Giải thích các bước giải:

Ta có:

\(\begin{array}{l}
a,\\
A = \left( {\dfrac{1}{{\sqrt x  - 1}} - \dfrac{{\sqrt x }}{{x - 1}}} \right):\dfrac{1}{{\sqrt x  + 1}}\,\,\,\,\,\,\,\,\,\,\,\,\,\left( \begin{array}{l}
x \ge 0\\
x \ne 1
\end{array} \right)\\
 = \left( {\dfrac{1}{{\sqrt x  - 1}} - \dfrac{{\sqrt x }}{{\left( {\sqrt x  - 1} \right)\left( {\sqrt x  + 1} \right)}}} \right):\dfrac{1}{{\sqrt x  + 1}}\\
 = \dfrac{{\left( {\sqrt x  + 1} \right) - \sqrt x }}{{\left( {\sqrt x  - 1} \right)\left( {\sqrt x  + 1} \right)}}:\dfrac{1}{{\sqrt x  + 1}}\\
 = \dfrac{1}{{\left( {\sqrt x  - 1} \right)\left( {\sqrt x  + 1} \right)}}.\left( {\sqrt x  + 1} \right)\\
 = \dfrac{1}{{\sqrt x  - 1}}\\
b,\\
A < 0 \Leftrightarrow \dfrac{1}{{\sqrt x  - 1}} < 0 \Leftrightarrow \sqrt x  - 1 < 0\\
 \Leftrightarrow \sqrt x  < 1 \Leftrightarrow 0 \le x < 1
\end{array}\) 

CHÚC BẠN HỌC TỐT!!!

$A=\bigg{(}\dfrac{1}{\sqrt{x}-1}-\dfrac{\sqrt{x}}{x-1}\bigg{)}:\dfrac{1}{\sqrt{x}+1}$ $(x\geq0;x\neq1)$.

$=\bigg{(}\dfrac{\sqrt{x}+1}{x-1}-\dfrac{\sqrt{x}}{x-1}\bigg{)}.(\sqrt{x}+1)$

$=\dfrac{1.(\sqrt{x+1}+1)}{(\sqrt{x}-1)(\sqrt{x}+1)}$

$=\dfrac{1}{\sqrt{x}-1}$