Đáp án:

 e) \(\left[ \begin{array}{l}
x = 9\\
x = 4
\end{array} \right.\)

Giải thích các bước giải:

\(\begin{array}{l}
a)A = \left( {\dfrac{{x - \sqrt x }}{{\sqrt x  - 1}} + 1} \right):\left( {\dfrac{{x + 2\sqrt x }}{{\sqrt x  + 2}} - 1} \right)\\
 = \left[ {\dfrac{{\sqrt x \left( {\sqrt x  - 1} \right)}}{{\sqrt x  - 1}} + 1} \right]:\left[ {\dfrac{{\sqrt x \left( {\sqrt x  + 2} \right)}}{{\sqrt x  + 2}} - 1} \right]\\
 = \left( {\sqrt x  + 1} \right):\left( {\sqrt x  - 1} \right)\\
 = \dfrac{{\sqrt x  + 1}}{{\sqrt x  - 1}}\\
b)Thay:x = 9\\
 \to A = \dfrac{{\sqrt 9  + 1}}{{\sqrt 9  - 1}} = \dfrac{4}{2} = 2\\
c)A = 5\\
 \to \dfrac{{\sqrt x  + 1}}{{\sqrt x  - 1}} = 5\\
 \to \sqrt x  + 1 = 5\sqrt x  - 5\\
 \to 4\sqrt x  = 6\\
 \to \sqrt x  = \dfrac{3}{2}\\
 \to x = \dfrac{9}{4}\\
d)A < 5\\
 \to \dfrac{{\sqrt x  + 1}}{{\sqrt x  - 1}} < 5\\
 \to \dfrac{{\sqrt x  + 1 - 5\sqrt x  + 5}}{{\sqrt x  - 1}} < 0\\
 \to \dfrac{{6 - 4\sqrt x }}{{\sqrt x  - 1}} < 0\\
 \to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
6 - 4\sqrt x  > 0\\
\sqrt x  - 1 < 0
\end{array} \right.\\
\left\{ \begin{array}{l}
6 - 4\sqrt x  < 0\\
\sqrt x  - 1 > 0
\end{array} \right.
\end{array} \right. \to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
\dfrac{3}{2} > \sqrt x \\
1 > \sqrt x 
\end{array} \right.\\
\left\{ \begin{array}{l}
\dfrac{3}{2} < \sqrt x \\
1 < \sqrt x 
\end{array} \right.
\end{array} \right.\\
 \to \left[ \begin{array}{l}
1 > \sqrt x \\
\dfrac{3}{2} < \sqrt x 
\end{array} \right. \to \left[ \begin{array}{l}
1 > x > 0\\
x > \dfrac{9}{4}
\end{array} \right.\\
e)A = \dfrac{{\sqrt x  + 1}}{{\sqrt x  - 1}} = \dfrac{{\sqrt x  - 1 + 2}}{{\sqrt x  - 1}} = 1 + \dfrac{2}{{\sqrt x  - 1}}\\
A \in Z \Leftrightarrow \dfrac{2}{{\sqrt x  - 1}} \in Z\\
 \Leftrightarrow \sqrt x  - 1 \in U\left( 2 \right)\\
 \to \left[ \begin{array}{l}
\sqrt x  - 1 = 2\\
\sqrt x  - 1 =  - 2\left( l \right)\\
\sqrt x  - 1 = 1\\
\sqrt x  - 1 =  - 1
\end{array} \right. \to \left[ \begin{array}{l}
\sqrt x  = 3\\
\sqrt x  = 2\\
\sqrt x  = 0\left( l \right)
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x = 9\\
x = 4
\end{array} \right.
\end{array}\)