Đáp án:
\(9 > x > 0;x \ne 1\)
Giải thích các bước giải:
\(\begin{array}{l}
B3:\\
DK:x > 0;x \ne 1\\
B = \dfrac{{\sqrt x - 1 + \sqrt x + 1}}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}.\dfrac{{\sqrt x - 1}}{{\sqrt x }}\\
= \dfrac{{2\sqrt x }}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}.\dfrac{{\sqrt x - 1}}{{\sqrt x }}\\
= \dfrac{2}{{\sqrt x + 1}}\\
B > \dfrac{1}{2}\\
\to \dfrac{2}{{\sqrt x + 1}} > \dfrac{1}{2}\\
\to \dfrac{{4 - \sqrt x - 1}}{{2\left( {\sqrt x + 1} \right)}} > 0\\
\to 3 - \sqrt x > 0\left( {do:\sqrt x + 1 > 0\forall x > 0} \right)\\
\to 9 > x > 0;x \ne 1
\end{array}\)
