Đáp án:
c) \(9 > x \ge 0\)
Giải thích các bước giải:
\(\begin{array}{l}
B1:\\
1)Thay:x = \dfrac{1}{4}\\
\to A = \dfrac{{\sqrt {\dfrac{1}{4}} + 1}}{{\sqrt {\dfrac{1}{4}} - 3}} = - \dfrac{3}{5}\\
2)B = \dfrac{{2\sqrt x \left( {\sqrt x - 3} \right) + \sqrt x \left( {\sqrt x + 3} \right) - 3x - 3}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}}\\
= \dfrac{{2x - 6\sqrt x + x + 3\sqrt x - 3x - 3}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}}\\
= \dfrac{{ - 3\sqrt x - 3}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}}\\
3)\dfrac{B}{A} = \dfrac{{ - 3\sqrt x - 3}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}}:\dfrac{{\sqrt x + 1}}{{\sqrt x - 3}}\\
= \dfrac{{ - 3\left( {\sqrt x + 1} \right)}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}}.\dfrac{{\sqrt x - 3}}{{\sqrt x + 1}}\\
= - \dfrac{3}{{\sqrt x + 3}}\\
\dfrac{B}{A} < - \dfrac{1}{2}\\
\to - \dfrac{3}{{\sqrt x + 3}} < - \dfrac{1}{2}\\
\to \dfrac{3}{{\sqrt x + 3}} > \dfrac{1}{2}\\
\to \dfrac{{6 - \sqrt x - 3}}{{2\left( {\sqrt x + 3} \right)}} > 0\\
\to 3 - \sqrt x > 0\left( {do:\sqrt x + 3 > 0\forall x \ge 0} \right)\\
\to 9 > x \ge 0
\end{array}\)
