Đáp án:
$\begin{array}{l}
f\left( x \right) = 2{x^{2011}} - 3{x^{2010}} + 2{x^{2009}} - 3{x^{2008}}\\
+ ... + 2{x^3} - 3{x^2} + 2x - 3\\
= {x^{2010}}\left( {2x - 3} \right) + {x^{2008}}\left( {2x - 3} \right) + ... + {x^2}\left( {2x - 3} \right) + \left( {2x - 3} \right)\\
= \left( {2x - 3} \right)\left( {{x^{2010}} + {x^{2008}} + ... + {x^2} + 1} \right)\\
Do:2{a^2} = 3a\\
\Rightarrow a.\left( {2a - 3} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
a = 0\\
2a - 3 = 0
\end{array} \right.\\
+ Khi:a = 0\\
\Rightarrow f\left( a \right) = \left( {2a - 3} \right).\left( {{a^{2010}} + {a^{2008}} + ... + {a^2} + 1} \right)\\
= \left( { - 3} \right).1 = - 3\\
+ Khi:2a - 3 = 0\\
\Rightarrow f\left( a \right) = 0\\
\text{Vậy}\,f\left( a \right) = - 3\,\text{hoặc}\,f\left( a \right) = 0
\end{array}$
