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Ta có : \(B=\frac{\frac{1}{39}-\frac{1}{6}-\frac{1}{51}}{\frac{1}{8}-\frac{1}{52}+\frac{1}{68}}:5\frac{1}{6}\)
=> \(B=\frac{\frac{\frac{1}{13}-\frac{1}{2}-\frac{1}{17}}{3}}{\frac{\frac{1}{2}-\frac{1}{13}+\frac{1}{17}}{4}}:5\frac{1}{6}\)
=> \(B=\frac{\frac{\frac{1}{13}-\frac{1}{2}-\frac{1}{17}}{3}}{\frac{\frac{1}{13}-\frac{1}{2}-\frac{1}{17}}{-4}}:5\frac{1}{6}\)
=> \(B=\frac{-4}{3}:5\frac{1}{6}\)
=> \(B=\frac{-8}{6}:\frac{31}{6}=-\frac{8}{6}.\frac{6}{31}=-\frac{8}{31}\)
\(\text{Đặt }A=\frac{\frac{1}{6}+\frac{1}{51}+\frac{1}{39}}{\frac{1}{8}-\frac{1}{52}+\frac{1}{68}}\)
\(\Rightarrow\frac{1}{A}=\frac{6+51+39}{8-52+68}=4\)
\(\text{Vậy }A=\frac{1}{4}\)
Ta có: \(\frac{\frac{1}{6}-\frac{1}{39}+\frac{1}{51}}{\frac{1}{8}-\frac{1}{52}+\frac{1}{68}}=\frac{\left(\frac{1}{6}-\frac{1}{39}+\frac{1}{51}\right).5304}{\left(\frac{1}{8}-\frac{1}{52}+\frac{1}{68}\right).5304}\)\(=\frac{136-884+104}{663-102+78}=-\frac{664}{639}\)
\(=\frac{\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{13}+\frac{1}{17}\right)}{\frac{1}{4}.\left(\frac{1}{2}-\frac{1}{13}+\frac{1}{17}\right)}=\frac{4}{3}\)
\(\frac{\frac{1}{6}-\frac{1}{39}+\frac{1}{51}}{\frac{1}{8}-\frac{1}{52}+\frac{1}{68}}=\frac{\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{13}+\frac{1}{17}\right)}{\frac{1}{4}.\left(\frac{1}{2}-\frac{1}{13}+\frac{1}{17}\right)}=\frac{4}{3}\)
\(A=\frac{1}{90}-\frac{1}{72}-\frac{1}{56}-\frac{1}{42}-\frac{1}{30}-\frac{1}{20}-\frac{1}{12}-\frac{1}{6}-\frac{1}{2}\)
\(A=\frac{1}{10.9}-\frac{1}{9.8}-\frac{1}{8.7}-\frac{1}{7.6}-\frac{1}{6.5}-\frac{1}{5.4}-\frac{1}{4.3}-\frac{1}{3.2}-\frac{1}{2.1}\)
\(-A=\left(\frac{1}{10.9}+\frac{1}{9.8}+\frac{1}{8.7}+\frac{1}{7.6}+...+\frac{1}{3.2}+\frac{1}{2.1}\right)\)
\(-A=\frac{1}{10}-\frac{1}{9}+\frac{1}{9}-\frac{1}{8}+\frac{1}{8}-\frac{1}{7}+...+\frac{1}{3}-\frac{1}{2}+\frac{1}{2}-1\)
\(-A=\frac{1}{10}-1=\frac{-9}{10}\Rightarrow A=\frac{9}{10}\)
\(A=\frac{1}{90}-\frac{1}{72}-\frac{1}{56}-\frac{1}{42}-\frac{1}{30}-\frac{1}{20}-\frac{1}{12}-\frac{1}{6}-\frac{1}{2}\)
\(=\frac{1}{90}-\left(\frac{1}{72}+\frac{1}{56}+\frac{1}{42}+\frac{1}{30}+\frac{1}{20}+\frac{1}{12}+\frac{1}{6}+\frac{1}{2}\right)\)
\(=\frac{1}{90}-\left(\frac{1}{8.9}+\frac{1}{7.8}+\frac{1}{6.7}+\frac{1}{5.6}+\frac{1}{4.5}+\frac{1}{3.4}+\frac{1}{2.3}+\frac{1}{1.2}\right)\)
\(=\frac{1}{90}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}\right)\)
\(=\frac{1}{90}-\left(1-\frac{1}{9}\right)=\frac{1}{90}-\frac{8}{9}=-\frac{79}{90}\)
Vậy A=-79/90
