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Ta có: \(1+2+...+n=\frac{n\left(n+1\right)}{2}\) áp dụng vào bài toán ta được
\(B=\frac{1}{3}+\frac{1}{6}\left(1+2\right)+\frac{1}{9}\left(1+2+3\right)+...+\frac{1}{6045}\left(1+2+...+2015\right)\)
\(=\frac{1}{3}+\frac{1}{2.3}.\frac{2.3}{2}+\frac{1}{3.3}.\frac{3.4}{2}+...+\frac{1}{2015.3}.\frac{2015.2016}{2}\)
\(=\frac{1}{3}\left(1+\frac{3}{2}+\frac{4}{2}+...+\frac{2016}{2}\right)\)
\(=\frac{1}{6}\left(2+3+4+...+2016\right)=\frac{1}{6}.\frac{2015.2018}{2}=\frac{2033135}{6}\)
CMCT : ( tự CM )
Áp dụng bài toán trên ta có : \(B=\frac{1}{3}+\frac{1}{6}\vec{\left(\frac{\left(x+1\right).2}{2}\right)+\frac{1}{9}\left(\frac{\left(1+3\right).3}{2}\right)+...+}\frac{1}{6045}\left(\frac{\left(1+2015\right).2}{2}\right)\)
\(B=\frac{1}{3}+\frac{1}{2}+\frac{2}{3}+....+....\) ( tự lm tiếp )
\(=\frac{16}{5}.\frac{15}{16}-\left(\frac{3}{4}+\frac{2}{7}\right):\left(\frac{-29}{28}\right)\)
\(=3-\left(\frac{21}{28}+\frac{8}{28}\right):\left(\frac{-29}{28}\right)\)
\(=3-\left(\frac{29}{28}\right).\left(\frac{-28}{29}\right)\)
\(=3-\left(-1\right)\)
\(=4\)
b) \(=\left(\frac{1}{4}+\frac{25}{2}-\frac{5}{16}\right):\left(12-\frac{7}{12}:\left(\frac{3}{8}-\frac{1}{12}\right)\right)\)
\(=\left(\frac{4}{16}+\frac{200}{16}-\frac{5}{16}\right):\left(12-\frac{7}{12}:\left(\frac{3.3}{2.3.4}-\frac{2}{2.3.4}\right)\right)\)
\(=\left(\frac{199}{16}\right):\left(12-\frac{7}{12}:\left(\frac{9}{24}-\frac{2}{24}\right)\right)\)
\(=\frac{199}{16}:\left(12-\frac{7}{12}.\frac{24}{7}\right)\)
\(=\frac{199}{16}:\left(12-2\right)\)
\(=\frac{199}{16}:10\)
\(=\frac{199}{160}\)
c) \(\left(\frac{-3}{5}+\frac{5}{11}\right):\frac{-3}{7}+\left(\frac{-2}{5}+\frac{6}{5}\right):\frac{-3}{7}\)
\(\left(\frac{-33}{55}+\frac{25}{55}\right):\frac{-3}{7}+\left(\frac{4}{5}\right):\frac{-3}{7}\)
\(\left(\frac{-8}{55}\right).\frac{-7}{3}+\frac{4}{5}.\frac{-7}{3}\)
\(\frac{-7}{3}\left(\frac{-8}{55}+\frac{4}{5}\right)\)
\(\frac{-7}{3}.\frac{36}{55}=\frac{-84}{55}\)