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\(A=\dfrac{\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{10}}}{\dfrac{2}{2}+\dfrac{2}{2^2}+...+\dfrac{2}{2^{10}}}=\dfrac{\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{10}}}{2\left(\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{10}}\right)}=\dfrac{1}{2}\)
A = \(\dfrac{\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+....+\dfrac{1}{2^{10}}}{\dfrac{2}{2}+\dfrac{2}{2^2}+\dfrac{2}{2^3}+...+\dfrac{2}{2^{10}}}\)
= \(\dfrac{\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{10}}}{2\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{10}}\right)}\)
= \(\dfrac{1}{2}\)
A1 là trung điểm của AB => AA1 = \(\frac{AB}{2}=\frac{AB}{2^1}\) => \(\frac{AA_1}{AB}=\frac{1}{2^1}\)
A2 là trung điểm của AA1 => AA2 = \(\frac{AA_1}{2}=\frac{AB}{2^2}\)=> \(\frac{AA_2}{AB}=\frac{1}{2^2}\)
...
A10 là trung điểm của AA9 => AA10 = \(\frac{AA_9}{2}=\frac{AB}{2^{10}}\)=> \(\frac{AA_{10}}{AB}=\frac{1}{2^{10}}\)
=> S = \(\frac{1}{2^1}+\frac{1}{2^2}+...+\frac{1}{2^{10}}\)
=> 2.S = \(1+\frac{1}{2^1}+\frac{1}{2^2}+...+\frac{1}{2^9}\)
=> 2S - S = S = \(1-\frac{1}{2^{10}}\)
\(A=\dfrac{1}{2}+\dfrac{1}{2}^2+\dfrac{1}{2}^3+...+\dfrac{1}{2}\)10
\(= (\dfrac{1}{2}+\dfrac{1}{2}^9)+(\dfrac{1}{2}^2+\dfrac{1}{2}^8)+(\dfrac{1}{2}^3+\dfrac{1}{2}^7)+\dfrac{1}{2}\)10
\(= \dfrac{257}{512}+\dfrac{65}{256}+\dfrac{17}{128}+\dfrac{1}{1024}\)
\(=( \dfrac{514}{1024}+\dfrac{136}{1024})+(\dfrac{260}{1024}+\dfrac{1}{1024})\)
\(=\dfrac{650}{1024}+\dfrac{261}{1024}\)
\(=\dfrac{911}{1024}\)