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19 tháng 8 2016

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

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

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

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

2 tháng 9 2015

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 = \(\frac{32}{99}\)

14 tháng 3 2016

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

21 tháng 3 2019

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

21 tháng 3 2019

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

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

21 tháng 3 2016

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

11 tháng 3 2015

 

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

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

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

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

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

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

14 tháng 7 2017

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

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

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

5 tháng 5 2016

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

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

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

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

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

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

8 tháng 5 2016

lớp 6ha

 

15 tháng 7 2017

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

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

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

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

15 tháng 7 2017

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

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

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

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

\(\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+....+\frac{4}{103.107}\)

=\(\frac{1}{3.7}+\frac{1}{7.11}+\frac{1}{11.15}+...+\frac{1}{103.107}\)

=\(\frac{1}{3.107}\)

=\(\frac{1}{321}\)

k mk nha bn

=

1 tháng 8 2017

\(\frac{1}{3.7}+\frac{1}{7.11}+\frac{1}{11.15}+...+\frac{1}{x\left(x+4\right)}=\frac{43}{552}\)

\(\Leftrightarrow\frac{1}{4}.\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...+\frac{1}{x}-\frac{1}{x+4}\right)=\frac{43}{552}\)

\(\Leftrightarrow\frac{1}{4}.\left(\frac{1}{3}-\frac{1}{x+4}\right)=\frac{43}{552}\)

\(\Leftrightarrow\frac{1}{3}-\frac{1}{x+4}=\frac{43}{552}\div\frac{1}{4}\)

\(\Leftrightarrow\frac{1}{3}-\frac{1}{x+4}=\frac{43}{138}\Leftrightarrow\frac{1}{x+4}=\frac{1}{3}-\frac{43}{138}\)

\(\Leftrightarrow\frac{1}{x+4}=\frac{1}{46}\Leftrightarrow x+4=46\Rightarrow x=46-4=42\)

Vậy x = 42 

1 tháng 8 2017

  \(s=\frac{1}{3.7}+\frac{1}{7.11}+...+\frac{1}{x\left(x+4\right)}=\)\(\frac{43}{552}\)

\(\Rightarrow S=\frac{4}{4}\left(\frac{1}{3.7}+\frac{1}{7.11}+...+\frac{1}{x\left(x+4\right)}\right)=\frac{43}{552}\)

\(\Rightarrow S=\frac{1}{4}\left(\frac{4}{3.7}+\frac{4}{7.11}+...+\frac{4}{x\left(x+4\right)}\right)=\frac{43}{552}\)

\(\Rightarrow S=\frac{1}{4}\left(\frac{4}{3}-\frac{4}{7}+\frac{4}{7}-\frac{4}{11}+...+\frac{4}{x}-\frac{4}{x+4}\right)=\frac{43}{552}\)

\(\Rightarrow S=\frac{1}{4}\left(\frac{4}{3}-\frac{4}{x+4}\right)=\frac{43}{552}\)

\(\Rightarrow\frac{4}{3}-\frac{4}{x+4}=\frac{43}{552}:\frac{1}{4}\)

\(\frac{\Rightarrow4}{3}-\frac{4}{x+4}=\frac{43}{138}\)

\(\frac{\Rightarrow4}{x+4}=\frac{4}{3}-\frac{43}{138}=\frac{47}{46}\)

\(\Rightarrow x+4=4:\frac{47}{46}=\frac{184}{47}\)

\(\Rightarrow x=\frac{184}{47}-4=\frac{-4}{47}\)