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5 tháng 6 2020

\(\frac{F}{2}=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{380}\)

\(\frac{F}{2}=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{19.20}\)

\(\frac{F}{2}=\frac{3-2}{2.3}+\frac{4-3}{3.5}+\frac{5-4}{4.5}+...+\frac{20-19}{19.20}\)

\(\frac{F}{2}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{19}-\frac{1}{20}\)

\(\frac{F}{2}=\frac{1}{2}-\frac{1}{20}=\frac{9}{20}\Rightarrow F=\frac{18}{20}=\frac{9}{10}\)

1 tháng 6 2020

\(A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{49\cdot50}\)

\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)

\(A=\frac{1}{1}-\frac{1}{50}\)

\(A=\frac{49}{50}\)

\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)

\(=1-\frac{1}{50}=\frac{49}{50}\)

2 tháng 5 2015

1/2D=1/2(1/6+1/10+......+1/45)

1/2D=1/12+1/20+1/30+.....+1/90

1/2D=1/3.4+1/4.5+1/5.6+......+1/9.10

1/2D=1/3-1/4+1/4-1/5+1/5-1/6+....+1/9-1/10

1/2D=1/3-1/10

1/2D=7/30

D=7/30:1/2

D=7/15

2 tháng 5 2015

Ta có:\(D=\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+\frac{1}{45}\)

\(=\frac{2}{12}+\frac{2}{20}+\frac{2}{30}+\frac{2}{42}+\frac{2}{56}+\frac{2}{72}+\frac{2}{90}\)

\(=2.\left(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}\right)\)

\(=2.\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}\right)\)

\(=2.\left(\frac{1}{3}-\frac{1}{10}\right)=2.\frac{7}{30}=\frac{7}{15}\)

Vậy \(D=\frac{7}{15}\)

13 tháng 6 2020

(1-1/3)x(1-1/5)x(1-1/7)x(1-1/9)x(1-1/2)x(1-1/4)x(1-1/6)x(1-1/8)x(1-1/10)

=2/3x4/5x6/7x8/9x1/2x3/4x5/6x7/8x9/10

=2x4x6x8x1x3x5x7x9 /3x5x7x9x2x4x6x8x10

=1/10

6 tháng 2 2016

\(\Rightarrow C=\frac{10}{5}\left(\frac{1}{7.12}+\frac{1}{12.17}+\frac{1}{17.22}+...+\frac{1}{502.507}\right)\)

\(\Rightarrow C=2.\left(\frac{1}{7}-\frac{1}{12}+\frac{1}{12}-\frac{1}{17}+....+\frac{1}{507}-\frac{1}{507}\right)\)

\(\Rightarrow C=2.\left(\frac{1}{7}-\frac{1}{507}\right)=2.\frac{1}{7}-2.\frac{1}{507}=\frac{2}{7}-\frac{2}{507}\)

 

10 tháng 4 2017

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

= 3(\(\frac{1}{3.6}\)+\(\frac{1}{6.9}\)+...+\(\frac{1}{30.33}\))

\(\frac{3}{3.6}\)+\(\frac{3}{6.9}\)+...+\(\frac{3}{30.33}\)

= 1 -\(\frac{1}{6}\)+\(\frac{1}{6}\)-\(\frac{1}{9}\)+...+\(\frac{1}{30}\)-\(\frac{1}{33}\)

= 1-\(\frac{1}{33}\)

=\(\frac{32}{33}\)

10 tháng 4 2017

gợi ý :1/18 +1/54 + ... +1/990

         = 1/3*6 + 1/6*9 + 1/9*13 + ... +1/30*33