Giải thích các bước giải:

Ta có:

\(\begin{array}{l}
1,\\
x = \dfrac{{\sqrt 2  - 1}}{{\sqrt 2  + 1}} = \dfrac{{{{\left( {\sqrt 2  - 1} \right)}^2}}}{{\left( {\sqrt 2  + 1} \right)\left( {\sqrt 2  - 1} \right)}} = \dfrac{{{{\left( {\sqrt 2  - 1} \right)}^2}}}{{2 - 1}} = {\left( {\sqrt 2  - 1} \right)^2}\\
A = \dfrac{x}{{x - \sqrt x  + 1}} = \dfrac{{{{\left( {\sqrt 2  - 1} \right)}^2}}}{{{{\left( {\sqrt 2  - 1} \right)}^2} - \sqrt {{{\left( {\sqrt 2  - 1} \right)}^2}}  + 1}}\\
 = \dfrac{{3 - 2\sqrt 2 }}{{\left( {3 - 2\sqrt 2 } \right) - \left( {\sqrt 2  - 1} \right) + 1}}\\
 = \dfrac{{3 - 2\sqrt 2 }}{{5 - 3\sqrt 2 }} = \dfrac{{\left( {3 - 2\sqrt 2 } \right)\left( {5 + 3\sqrt 2 } \right)}}{{\left( {5 - 3\sqrt 2 } \right)\left( {5 + 3\sqrt 2 } \right)}}\\
 = \dfrac{{15 + 9\sqrt 2  - 10\sqrt 2  - 2\sqrt 2 .3\sqrt 2 }}{{25 - {{\left( {3\sqrt 2 } \right)}^2}}}\\
 = \dfrac{{3 - \sqrt 2 }}{7}\\
2,\\
B = \dfrac{{x + 2\sqrt x  + 1}}{{x\sqrt x  + 1}} = \dfrac{{{{\left( {\sqrt x  + 1} \right)}^2}}}{{\left( {\sqrt x  + 1} \right)\left( {x - \sqrt x  + 1} \right)}} = \dfrac{{\sqrt x  + 1}}{{x - \sqrt x  + 1}}\\
1 - A = 1 - \dfrac{x}{{x - \sqrt x  + 1}} = \dfrac{{\left( {x - \sqrt x  + 1} \right) - x}}{{x - \sqrt x  + 1}} = \dfrac{{ - \sqrt x  + 1}}{{x - \sqrt x  + 1}}\\
P = \dfrac{{1 - A}}{B} = \dfrac{{ - \sqrt x  + 1}}{{x - \sqrt x  + 1}}:\dfrac{{\sqrt x  + 1}}{{x - \sqrt x  + 1}} = \dfrac{{ - \sqrt x  + 1}}{{\sqrt x  + 1}}\\
\left( {x - 1} \right).P =  - 9\\
 \Leftrightarrow \left( {x - 1} \right).\dfrac{{ - \sqrt x  + 1}}{{\sqrt x  + 1}} =  - 9\\
 \Leftrightarrow \left( {\sqrt x  - 1} \right)\left( {\sqrt x  + 1} \right).\dfrac{{\sqrt x  - 1}}{{\sqrt x  + 1}} = 9\\
 \Leftrightarrow {\left( {\sqrt x  - 1} \right)^2} = 9\\
 \Leftrightarrow \left[ \begin{array}{l}
\sqrt x  - 1 = 3\\
\sqrt x  - 1 =  - 3
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\sqrt x  = 4\\
\sqrt x  =  - 2
\end{array} \right. \Leftrightarrow \sqrt x  = 4 \Leftrightarrow x = 16\\
3,\\
P > \dfrac{1}{2} \Leftrightarrow \dfrac{{ - \sqrt x  + 1}}{{\sqrt x  + 1}} - \dfrac{1}{2} > 0\\
 \Leftrightarrow \dfrac{{2.\left( { - \sqrt x  + 1} \right) - \left( {\sqrt x  + 1} \right)}}{{2\left( {\sqrt x  + 1} \right)}} > 0\\
 \Leftrightarrow \dfrac{{ - 3\sqrt x  + 1}}{{2\left( {\sqrt x  + 1} \right)}} > 0\\
 \Leftrightarrow  - 3\sqrt x  + 1 > 0\\
 \Leftrightarrow 3\sqrt x  - 1 < 0\\
 \Leftrightarrow \sqrt x  < \dfrac{1}{3}\\
 \Leftrightarrow 0 \le x < \dfrac{1}{9}
\end{array}\)