Đáp án:
\(a \in \emptyset \)
Giải thích các bước giải:
\(\begin{array}{l}
DK:a \ge 0;a \ne 1\\
Q = \left( {\dfrac{1}{{\sqrt a + 1}} - \dfrac{1}{{a + \sqrt a }}} \right):\dfrac{{\sqrt a - 1}}{{a + 2\sqrt a + 1}}\\
= \dfrac{{\sqrt a - 1}}{{\sqrt a \left( {\sqrt a + 1} \right)}}.\dfrac{{{{\left( {\sqrt a + 1} \right)}^2}}}{{\sqrt a - 1}}\\
= \dfrac{{\sqrt a + 1}}{{\sqrt a }} = 1 + \dfrac{1}{{\sqrt a }}\\
Q \in Z \to \dfrac{1}{{\sqrt a }} \in Z\\
\to \sqrt a \in U\left( 1 \right)\\
\to \sqrt a = 1\\
\to a = 1\left( l \right)\\
\to a \in \emptyset
\end{array}\)
