Đáp án:
$\begin{array}{l}
a)Dkxd:x - 1 > 0\\
\Rightarrow x > 1\\
b)Dkxd:x > 1\\
P = 2\left( {\dfrac{1}{{\sqrt {x - 1} }} - \dfrac{1}{{\sqrt {x - 1} + 1}}} \right):\dfrac{{\sqrt {x - 1} }}{{x + \sqrt {x - 1} - 1}}\\
= 2.\dfrac{{\sqrt {x - 1} + 1 - \sqrt {x - 1} }}{{\sqrt {x - 1} \left( {\sqrt {x - 1} + 1} \right)}}.\dfrac{{x - 1 + \sqrt {x - 1} }}{{\sqrt {x - 1} }}\\
= 2.\dfrac{1}{{\sqrt {x - 1} .\left( {\sqrt {x - 1} + 1} \right)}}.\dfrac{{\sqrt {x - 1} \left( {\sqrt {x - 1} + 1} \right)}}{{\sqrt {x - 1} }}\\
= \dfrac{2}{{\sqrt {x - 1} }}\\
c)P = \dfrac{2}{{\sqrt {x - 1} }} \in Z\\
\Rightarrow \left[ \begin{array}{l}
\sqrt {x - 1} = 1\\
\sqrt {x - 1} = 2
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x - 1 = 1\\
x - 1 = 4
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = 2\left( {tmdk} \right)\\
x = 5\left( {tmdk} \right)
\end{array} \right.
\end{array}$
Vậy x=2 hoặc x=5 thì P nguyên
