Đáp án:
\(\left[ \begin{array}{l}
x = 5\\
x = 2
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
DK:x > 1\\
P = 2\left( {\dfrac{{\sqrt {x - 1} + 1 - \sqrt {x - 1} }}{{\sqrt {x - 1} \left( {\sqrt {x - 1} + 1} \right)}}} \right):\dfrac{{\sqrt {x - 1} }}{{{{\left( {\sqrt {x - 1} } \right)}^2} + \sqrt {x - 1} }}\\
= \dfrac{2}{{\sqrt {x - 1} \left( {\sqrt {x - 1} + 1} \right)}}:\dfrac{{\sqrt {x - 1} }}{{\left( {\sqrt {x - 1} } \right)\left( {\sqrt {x - 1} + 1} \right)}}\\
= \dfrac{2}{{\sqrt {x - 1} \left( {\sqrt {x - 1} + 1} \right)}}.\dfrac{{\left( {\sqrt {x - 1} } \right)\left( {\sqrt {x - 1} + 1} \right)}}{{\sqrt {x - 1} }}\\
= \dfrac{2}{{\sqrt {x - 1} }}\\
P \in Z \to \dfrac{2}{{\sqrt {x - 1} }} \in Z\\
\to \sqrt {x - 1} \in U\left( 2 \right)\\
\to \left[ \begin{array}{l}
\sqrt {x - 1} = 2\\
\sqrt {x - 1} = 1
\end{array} \right.\\
\to \left[ \begin{array}{l}
x - 1 = 4\\
x - 1 = 1
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 5\\
x = 2
\end{array} \right.\left( {TM} \right)
\end{array}\)
