Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
DKXD:\,\,\,\left\{ \begin{array}{l}
x \ge 0\\
x \ne 1
\end{array} \right.\\
a,\\
P = \left( {\dfrac{{2x + 1}}{{x\sqrt x - 1}} - \dfrac{1}{{\sqrt x - 1}}} \right):\left( {1 - \dfrac{{x + 4}}{{x + \sqrt x + 1}}} \right)\\
= \left( {\dfrac{{2x + 1}}{{{{\sqrt x }^3} - {1^3}}} - \dfrac{1}{{\sqrt x - 1}}} \right):\left( {1 - \dfrac{{x + 4}}{{x + \sqrt x + 1}}} \right)\\
= \left( {\dfrac{{2x + 1}}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}} - \dfrac{1}{{\sqrt x - 1}}} \right):\dfrac{{\left( {x + \sqrt x + 1} \right) - \left( {x + 4} \right)}}{{x + \sqrt x + 1}}\\
= \dfrac{{\left( {2x + 1} \right) - \left( {x + \sqrt x + 1} \right)}}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}:\dfrac{{x + \sqrt x + 1 - x - 4}}{{x + \sqrt x + 1}}\\
= \dfrac{{2x + 1 - x - \sqrt x - 1}}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}:\dfrac{{\sqrt x - 3}}{{x + \sqrt x + 1}}\,\,\,\,\,\,\,\left( {x \ne 9} \right)\\
= \dfrac{{x - \sqrt x }}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}:\dfrac{{\sqrt x - 3}}{{x + \sqrt x + 1}}\\
= \dfrac{{\sqrt x \left( {\sqrt x - 1} \right)}}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}.\dfrac{{x + \sqrt x + 1}}{{\sqrt x - 3}}\\
= \dfrac{{\sqrt x }}{{\sqrt x - 3}}\\
b,\\
P \in {Z^ + } \Leftrightarrow \left\{ \begin{array}{l}
\dfrac{{\sqrt x }}{{\sqrt x - 3}} \in Z\\
\dfrac{{\sqrt x }}{{\sqrt x - 3}} > 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
\dfrac{{\left( {\sqrt x - 3} \right) + 3}}{{\sqrt x - 3}} \in Z\\
\sqrt x - 3 > 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
1 + \dfrac{3}{{\sqrt x - 3}} \in Z\\
\sqrt x > 3
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
\dfrac{3}{{\sqrt x - 3}} \in Z\\
x > 9
\end{array} \right.\\
x \in Z \Rightarrow \left\{ \begin{array}{l}
\left( {\sqrt x - 3} \right) \in \left\{ { \pm 1;\,\, \pm 3} \right\}\\
x > 9
\end{array} \right.\\
\Leftrightarrow \sqrt x - 3 \in \left\{ {1;3} \right\}\\
\Leftrightarrow \sqrt x \in \left\{ {4;6} \right\}\\
\Leftrightarrow x \in \left\{ {16;36} \right\}
\end{array}\)
