Đáp án:
$\begin{array}{l}
a)A = {3^1} + {3^2} + {3^3} + {3^4} + ... + {3^{99}} + {3^{100}}\\
= \left( {{3^1} + {3^2}} \right) + \left( {{3^3} + {3^4}} \right) + .. + \left( {{3^{99}} + {3^{100}}} \right)\\
= 3.\left( {1 + 3} \right) + {3^3}.\left( {1 + 3} \right) + ... + {3^{99}}.\left( {1 + 3} \right)\\
= 3.4 + {3^3}.4 + ... + {3^{99}}.4\\
= \left( {3 + {3^3} + ... + {3^{99}}} \right).4 \vdots 4\\
b)\left| {a - 2010} \right| = 2009\\
\Rightarrow \left[ \begin{array}{l}
a - 2010 = 2009\\
a - 2010 = - 2009
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
a = 4019\\
a = 1
\end{array} \right.\\
Vậy\,a = 4019;a = 1\\
c)\dfrac{a}{b} = \dfrac{b}{c} = \dfrac{c}{a} = k\\
\Rightarrow \left\{ \begin{array}{l}
a = b.k\\
b = c.k\\
c = a.k
\end{array} \right.\\
\Rightarrow a = b.k = c.k.k = a.k.k.k = a.{k^3}\\
\Rightarrow {k^3} = 1\\
\Rightarrow k = 1\\
\Rightarrow a = b = c\\
M = \dfrac{{{a^6}.{b^6}.{c^{2006}}}}{{{b^{2018}}}}\\
= \dfrac{{{b^6}.{b^6}.{b^{2006}}}}{{{b^{2018}}}}\\
= {b^{6 + 6 + 2006 - 2018}}\\
= {b^0}\\
= 1
\end{array}$
