Giải thích các bước giải:

Ta có:

\(\begin{array}{l}
7,\\
a,\,\,\,\,DKXD:\,\,\left\{ \begin{array}{l}
x \ge 0\\
3 + \sqrt x  \ne 0\\
9 - x \ne 0\\
x - 3\sqrt x  \ne 0\\
\sqrt x  \ne 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x > 0\\
x \ne 9
\end{array} \right.\\
b,\,\,\,\,C = \left( {\dfrac{{\sqrt x }}{{3 + \sqrt x }} + \dfrac{{x + 9}}{{9 - x}}} \right):\left( {\dfrac{{3\sqrt x  + 1}}{{x - 3\sqrt x }} - \dfrac{1}{{\sqrt x }}} \right)\\
 = \left( {\dfrac{{\sqrt x }}{{3 + \sqrt x }} + \dfrac{{x + 9}}{{\left( {3 - \sqrt x } \right)\left( {3 + \sqrt x } \right)}}} \right):\left( {\dfrac{{3\sqrt x  + 1}}{{\sqrt x \left( {\sqrt x  - 3} \right)}} - \dfrac{1}{{\sqrt x }}} \right)\\
 = \dfrac{{\sqrt x .\left( {3 - \sqrt x } \right) + x + 9}}{{\left( {3 - \sqrt x } \right)\left( {3 + \sqrt x } \right)}}:\dfrac{{\left( {3\sqrt x  + 1} \right) - 1.\left( {\sqrt x  - 3} \right)}}{{\sqrt x \left( {\sqrt x  - 3} \right)}}\\
 = \dfrac{{3\sqrt x  - x + x + 9}}{{\left( {3 - \sqrt x } \right)\left( {3 + \sqrt x } \right)}}:\dfrac{{2\sqrt x  + 4}}{{\sqrt x \left( {\sqrt x  - 3} \right)}}\\
 = \dfrac{{3\sqrt x  + 9}}{{\left( {3 - \sqrt x } \right)\left( {3 + \sqrt x } \right)}}.\dfrac{{\sqrt x \left( {\sqrt x  - 3} \right)}}{{2\sqrt x  + 4}}\\
 = \dfrac{{3.\left( {\sqrt x  + 3} \right)}}{{\left( {3 - \sqrt x } \right)\left( {3 + \sqrt x } \right)}}.\dfrac{{\sqrt x .\left( {\sqrt x  - 3} \right)}}{{2\sqrt x  + 4}}\\
 = \dfrac{{ - 3\sqrt x }}{{2\sqrt x  + 4}}\\
c,\,\,\,C <  - 1\\
 \Leftrightarrow \dfrac{{ - 3\sqrt x }}{{2\sqrt x  + 4}} <  - 1\\
 \Leftrightarrow \dfrac{{ - 3\sqrt x }}{{2\sqrt x  + 4}} + 1 < 0\\
 \Leftrightarrow \dfrac{{ - 3\sqrt x  + 2\sqrt x  + 4}}{{2\sqrt x  + 4}} < 0\\
 \Leftrightarrow \dfrac{{4 - \sqrt x }}{{2\sqrt x  + 4}} < 0\\
 \Leftrightarrow 4 - \sqrt x  < 0\\
 \Leftrightarrow \sqrt x  > 4\\
 \Leftrightarrow x > 16\\
8,\,\,\,DKXD:\,\,\,\left\{ \begin{array}{l}
x \ge 0\\
x \ne 1
\end{array} \right.\\
a,\,\,\,P = \left( {\dfrac{{x + 2}}{{\sqrt x  + 1}} - \sqrt x } \right):\left( {\dfrac{{\sqrt x  - 4}}{{1 - x}} - \dfrac{{\sqrt x }}{{\sqrt x  + 1}}} \right)\\
 = \dfrac{{\left( {x + 2} \right) - \sqrt x \left( {\sqrt x  + 1} \right)}}{{\sqrt x  + 1}}:\left( {\dfrac{{\sqrt x  - 4}}{{\left( {1 - \sqrt x } \right)\left( {1 + \sqrt x } \right)}} - \dfrac{{\sqrt x }}{{\sqrt x  + 1}}} \right)\\
 = \dfrac{{x + 2 - x - \sqrt x }}{{\sqrt x  + 1}}:\dfrac{{\sqrt x  - 4 - \sqrt x .\left( {1 - \sqrt x } \right)}}{{\left( {1 - \sqrt x } \right)\left( {1 + \sqrt x } \right)}}\\
 = \dfrac{{2 - \sqrt x }}{{\sqrt x  + 1}}:\dfrac{{\sqrt x  - 4 - \sqrt x  + x}}{{\left( {1 - \sqrt x } \right)\left( {1 + \sqrt x } \right)}}\\
 = \dfrac{{2 - \sqrt x }}{{\sqrt x  + 1}}:\dfrac{{x - 4}}{{\left( {1 - \sqrt x } \right)\left( {1 + \sqrt x } \right)}}\\
 = \dfrac{{2 - \sqrt x }}{{\sqrt x  + 1}}:\dfrac{{\left( {\sqrt x  - 2} \right)\left( {\sqrt x  + 2} \right)}}{{\left( {1 - \sqrt x } \right)\left( {1 + \sqrt x } \right)}}\\
 = \dfrac{{2 - \sqrt x }}{{\sqrt x  + 1}}.\dfrac{{\left( {1 - \sqrt x } \right)\left( {1 + \sqrt x } \right)}}{{\left( {\sqrt x  - 2} \right)\left( {\sqrt x  + 2} \right)}} = \dfrac{{\sqrt x  - 1}}{{\sqrt x  + 2}}\\
b,\\
Q < 1 \Leftrightarrow \dfrac{{\sqrt x  - 1}}{{\sqrt x  + 2}} - 1 < 0\\
 \Leftrightarrow \dfrac{{\sqrt x  - 1 - \left( {\sqrt x  + 2} \right)}}{{\sqrt x  + 2}} < 0\\
 \Leftrightarrow \dfrac{{ - 3}}{{\sqrt x  + 2}} < 0,\,\,\,\forall x \ge 0,x \ne 1\\
c,\\
P = \dfrac{{\sqrt x  - 1}}{{\sqrt x  + 2}} = \dfrac{{\left( {\sqrt x  + 2} \right) - 3}}{{\sqrt x  + 2}} = 1 - \dfrac{3}{{\sqrt x  + 2}}\\
\sqrt x  \ge 0 \Rightarrow \sqrt x  + 2 \ge 2 \Rightarrow \dfrac{3}{{\sqrt x  + 2}} \le \dfrac{3}{2}\\
 \Rightarrow P = 1 - \dfrac{3}{{\sqrt x  + 2}} \ge 1 - \dfrac{3}{2} =  - \dfrac{1}{2}\\
 \Rightarrow {P_{\min }} =  - \dfrac{1}{2} \Leftrightarrow x = 0
\end{array}\)