a: Đặt \(A=\frac12+\frac14+\frac18+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}\)
=>2A=\(1+\frac12+\frac14+\frac18+\frac{1}{16}+\frac{1}{32}\)
=>2A-A=\(1+\frac12+\frac14+\frac18+\frac{1}{16}+\frac{1}{32}-\frac12-\frac14-\frac18-\frac{1}{16}-\frac{1}{32}-\frac{1}{64}\)
=>A=\(1-\frac{1}{64}=\frac{63}{64}\)
Ta có: \(27\left(\frac12+\frac14+\frac18+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}\right)\cdot x=\frac{63}{64}\)
=>\(27\cdot x\cdot\frac{63}{64}=\frac{63}{64}\)
=>27x=1
=>\(x=\frac{1}{27}\)
b: Đặt \(B=\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+\frac{3}{11\cdot14}\)
\(=\frac12-\frac15+\frac15-\frac18+\frac18-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}\)
\(=\frac12-\frac{1}{14}=\frac{6}{14}=\frac37\)
Ta có: \(28\left(\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+\frac{3}{11\cdot14}\right):x=\frac37\)
=>\(28\cdot\frac37:x=\frac37\)
=>\(\frac{12}{x}=\frac37=\frac{12}{28}\)
=>x=28
