Đáp án:$\dfrac{5}{{{a^2} + 5a}}$
Giải thích các bước giải:
$\begin{array}{l}
\dfrac{1}{{x\left( {x + 1} \right)}} = \dfrac{{x + 1 - x}}{{x\left( {x + 1} \right)}} = \dfrac{1}{x} - \dfrac{1}{{x + 1}}\\
\left\{ \begin{array}{l}
\dfrac{1}{{{a^2} + a}} = \dfrac{1}{{a\left( {a + 1} \right)}} = \dfrac{1}{a} - \dfrac{1}{{a + 1}}\\
\dfrac{1}{{{a^2} + 3a + 2}} = \dfrac{1}{{\left( {a + 1} \right)\left( {a + 2} \right)}} = \dfrac{1}{{a + 1}} - \dfrac{1}{{a + 2}}\\
....\dfrac{1}{{{a^2} + 9a + 20}} = \dfrac{1}{{\left( {a + 4} \right)\left( {a + 5} \right)}} = \dfrac{1}{{a + 4}} - \dfrac{1}{{a + 5}}
\end{array} \right.\\
\Leftrightarrow \dfrac{1}{{{a^2} + a}} + \dfrac{1}{{{a^2} + 3a + 2}} + ... + \dfrac{1}{{{a^2} + 9a + 20}}\\
= \dfrac{1}{a} - \dfrac{1}{{a + 1}} + \dfrac{1}{{a + 1}} - \dfrac{1}{{a + 2}} + ... + \dfrac{1}{{a + 4}} - \dfrac{1}{{a + 5}}\\
= \dfrac{1}{a} - \dfrac{1}{{a + 5}}\\
= \dfrac{{a + 5 - a}}{{a\left( {a + 5} \right)}}\\
= \dfrac{5}{{{a^2} + 5a}}
\end{array}$
