Đáp án: A<B
Giải thích các bước giải:
$\begin{array}{l}
A = \dfrac{1}{2} + \dfrac{1}{{{2^2}}} + \dfrac{1}{{{2^3}}} + ... + \dfrac{1}{{{2^{10}}}}\\
\Rightarrow 2.A = 1 + \dfrac{1}{2} + \dfrac{1}{{{2^2}}} + ... + \dfrac{1}{{{2^9}}}\\
\Rightarrow 2A - A = A = 1 - \dfrac{1}{{{2^{10}}}}\\
\Rightarrow A = \dfrac{1}{2}.\dfrac{{{2^{11}} - 1}}{{{2^9}}}\\
B = 1 + \dfrac{1}{3} + \dfrac{1}{{{3^2}}} + \dfrac{1}{{{3^3}}} + \dfrac{1}{{{3^4}}} + \dfrac{1}{{{3^5}}}\\
\Rightarrow 3.B = 3 + 1 + \dfrac{1}{3} + \dfrac{1}{{{3^2}}} + \dfrac{1}{{{3^3}}} + \dfrac{1}{{{3^4}}}\\
\Rightarrow 3B - B = 2B = 3 - \dfrac{1}{{{3^5}}}\\
\Rightarrow B = \dfrac{3}{2} - \dfrac{1}{{{{2.3}^5}}} = \dfrac{1}{2}.\dfrac{{{3^6} - 1}}{{{3^5}}}\\
Do:\dfrac{{{3^6} - 1}}{{{3^5}}} > \dfrac{{{2^{11}} - 1}}{{{2^9}}}\\
\Rightarrow B > A
\end{array}$
