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29 tháng 5 2020

A=\(\frac{4}{2.4}+\frac{4}{4.6}+\frac{4}{6.8}+...+\frac{4}{2018.2020}\)

\(\frac{1}{2}\)A= \(\frac{1}{2}.\left(\frac{4}{2.4}+\frac{4}{4.6}+\frac{4}{6.8}+...+\frac{4}{2018.2020}\right)\)

\(\frac{1}{2}A\)\(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{2018.2020}\)

\(\frac{1}{2}A\)\(\frac{4-2}{2.4}+\frac{6-4}{4.6}+\frac{8-6}{6.8}+...+\frac{2020-2018}{2018.2020}\)

\(\frac{1}{2}A\)\(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{2018}-\frac{1}{2020}\)

\(\frac{1}{2}A\)\(\frac{1}{2}-\frac{1}{2020}\)

\(\frac{1}{2}A=\frac{1009}{2020}\)

\(A=\frac{1009}{2020}:\frac{1}{2}\)

\(A=\frac{1009}{1010}\)

29 tháng 5 2020

a) Ta có 

A= 4/2*4+4/4*6+....+4/2018*2020

=> A= 2*(2/2*4+2/4*6+...+2*(2018*2020)

=> A= 2*(1/2-1/4+1/4-1/6+...+1/2018-1/2020)

=> A= 2*(1/2-1/2020)

=> A= 2* 1009/2020

=> A= 1009/1010

b) B= 1/18+1/54+1/108+...+1/990

=> B= 3/3*(1/18+1/54+1/108+..+1/990)

=> B= 1/3*( 3/3*6+3/6*9+...+3/30*33)

=> B= 1/3*(1/3-1/6+1/6-1/9+1/9-1/12+...+1/30-1/33)

=> B= 1/3*( 1/3-1/33)

=> B=1/3*10/33

=> B=10/99

21 tháng 8 2016

\(A=\frac{4}{2.4}+\frac{4}{4.6}+\frac{4}{6.8}+.....+\frac{4}{2008.2010}\)

\(\Rightarrow A=4\left(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+.....+\frac{1}{2008.2010}\right)\)

\(\Rightarrow A=4\left[\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+....+\frac{1}{2008}-\frac{1}{2010}\right)\right]\)

\(\Rightarrow A=4\left[\frac{1}{2}\left(\frac{1}{2}-\frac{1}{2010}\right)\right]\Rightarrow A=4\left(\frac{1}{2}.\frac{502}{1005}\right)\Rightarrow A=4.\frac{251}{1005}\Rightarrow A=\frac{1004}{1005}\)

21 tháng 8 2016

\(B=\frac{1}{18}+\frac{1}{54}+\frac{1}{108}+....+\frac{1}{990}\)

\(\Rightarrow B=\frac{1}{3.6}+\frac{1}{6.9}+\frac{1}{9.12}+....+\frac{1}{30.33}\)

\(\Rightarrow B=\frac{1}{3}\left(\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+\frac{1}{9}-\frac{1}{12}+.....+\frac{1}{30}-\frac{1}{33}\right)\)

\(\Rightarrow B=\frac{1}{3}.\left(\frac{1}{3}-\frac{1}{33}\right)\Rightarrow B=\frac{1}{3}.\frac{10}{33}\Rightarrow B=\frac{10}{99}\)

23 tháng 2 2020

Ta thấy : \(\frac{1}{11}>\frac{1}{100},\frac{1}{12}>\frac{1}{100},...,\frac{1}{100}=\frac{1}{100}\)

\(\Rightarrow\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+...+\frac{1}{100}>\frac{1}{100}+\frac{1}{100}+\frac{1}{100}+...+\frac{1}{100}=\frac{90}{100}=\frac{9}{10}\)

\(\Rightarrow\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+...+\frac{1}{100}>\frac{9}{10}+\frac{1}{10}=1\)

Do đó : \(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+...+\frac{1}{100}>1\)

1 tháng 8 2019

a, \(A=\frac{6}{10.11}+\frac{6}{11.12}+\frac{6}{12.13}+...+\frac{6}{69.70}\)

\(A=\frac{6}{10}-\frac{6}{11}+\frac{6}{11}-\frac{6}{12}+\frac{6}{12}-\frac{6}{13}+...+\frac{6}{69}-\frac{6}{70}\)

\(A=\frac{6}{10}-\frac{6}{70}\)

\(A=\frac{18}{35}\)

b, \(B=\frac{4}{2.4}+\frac{4}{4.6}+\frac{4}{6.8}+...+\frac{4}{2018.2020}\)

\(B=\frac{4}{2}.\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{2018.2020}\right)\)

\(B=2.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{2018}-\frac{1}{2020}\right)\)

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

\(B=2.\frac{1009}{2020}\)

\(B=\frac{1009}{1010}\)

Chúc bạn học tốt thanghoa

6 tháng 10 2019

vuiHơi thắc mắc câu B cậu oi!!!Gỉai thích cho mk vs ạ!!Thanks

14 tháng 8 2016

dễ mà bạn làm từ câu a nếu ra thì các câu khác cũng dễ thôi

14 tháng 8 2016

\(A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+....+\frac{1}{2009\cdot2010}\)

\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2009}-\frac{1}{2010}\)

\(A=1-\frac{1}{2010}\)

\(A=\frac{2009}{2010}\)

8 tháng 2 2019

\(\frac{1}{18}+\frac{1}{54}+\frac{1}{108}+...+\frac{1}{990}\)

\(=\frac{1}{3\cdot6}+\frac{1}{6\cdot9}+\frac{1}{9\cdot12}+...+\frac{1}{30\cdot33}\)

\(=\frac{1}{3}\left(\frac{3}{3\cdot6}+\frac{3}{6\cdot9}+\frac{3}{9\cdot12}+...+\frac{3}{30\cdot33}\right)\)

\(=\frac{1}{3}\left(\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+\frac{1}{9}-\frac{1}{12}+...+\frac{1}{30}-\frac{1}{33}\right)\)

\(=\frac{1}{3}\left(\frac{1}{3}-\frac{1}{33}\right)\)

\(=\frac{1}{3}\cdot\frac{10}{33}=\frac{10}{99}\)

\(\frac{1}{18}+\frac{1}{54}+\frac{1}{108}+...+\frac{1}{990}\)

\(=\frac{1}{3.6}+\frac{1}{6.9}+\frac{1}{9.12}+...+\frac{1}{30.33}\)

\(=\frac{1}{3}\left(\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+\frac{1}{9}-\frac{1}{12}+...+\frac{1}{30}-\frac{1}{33}\right)\)

\(=\frac{1}{3}\left(\frac{1}{3}-\frac{1}{33}\right)\)

\(=\frac{1}{3}.\frac{10}{33}\)

\(=\frac{10}{99}\)

27 tháng 7 2017

a, 11 1/4-(2 5/7+5 1/4)

= 45/4-(19/7+21/4)

= 45/4-223/28

=23/7

b, (8 5/11+3 5/8)-3 5/11

=(93/11+29/8)-38/11

=1063/88-38/11

=69/8

27 tháng 7 2017

a,  =\(11\frac{1}{4}-2\frac{5}{7}-5\frac{1}{4}\)

\(=\left(11\frac{1}{4}-5\frac{1}{4}\right)-2\frac{5}{7}\)

\(=6-2\frac{5}{7}\)

\(=\frac{23}{7}\)

b,  \(=8\frac{5}{11}+3\frac{5}{8}-3\frac{5}{11}\)

\(=\left(8\frac{5}{11}-3\frac{5}{11}\right)+3\frac{5}{8}\)

\(=5+3\frac{5}{8}\)

\(=\frac{69}{8}\)