1/7+1/91+1/247+1/475+1/775+1/1147=? (1)
ta có: (1) <=>: 1/(1.7)+1/(7.13)+1/(13.19)+1/(19.25)+1/(25.31)+1/(31.37)
=1/6.(1-1/7+1/7-1/13+1/13-1/19+1/19-1/25+1/25-1/31+1/31-1/37)
=1/6.(1-1/37)=6/37

ta có: 

\(S=\frac{4}{6}\left(\frac{6}{1.7}+\frac{6}{7.13}+...+\frac{6}{43.49}\right)\)\(=\frac{4}{6}\left(\frac{7-1}{1.7}+\frac{13-7}{7.13}+...+\frac{49-43}{43.49}\right)=\frac{4}{6}\left(1-\frac{1}{7}+\frac{1}{7}-\frac{1}{13}+...+\frac{1}{43}-\frac{1}{49}\right)\)

\(\frac{4}{6}\left(1-\frac{1}{49}\right)=\frac{4.48}{6.49}=\frac{32}{49}\)

đặt biểu thức trên =A

ta có: A=2/13.19+2/19.25+2/25.31+...+2/613.619

=(2/13-2/19)+(2/19-2/25)+(2/25-2/31)+...+(2/613-2/619)

=2/13-2/19+2/19-2/25+2/25-2/31+...+2/613-2/619

=2/13+(2/19-2/19)+(2/25-2/25)+(2/31-2/31)+...+(2/613-2/613)-2/619

=2/13-2/619=1212/8047

= 2 .[ \(\dfrac{1}{13.19}+\dfrac{1}{19.25}+\dfrac{1}{25.31}+........+\dfrac{1}{613.619}\)]

= 2 .[\(\dfrac{1}{6}\).(\(\dfrac{1}{13}-\dfrac{1}{19}\))+\(\dfrac{1}{6}\). (\(\dfrac{1}{19}-\dfrac{1}{25}\))+\(\dfrac{1}{6}\). (\(\dfrac{1}{25}-\dfrac{1}{31}\)) + ......+\(\dfrac{1}{6}\). (\(\dfrac{1}{613}-\dfrac{1}{619}\) ) ]

= 2. [\(\dfrac{1}{6}\).( \(\dfrac{1}{13}-\dfrac{1}{19}\)+\(\dfrac{1}{19}-\dfrac{1}{25}\)+\(\dfrac{1}{25}-\dfrac{1}{31}\)+ ......+\(\dfrac{1}{613}-\dfrac{1}{619}\))]

= 2. [\(\dfrac{1}{6}\).\(\dfrac{1}{619}\)]

=2 . \(\dfrac{1}{3714}\)

= \(\dfrac{1}{1857}\)

có j sai mọi người cứ nói

= 2 .[

= 2 .[.()+. ()+. () + ......+. () ]

= 2. [ 1/6+1/6. 1/6.1+ 1/6. 1/6. + ......+ 1/6. 1/6. )]

= 2. [.1/6.]

=2 . [1/6.(1/3 - 1/619]

= 2. [1/6 .616/1857]

= 2.308/5571

=616/5571

\(\frac{5}{1.7}+\frac{5}{7.13}+\frac{5}{13.19}+...+\frac{5}{2017.2023}\)

\(=5.\frac{1}{6}.\left(\frac{6}{1.7}+\frac{6}{7.13}+\frac{6}{13.19}+...+\frac{6}{2017.2023}\right)\)

\(=\frac{5}{6}.\left(1-\frac{1}{7}+\frac{1}{7}-\frac{1}{13}+\frac{1}{13}-\frac{1}{19}+...+\frac{1}{2017}-\frac{1}{2023}\right)\)

\(=\frac{5}{6}.\left(1-\frac{1}{2023}\right)\)

\(=\frac{5}{6}.\frac{2022}{2023}\)

\(=\frac{1685}{2023}\)

\(\frac{5}{1.7}+\frac{5}{7.13}+\frac{5}{13.19}+...+\frac{5}{2017.2023}\)

\(=\frac{5.6}{1.7.6}+\frac{5.6}{7.13.6}+\frac{5.6}{13.19.6}+.....+\frac{5.6}{2017.2023.6}\)

\(=\frac{5}{6}.\left(\frac{6}{1.7}+\frac{6}{7.13}+\frac{6}{13.19}+...+\frac{6}{2017.2023}\right)\)

\(=\frac{5}{6}.\left(1-\frac{1}{7}+\frac{1}{7}-\frac{1}{13}+\frac{1}{13}-\frac{1}{19}+...+\frac{1}{2017}-\frac{1}{2023}\right)\)

\(=\frac{5}{6}.\left(1-\frac{1}{2023}\right)\)

\(=\frac{5}{6}.\frac{2022}{2023}\)

\(=\frac{1685}{2023}\)

\(=\dfrac{1}{7}+\dfrac{4}{6}\left(\dfrac{6}{7\cdot13}+\dfrac{6}{13\cdot19}+...+\dfrac{6}{43\cdot49}\right)\)

\(=\dfrac{1}{7}+\dfrac{2}{3}\left(\dfrac{1}{7}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{19}+...+\dfrac{1}{43}-\dfrac{1}{49}\right)\)

\(=\dfrac{1}{7}+\dfrac{2}{3}\cdot\dfrac{6}{49}=\dfrac{1}{7}+\dfrac{4}{49}=\dfrac{11}{49}\)

a/\(C=\dfrac{2}{1.7}+\dfrac{2}{7.13}+\dfrac{2}{13.19}+...+\dfrac{2}{1013.1019}\)
\(=\dfrac{1}{3}\left(\dfrac{6}{1.7}+\dfrac{6}{7.13}+\dfrac{6}{13.19}+...+\dfrac{6}{1013.1019}\right)\)
\(=\dfrac{1}{3}\left(1-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{19}+...+\dfrac{1}{1013}-\dfrac{1}{1019}\right)\)
\(=\dfrac{1}{3}\left(1-\dfrac{1}{1019}\right)\)
\(=\dfrac{1}{3}\cdot\dfrac{1018}{1019}\)
\(=\dfrac{1018}{3057}\)
b/\(D=\dfrac{7}{1.9}+\dfrac{7}{9.17}+\dfrac{7}{17.25}+...+\dfrac{7}{2011.2019}\)
\(=\dfrac{7}{8}\left(\dfrac{8}{1.9}+\dfrac{8}{9.17}+\dfrac{8}{17.25}+...+\dfrac{8}{2011.2019}\right)\)
\(=\dfrac{7}{8}\left(1-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{17}+\dfrac{1}{17}-\dfrac{1}{25}+...+\dfrac{1}{2011}-\dfrac{1}{2019}\right)\)
\(=\dfrac{7}{8}\left(1-\dfrac{1}{2019}\right)\)
\(=\dfrac{7}{8}\cdot\dfrac{2018}{2019}\)
\(=\dfrac{7063}{8076}\)

a) Ta có :

\(C=\dfrac{2}{1.7}+\dfrac{2}{7.13}+\dfrac{2}{13.19}+...+\dfrac{2}{1013.1019}\)

\(\Rightarrow C=\dfrac{1}{3}.\left(\dfrac{6}{1.7}+\dfrac{6}{7.13}+\dfrac{6}{13.19}+...+\dfrac{6}{1013.1019}\right)\)

\(\Rightarrow C=\dfrac{1}{3}.\left(\dfrac{1}{1}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{19}+...+\dfrac{1}{1013}-\dfrac{1}{1019}\right)\)

\(\Rightarrow C=\dfrac{1}{3}.\left(\dfrac{1}{1}-\dfrac{1}{1019}\right)\)

\(\Rightarrow C=\dfrac{1}{3}.\dfrac{1018}{1019}\)

\(\Rightarrow C=\dfrac{1018}{3057}\)

b) Ta có:

\(D=\dfrac{7}{1.9}+\dfrac{7}{9.17}+\dfrac{7}{17.25}+...+\dfrac{7}{2011.2019}\)

\(\Rightarrow D=\dfrac{7}{8}.\left(\dfrac{8}{1.9}+\dfrac{8}{9.17}+\dfrac{8}{17.25}+...+\dfrac{8}{2011.2019}\right)\)

\(\Rightarrow D=\dfrac{7}{8}.\left(\dfrac{1}{1}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{17}+\dfrac{1}{17}-\dfrac{1}{25}+...+\dfrac{1}{2011}-\dfrac{1}{2019}\right)\)

\(\Rightarrow D=\dfrac{7}{8}.\left(\dfrac{1}{1}-\dfrac{1}{2019}\right)\)

\(\Rightarrow D=\dfrac{7}{8}.\dfrac{2018}{2019}\)

\(\Rightarrow D=\dfrac{7063}{8076}\)