Giải thích các bước giải:

Ta có:

\(\begin{array}{l}
2,\\
a,\\
P = \left( {\dfrac{1}{{\sqrt x }} + \dfrac{{\sqrt x }}{{\sqrt x  + 1}}} \right):\dfrac{{\sqrt x }}{{x + \sqrt x }}\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {x > 0} \right)\\
 = \dfrac{{\left( {\sqrt x  + 1} \right) + {{\sqrt x }^2}}}{{\sqrt x .\left( {\sqrt x  + 1} \right)}}:\dfrac{{\sqrt x }}{{\sqrt x \left( {\sqrt x  + 1} \right)}}\\
 = \dfrac{{x + \sqrt x  + 1}}{{\sqrt x \left( {\sqrt x  + 1} \right)}}:\dfrac{1}{{\sqrt x  + 1}}\\
 = \dfrac{{x + \sqrt x  + 1}}{{\sqrt x \left( {\sqrt x  + 1} \right)}}.\left( {\sqrt x  + 1} \right)\\
 = \dfrac{{x + \sqrt x  + 1}}{{\sqrt x }}\\
b,\\
x = 4 \Rightarrow P = \dfrac{{4 + \sqrt 4  + 1}}{{\sqrt 4 }} = \dfrac{7}{2}\\
c,\\
P = \dfrac{{13}}{3} \Leftrightarrow \dfrac{{x + \sqrt x  + 1}}{{\sqrt x }} = \dfrac{{13}}{3}\\
 \Leftrightarrow 3x + 3\sqrt x  + 3 = 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}
3\sqrt x  - 1 = 0\\
\sqrt x  - 3 = 0
\end{array} \right.\\
 \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.\\
3,\\
a,\\
x = 9 \Rightarrow A = \dfrac{{\sqrt 9  + 4}}{{\sqrt 9  - 1}} = \dfrac{{3 + 4}}{{3 - 1}} = \dfrac{7}{2}\\
b,\\
B = \dfrac{{3\sqrt x  + 1}}{{x + 2\sqrt x  - 3}} - \dfrac{2}{{\sqrt x  + 3}}\\
 = \dfrac{{3\sqrt x  + 1}}{{\left( {\sqrt x  + 3} \right)\left( {\sqrt x  - 1} \right)}} - \dfrac{2}{{\sqrt x  + 3}}\\
 = \dfrac{{\left( {3\sqrt x  + 1} \right) - 2.\left( {\sqrt x  - 1} \right)}}{{\left( {\sqrt x  + 3} \right)\left( {\sqrt x  - 1} \right)}}\\
 = \dfrac{{\sqrt x  + 3}}{{\left( {\sqrt x  + 3} \right)\left( {\sqrt x  - 1} \right)}}\\
 = \dfrac{1}{{\sqrt x  - 1}}\\
c,\\
\dfrac{A}{B} = \dfrac{{\sqrt x  + 4}}{{\sqrt x  - 1}}:\dfrac{1}{{\sqrt x  - 1}} = \dfrac{{\sqrt x  + 4}}{{\sqrt x  - 1}}.\left( {\sqrt x  - 1} \right) = \sqrt x  + 4\\
\dfrac{A}{B} > \dfrac{x}{4} + 5\\
 \Leftrightarrow \sqrt x  + 4 > \dfrac{x}{4} + 5\\
 \Leftrightarrow \dfrac{x}{4} - \sqrt x  + 1 > 0\\
 \Leftrightarrow x - 4\sqrt x  + 4 > 0\\
 \Leftrightarrow {\left( {\sqrt x  - 2} \right)^2} > 0\\
 \Rightarrow \sqrt x  - 2 \ne 0 \Rightarrow \sqrt x  \ne 2 \Rightarrow x \ne 4\\
 \Rightarrow \left\{ \begin{array}{l}
x \ge 0\\
x \ne 1\\
x \ne 4
\end{array} \right.
\end{array}\)