Ta có:
\(S=\dfrac{2}{3}+\dfrac{2}{6}+\dfrac{2}{12}+\dfrac{2}{24}+\dfrac{2}{48}+\dfrac{2}{96}+\dfrac{2}{192}+\dfrac{2}{384}\\ =\dfrac{2}{3}+\dfrac{2}{2\times3}+\dfrac{2}{2\times6}+\dfrac{2}{2\times12}+\dfrac{2}{2\times24}+\dfrac{2}{2\times48}+\dfrac{2}{2\times96}+\dfrac{2}{2\times192}\\ =\dfrac{2}{3}+\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{24}+\dfrac{1}{48}+\dfrac{1}{96}+\dfrac{1}{192}\\ \)
\(\dfrac{S}{2}=\dfrac{1}{2}\left(\dfrac{2}{3}+\dfrac{2}{6}+\dfrac{2}{12}+\dfrac{2}{24}+\dfrac{2}{48}+\dfrac{2}{96}+\dfrac{2}{192}+\dfrac{2}{384}\right)\\ =\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{24}+\dfrac{1}{48}+\dfrac{1}{96}+\dfrac{1}{192}+\dfrac{1}{384}\)
\(S-\dfrac{S}{2}=\dfrac{2}{3}+\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{24}+\dfrac{1}{48}+\dfrac{1}{96}+\dfrac{1}{192}-\left(\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{24}+\dfrac{1}{48}+\dfrac{1}{96}+\dfrac{1}{192}+\dfrac{1}{384}\right)\\ =\dfrac{2}{3}-\dfrac{1}{384}=\dfrac{2\times128-1}{384}\\ =\dfrac{85}{128}\\ \Rightarrow S=\dfrac{85}{128}\times2=\dfrac{85}{64}\)
\(A=\dfrac{2}{3}+\dfrac{2}{6}+\dfrac{2}{12}+\dfrac{2}{24}+\dfrac{2}{48}+\dfrac{2}{96}+\dfrac{2}{192}+\dfrac{2}{384}\)
\(A.2=\dfrac{4}{3}+\dfrac{2}{3}+\dfrac{2}{6}+\dfrac{2}{12}+\dfrac{2}{24}+\dfrac{2}{48}+\dfrac{2}{96}+\dfrac{2}{192}\)
\(A=A.2-A=\dfrac{4}{3}-\dfrac{2}{384}=\dfrac{127}{96}\)
Đáp án cuối là: \(\dfrac{85}{64}\) nhé nãy mình tính nhầm