\(A=\frac{4}{3.7}+\frac{4}{7.11}+....+\frac{4}{95.99}\)

\(=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{95}-\frac{1}{99}\)

\(=\frac{1}{3}-\frac{1}{99}\)

\(=\frac{32}{99}\)

A = \(\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+...+\frac{4}{95.99}\)

A = \(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...+\frac{1}{95}-\frac{1}{99}\)

A = \(\frac{1}{3}-\frac{1}{99}\)

A = \(\frac{32}{99}\)

A=1/3*7+1/7*11+..+1/95*99

=> 4A=4/3*7+4/7*11+..+4/95*99

=>4A=1/3-1/7+1/7-1/11+...+1/95-1/99=1/3-1/99=32/99

=>A=8/99

\(=\frac{1}{4}\left(\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+.......+\frac{4}{95.99}\right)=\frac{1}{4}\left(\frac{1}{3}-\frac{1}{99}\right)\)

\(=\frac{1}{4}.\frac{32}{99}=\frac{8}{99}\)

\(\frac{1}{3.7}+\frac{1}{7.11}+\frac{1}{11.15}+...+\frac{1}{95.99}\)

\(=\frac{1}{4}\left(\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+...+\frac{4}{95.99}\right)\)

\(=\frac{1}{4}\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{95}-\frac{1}{99}\right)\)

\(=\frac{1}{4}\left(\frac{1}{3}-\frac{1}{99}\right)\)

\(=\frac{1}{4}\cdot\frac{32}{99}=\frac{8}{99}\)

Ta thấy \(\frac{1}{3}-\frac{1}{7}=\frac{7-3}{3.7}=\frac{4}{3.7}\) 

             \(\frac{1}{7}-\frac{1}{11}=\frac{11-7}{7.11}=\frac{4}{7.11}\)

             ..........................

             \(\frac{1}{1023}-\frac{1}{1027}=\frac{1027-1023}{1023.1027}=\frac{4}{1023.1027}\)

=> \(\frac{4}{3.7}+\frac{4}{7.11}+....+\frac{4}{1023.1027}=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+....+\frac{1}{1023}-\frac{1}{1027}\)

=>                                                       =\(\frac{1}{3}-\frac{1}{1027}=\frac{1024}{3.1027}\)

Ta có: \(\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+...+\frac{4}{1023.1027}\)

\(=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...+\frac{1}{1023}-\frac{1}{1027}\)

\(=\frac{1}{3}-\frac{1}{1027}=\frac{1024}{3081}\)

\(\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+....+\frac{4}{1023.1027}\)

\(=\frac{1}{3.7}+\frac{1}{7.11}+\frac{1}{11.15}+...+\frac{1}{1023.1027}\)

\(=\frac{1}{3}-\frac{1}{1027}\)

\(=\frac{1024}{3081}\)

ai k mik mik k lại nha

\(\Leftrightarrow\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...+\frac{1}{107}-\frac{1}{111}\)

\(\Rightarrow=\frac{1}{3}-\frac{1}{111}\)

\(=\frac{12}{37}\)

k nha

\(=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...+\frac{1}{107}-\frac{1}{111}\)

\(=\frac{1}{3}-\frac{1}{111}\)

\(=\frac{108}{333}=\frac{12}{37}\)

ten gia mao nobita 

nobi k the gioi nhu the 

do gia :v

4x(\(\frac{1}{3.7}+...+\frac{1}{107.111}\) )

4(\(\frac{1}{3}-\frac{1}{7}+...+\frac{1}{107}-\frac{1}{111}\))

4(\(\frac{1}{3}-\frac{1}{111}\))

4.\(\frac{12}{37}\)

48/37

\(\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+...+\frac{4}{23.27}\)

\(=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{23}-\frac{1}{27}\)

\(=\frac{1}{3}-\frac{1}{27}+0+0+0+0\)

\(=\frac{8}{27}\)

Ta có : \(\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+\frac{4}{15.19}+\frac{4}{19.23}+\frac{4}{23.27}\)

\(=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+.....+\frac{1}{23}-\frac{1}{27}\)

\(=\frac{1}{3}-\frac{1}{27}\)

\(=\frac{8}{27}\)

\(\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+\frac{4}{15.19}+\frac{4}{19.23}+\frac{4}{23.27}\)

\(=\frac{7-3}{3.7}+\frac{11-7}{7.11}+.....+\frac{27-23}{23.27}\)

\(=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+....+\frac{1}{23}-\frac{1}{27}\)

\(=\frac{1}{3}-\frac{1}{27}=\frac{8}{27}\)

Ta có A = \(\frac{4}{3.7}+\frac{4}{7.11}+..............+\frac{4}{107.111}\)

=> A = \(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+.............+\frac{1}{107}-\frac{1}{111}\)

A = \(\frac{1}{3}-\frac{1}{111}=\frac{12}{37}\)

k nha bạn

\(A=\frac{4^2}{3.7}+\frac{4^2}{7.11}+\frac{4^2}{11.15}+...+\frac{4^2}{107.111}\)

\(A=\) \(4\left(\frac{1}{3.7}+\frac{1}{7.11}+\frac{1}{11.15}+...+\frac{1}{107.111}\right)\)

\(A=4\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...+\frac{1}{107}-\frac{1}{111}\right)\)

\(A=4\left(\frac{1}{3}-\frac{1}{111}\right)\)

\(A=4.\frac{12}{37}\)

\(A=\frac{48}{37}\)

lớp 6