Đáp án:

\(\begin{array}{l}
4,\\
a,\\
DKXD:\,\,\,\left\{ \begin{array}{l}
x > 0\\
x \ne 4
\end{array} \right.\\
b,\\
C = \sqrt x  - 1\\
c,\\
C = \dfrac{{\sqrt 5  - 3}}{2}\\
5,\\
DKXD:\,\,\,\left\{ \begin{array}{l}
x \ge 0\\
x \ne 1
\end{array} \right.
\end{array}\)

Giải thích các bước giải:

 Ta có:

\(\begin{array}{l}
4,\\
a,\\
DKXD:\,\,\,\left\{ \begin{array}{l}
x \ge 0\\
\sqrt x  - 2 \ne 0\\
2\sqrt x  - x \ne 0\\
\sqrt x  \ne 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ge 0\\
\sqrt x  \ne 2\\
\sqrt x \left( {2 - \sqrt x } \right) \ne 0\\
\sqrt x  \ne 0
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
x \ge 0\\
\sqrt x  \ne 2\\
\sqrt x  \ne 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ge 0\\
x \ne 4\\
x \ne 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x > 0\\
x \ne 4
\end{array} \right.\\
b,\\
C = \left( {\dfrac{1}{{\sqrt x  - 2}} + \dfrac{{5\sqrt x  - 4}}{{2\sqrt x  - x}}} \right):\left( {\dfrac{{2 + \sqrt x }}{{\sqrt x }} - \dfrac{{\sqrt x }}{{\sqrt x  - 2}}} \right)\\
 = \left( {\dfrac{1}{{\sqrt x  - 2}} - \dfrac{{5\sqrt x  - 4}}{{x - 2\sqrt x }}} \right):\left( {\dfrac{{\sqrt x  + 2}}{{\sqrt x }} - \dfrac{{\sqrt x }}{{\sqrt x  - 2}}} \right)\\
 = \left( {\dfrac{1}{{\sqrt x  - 2}} - \dfrac{{5\sqrt x  - 4}}{{\sqrt x \left( {\sqrt x  - 2} \right)}}} \right):\dfrac{{\left( {\sqrt x  + 2} \right)\left( {\sqrt x  - 2} \right) - {{\sqrt x }^2}}}{{\sqrt x \left( {\sqrt x  - 2} \right)}}\\
 = \dfrac{{\sqrt x  - \left( {5\sqrt x  - 4} \right)}}{{\sqrt x \left( {\sqrt x  - 2} \right)}}:\dfrac{{\left( {x - 4} \right) - x}}{{\sqrt x \left( {\sqrt x  - 2} \right)}}\\
 = \dfrac{{\sqrt x  - 5\sqrt x  + 4}}{{\sqrt x \left( {\sqrt x  - 2} \right)}}:\dfrac{{ - 4}}{{\sqrt x \left( {\sqrt x  - 2} \right)}}\\
 = \dfrac{{ - 4\sqrt x  + 4}}{{\sqrt x \left( {\sqrt x  - 2} \right)}}.\dfrac{{\sqrt x \left( {\sqrt x  - 2} \right)}}{{ - 4}}\\
 = \dfrac{{ - 4\sqrt x  + 4}}{{ - 4}}\\
 = \sqrt x  - 1\\
c,\\
x = \dfrac{{3 - \sqrt 5 }}{2} = \dfrac{{6 - 2\sqrt 5 }}{4} = \dfrac{{5 - 2.\sqrt 5 .1 + 1}}{4} = \dfrac{{{{\left( {\sqrt 5  - 1} \right)}^2}}}{4}\\
 \Rightarrow \sqrt x  = \sqrt {\dfrac{{{{\left( {\sqrt 5  - 1} \right)}^2}}}{4}}  = \dfrac{{\left| {\sqrt 5  - 1} \right|}}{2} = \dfrac{{\sqrt 5  - 1}}{2}\\
 \Rightarrow C = \dfrac{{\sqrt 5  - 1}}{2} - 1 = \dfrac{{\sqrt 5  - 1 - 2}}{2} = \dfrac{{\sqrt 5  - 3}}{2}\\
5,\\
DKXD:\,\,\,\left\{ \begin{array}{l}
x \ge 0\\
\sqrt x  + 1 \ne 0\\
\sqrt x  - 1 \ne 0\\
x - 1 \ne 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ge 0\\
\sqrt x  \ne  - 1\\
\sqrt x  \ne 1\\
x \ne 1
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
x \ge 0\\
\sqrt x  \ne 1\\
x \ne 1
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ge 0\\
x \ne 1
\end{array} \right.
\end{array}\)