Giải:
\(\dfrac{1}{3}-\dfrac{1}{12}-\dfrac{1}{20}-...-\dfrac{1}{90}-\dfrac{1}{110}=x-\dfrac{5}{13}\)
\(\Leftrightarrow\dfrac{1}{1.3}-\dfrac{1}{3.4}-\dfrac{1}{4.5}-...-\dfrac{1}{9.10}-\dfrac{1}{10.11}=x-\dfrac{5}{13}\)
\(\Leftrightarrow\dfrac{-1}{2}\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{9}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{11}\right)=x-\dfrac{5}{13}\)
\(\Leftrightarrow\dfrac{-1}{2}\left(\dfrac{1}{1}-\dfrac{1}{11}\right)=x-\dfrac{5}{13}\)
\(\Leftrightarrow\dfrac{-1}{2}.\dfrac{10}{11}=x-\dfrac{5}{13}\)
\(\Leftrightarrow\dfrac{-5}{11}=x-\dfrac{5}{13}\)
\(\Leftrightarrow\dfrac{-5}{11}+\dfrac{5}{13}=x\)
\(\Leftrightarrow x=\dfrac{-10}{143}\)
Vậy ...
Ta có:
\(\dfrac{1}{3}-\dfrac{1}{12}-\dfrac{1}{20}-\dfrac{1}{30}-...-\dfrac{1}{110}=x-\dfrac{5}{13}\\ -\dfrac{1}{12}-\dfrac{1}{20}-\dfrac{1}{30}-...-\dfrac{1}{110}=x-\dfrac{5}{13}-\dfrac{1}{3}\\ -\dfrac{1}{12}-\dfrac{1}{20}-\dfrac{1}{30}-...-\dfrac{1}{110}=x-\dfrac{28}{39}\\ -\left(\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+...+\dfrac{1}{110}\right)=x-\dfrac{28}{39}\\ -\left(\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+...+\dfrac{1}{10.11}\right)=x-\dfrac{28}{39}\\ -\left(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{10}-\dfrac{1}{11}\right)=x-\dfrac{28}{39}\\ -\left(\dfrac{1}{3}-\dfrac{1}{11}\right)=x-\dfrac{28}{39}\\ -\dfrac{8}{33}=x-\dfrac{28}{39}\\ x=-\dfrac{8}{33}+\dfrac{28}{39}\\ x=\dfrac{68}{143}\)
Vậy \(x=\dfrac{68}{143}\)
Giải
⇔ \(\dfrac{1}{1.3}\) - \(\dfrac{1}{3.4}\) - .... - \(\dfrac{1}{9.10}\)- \(\dfrac{1}{10.11}\)= X - \(\dfrac{5}{13}\)
⇔\(\dfrac{-1}{2}\)( \(\dfrac{1}{1}\)- \(\dfrac{1}{3}\) + \(\dfrac{1}{3}\) - \(\dfrac{1}{4}\)+ .........+ \(\dfrac{1}{9}\) - \(\dfrac{1}{10}\) + \(\dfrac{1}{10}\) - \(\dfrac{1}{11}\)) = X - \(\dfrac{5}{13}\)
⇔\(\dfrac{-1}{2}\)( \(\dfrac{1}{1}\) - \(\dfrac{1}{11}\)) = X - \(\dfrac{5}{13}\)
⇔ \(\dfrac{-1}{2}\) . \(\dfrac{10}{11}\)= X - \(\dfrac{5}{13}\)
⇔ \(\dfrac{-5}{11}\) + \(\dfrac{5}{13}\) = X
⇔ \(\dfrac{-10}{143}\) = X
Vậy X = \(\dfrac{-10}{143}\)
