Đáp án:
$\begin{array}{l}
1)x = \dfrac{{\sqrt 2 - 1}}{{\sqrt 2 + 1}} = \dfrac{{\left( {\sqrt 2 - 1} \right)\left( {\sqrt 2 - 1} \right)}}{{2 - 1}}\\
= {\left( {\sqrt 2 - 1} \right)^2}\\
\Rightarrow \sqrt x = \sqrt 2 - 1\\
A = \dfrac{x}{{x - \sqrt x + 1}} = \dfrac{{{{\left( {\sqrt 2 - 1} \right)}^2}}}{{{{\left( {\sqrt 2 - 1} \right)}^2} - \sqrt 2 - 1 + 1}}\\
= \dfrac{{3 - 2\sqrt 2 }}{{3 - 2\sqrt 2 - \sqrt 2 }}\\
= \dfrac{{3 - 2\sqrt 2 }}{{3 - 3\sqrt 2 }}\\
= \dfrac{{\left( {3 - 2\sqrt 2 } \right)\left( {1 + \sqrt 2 } \right)}}{{3.\left( {1 - \sqrt 2 } \right)\left( {1 + \sqrt 2 } \right)}}\\
= \dfrac{{3 + 3\sqrt 2 - 2\sqrt 2 - 8}}{{3.\left( { - 1} \right)}}\\
= \dfrac{{5 - \sqrt 2 }}{3}\\
2)B = \dfrac{{x + 2\sqrt x + 1}}{{x\sqrt x + 1}}\\
= \dfrac{{{{\left( {\sqrt x + 1} \right)}^2}}}{{\left( {\sqrt x + 1} \right)\left( {x - \sqrt x + 1} \right)}}\\
= \dfrac{{\sqrt x + 1}}{{x - \sqrt x + 1}}\\
P = \dfrac{{1 - A}}{B}\\
= \left( {1 - \dfrac{x}{{x - \sqrt x + 1}}} \right):\dfrac{{\sqrt x + 1}}{{x - \sqrt x + 1}}\\
= \dfrac{{ - \sqrt x - 1}}{{x - \sqrt x + 1}}.\dfrac{{x - \sqrt x + 1}}{{\sqrt x + 1}}\\
= - 1\\
\left( {x - 1} \right).P = - 9\\
\Rightarrow \left( {x - 1} \right).\left( { - 1} \right) = - 9\\
\Rightarrow x - 1 = 9\\
\Rightarrow x = 10\left( {tmdk} \right)\\
3)P > \dfrac{1}{2}\\
\Rightarrow - 1 > \dfrac{1}{2}\left( l \right)
\end{array}$
Vậy ko có giá trị của x để P>1/2
