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