Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
DKXD:\,\,\,\left\{ \begin{array}{l}
a \ge 0\\
a + 2\sqrt a \ne 0\\
\sqrt a + 2 \ne 0\\
a\sqrt a + 4a + 4\sqrt a \ne 0
\end{array} \right. \Leftrightarrow a > 0\\
b,\\
A = \left( {\dfrac{2}{{a + 2\sqrt a }} + \dfrac{1}{{\sqrt a + 2}}} \right):\dfrac{{\sqrt a + 2}}{{a\sqrt a + 4a + 4\sqrt a }}\\
= \left( {\dfrac{2}{{\sqrt a \left( {\sqrt a + 2} \right)}} + \dfrac{1}{{\sqrt a + 2}}} \right):\dfrac{{\sqrt a + 2}}{{\sqrt a .\left( {a + 4\sqrt a + 4} \right)}}\\
= \dfrac{{2 + \sqrt a }}{{\sqrt a .\left( {\sqrt a + 2} \right)}}:\dfrac{{\sqrt a + 2}}{{\sqrt a {{\left( {\sqrt a + 2} \right)}^2}}}\\
= \dfrac{1}{{\sqrt a }}:\dfrac{1}{{\sqrt a \left( {\sqrt a + 2} \right)}}\\
= \dfrac{1}{{\sqrt a }}.\sqrt a \left( {\sqrt a + 2} \right)\\
= \sqrt a + 2\\
c,\\
A \in Z \Rightarrow \left( {\sqrt a + 2} \right) \in Z \Rightarrow \sqrt a \in Z
\end{array}\)
Suy ra \(x\) là số chính phương.
