Đáp án:
$\begin{array}{l}
a)A = \left( {a + b} \right)\left( {{a^2} + {b^2}} \right)\left( {{a^4} + {b^4}} \right)...\left( {{a^{32}} + {b^{32}}} \right)\\
= \dfrac{1}{{a - b}}.\left( {a - b} \right).\left( {a + b} \right)\left( {{a^2} + {b^2}} \right)\left( {{a^4} + {b^4}} \right)...\left( {{a^{32}} + {b^{32}}} \right)\\
= \dfrac{1}{{a - b}}.\left( {{a^2} - {b^2}} \right)\left( {{a^2} + {b^2}} \right)\left( {{a^4} + {b^4}} \right)...\left( {{a^{32}} + {b^{32}}} \right)\\
= \dfrac{1}{{a - b}}.\left( {{a^4} - {b^4}} \right)\left( {{a^4} + {b^4}} \right)...\left( {{a^{32}} + {b^{32}}} \right)\\
= \dfrac{1}{{a - b}}.\left( {{a^{32}} - {b^{32}}} \right)\left( {{a^{32}} + {b^{32}}} \right)\\
= \dfrac{1}{{a - b}}.\left( {{a^{64}} - {b^{64}}} \right)\\
= {a^{64}} - {b^{64}}\left( {do:a = b + 1 \Leftrightarrow a - b = 1} \right)\\
Vay\,A = B\\
b)A = 2005.7.\left( {{8^{2003}} + {8^{2002}} + ... + {8^2} + 8 + 1} \right) + 2005\\
a = \left( {{8^{2003}} + {8^{2002}} + ... + {8^2} + 8 + 1} \right)\\
\Leftrightarrow 8a = {8^{2004}} + {8^{2003}} + ... + {8^3} + {8^2} + 8\\
\Leftrightarrow 8a - a = {8^{2004}} - 1\\
\Leftrightarrow a = \dfrac{{{8^{2004}} - 1}}{7}\\
\Leftrightarrow A = 2005.7.\dfrac{{{8^{2004}} - 1}}{7} + 2005\\
= 2005.\left( {{8^{2004}} - 1} \right) + 2005\\
= {2005.8^{2004}}\\
\Leftrightarrow A = B
\end{array}$
