Giải thích các bước giải:

Ta có:

\(\begin{array}{l}
E = \dfrac{{x + \sqrt x }}{{x - 2\sqrt x  + 1}}:\left( {\dfrac{{\sqrt x  + 1}}{{\sqrt x }} - \dfrac{1}{{1 - \sqrt x }} + \dfrac{{2 - x}}{{x - \sqrt x }}} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\left( \begin{array}{l}
x > 0\\
x \ne 1
\end{array} \right)\\
 = \dfrac{{\sqrt x \left( {\sqrt x  + 1} \right)}}{{{{\left( {\sqrt x  - 1} \right)}^2}}}:\left( {\dfrac{{\sqrt x  + 1}}{{\sqrt x }} + \dfrac{1}{{\sqrt x  - 1}} + \dfrac{{2 - x}}{{\sqrt x \left( {\sqrt x  - 1} \right)}}} \right)\\
 = \dfrac{{\sqrt x \left( {\sqrt x  + 1} \right)}}{{{{\left( {\sqrt x  - 1} \right)}^2}}}:\dfrac{{\left( {\sqrt x  + 1} \right).\left( {\sqrt x  - 1} \right) + \sqrt x  + \left( {2 - x} \right)}}{{\sqrt x \left( {\sqrt x  - 1} \right)}}\\
 = \dfrac{{\sqrt x \left( {\sqrt x  + 1} \right)}}{{{{\left( {\sqrt x  - 1} \right)}^2}}}:\dfrac{{\left( {x - 1} \right) + \sqrt x  + 2 - x}}{{\sqrt x \left( {\sqrt x  - 1} \right)}}\\
 = \dfrac{{\sqrt x \left( {\sqrt x  + 1} \right)}}{{{{\left( {\sqrt x  - 1} \right)}^2}}}:\dfrac{{\sqrt x  + 1}}{{\sqrt x \left( {\sqrt x  - 1} \right)}}\\
 = \dfrac{{\sqrt x .\left( {\sqrt x  + 1} \right)}}{{{{\left( {\sqrt x  - 1} \right)}^2}}}.\dfrac{{\sqrt x \left( {\sqrt x  - 1} \right)}}{{\sqrt x  + 1}}\\
 = \dfrac{x}{{\sqrt x  - 1}}\\
b,\\
E > 1 \Leftrightarrow \dfrac{x}{{\sqrt x  - 1}} > 1\\
 \Leftrightarrow \dfrac{x}{{\sqrt x  - 1}} - 1 > 0\\
 \Leftrightarrow \dfrac{{x - \left( {\sqrt x  - 1} \right)}}{{\sqrt x  - 1}} > 0\\
 \Leftrightarrow \dfrac{{x - \sqrt x  + 1}}{{\sqrt x  - 1}} > 0\\
x - \sqrt x  + 1 = \left( {x - \sqrt x  + \dfrac{1}{4}} \right) + \dfrac{3}{4} = {\left( {\sqrt x  - \dfrac{1}{2}} \right)^2} + \dfrac{3}{4} \ge \dfrac{3}{4} > 0,\,\,\,\forall x \ge 0\\
 \Rightarrow \sqrt x  - 1 > 0\\
 \Leftrightarrow \sqrt x  > 1\\
 \Leftrightarrow x > 1\\
c,\\
E = \dfrac{x}{{\sqrt x  - 1}} = \dfrac{{\left( {x - 1} \right) + 1}}{{\sqrt x  - 1}} = \dfrac{{\left( {\sqrt x  - 1} \right)\left( {\sqrt x  + 1} \right) + 1}}{{\sqrt x  - 1}}\\
 = \left( {\sqrt x  + 1} \right) + \dfrac{1}{{\sqrt x  - 1}} = \left[ {\left( {\sqrt x  - 1} \right) + \dfrac{1}{{\sqrt x  - 1}}} \right] + 2\\
 \ge 2.\sqrt {\left( {\sqrt x  - 1} \right).\dfrac{1}{{\sqrt x  - 1}}}  + 2 = 2.\sqrt 1  + 2 = 4\\
 \Rightarrow {E_{\min }} = 4 \Leftrightarrow \sqrt x  - 1 = \dfrac{1}{{\sqrt x  - 1}} \Leftrightarrow x = 4\\
d,\\
E = \sqrt x  + 1 + \dfrac{1}{{\sqrt x  - 1}}\\
E \in Z \Leftrightarrow \dfrac{1}{{\sqrt x  - 1}} \in Z \Leftrightarrow \left( {\sqrt x  - 1} \right) \in \left\{ { \pm 1} \right\}\\
 \Rightarrow \sqrt x  \in \left\{ {0;2} \right\}\\
x \ne 0 \Rightarrow x = 4\\
e,\\
\left| {2x + 1} \right| = 5 \Leftrightarrow \left[ \begin{array}{l}
2x + 1 = 5\\
2x + 1 =  - 5
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
2x = 4\\
2x =  - 6
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 2\\
x =  - 3
\end{array} \right.\\
x > 0,\,\,x \ne 1 \Rightarrow x = 2\\
 \Rightarrow E = \dfrac{2}{{\sqrt 2  - 1}} = \dfrac{{2.\left( {\sqrt 2  + 1} \right)}}{{\left( {\sqrt 2  - 1} \right)\left( {\sqrt 2  + 1} \right)}} = \dfrac{{2\left( {\sqrt 2  + 1} \right)}}{{2 - 1}} = 2\sqrt 2  + 2\\
g,\\
E = \dfrac{9}{2} \Leftrightarrow \dfrac{x}{{\sqrt x  - 1}} = \dfrac{9}{2}\\
 \Leftrightarrow 2x = 9\left( {\sqrt x  - 1} \right)\\
 \Leftrightarrow 2x - 9\sqrt x  + 9 = 0\\
 \Leftrightarrow \left( {\sqrt x  - 3} \right)\left( {2\sqrt x  - 3} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
\sqrt x  = 3\\
\sqrt x  = \dfrac{3}{2}
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 9\\
x = \dfrac{9}{4}
\end{array} \right.
\end{array}\)