
`<=>`$\sqrt{x}(\sqrt{x}-1)=0$
`<=>`\(\left[ \begin{array}{l}\sqrt{x}=0\\\sqrt{x}-1=0\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}x=0(loại)\\x=1(t/m)\end{array} \right.\)
Vậy $x=1$ thì $\dfrac{1}{P}$ có giá trị nguyên
