Đáp án:

\(\begin{array}{l}
3,\\
a,\\
P = \dfrac{{x + 8}}{{\sqrt x  + 1}}\\
b,\\
P = \dfrac{{58 - 2\sqrt 5 }}{{11}}\\
c,\\
{P_{\min }} = 4 \Leftrightarrow x = 4\\
4,\\
a,\\
M = \dfrac{{x + 2\sqrt x  + 3}}{{\sqrt x  + 1}}\\
b,\\
x = 9\\
5,\\
a,\\
K = \dfrac{1}{{\sqrt x  - 2}}\\
b,\\
Phương\,\,\,trình\,\,vô\,\,\,nghiệm
\end{array}\)

Giải thích các bước giải:

 Ta có:

\(\begin{array}{l}
3,\\
DKXD:\,\,\,\,\left\{ \begin{array}{l}
x \ge 0\\
x - 2\sqrt x  - 3 \ne 0\\
\sqrt x  + 1 \ne 0\\
3 - \sqrt x  \ne 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ge 0\\
x \ne 9
\end{array} \right.\\
a,\\
P = \dfrac{{x\sqrt x  - 3}}{{x - 2\sqrt x  - 3}} - \dfrac{{2\left( {\sqrt x  - 3} \right)}}{{\sqrt x  + 1}} + \dfrac{{\sqrt x  + 3}}{{3 - \sqrt x }}\\
 = \dfrac{{x\sqrt x  - 3}}{{\left( {x - 3\sqrt x } \right) + \left( {\sqrt x  - 3} \right)}} - \dfrac{{2\left( {\sqrt x  - 3} \right)}}{{\sqrt x  + 1}} - \dfrac{{\sqrt x  + 3}}{{\sqrt x  - 3}}\\
 = \dfrac{{x\sqrt x  - 3}}{{\sqrt x \left( {\sqrt x  - 3} \right) + \left( {\sqrt x  - 3} \right)}} - \dfrac{{2\left( {\sqrt x  - 3} \right)}}{{\sqrt x  + 1}} - \dfrac{{\sqrt x  + 3}}{{\sqrt x  - 3}}\\
 = \dfrac{{x\sqrt x  - 3}}{{\left( {\sqrt x  - 3} \right)\left( {\sqrt x  + 1} \right)}} - \dfrac{{2\left( {\sqrt x  - 3} \right)}}{{\sqrt x  + 1}} - \dfrac{{\sqrt x  + 3}}{{\sqrt x  - 3}}\\
 = \dfrac{{\left( {x\sqrt x  - 3} \right) - 2.{{\left( {\sqrt x  - 3} \right)}^2} - \left( {\sqrt x  + 3} \right)\left( {\sqrt x  + 1} \right)}}{{\left( {\sqrt x  - 3} \right)\left( {\sqrt x  + 1} \right)}}\\
 = \dfrac{{\left( {x\sqrt x  - 3} \right) - 2.\left( {x - 6\sqrt x  + 9} \right) - \left( {x + 4\sqrt x  + 3} \right)}}{{\left( {\sqrt x  - 3} \right)\left( {\sqrt x  + 1} \right)}}\\
 = \dfrac{{x\sqrt x  - 3 - 2x + 12\sqrt x  - 18 - x - 4\sqrt x  - 3}}{{\left( {\sqrt x  - 3} \right)\left( {\sqrt x  + 1} \right)}}\\
 = \dfrac{{x\sqrt x  - 3x + 8\sqrt x  - 24}}{{\left( {\sqrt x  - 3} \right)\left( {\sqrt x  + 1} \right)}}\\
 = \dfrac{{x.\left( {\sqrt x  - 3} \right) + 8.\left( {\sqrt x  - 3} \right)}}{{\left( {\sqrt x  - 3} \right)\left( {\sqrt x  + 1} \right)}}\\
 = \dfrac{{\left( {\sqrt x  - 3} \right)\left( {x + 8} \right)}}{{\left( {\sqrt x  - 3} \right)\left( {\sqrt x  + 1} \right)}}\\
 = \dfrac{{x + 8}}{{\sqrt x  + 1}}\\
b,\\
x = 14 - 6\sqrt 5  = 9 - 2.3.\sqrt 5  + 5 = {3^2} - 2.3.\sqrt 5  + {\sqrt 5 ^2} = {\left( {3 - \sqrt 5 } \right)^2}\\
 \Rightarrow \sqrt x  = \sqrt {{{\left( {3 - \sqrt 5 } \right)}^2}}  = \left| {3 - \sqrt 5 } \right| = 3 - \sqrt 5 \\
 \Rightarrow P = \dfrac{{\left( {14 - 6\sqrt 5 } \right) + 8}}{{\left( {3 - \sqrt 5 } \right) + 1}} = \dfrac{{22 - 6\sqrt 5 }}{{4 - \sqrt 5 }}\\
 = \dfrac{{\left( {22 - 6\sqrt 5 } \right)\left( {4 + \sqrt 5 } \right)}}{{\left( {4 - \sqrt 5 } \right)\left( {4 + \sqrt 5 } \right)}}\\
 = \dfrac{{88 + 22\sqrt 5  - 24\sqrt 5  - 30}}{{16 - 5}}\\
 = \dfrac{{58 - 2\sqrt 5 }}{{11}}\\
c,\\
P = \dfrac{{x + 8}}{{\sqrt x  + 1}}\\
P - 4 = \dfrac{{x + 8}}{{\sqrt x  + 1}} - 4 = \dfrac{{x + 8 - 4\left( {\sqrt x  + 1} \right)}}{{\sqrt x  + 1}}\\
 = \dfrac{{x + 8 - 4\sqrt x  - 4}}{{\sqrt x  + 1}} = \dfrac{{x - 4\sqrt x  + 4}}{{\sqrt x  + 1}}\\
 = \dfrac{{{{\left( {\sqrt x  - 2} \right)}^2}}}{{\sqrt x  + 1}}\\
{\left( {\sqrt x  - 2} \right)^2} \ge 0,\,\,\,\forall x \ge 0\\
\sqrt x  + 1 \ge 1 > 0,\,\,\,\forall x \ge 0\\
 \Rightarrow P - 4 \ge 0,\,\,\,\,\forall x \ge 0,x \ne 9\\
 \Rightarrow P \ge 4,\,\,\,\,\forall x \ge 0,x \ne 9\\
 \Rightarrow {P_{\min }} = 4 \Leftrightarrow {\left( {\sqrt x  - 2} \right)^2} = 0 \Leftrightarrow \sqrt x  = 2 \Leftrightarrow x = 4\\
Vậy\,\,\,{P_{\min }} = 4 \Leftrightarrow x = 4\\
4,\\
DKXD:\,\,\,\left\{ \begin{array}{l}
x \ge 0\\
x - \sqrt x  \ne 0\\
x + \sqrt x  \ne 0\\
\sqrt x  + 1 \ne 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x > 0\\
x \ne 1
\end{array} \right.\\
a,\\
M = \dfrac{{x\sqrt x  - 1}}{{x - \sqrt x }} - \dfrac{{x\sqrt x  + 1}}{{x + \sqrt x }} + \dfrac{{x + 1}}{{\sqrt x  + 1}}\\
 = \dfrac{{{{\sqrt x }^3} - {1^3}}}{{\sqrt x \left( {\sqrt x  - 1} \right)}} - \dfrac{{{{\sqrt x }^3} + {1^3}}}{{\sqrt x \left( {\sqrt x  + 1} \right)}} + \dfrac{{x + 1}}{{\sqrt x  + 1}}\\
 = \dfrac{{\left( {\sqrt x  - 1} \right)\left( {x + \sqrt x  + 1} \right)}}{{\sqrt x \left( {\sqrt x  - 1} \right)}} - \dfrac{{\left( {\sqrt x  + 1} \right)\left( {x - \sqrt x  + 1} \right)}}{{\sqrt x \left( {\sqrt x  + 1} \right)}} + \dfrac{{x + 1}}{{\sqrt x  + 1}}\\
 = \dfrac{{x + \sqrt x  + 1}}{{\sqrt x }} - \dfrac{{x - \sqrt x  + 1}}{{\sqrt x }} + \dfrac{{x + 1}}{{\sqrt x  + 1}}\\
 = \dfrac{{\left( {x + \sqrt x  + 1} \right) - \left( {x - \sqrt x  + 1} \right)}}{{\sqrt x }} + \dfrac{{x + 1}}{{\sqrt x  + 1}}\\
 = \dfrac{{2\sqrt x }}{{\sqrt x }} + \dfrac{{x + 1}}{{\sqrt x  + 1}}\\
 = 2 + \dfrac{{x + 1}}{{\sqrt x  + 1}}\\
 = \dfrac{{2.\left( {\sqrt x  + 1} \right) + x + 1}}{{\sqrt x  + 1}}\\
 = \dfrac{{x + 2\sqrt x  + 3}}{{\sqrt x  + 1}}\\
b,\\
M = \dfrac{9}{2}\\
 \Leftrightarrow \dfrac{{x + 2\sqrt x  + 3}}{{\sqrt x  + 1}} = \dfrac{9}{2}\\
 \Leftrightarrow 2.\left( {x + 2\sqrt x  + 3} \right) = 9.\left( {\sqrt x  + 1} \right)\\
 \Leftrightarrow 2x + 4\sqrt x  + 6 = 9\sqrt x  + 9\\
 \Leftrightarrow 2x - 5\sqrt x  - 3 = 0\\
 \Leftrightarrow \left( {2x - 6\sqrt x } \right) + \left( {\sqrt x  - 3} \right) = 0\\
 \Leftrightarrow 2\sqrt x \left( {\sqrt x  - 3} \right) + \left( {\sqrt x  - 3} \right) = 0\\
 \Leftrightarrow \left( {\sqrt x  - 3} \right)\left( {2\sqrt x  + 1} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
\sqrt x  - 3 = 0\\
2\sqrt x  + 1 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\sqrt x  = 3\\
\sqrt x  =  - \dfrac{1}{2}\,\,\,\left( L \right)
\end{array} \right.\\
 \Rightarrow \sqrt x  = 3 \Rightarrow x = 9\\
5,\\
a,\\
K = \left( {1 - \dfrac{{x - 3\sqrt x }}{{x - 9}}} \right):\left( {\dfrac{{\sqrt x  - 3}}{{2 - \sqrt x }} + \dfrac{{\sqrt x  - 2}}{{3 + \sqrt x }} - \dfrac{{9 - x}}{{x + \sqrt x  - 6}}} \right)\\
 = \left( {1 - \dfrac{{\sqrt x \left( {\sqrt x  - 3} \right)}}{{\left( {\sqrt x  - 3} \right)\left( {\sqrt x  + 3} \right)}}} \right):\left( { - \dfrac{{\sqrt x  - 3}}{{\sqrt x  - 2}} + \dfrac{{\sqrt x  - 2}}{{\sqrt x  + 3}} - \dfrac{{9 - x}}{{\left( {\sqrt x  + 3} \right)\left( {\sqrt x  - 2} \right)}}} \right)\\
 = \left( {1 - \dfrac{{\sqrt x }}{{\sqrt x  + 3}}} \right):\dfrac{{ - \left( {\sqrt x  - 3} \right)\left( {\sqrt x  + 3} \right) + {{\left( {\sqrt x  - 2} \right)}^2} - \left( {9 - x} \right)}}{{\left( {\sqrt x  - 2} \right)\left( {\sqrt x  + 3} \right)}}\\
 = \dfrac{{\left( {\sqrt x  + 1} \right) - \sqrt x }}{{\sqrt x  + 3}}:\dfrac{{ - \left( {x - 9} \right) + {{\left( {\sqrt x  - 2} \right)}^2} + \left( {x - 9} \right)}}{{\left( {\sqrt x  - 2} \right)\left( {\sqrt x  + 3} \right)}}\\
 = \dfrac{1}{{\sqrt x  + 3}}:\dfrac{{{{\left( {\sqrt x  - 2} \right)}^2}}}{{\left( {\sqrt x  - 2} \right)\left( {\sqrt x  + 3} \right)}}\\
 = \dfrac{1}{{\sqrt x  + 3}}:\dfrac{{\sqrt x  - 2}}{{\sqrt x  + 3}}\\
 = \dfrac{1}{{\sqrt x  - 2}}\\
b,\\
K = 1 \Leftrightarrow \dfrac{1}{{\sqrt x  - 2}} = 1\\
 \Leftrightarrow \sqrt x  - 2 = 1\\
 \Leftrightarrow \sqrt x  = 3\\
 \Leftrightarrow x = 9\,\,\,\,\left( {không\,\,\,thoả\,\,\,mãn\,\,\,ĐKXĐ} \right)\\
Vậy\,\,\,phương\,\,\,trình\,\,\,đã\,\,\,cho\,\,\,vô\,\,\,nghiệm
\end{array}\)