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14 tháng 8 2015

\(a=3\left(\frac{1}{5.10}+\frac{1}{10.15}+...+\frac{1}{45.50}\right)\)

\(a=\frac{3}{5}\left(\frac{5}{5.10}+\frac{5}{10.15}+...+\frac{5}{45.50}\right)\)

\(a=\frac{3}{5}\left(\frac{1}{5}-\frac{1}{10}+\frac{1}{10}-\frac{1}{15}+...+\frac{1}{45}-\frac{1}{50}\right)\)

\(a=\frac{3}{5}.\left(\frac{1}{5}-\frac{1}{50}\right)\)

\(a=\frac{3}{5}\cdot\frac{9}{50}\)

\(a=\frac{27}{250}\)

27 tháng 6 2018

\(1-\frac{1}{5.10}-\frac{1}{10.15}-\frac{1}{15.20}-...-\frac{1}{95.100}\)

\(=1-\left(\frac{1}{5.10}+\frac{1}{10.15}+\frac{1}{15.20}+...+\frac{1}{95.100}\right)\)

\(=1-\frac{1}{5}.\left(\frac{1}{5}-\frac{1}{10}+\frac{1}{10}-\frac{1}{15}+\frac{1}{15}-\frac{1}{20}+...+\frac{1}{95}-\frac{1}{100}\right)\)

\(=1-\frac{1}{5}.\left(\frac{1}{5}-\frac{1}{100}\right)\)

\(=1-\frac{1}{5}.\frac{19}{100}\)

\(=1-\frac{19}{500}\)

\(=\frac{481}{500}\)

27 tháng 6 2018

\(1-\frac{1}{5.10}-\frac{1}{10.15}-\frac{1}{15.20}-.....-\frac{1}{95.100}\)

\(=1-\left(\frac{1}{5.10}+\frac{1}{10.15}+\frac{1}{15.20}+...+\frac{1}{95.100}\right)\)

Đặt \(C=\frac{1}{5.10}+\frac{1}{10.15}+\frac{1}{15.20}+....+\frac{1}{95.100}\)

\(\Rightarrow C=\frac{1}{5}.\left(\frac{5}{5.10}+\frac{5}{10.15}+\frac{5}{15.20}+....+\frac{5}{95.100}\right)\)

           \(=\frac{1}{5}.\left(\frac{1}{5}-\frac{1}{10}+\frac{1}{10}-\frac{1}{15}+....+\frac{1}{95}-\frac{1}{100}\right)\)

             \(=\frac{1}{5}.\left(\frac{1}{5}-\frac{1}{100}\right)=\frac{1}{5}.\frac{19}{100}=\frac{19}{500}\)

\(\Rightarrow1-C=1-\frac{19}{500}=\frac{481}{500}\)

Chúc bạn học tốt

6 tháng 7 2016

\(\frac{2}{5.10}+\frac{2}{10.15}+\frac{2}{15.20}+...+\frac{2}{2015.2020}\)

\(=2.\left(\frac{1}{5.10}+\frac{1}{10.15}+\frac{1}{15.20}+...+\frac{1}{2015.2020}\right)\)

\(=2.\left(\frac{1}{5}-\frac{1}{10}+\frac{1}{10}-\frac{1}{15}+\frac{1}{15}-\frac{1}{20}+...+\frac{1}{2015}-\frac{1}{2020}\right)\)

\(=2.\left(\frac{1}{5}-\frac{1}{2020}\right)\)

\(=2.\frac{403}{2020}=\frac{403}{1010}\)

6 tháng 7 2016

\(\frac{2}{5.10}+\frac{2}{10.15}+\frac{2}{15.20}+...+\frac{2}{2015.2020}\)

=\(\frac{2}{5}\left(\frac{5}{5.10}+\frac{5}{10.15}+\frac{5}{15.20}+...+\frac{5}{2015.2020}\right)\)

=\(\frac{2}{5}\left(\frac{1}{5}-\frac{1}{10}+\frac{1}{10}-\frac{1}{15}+\frac{1}{15}-\frac{1}{20}+...+\frac{1}{2015}-\frac{1}{2016}\right)\)

=\(\frac{2}{5}.\left(\frac{1}{5}-\frac{1}{2020}\right)\)

=\(\frac{2}{5}.\frac{403}{2020}\)

=\(\frac{403}{5005}\)

11 tháng 5 2016

kho nhi

4 tháng 9 2016

mih biet lam nhung nhac qua

11 tháng 5 2016

mk chỉ làm thôi nhé

Ta tách biểu thức trên thành hai phần A và B

\(A=\frac{12-\frac{12}{7}-\frac{12}{289}-\frac{12}{85}}{4-\frac{4}{7}-\frac{4}{289}-\frac{12}{85}}\)

\(A=\frac{12.\left(1-\frac{1}{7}-\frac{1}{289}-\frac{1}{85}\right)}{4.\left(1-\frac{1}{7}-\frac{1}{289}-\frac{1}{85}\right)}\)

\(A=3\)

\(B=\frac{3+\frac{3}{13}+\frac{3}{169}+\frac{3}{91}}{7+\frac{7}{13}+\frac{7}{169}+\frac{7}{91}}\)

\(B=\frac{3.\left(1+\frac{1}{13}+\frac{1}{169}+\frac{1}{91}\right)}{7.\left(1+\frac{1}{13}+\frac{1}{169}+\frac{1}{91}\right)}\)

\(B=\frac{3}{7}\)

=> A:B=7

=>b=7

11 tháng 5 2016

k mình nha

15 tháng 7 2019

E = 2/5.10 + 2/10.15 + ... + 2/35.40

E = 2/5.(1/5 - 1/10 + 1/10 - 1/15 + ... + 1/35 - 1/40)

E = 2/5.(1/5 - 1/40)

E = 2/5.7/40

E = 7/100

15 tháng 7 2019

E = \(\frac{2}{5.10}+\frac{2}{10.15}+...+\frac{2}{35.40}\)

   = \(\frac{2}{5}.\left(\frac{5}{5.10}+\frac{5}{10.15}+...+\frac{5}{35.40}\right)\)

   = \(\frac{2}{5}.\left(\frac{1}{5}-\frac{1}{10}+\frac{1}{10}-\frac{1}{15}+...+\frac{1}{35}-\frac{1}{40}\right)\)

   = \(\frac{2}{5}.\left(\frac{1}{5}-\frac{1}{40}\right)\)

   = \(\frac{2}{5}.\frac{7}{40}\)

   = \(\frac{7}{100}\)

31 tháng 8 2016

\(\frac{1.2+3.6+5.10+7.14}{2.3+6.9+10.15+14.21}\)

\(=\frac{1.2+3.6+5.10+7.14}{1.2.3+3.6.3+5.10.3+7.14.3}\)

\(=\frac{1.2+3.6+5.10+7.14}{3.\left(1.2+3.6+5.10+7.14\right)}\)

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

9 tháng 7 2016

\(A=\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{29.32}+\frac{1}{32.35}\)

\(3A=\frac{3}{5.8}+\frac{3}{8.11}+....+\frac{3}{32.35}\)

\(3A=\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{32}-\frac{1}{35}\)

\(3A=\frac{1}{5}-\frac{1}{35}=\frac{6}{35}\)

\(A=\frac{6}{35}.\frac{1}{3}=\frac{2}{35}\)

\(B=\frac{1}{5.10}+\frac{1}{10.15}+...+\frac{1}{30.35}\)

\(5B=\frac{5}{5.10}+\frac{5}{10.15}+....+\frac{5}{30.35}\)

\(5B=\frac{1}{5}-\frac{1}{10}+\frac{1}{10}-\frac{1}{15}+...+\frac{1}{30}-\frac{1}{35}\)

\(5B=\frac{1}{5}-\frac{1}{35}=\frac{6}{35}\)

\(B=\frac{6}{35}.\frac{1}{5}=\frac{6}{175}\)

\(\frac{\left(\frac{6}{5}-\frac{4}{9}\right).\left(3^{22}+4.3^{18}\right)}{9^8.\left(x-2\right)}=\frac{\frac{2}{35}}{\frac{6}{175}}\)

\(\frac{\frac{34}{45}.3^{18}.\left(3^4+4\right)}{\left(3^2\right)^8.\left(x-2\right)}=\frac{5}{3}\)

\(\frac{\frac{34}{45}.85.3^{18}}{3^{16}.\left(x-2\right)}=\frac{5}{3}\)

\(\frac{\frac{578}{9}.3^2}{x-2}=\frac{5}{3}\)

\(\frac{578}{x-2}=\frac{5}{3}\)

\(\Rightarrow578.3=5x-10\)

\(\Rightarrow1734+10=5x\)

\(\Rightarrow x=\frac{1744}{5}=348,8\)

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