Giải thích các bước giải:
\(\begin{array}{l}
a,\\
DKXD:\,\,\,\,\left\{ \begin{array}{l}
x \ge 0\\
x - \sqrt x \ne 0\\
x + \sqrt x \ne 0\\
2.\left( {x - 2\sqrt x + 1} \right) \ne 0\\
x - 1 \ne 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x > 0\\
x \ne 1
\end{array} \right.\\
B = \left( {\dfrac{{x\sqrt x - 1}}{{x - \sqrt x }} - \dfrac{{x\sqrt x + 1}}{{x + \sqrt x }}} \right):\dfrac{{2.\left( {x - 2\sqrt x + 1} \right)}}{{x - 1}}\\
= \left( {\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)}}} \right):\dfrac{{2.{{\left( {\sqrt x - 1} \right)}^2}}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}\\
= \left( {\dfrac{{x + \sqrt x + 1}}{{\sqrt x }} - \dfrac{{x - \sqrt x + 1}}{{\sqrt x }}} \right):\dfrac{{2\left( {\sqrt x - 1} \right)}}{{\sqrt x + 1}}\\
= \dfrac{{2\sqrt x }}{{\sqrt x }}:\dfrac{{2.\left( {\sqrt x - 1} \right)}}{{\sqrt x + 1}}\\
= 2.\dfrac{{\sqrt x + 1}}{{2.\left( {\sqrt x - 1} \right)}}\\
= \dfrac{{\sqrt x + 1}}{{\sqrt x - 1}}\\
b,\\
B = \dfrac{{\sqrt x + 1}}{{\sqrt x - 1}} = \dfrac{{\left( {\sqrt x - 1} \right) + 2}}{{\sqrt x - 1}} = 1 + \dfrac{2}{{\sqrt x - 1}}\\
B \in Z \Leftrightarrow \dfrac{2}{{\sqrt x - 1}} \in Z\\
\Leftrightarrow \left( {\sqrt x - 1} \right) \in \left\{ { \pm 1; \pm 2} \right\}\\
\Rightarrow \sqrt x \in \left\{ { - 1;0;2;3} \right\}\\
\Rightarrow x \in \left\{ {0;4;9} \right\}\\
x > 0,x \ne 1 \Rightarrow x \in \left\{ {4;9} \right\}\\
c,\\
P < \dfrac{1}{3} \Leftrightarrow \dfrac{{\sqrt x + 1}}{{\sqrt x - 1}} < \dfrac{1}{3}\\
\Leftrightarrow \dfrac{{\sqrt x + 1}}{{\sqrt x - 1}} - \dfrac{1}{3} < 0\\
\Leftrightarrow \dfrac{{3.\left( {\sqrt x + 1} \right) - \left( {\sqrt x - 1} \right)}}{{3.\left( {\sqrt x - 1} \right)}} < 0\\
\Leftrightarrow \dfrac{{2\sqrt x + 4}}{{3.\left( {\sqrt x - 1} \right)}} < 0\\
\Leftrightarrow \dfrac{{2.\left( {\sqrt x + 2} \right)}}{{3.\left( {\sqrt x - 1} \right)}} < 0\\
\Leftrightarrow \sqrt x < 1\\
\Leftrightarrow 0 < x < 1
\end{array}\)
