Đáp án:
\(\begin{array}{l}
a)A = \dfrac{{\sqrt x - 1}}{{\sqrt x }}\\
b)0 < x < 4;x \ne 1
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)A = \dfrac{{1 + \sqrt x }}{{\sqrt x \left( {\sqrt x - 1} \right)}}:\dfrac{{\sqrt x + 1}}{{{{\left( {\sqrt x - 1} \right)}^2}}}\\
= \dfrac{{1 + \sqrt x }}{{\sqrt x \left( {\sqrt x - 1} \right)}}.\dfrac{{{{\left( {\sqrt x - 1} \right)}^2}}}{{\sqrt x + 1}}\\
= \dfrac{{\sqrt x - 1}}{{\sqrt x }}\\
b)A < \dfrac{1}{2}\\
\to \dfrac{{\sqrt x - 1}}{{\sqrt x }} < \dfrac{1}{2}\\
\to \dfrac{{2\sqrt x - 2 - \sqrt x }}{{2\sqrt x }} < 0\\
\to \dfrac{{\sqrt x - 2}}{{2\sqrt x }} < 0\\
\to \sqrt x - 2 < 0\left( {do:\sqrt x > 0\forall x > 0} \right)\\
\to 0 < x < 4;x \ne 1
\end{array}\)
