Đáp án:
b) \(0 < a < 1\)
Giải thích các bước giải:
\(\begin{array}{l}
C4:\\
a)A = \left( {\dfrac{{\sqrt a }}{{\sqrt a - 1}} - \dfrac{{\sqrt a }}{{a - \sqrt a }}} \right):\left( {\dfrac{{\sqrt a + 1}}{{a - 1}}} \right)\\
= \left[ {\dfrac{{\sqrt a .\sqrt a - \sqrt a }}{{\sqrt a \left( {\sqrt a - 1} \right)}}} \right].\dfrac{{\left( {\sqrt a + 1} \right)\left( {\sqrt a - 1} \right)}}{{\sqrt a + 1}}\\
= \dfrac{{a - \sqrt a }}{{\sqrt a \left( {\sqrt a - 1} \right)}}.\dfrac{{\left( {\sqrt a + 1} \right)\left( {\sqrt a - 1} \right)}}{{\sqrt a + 1}}\\
= \dfrac{{\sqrt a \left( {\sqrt a - 1} \right)}}{{\sqrt a \left( {\sqrt a - 1} \right)}}.\dfrac{{\left( {\sqrt a + 1} \right)\left( {\sqrt a - 1} \right)}}{{\sqrt a + 1}}\\
= \sqrt a - 1\\
b)A < 0\\
\to \sqrt a - 1 < 0\\
\to \sqrt a < 1\\
\to 0 < a < 1
\end{array}\)
