Đáp án:
\[0 \le x < \frac{1}{4}\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
B = \left( {\frac{{x\sqrt x + x + \sqrt x }}{{x\sqrt x - 1}} - \frac{{\sqrt x + 3}}{{1 - \sqrt x }}} \right).\frac{{x - 1}}{{2x + \sqrt x - 1}}\\
= \left( {\frac{{\sqrt x \left( {x + \sqrt x + 1} \right)}}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}} + \frac{{\sqrt x + 3}}{{\sqrt x - 1}}} \right).\frac{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}{{\left( {2\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}\\
= \left( {\frac{{\sqrt x }}{{\sqrt x - 1}} + \frac{{\sqrt x + 3}}{{\sqrt x - 1}}} \right).\frac{{\sqrt x - 1}}{{2\sqrt x - 1}}\\
= \frac{{2\sqrt x + 3}}{{\sqrt x - 1}}.\frac{{\sqrt x - 1}}{{2\sqrt x - 1}}\\
= \frac{{2\sqrt x + 3}}{{2\sqrt x - 1}}\\
2\sqrt x + 3 > 0,\,\,\,\forall x \Rightarrow B < 0 \Leftrightarrow 2\sqrt x - 1 < 0 \Leftrightarrow 0 \le x < \frac{1}{4}
\end{array}\)
