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1: \(=\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{11}-\dfrac{1}{13}\right)\)
=1/2*10/39
=5/39
2: \(=\dfrac{5}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{9}-\dfrac{1}{11}\right)=\dfrac{5}{2}\cdot\dfrac{10}{11}=\dfrac{50}{22}=\dfrac{25}{11}\)
\(B=\dfrac{1}{15}+\dfrac{1}{35}+\dfrac{1}{63}+\dfrac{1}{99}+\dfrac{1}{143}\)
\(=\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+\dfrac{1}{9.11}+\dfrac{1}{11.13}\)
\(=\dfrac{1}{2}.\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{13}\right)\)
\(=\dfrac{1}{2}.\left(\dfrac{1}{3}-\dfrac{1}{13}\right)\)
\(=\dfrac{1}{2}.\dfrac{10}{39}=\dfrac{5}{39}\)
Vậy \(B=\dfrac{5}{39}\)
`1/15+1/35+1/63+1/99+1/143`
`=1/[3.5]+1/[5.7]+1/[7.9]+1/[9.11]+1/[11.13]`
`=1/2(2/[3.5]+2/[5.7]+2/[7.9]+2/[9.11]+2/[11.13])`
`=1/2.(1/3-1/5+1/5-1/7+...+1/11-1/13)`
`=1/2.(1/3-1/13)`
`=1/2 . 10/39`
`=5/39`
\(=\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{13}\right)\)
\(=\dfrac{1}{2}\cdot\left(\dfrac{1}{3}-\dfrac{1}{13}\right)=\dfrac{1}{2}\cdot\dfrac{10}{39}=\dfrac{5}{39}\)
\(\dfrac{1}{15}+\dfrac{1}{35}+\dfrac{1}{63}+\dfrac{1}{99}+\dfrac{1}{143}+\dfrac{1}{195}\)
= \(\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+\dfrac{1}{9.11}+\dfrac{1}{11.13}+\dfrac{1}{13.15}\)
= 2(\(\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+\dfrac{1}{9.11}+\dfrac{1}{11.13}+\dfrac{1}{13.15}\)) :2
= (\(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+\dfrac{2}{9.11}+\dfrac{2}{11.13}+\dfrac{2}{13.15}\)) : 2
= (\(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}+\dfrac{1}{11}\) \(-\dfrac{1}{13}+\dfrac{1}{13}\)\(-\dfrac{1}{15}\)):2
= (\(\dfrac{1}{3}-\dfrac{1}{15}\)) :2
= \(\dfrac{4}{15}\): 2 = \(\dfrac{2}{15}\)
a)\(=\dfrac{211}{180}\)
b)\(=\dfrac{5}{39}\)
c)=\(=-\dfrac{65}{168}\)
\(B=\dfrac{1}{15}+\dfrac{1}{35}+\dfrac{1}{63}+\dfrac{1}{99}+\dfrac{1}{143}\)
=> \(2B=2\left(\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+\dfrac{1}{9.11}+\dfrac{1}{11.13}\right)\) => \(2B=\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+\dfrac{2}{9.11}+\dfrac{2}{11.13}\) => \(2B=\dfrac{5-3}{3.5}+\dfrac{7-5}{5.7}+\dfrac{9-7}{7.9}+\dfrac{11-9}{9.11}+\dfrac{13-11}{11.13}\) => \(2B=\dfrac{5}{3.5}-\dfrac{3}{3.5}+\dfrac{7}{5.7}-\dfrac{5}{5.7}+\dfrac{9}{7.9}-\dfrac{7}{7.9}+\dfrac{11}{9.11}-\dfrac{9}{9.11}+\dfrac{13}{11.13}-\dfrac{11}{11.13}\) => \(2B=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{13}\) => \(2B=\dfrac{1}{3}-\dfrac{1}{13}\)
=> \(B=\left(\dfrac{13}{39}-\dfrac{3}{39}\right):2\)
=> \(B=\dfrac{10}{39}.\dfrac{1}{2}\)
=> \(B=\dfrac{10}{39.2}\)
=> \(B=\dfrac{5}{39}\)
Vậy \(B=\dfrac{5}{39}\)
\(B=\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+\dfrac{1}{9.11}+\dfrac{1}{11.13}\)
\(B=\dfrac{1}{2}\left(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+\dfrac{2}{9.11}+\dfrac{2}{11.13}\right)\)
\(B=\dfrac{1}{2}\left(\dfrac{5-3}{3.5}+\dfrac{7-5}{5.7}+\dfrac{9-7}{7.9}+\dfrac{11-9}{9.11}+\dfrac{13-11}{11.13}\right)\)
\(B=\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{13}\right)\)
\(B=\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{13}\right)\)
\(B=\dfrac{1}{2}.\dfrac{10}{39}=\dfrac{5}{39}\)