Đáp án:
$\begin{array}{l}
B = \dfrac{{a - \sqrt a }}{{a - 1}} + \dfrac{1}{{\sqrt a + 1}}\left( {a \ge 0;a \ne 1} \right)\\
= \dfrac{{\sqrt a \left( {\sqrt a - 1} \right)}}{{\left( {\sqrt a + 1} \right)\left( {\sqrt a - 1} \right)}} + \dfrac{1}{{\sqrt a + 1}}\\
= \dfrac{{\sqrt a }}{{\sqrt a + 1}} + \dfrac{1}{{\sqrt a + 1}}\\
= \dfrac{{\sqrt a + 1}}{{\sqrt a + 1}}\\
= 1
\end{array}$