Ta có:\(B=\dfrac{1}{6}+\dfrac{1}{24}+\dfrac{1}{60}+...+\dfrac{1}{990}\)

\(2B=\dfrac{2}{6}+\dfrac{2}{24}+\dfrac{2}{60}+...+\dfrac{2}{990}\)

\(2B=\dfrac{2}{1\cdot2\cdot3}+\dfrac{2}{2\cdot3\cdot4}+\dfrac{2}{3\cdot4\cdot5}+...+\dfrac{2}{9\cdot10\cdot11}\)

\(2B=\dfrac{1}{1\cdot2}-\dfrac{1}{2\cdot3}+\dfrac{1}{2\cdot3}-\dfrac{1}{3\cdot4}+\dfrac{1}{3\cdot4}-\dfrac{1}{4\cdot5}+...+\dfrac{1}{9\cdot10}-\dfrac{1}{10\cdot11}\)

\(2B=\dfrac{1}{1\cdot2}-\dfrac{1}{10\cdot11}\)

\(2B=\dfrac{27}{55}\)

\(B=\dfrac{27}{55}:2\)

\(B=\dfrac{27}{110}\)

đó bạn

a: \(=\dfrac{99}{100}:\left(\dfrac{3}{12}-\dfrac{1}{12}+\dfrac{4}{12}\right)-\dfrac{49}{25}\)

\(=\dfrac{99}{100}:\dfrac{1}{2}-\dfrac{49}{25}\)

\(=\dfrac{99}{50}-\dfrac{98}{50}=\dfrac{1}{50}\)

b: \(=\dfrac{13}{15}\cdot\dfrac{1}{4}\cdot3+\left(\dfrac{32}{60}-1-\dfrac{19}{60}\right):\dfrac{47}{24}\)

\(=\dfrac{39}{60}+\dfrac{-19}{60}\cdot\dfrac{24}{47}\)

=459/940

\(F=\dfrac{1}{18}+\dfrac{1}{54}+\dfrac{1}{108}+...+\dfrac{1}{990}\)

\(F=\dfrac{1}{3.6}+\dfrac{1}{6.9}+\dfrac{1}{9.12}+....+\dfrac{1}{30.33}\)

\(F=\dfrac{1}{3}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{12}+...+\dfrac{1}{30}-\dfrac{1}{33}\)

\(F=\dfrac{1}{3}-\dfrac{1}{30}\)

\(F=\dfrac{3}{10}\)

Thử làm xem , không biết đúng không nhé !

\(F=\dfrac{1}{18}+\dfrac{1}{54}+\dfrac{1}{108}+...+\dfrac{1}{990}\)

\(F=\dfrac{1}{3.6}+\dfrac{1}{6.9}+\dfrac{1}{9.12}+...+\dfrac{1}{30.33}\)

\(F=\dfrac{1}{3}\left(\dfrac{1}{3}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{12}+...+\dfrac{1}{30}-\dfrac{1}{33}\right)\)

\(F=\dfrac{1}{3}\left(\dfrac{1}{3}-\dfrac{1}{33}\right)\)

\(F=\dfrac{1}{3}.\dfrac{10}{33}\)

\(F=\dfrac{10}{99}\)

\(F=\dfrac{1}{18}+\dfrac{1}{54}+\dfrac{1}{108}+.....+\dfrac{1}{990}\)

\(F=\dfrac{1}{3.6}+\dfrac{1}{6.9}+\dfrac{1}{9.12}+......+\dfrac{1}{30.33}\)

\(F=\dfrac{1}{3}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{12}+.....+\dfrac{1}{30}-\dfrac{1}{33}\)

\(F=\dfrac{1}{3}-\dfrac{1}{33}\)

\(F=\dfrac{10}{33}\)

\(B=\dfrac{1}{18}+\dfrac{1}{54}+...+\dfrac{1}{990}\)

\(\Rightarrow B=\dfrac{1}{3.6}+\dfrac{1}{6.9}+...+\dfrac{1}{30.33}\)

\(\Rightarrow B=\dfrac{1}{3}\left(\dfrac{3}{3.6}+\dfrac{3}{6.9}+...+\dfrac{3}{30.33}\right)\)

\(\Rightarrow B=\dfrac{1}{3}\left(\dfrac{1}{3}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{9}+...+\dfrac{1}{30}-\dfrac{1}{33}\right)\)

\(\Rightarrow B=\dfrac{1}{3}\left(\dfrac{1}{3}-\dfrac{1}{33}\right)\)

\(\Rightarrow B=\dfrac{1}{3}.\dfrac{10}{33}\)

\(\Rightarrow B=\dfrac{10}{99}\)

Vậy...

\(B=\dfrac{1}{18}+\dfrac{1}{54}+\dfrac{1}{108}+...+\dfrac{1}{990}\)

\(\Leftrightarrow B=\dfrac{1}{3.6}+\dfrac{1}{6.9}+\dfrac{1}{9.12}+..+\dfrac{1}{30.33}\)

\(\Leftrightarrow B=\left(\dfrac{1}{3}-\dfrac{1}{6}\right)+\left(\dfrac{1}{6}-\dfrac{1}{9}\right)+...+\left(\dfrac{1}{30}-\dfrac{1}{33}\right)\)

\(\Leftrightarrow B=\dfrac{1}{3}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{9}+\dfrac{1}{9}-...+\dfrac{1}{30}-\dfrac{1}{33}\)

\(\Leftrightarrow B=\dfrac{1}{3}-\dfrac{1}{33}\)

\(\Leftrightarrow B=\dfrac{10}{33}\).

\(1\dfrac{13}{15}.\left(0,5\right)^2.3+\left(\dfrac{8}{15}-1\dfrac{19}{60}\right):1\dfrac{23}{24}\\ =\dfrac{28}{15}.0,25.3+\left(\dfrac{8}{15}-\dfrac{79}{60}\right):\dfrac{47}{24}\\ =\dfrac{7}{15}.3+\dfrac{-47}{60}:\dfrac{47}{24}\\ =\dfrac{7}{5}+\dfrac{-2}{5}\\ =\dfrac{5}{5}=1\)

\(1\dfrac{13}{15}\cdot\left(0,5\right)^2+3\cdot\left(\dfrac{8}{15}+1\dfrac{19}{60}\right):1\dfrac{23}{24}\)

\(=\dfrac{28}{15}\cdot\dfrac{1}{4}+3\cdot\left(\dfrac{8}{15}+\dfrac{79}{60}\right):\dfrac{47}{24}\)

\(=\dfrac{7}{15}+3\cdot\dfrac{37}{20}\cdot\dfrac{24}{47}\)

\(=\dfrac{7}{15}+\dfrac{666}{235}=\dfrac{2327}{705}\)

\(1\dfrac{13}{15}.\left(0,5\right)^2+3.\left(\dfrac{8}{15}+1\dfrac{19}{60}\right):1\dfrac{23}{24}\) 

\(=\dfrac{28}{15}.\dfrac{1}{4}+3.\left(\dfrac{8}{15}+\dfrac{79}{60}\right):\dfrac{47}{24}\) 

\(=\dfrac{7}{15}+3.\dfrac{37}{20}:\dfrac{47}{24}\) 

\(=\dfrac{7}{15}+\dfrac{666}{235}\) 

\(=\dfrac{2327}{705}\)