Giải thích các bước giải:

 a,

ĐKXĐ:

\(\left\{ \begin{array}{l}
{a^2} - a \ne 0\\
{a^2} - 3a + 2 \ne 0\\
{a^2} - 5a + 6 \ne 0\\
{a^2} - 7a + 12 \ne 0\\
{a^2} - 9a + 20 \ne 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
a\left( {a - 1} \right) \ne 0\\
\left( {a - 1} \right)\left( {a - 2} \right) \ne 0\\
\left( {a - 2} \right)\left( {a - 3} \right) \ne 0\\
\left( {a - 3} \right)\left( {a - 4} \right) \ne 0\\
\left( {a - 4} \right)\left( {a - 5} \right) \ne 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
a \ne 0\\
a \ne 1\\
a \ne 2\\
a \ne 3\\
a \ne 4\\
a \ne 5
\end{array} \right.\)

b,

Ta có:

\(\begin{array}{l}
P = \dfrac{1}{{{a^2} - a}} + \dfrac{1}{{{a^2} - 3a + 2}} + \dfrac{1}{{{a^2} - 5a + 6}} + \dfrac{1}{{{a^2} - 7a + 12}} + \dfrac{1}{{{a^2} - 9a + 20}}\\
 = \dfrac{1}{{a\left( {a - 1} \right)}} + \dfrac{1}{{\left( {a - 1} \right)\left( {a - 2} \right)}} + \dfrac{1}{{\left( {a - 2} \right)\left( {a - 3} \right)}} + \dfrac{1}{{\left( {a - 3} \right)\left( {a - 4} \right)}} + \dfrac{1}{{\left( {a - 4} \right)\left( {a - 5} \right)}}\\
 = \dfrac{{a - \left( {a - 1} \right)}}{{a\left( {a - 1} \right)}} + \dfrac{{\left( {a - 1} \right) - \left( {a - 2} \right)}}{{\left( {a - 1} \right)\left( {a - 2} \right)}} + \dfrac{{\left( {a - 2} \right) - \left( {a - 3} \right)}}{{\left( {a - 2} \right)\left( {a - 3} \right)}} + \dfrac{{\left( {a - 3} \right) - \left( {a - 4} \right)}}{{\left( {a - 3} \right)\left( {a - 4} \right)}} + \dfrac{{\left( {a - 4} \right) - \left( {a - 5} \right)}}{{\left( {a - 4} \right)\left( {a - 5} \right)}}\\
 = \left( {\dfrac{1}{{a - 1}} - \dfrac{1}{a}} \right) + \left( {\dfrac{1}{{a - 2}} - \dfrac{1}{{a - 1}}} \right) + \left( {\dfrac{1}{{a - 3}} - \dfrac{1}{{a - 2}}} \right) + \left( {\dfrac{1}{{a - 4}} - \dfrac{1}{{a - 3}}} \right) + \left( {\dfrac{1}{{a - 5}} - \dfrac{1}{{a - 4}}} \right)\\
 =  - \dfrac{1}{a} + \dfrac{1}{{a - 5}}\\
 = \dfrac{1}{{a - 5}} - \dfrac{1}{a}\\
 = \dfrac{{a - \left( {a - 5} \right)}}{{a\left( {a - 5} \right)}}\\
 = \dfrac{5}{{a\left( {a - 5} \right)}}\\
c,\\
{a^3} - {a^2} + 2 = 0\\
 \Leftrightarrow \left( {{a^3} + {a^2}} \right) - \left( {2{a^2} + 2a} \right) + \left( {2a + 2} \right) = 0\\
 \Leftrightarrow {a^2}\left( {a + 1} \right) - 2a\left( {a + 1} \right) + 2\left( {a + 1} \right) = 0\\
 \Leftrightarrow \left( {a + 1} \right)\left( {{a^2} - 2a + 2} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
a + 1 = 0\\
{a^2} - 2a + 2 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
a =  - 1\\
{\left( {a - 1} \right)^2} + 1 = 0
\end{array} \right. \Leftrightarrow a =  - 1\\
 \Rightarrow P = \dfrac{5}{{\left( { - 1} \right).\left( { - 1 - 5} \right)}} = \dfrac{5}{6}
\end{array}\)