1: \(P=\dfrac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}\cdot\dfrac{\sqrt{x}+1}{2}=\dfrac{\sqrt{x}+1}{2\sqrt{x}}\)
2: Để P là số nguyên thì \(2\sqrt{x}+2⋮2\sqrt{x}\)
\(\Leftrightarrow2\sqrt{x}=2\)
hay x=1(nhận)
3: \(P-\dfrac{1}{2}=\dfrac{\sqrt{x}+1}{2\sqrt{x}}-\dfrac{1}{2}=\dfrac{2\sqrt{x}+2-\sqrt{x}}{2\sqrt{x}}=\dfrac{\sqrt{x}+2}{2\sqrt{x}}>0\)
=>P>1/2