Giải thích các bước giải:
\(\begin{array}{l}
1.A = \frac{{x - 1}}{{\sqrt x \left( {\sqrt x - 1} \right)}} = \frac{{\sqrt x + 1}}{{\sqrt x }}\\
B = \frac{{\sqrt x + 1 + 2}}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}} = \frac{{\sqrt x + 3}}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}\\
2.A = \frac{{\sqrt x + 1}}{{\sqrt x }} = 1 + \frac{1}{{\sqrt x }}\\
Do:1 + \frac{1}{{\sqrt x }} > 1\\
\to A > 1\\
3.B < 0\\
\to \frac{{\sqrt x + 3}}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}} < 0\\
Do:\sqrt x + 3 > 0\forall x \ge 0\\
\to \frac{{\sqrt x + 3}}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}} < 0\\
\Leftrightarrow \left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right) < 0\\
\Leftrightarrow \left\{ \begin{array}{l}
\sqrt x + 1 > 0\left( {ld} \right)\\
\sqrt x - 1 < 0
\end{array} \right. \to 0 \le x < 1
\end{array}\)
