Giải thích các bước giải:

Ta có:

\(\begin{array}{l}
5,\\
a,\\
\sqrt {x + 1}  > 3\,\,\,\,\,\,\,\,\,\,\,\left( {x \ge  - 1} \right)\\
 \Leftrightarrow x + 1 > {3^2}\\
 \Leftrightarrow x > 8\\
b,\\
\sqrt {2x}  < 1 + \sqrt 3 \,\,\,\,\,\,\,\,\,\,\left( {x \ge 0} \right)\\
 \Leftrightarrow 2x < {\left( {1 + \sqrt 3 } \right)^2}\\
 \Leftrightarrow 2x < 4 + 2\sqrt 3 \\
 \Leftrightarrow x < 2 + \sqrt 3 \\
 \Rightarrow 0 \le x < 2 + \sqrt 3 \\
c,\\
x - 4\sqrt 3 .\sqrt x  + 12 = 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {x \ge 0} \right)\\
 \Leftrightarrow {\sqrt x ^2} - 2.\sqrt x .2\sqrt 3  + {\left( {2\sqrt 3 } \right)^2} = 0\\
 \Leftrightarrow {\left( {\sqrt x  - 2\sqrt 3 } \right)^2} = 0\\
 \Leftrightarrow \sqrt x  - 2\sqrt 3  = 0\\
 \Leftrightarrow x = 12\\
d,\\
2x = \sqrt {2x - 1}  + 1\,\,\,\,\,\,\,\left( {x \ge \dfrac{1}{2}} \right)\\
 \Leftrightarrow 2x - 1 - \sqrt {2x - 1}  = 0\\
 \Leftrightarrow {\sqrt {2x - 1} ^2} - \sqrt {2x - 1}  = 0\\
 \Leftrightarrow \sqrt {2x - 1} \left( {\sqrt {2x - 1}  - 1} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
\sqrt {2x - 1}  = 0\\
\sqrt {2x - 1}  - 1 = 0
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{1}{2}\\
x = 1
\end{array} \right.\\
6,\\
a,\\
\dfrac{{x + \sqrt x  + 1}}{{\sqrt x }} = \dfrac{{13}}{3}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {x > 0} \right)\\
 \Leftrightarrow 3.\left( {x + \sqrt x  + 1} \right) = 13\sqrt x \\
 \Leftrightarrow 3x - 10\sqrt x  + 3 = 0\\
 \Leftrightarrow \left( {3\sqrt x  - 1} \right)\left( {\sqrt x  - 3} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
\sqrt x  = \dfrac{1}{3}\\
\sqrt x  = 3
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{1}{9}\\
x = 9
\end{array} \right.\\
b,\\
\dfrac{x}{{\sqrt x  - 5}} =  - \dfrac{4}{3}\,\,\,\,\,\,\,\left( {x \ge 0,\,\,\,x \ne 25} \right)\\
 \Leftrightarrow 3x =  - 4.\left( {\sqrt x  - 5} \right)\\
 \Leftrightarrow 3x + 4\sqrt x  - 20 = 0\\
 \Leftrightarrow \left( {3\sqrt x  + 10} \right)\left( {\sqrt x  - 2} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
\sqrt x  =  - \dfrac{{10}}{3}\,\,\,\,\,\left( L \right)\\
\sqrt x  = 2
\end{array} \right. \Leftrightarrow x = 4\\
c,\\
\dfrac{{\sqrt x  + 1}}{{3.\left( {\sqrt x  - 1} \right)}} > 1\,\,\,\,\,\,\,\,\,\,\,\,\left( {x \ge 0;\,\,x \ne 1} \right)\\
 \Leftrightarrow \dfrac{{\sqrt x  + 1}}{{3\left( {\sqrt x  - 1} \right)}} - 1 > 0\\
 \Leftrightarrow \dfrac{{\sqrt x  + 1 - 3.\left( {\sqrt x  - 1} \right)}}{{3\left( {\sqrt x  - 1} \right)}} > 0\\
 \Leftrightarrow \dfrac{{ - 2\sqrt x  + 4}}{{3.\left( {\sqrt x  - 1} \right)}} > 0\\
 \Leftrightarrow \dfrac{{2.\left( {\sqrt x  - 2} \right)}}{{3.\left( {\sqrt x  - 1} \right)}} < 0\\
 \Leftrightarrow 1 < \sqrt x  < 2\\
 \Leftrightarrow 1 < x < 2\\
d,\\
\dfrac{{\sqrt x  - 5}}{{\sqrt x  + 5}} < \dfrac{1}{3}\,\,\,\,\,\,\,\left( {x \ge 0} \right)\\
 \Leftrightarrow 3\sqrt x  - 15 < \sqrt x  + 5\\
 \Leftrightarrow 2\sqrt x  < 20\\
 \Leftrightarrow x < 100
\end{array}\)