Đáp án:
\(\dfrac{{\left( {\sqrt x - 1} \right)\left( {x - \sqrt x + 1} \right)}}{{{{\left( {\sqrt x + 1} \right)}^2}}}\)
Giải thích các bước giải:
\(\begin{array}{l}
DK:x \ge 0\\
A = \left( {\dfrac{{x + 1 - \sqrt x }}{{x + 1}}} \right):\left[ {\dfrac{1}{{\sqrt x - 1}} + \dfrac{{2\sqrt x }}{{x\left( {\sqrt x - 1} \right) + \left( {\sqrt x - 1} \right)}}} \right]\\
= \dfrac{{x - \sqrt x + 1}}{{x + 1}}:\left[ {\dfrac{1}{{\sqrt x - 1}} + \dfrac{{2\sqrt x }}{{\left( {x + 1} \right)\left( {\sqrt x - 1} \right)}}} \right]\\
= \dfrac{{x - \sqrt x + 1}}{{x + 1}}:\dfrac{{x + 1 + 2\sqrt x }}{{\left( {x + 1} \right)\left( {\sqrt x - 1} \right)}}\\
= \dfrac{{x - \sqrt x + 1}}{{x + 1}}.\dfrac{{\left( {x + 1} \right)\left( {\sqrt x - 1} \right)}}{{{{\left( {\sqrt x + 1} \right)}^2}}}\\
= \dfrac{{\left( {\sqrt x - 1} \right)\left( {x - \sqrt x + 1} \right)}}{{{{\left( {\sqrt x + 1} \right)}^2}}}
\end{array}\)
