Đáp án:
b) \(9 > x \ge 0;x \ne 1\)
Giải thích các bước giải:
\(\begin{array}{l}
B2:\\
C1:\\
A = \dfrac{{2x - 4\sqrt x + 2 - \left( {2\sqrt x - 1} \right)\left( {\sqrt x - 2} \right)}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}.\dfrac{{\sqrt x - 2}}{{\sqrt x }}\\
= \dfrac{{2x - 4\sqrt x + 2 - 2x + 5\sqrt x - 2}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}.\dfrac{{\sqrt x - 2}}{{\sqrt x }}\\
= \dfrac{{\sqrt x }}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}.\dfrac{{\sqrt x - 2}}{{\sqrt x }}\\
= \dfrac{1}{{\sqrt x + 2}}\\
C2:\\
a)B = \left[ {\dfrac{{\sqrt x + 3 - 3 + \sqrt x }}{{\left( {\sqrt x + 3} \right)\left( {3 - \sqrt x } \right)}}} \right].\dfrac{{\sqrt x + 3}}{{\sqrt x }}\\
= \dfrac{{2\sqrt x }}{{\left( {\sqrt x + 3} \right)\left( {3 - \sqrt x } \right)}}.\dfrac{{\sqrt x + 3}}{{\sqrt x }}\\
= \dfrac{2}{{3 - \sqrt x }}\\
b)B > \dfrac{1}{2}\\
\to \dfrac{2}{{3 - \sqrt x }} > \dfrac{1}{2}\\
\to \dfrac{{4 - 3 + \sqrt x }}{{2\left( {3 - \sqrt x } \right)}} > 0\\
\to \dfrac{{\sqrt x + 1}}{{2\left( {3 - \sqrt x } \right)}} > 0\\
\to 3 - \sqrt x > 0\left( {do:\sqrt x + 1 > 0\forall x \ge 0} \right)\\
\to 9 > x \ge 0;x \ne 1
\end{array}\)
