Giải thích các bước giải:
1,
a, Ta có:
\(\begin{array}{l}
P = \left( {\frac{{2x + 1}}{{x\sqrt x - 1}} - \frac{{\sqrt x }}{{x + \sqrt x + 1}}} \right).\left( {\frac{{x\sqrt x - 1}}{{\sqrt x - 1}} - \sqrt x } \right)\\
= \left( {\frac{{2x + 1}}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}} - \frac{{\sqrt x \left( {\sqrt x - 1} \right)}}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}} \right).\left( {\frac{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}{{\sqrt x - 1}} - \sqrt x } \right)\\
= \left( {\frac{{2x + 1 - x + \sqrt x }}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}} \right).\left( {x + \sqrt x + 1 - \sqrt x } \right)\\
= \frac{{x + \sqrt x + 1}}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}.\left( {x + 1} \right)\\
= \frac{{x + 1}}{{\sqrt x - 1}}
\end{array}\)
b,
\(P.\sqrt {1 - x} < 0 \Leftrightarrow \frac{{x + 1}}{{\sqrt x - 1}}.\sqrt {1 - x} < 0 \Leftrightarrow 0 < x < 1\)