Đáp án: $B = \dfrac{{\sqrt x }}{{x + \sqrt x + 1}}$
Giải thích các bước giải:
$\begin{array}{l}
b)B = \dfrac{{x + 2}}{{x\sqrt x - 1}} + \dfrac{{\sqrt x + 1}}{{x + \sqrt x + 1}} - \dfrac{1}{{\sqrt x - 1}}\\
= \dfrac{{x + 2 + \left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right) - \left( {x + \sqrt x + 1} \right)}}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}\\
= \dfrac{{x + 2 + x - 1 - x - \sqrt x - 1}}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}\\
= \dfrac{{x - \sqrt x }}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}\\
= \dfrac{{\sqrt x \left( {\sqrt x - 1} \right)}}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}\\
= \dfrac{{\sqrt x }}{{x + \sqrt x + 1}}
\end{array}$
