\(P=\dfrac{x-1}{\sqrt{x}}\)(ĐK:x>0)
\(=\dfrac{\left(\sqrt{x}\right)^2-1}{\sqrt{x}}=\sqrt{x}-\dfrac{1}{\sqrt{x}}\)
Vậy để P nguyên thì \(\left\{{}\begin{matrix}\sqrt{x}\in Z\\\sqrt{x}\inƯ\left(1\right)\in\left(\pm1\right)\end{matrix}\right.\)
Vì \(\sqrt{x}>0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x}\in Z\\\sqrt{x}=1\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x}\in Z\\x=1\end{matrix}\right.\)(tm)
Vậy x=1 thì P\(\in Z\)