Câu trả lời:
+ \(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}\)
=> \(2A=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}\)
=> \(A=2A-A=1-\frac{1}{16}=\frac{15}{16}\)
+ \(B=\frac{1}{1x2}+\frac{1}{2x3}+\frac{1}{3x4}+...+\frac{1}{11x12}\)
\(B=\frac{2-1}{1x2}+\frac{3-2}{2x3}+\frac{4-3}{3x4}+...+\frac{12-11}{11x12}\)
\(B=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{11}-\frac{1}{12}\)
\(B=1-\frac{1}{12}=\frac{11}{12}\)
\(A:x=B\Rightarrow x=A:B=\frac{15}{16}:\frac{11}{12}=\frac{15}{16}x\frac{12}{11}=\frac{45}{44}=1\frac{1}{44}\)