Đáp án:
a) \(\dfrac{{\sqrt a - 1}}{{\sqrt a }}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)P = \left[ {\dfrac{{1 + \sqrt a }}{{\sqrt a \left( {\sqrt a - 1} \right)}}} \right].\dfrac{{{{\left( {\sqrt a - 1} \right)}^2}}}{{\sqrt a + 1}}\\
= \dfrac{{\sqrt a - 1}}{{\sqrt a }}\\
b)P = \dfrac{{\sqrt a - 1}}{{\sqrt a }} = 1 - \dfrac{1}{{\sqrt a }}\\
P \in Z \Leftrightarrow \dfrac{1}{{\sqrt a }} \in Z\\
\to \sqrt a \in U\left( 1 \right)\\
\to \left[ \begin{array}{l}
\sqrt a = 1\\
\sqrt a = - 1\left( l \right)
\end{array} \right.\\
\to a = 1\left( l \right)\\
\to a \in \emptyset
\end{array}\)
