Đáp án:
$\begin{array}{l}
Dkxd:a \ne 0;a \ne - 1\\
B = \left( {\dfrac{1}{{{a^2} + a}} - \dfrac{1}{{a + 1}}} \right):\dfrac{{1 - a}}{{{a^2} + 2a + 1}}\\
= \left( {\dfrac{1}{{a\left( {a + 1} \right)}} - \dfrac{1}{{a + 1}}} \right):\dfrac{{1 - a}}{{{{\left( {a + 1} \right)}^2}}}\\
= \dfrac{{1 - a}}{{a\left( {a + 1} \right)}}.\dfrac{{{{\left( {a + 1} \right)}^2}}}{{1 - a}}\\
= \dfrac{{a + 1}}{a}
\end{array}$