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23 tháng 1 2022

\(\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{79.80}\)

\(=\)\(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{79}-\frac{1}{80}\)

\(=\frac{1}{4}-\frac{1}{80}\)

\(=\frac{20}{80}-\frac{1}{80}\)

\(=\frac{19}{80}\)

DD
23 tháng 1 2022

\(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{78.79}+\frac{1}{79.80}\)

\(=\frac{4-3}{3.4}+\frac{5-4}{4.5}+\frac{6-5}{5.6}+...+\frac{79-78}{78.79}+\frac{80-79}{79.80}\)

\(=\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{78}-\frac{1}{79}+\frac{1}{79}-\frac{1}{80}\)

\(=\frac{1}{3}-\frac{1}{80}=\frac{77}{240}\)

4 tháng 5 2016

\(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}\)

=\(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}\)

=\(\frac{1}{3}-\frac{1}{8}\)

=\(\frac{8}{24}-\frac{3}{24}\)

=\(\frac{5}{24}\)

4 tháng 5 2016

bằng 853/ 140 nhé bn!

6 tháng 2 2023

A=1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7+1/7-1/8+1/8-1/9

A=1/3-1/9

A=2/9

6 tháng 2 2023

các câu 2;3 còn lại giống câu 1 bạn nhé

bạn thay số vào rồi làm tương tự

27 tháng 4 2018

Ta có : \(\frac{1}{1.2}\)+  \(\frac{1}{2.3}\)+\(\frac{1}{3.4}\)+\(\frac{1}{4.5}\)+\(\frac{1}{5.6}\)+\(\frac{1}{6.7}\)

= 1  - \(\frac{1}{2}\)+\(\frac{1}{2}\)-\(\frac{1}{3}\)+\(\frac{1}{3}\)-\(\frac{1}{4}\)+\(\frac{1}{4}\)-\(\frac{1}{5}\)+\(\frac{1}{5}\)-\(\frac{1}{6}\)+\(\frac{1}{6}\)-\(\frac{1}{7}\)

= 1 - \(\frac{1}{7}\)=  \(\frac{6}{7}\)

27 tháng 4 2018

=1-1/2+1/2-1/3+1/3-1/4+...+1/6-1/7=1-1/7=6/7

3 tháng 3 2022

\(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}\\ =\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}\\ =\dfrac{1}{2}-\dfrac{1}{7}\\ =\dfrac{5}{14}\)

15 tháng 10 2023

Hh

 

DT
15 tháng 10 2023

P = 1/2.3 + 1/3.4 + 1/4.5 + ... + 1/8.9

= 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 +  ... + 1/8 - 1/9

= 1/2 - 1/9

= 7/18

25 tháng 3 2022

\(M=\dfrac{1}{2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}\)

\(M=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}\)

\(M=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}\)

\(M=1-\dfrac{1}{7}\)

\(M=\dfrac{6}{7}\)

 

25 tháng 3 2022

tham khảo

https://hoc24.vn/cau-hoi/123134145156167.5003535458609#:~:text=l%C3%BAc%2021%3A02-,1,14,-12.3%2B13.4%2B14.5

vào đi 

`@` `\text {Ans}`

`\downarrow`

\(\dfrac{1}{3\cdot4}-\dfrac{1}{4\cdot5}-\dfrac{1}{5\cdot6}-\dfrac{1}{6\cdot7}-\dfrac{1}{7\cdot8}-\dfrac{1}{8\cdot9}\)

`=`\(\dfrac{1}{3}-\left(\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{8}-\dfrac{1}{9}\right)\)

`=`\(\dfrac{1}{3}-\left(\dfrac{1}{2}-\dfrac{1}{9}\right)\)

`=`\(\dfrac{1}{3}-\dfrac{7}{18}=-\dfrac{1}{18}\)

`@` `\text {Ans}`

`\downarrow`

`a)`

\(A=\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}+\dfrac{1}{7\cdot8}+\dfrac{1}{8\cdot9}\)

`=`\(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{8}-\dfrac{1}{9}\)

`=`\(\dfrac{1}{3}-\left(\dfrac{1}{4}-\dfrac{1}{4}\right)-\left(\dfrac{1}{5}-\dfrac{1}{5}\right)-...-\dfrac{1}{9}\)

`=`\(\dfrac{1}{3}-\dfrac{1}{9}\)

`=`\(\dfrac{2}{9}\)

Vậy, \(A=\dfrac{2}{9}\)

`b)`

\(B=\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}+\dfrac{1}{7\cdot8}+...+\dfrac{1}{23\cdot24}+\dfrac{1}{24\cdot25}\)

`=`\(\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{24}-\dfrac{1}{25}\)

`=`\(\dfrac{1}{5}-\left(\dfrac{1}{6}-\dfrac{1}{6}\right)-\left(\dfrac{1}{7}-\dfrac{1}{7}\right)-...-\dfrac{1}{25}\)

`=`\(\dfrac{1}{5}-\dfrac{1}{25}=\dfrac{4}{25}\)

Vậy, \(B=\dfrac{4}{25}\)

`c)`

\(C=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{98\cdot99}+\dfrac{1}{99\cdot100}\)

`=`\(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\)

`=`\(1-\left(\dfrac{1}{2}-\dfrac{1}{2}\right)-\left(\dfrac{1}{3}-\dfrac{1}{3}\right)-...-\dfrac{1}{100}\)

`=`\(1-\dfrac{1}{100}=\dfrac{99}{100}\)

Vậy, \(C=\dfrac{99}{100}\)

26 tháng 5 2020

\(\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}=\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{7}-\frac{1}{8}\)

\(=\frac{1}{3}-\frac{1}{8}=\frac{5}{24}\)

Ez :))