Giải thích các bước giải:

Ta có:

\(\begin{array}{l}
a,\\
DKXD:\,\,\,\left\{ \begin{array}{l}
x \ge 0\\
x - 1 \ge 0\\
\sqrt x  - \sqrt {x - 1}  \ne 0\\
\sqrt {x - 1}  - \sqrt 2  \ne 0\\
\sqrt 2  - \sqrt x  \ne 0\\
\sqrt {2x}  - x \ne 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ge 0\\
x \ge 1\\
x \ne x - 1\\
x - 1 \ne 2\\
x \ne 2\\
x \ne 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ge 1\\
x \ne 2\\
x \ne 3
\end{array} \right.\\
b,\\
P = \left( {\dfrac{1}{{\sqrt x  - \sqrt {x - 1} }} - \dfrac{{x - 3}}{{\sqrt {x - 1}  - \sqrt 2 }}} \right)\left( {\dfrac{2}{{\sqrt 2  - \sqrt x }} - \dfrac{{\sqrt x  + \sqrt 2 }}{{\sqrt {2x}  - x}}} \right)\\
 = \left( {\dfrac{{\sqrt x  + \sqrt {x - 1} }}{{\left( {\sqrt x  - \sqrt {x - 1} } \right)\left( {\sqrt x  + \sqrt {x - 1} } \right)}} - \dfrac{{\left( {x - 3} \right)\left( {\sqrt {x - 1}  + \sqrt 2 } \right)}}{{\left( {\sqrt {x - 1}  - \sqrt 2 } \right)\left( {\sqrt {x - 1}  + \sqrt 2 } \right)}}} \right)\left( {\dfrac{2}{{\sqrt 2  - \sqrt x }} - \dfrac{{\sqrt x  + \sqrt 2 }}{{\sqrt x .\left( {\sqrt 2  - \sqrt x } \right)}}} \right)\\
 = \left( {\dfrac{{\sqrt x  + \sqrt {x - 1} }}{{x - \left( {x - 1} \right)}} - \dfrac{{\left( {x - 3} \right)\left( {\sqrt {x - 1}  + \sqrt 2 } \right)}}{{\left( {x - 1} \right) - 2}}} \right)\left( {\dfrac{{2\sqrt x  - \left( {\sqrt x  + \sqrt 2 } \right)}}{{\sqrt x .\left( {\sqrt 2  - \sqrt x } \right)}}} \right)\\
 = \left( {\dfrac{{\sqrt x  + \sqrt {x - 1} }}{1} - \dfrac{{\left( {x - 3} \right)\left( {\sqrt {x - 1}  + \sqrt 2 } \right)}}{{x - 3}}} \right)\dfrac{{\sqrt x  - \sqrt 2 }}{{\sqrt x \left( {\sqrt 2  - \sqrt x } \right)}}\\
 = \left( {\sqrt x  + \sqrt {x - 1}  - \left( {\sqrt {x - 1}  + \sqrt 2 } \right)} \right).\dfrac{{ - 1}}{{\sqrt x }}\\
 = \left( {\sqrt x  - \sqrt 2 } \right).\dfrac{{ - 1}}{{\sqrt x }}\\
 = \dfrac{{\sqrt 2  - \sqrt x }}{{\sqrt x }}\\
c,\\
x = 3 - 2\sqrt 2  = 2 - 2.\sqrt 2 .1 + 1 = {\left( {\sqrt 2  - 1} \right)^2}\\
 \Rightarrow \sqrt x  = \sqrt 2  - 1\\
 \Rightarrow P = \dfrac{{\sqrt 2  - \left( {\sqrt 2  - 1} \right)}}{{\sqrt 2  - 1}} = \dfrac{1}{{\sqrt 2  - 1}} = \dfrac{{\sqrt 2  + 1}}{{\left( {\sqrt 2  - 1} \right)\left( {\sqrt 2  + 1} \right)}} = \dfrac{{\sqrt 2  + 1}}{{2 - 1}} = \sqrt 2  + 1\\
d,\\
P = \dfrac{{\sqrt 2  - \sqrt x }}{{\sqrt x }} = \dfrac{{\sqrt 2 }}{{\sqrt x }} - 1\\
x \ge 1 \Rightarrow \sqrt x  \ge 1 \Rightarrow \dfrac{{\sqrt 2 }}{{\sqrt x }} \le \dfrac{{\sqrt 2 }}{1} = \sqrt 2 \\
 \Rightarrow P = \dfrac{{\sqrt 2 }}{{\sqrt x }} - 1 \le \sqrt 2  - 1\\
 \Rightarrow {P_{\max }} = \sqrt 2  - 1 \Leftrightarrow x = 1
\end{array}\)