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

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

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

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

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=

1 tháng 7 2015

\(\frac{1}{2}-\frac{1}{3.7}-\frac{1}{7.11}-...-\frac{1}{23.27}=\frac{1}{2}-\left(\frac{1}{3.7}+\frac{1}{7.11}+...+\frac{1}{23.27}\right)=\frac{1}{2}-\frac{1}{4}\cdot\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{23}-\frac{1}{27}\right)=\frac{1}{2}-\frac{1}{4}\cdot\left(\frac{1}{3}-\frac{1}{27}\right)=\frac{1}{2}-\frac{1}{4}\cdot\frac{8}{27}=\frac{23}{54}\)

24 tháng 4 2016

NHAN A VOI 3 RUI TU TINH

DỄ MÀ

24 tháng 4 2016

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

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

4A=\(\frac{1}{3}-\frac{1}{111}=\frac{12}{37}\)

A=\(\frac{12}{37}:4=\frac{12}{37}.\frac{1}{4}=\frac{3}{37}\)

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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