Đáp án:
a) \(\dfrac{{\sqrt x + 1}}{{\sqrt x - 1}}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)A = \dfrac{{\sqrt x }}{{\sqrt x - 1}} + \dfrac{{2\sqrt x }}{{x - 1}} - \dfrac{1}{{\sqrt x + 1}}\\
= \dfrac{{x + \sqrt x + 2\sqrt x - \sqrt x + 1}}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}\\
= \dfrac{{x + 2\sqrt x + 1}}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}\\
= \dfrac{{{{\left( {\sqrt x + 1} \right)}^2}}}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}} = \dfrac{{\sqrt x + 1}}{{\sqrt x - 1}}\\
b)A < \dfrac{1}{2}\\
\to \dfrac{{\sqrt x + 1}}{{\sqrt x - 1}} < \dfrac{1}{2}\\
\to \dfrac{{2\sqrt x + 2 - \sqrt x + 1}}{{2\left( {\sqrt x - 1} \right)}} < 0\\
\to \dfrac{{\sqrt x + 3}}{{2\left( {\sqrt x - 1} \right)}} < 0\\
\to \sqrt x - 1 < 0\left( {do:\sqrt x + 3 > 0\forall x \ge 0} \right)\\
\to 0 \le x < 1
\end{array}\)
