Đáp án:
\(0 \le x < \dfrac{1}{4}\)
Giải thích các bước giải:
\(\begin{array}{l}
B = \left[ {\dfrac{{x\sqrt x + x + \sqrt x + \left( {\sqrt x + 3} \right)\left( {x + \sqrt x + 1} \right)}}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}} \right].\dfrac{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}{{2x + \sqrt x - 1}}\\
= \dfrac{{x\sqrt x + x + \sqrt x + x\sqrt x + 3x + x + 3\sqrt x + \sqrt x + 3}}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}.\dfrac{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}{{2x + \sqrt x - 1}}\\
= \dfrac{{2x\sqrt x + 5x + 5\sqrt x + 3}}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}.\dfrac{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}{{\left( {2\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}\\
= \dfrac{{\left( {2\sqrt x + 3} \right)\left( {x + \sqrt x + 1} \right)}}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}.\dfrac{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}{{\left( {2\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}\\
= \dfrac{{2\sqrt x + 3}}{{2\sqrt x - 1}}\\
B < 0\\
\to \dfrac{{2\sqrt x + 3}}{{2\sqrt x - 1}} < 0\\
\to 2\sqrt x - 1 < 0\\
\to \sqrt x < \dfrac{1}{2}\\
\to x < \dfrac{1}{4}\\
KL:0 \le x < \dfrac{1}{4}
\end{array}\)
