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\(\text{Đặt }A=\frac{\frac{1}{6}+\frac{1}{51}+\frac{1}{39}}{\frac{1}{8}-\frac{1}{52}+\frac{1}{68}}\)
\(\Rightarrow\frac{1}{A}=\frac{6+51+39}{8-52+68}=4\)
\(\text{Vậy }A=\frac{1}{4}\)
\(=\frac{\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{13}+\frac{1}{17}\right)}{\frac{1}{4}.\left(\frac{1}{2}-\frac{1}{13}+\frac{1}{17}\right)}=\frac{4}{3}\)
\(\frac{\frac{1}{6}-\frac{1}{39}+\frac{1}{51}}{\frac{1}{8}-\frac{1}{52}+\frac{1}{68}}=\frac{\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{13}+\frac{1}{17}\right)}{\frac{1}{4}.\left(\frac{1}{2}-\frac{1}{13}+\frac{1}{17}\right)}=\frac{4}{3}\)
Ta có : \(B=\frac{\frac{1}{39}-\frac{1}{6}-\frac{1}{51}}{\frac{1}{8}-\frac{1}{52}+\frac{1}{68}}:5\frac{1}{6}\)
=> \(B=\frac{\frac{\frac{1}{13}-\frac{1}{2}-\frac{1}{17}}{3}}{\frac{\frac{1}{2}-\frac{1}{13}+\frac{1}{17}}{4}}:5\frac{1}{6}\)
=> \(B=\frac{\frac{\frac{1}{13}-\frac{1}{2}-\frac{1}{17}}{3}}{\frac{\frac{1}{13}-\frac{1}{2}-\frac{1}{17}}{-4}}:5\frac{1}{6}\)
=> \(B=\frac{-4}{3}:5\frac{1}{6}\)
=> \(B=\frac{-8}{6}:\frac{31}{6}=-\frac{8}{6}.\frac{6}{31}=-\frac{8}{31}\)
\(A=\frac{\frac{1}{6}-\frac{1}{39}+\frac{1}{51}}{\frac{1}{8}-\frac{1}{52}+\frac{1}{68}}=\frac{\frac{1}{3}\cdot\left(\frac{1}{2}-\frac{1}{13}+\frac{1}{17}\right)}{\frac{1}{4}\cdot\left(\frac{1}{2}-\frac{1}{13}+\frac{1}{17}\right)}=\frac{1}{3}:\frac{1}{4}=\frac{4}{3}\)
Ta có: \(\frac{\frac{1}{6}-\frac{1}{39}+\frac{1}{51}}{\frac{1}{8}-\frac{1}{52}+\frac{1}{68}}=\frac{\left(\frac{1}{6}-\frac{1}{39}+\frac{1}{51}\right).5304}{\left(\frac{1}{8}-\frac{1}{52}+\frac{1}{68}\right).5304}\)\(=\frac{136-884+104}{663-102+78}=-\frac{664}{639}\)