E cần gấp ạ .................

Đáp án:

$\begin{array}{l}
a)Dkxd:x > 0;x \ne 1\\
A = \left( {\dfrac{{x - \sqrt x }}{{\sqrt x  - 1}} - \dfrac{{\sqrt x  + 1}}{{\sqrt x  + x}}} \right):\dfrac{{\sqrt x  + 1}}{x}\\
 = \left( {\dfrac{{\sqrt x \left( {\sqrt x  - 1} \right)}}{{\sqrt x  - 1}} - \dfrac{{\sqrt x  + 1}}{{\sqrt x \left( {\sqrt x  + 1} \right)}}} \right).\dfrac{x}{{\sqrt x  + 1}}\\
 = \left( {\sqrt x  - \dfrac{1}{{\sqrt x }}} \right).\dfrac{x}{{\sqrt x  + 1}}\\
 = \dfrac{{x - 1}}{{\sqrt x }}.\dfrac{x}{{\sqrt x  + 1}}\\
 = \dfrac{{\left( {\sqrt x  + 1} \right)\left( {\sqrt x  - 1} \right).\sqrt x }}{{\sqrt x  + 1}}\\
 = \sqrt x \left( {\sqrt x  - 1} \right)\\
 = x - \sqrt x \\
b)A =  - \dfrac{1}{4}\\
 \Leftrightarrow x - \sqrt x  =  - \dfrac{1}{4}\\
 \Leftrightarrow x - \sqrt x  + \dfrac{1}{4} = 0\\
 \Leftrightarrow x - 2.\sqrt x .\dfrac{1}{2} + \dfrac{1}{4} = 0\\
 \Leftrightarrow {\left( {\sqrt x  - \dfrac{1}{2}} \right)^2} = 0\\
 \Leftrightarrow \sqrt x  = \dfrac{1}{2}\\
 \Leftrightarrow x = \dfrac{1}{4}\left( {tmdk} \right)\\
Vậy\,x = \dfrac{1}{4}
\end{array}$