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8 tháng 4 2016

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

=\(\frac{1}{3}\).(3.( \(\frac{1}{3.6}\) + \(\frac{1}{6.9}\) + \(\frac{1}{9.12}\) + ... + \(\frac{1}{30.33}\) ))

\(\frac{1}{3}\). (\(\frac{3}{3.6}\) + \(\frac{3}{6.9}\) + \(\frac{3}{9.12}\) + ... + \(\frac{3}{30.33}\) )

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

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

\(\frac{1}{3}\) . \(\frac{10}{33}\)

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

Nhớ k cho mình nhé!!!

8 tháng 4 2016

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

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

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

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

\(=\frac{1}{3}.\left(\frac{11}{33}-\frac{1}{33}\right)\)

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

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

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

1 tháng 4 2017

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

\(=\frac{1}{3}.\left(\frac{3}{3.6}+\frac{3}{6.9}+\frac{3}{9.12}+...+\frac{3}{30.33}\right)\)

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

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

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

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

Đúng không Bạch Dương ? 

1 tháng 4 2017

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

\(=\frac{1}{2.9}+\frac{1}{6.9}+\frac{1}{12.9}+...+\frac{1}{110.9}\)

\(=\frac{1}{9}\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{110}\right)\)

\(=\frac{1}{9}\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{10.11}\right)\)

\(=\frac{1}{9}\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{10}-\frac{1}{11}\right)\)

\(=\frac{1}{9}\left(\frac{1}{1}-\frac{1}{11}\right)\)

\(=\frac{1}{9}.\frac{10}{11}\)

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

                Vậy \(\frac{1}{18}+\frac{1}{54}+\frac{1}{108}+...+\frac{1}{990}=\frac{10}{99}\)

25 tháng 4 2017

A=\(\frac{1}{18}+\frac{1}{54}+\frac{1}{108}+....+\frac{1}{990}\) =\(\frac{1}{3.6}+\frac{1}{6.9}+\frac{1}{9.12}+...+\frac{1}{30.33}\)                                                                    =\(\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{33}\right)=\frac{1}{2}.\frac{10}{33}=\frac{5}{33}\)

25 tháng 4 2017

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

\(A=\frac{1}{9}\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{110}\right)\)

\(A=\frac{1}{9}\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{10.11}\right)\)

\(A=\frac{1}{9}\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...-\frac{1}{11}\right)\)

\(A=\frac{1}{9}.\frac{10}{11}=\frac{10}{99}\)

17 tháng 4 2016

K hiểu cái quy luật là j luôn

17 tháng 4 2016

Trần Quang Đài vẫn có quy luật mà

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

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

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

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

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

17 tháng 4 2016

=1/3x6+1/6x9+1/9x12+...+1/30x33

=1/3-1/6+1/6-1/9+1/9-1/12+...+1/30-1/33

=1/3-1/33

=10/33

17 tháng 4 2016

\(\frac{1}{18}+\frac{1}{54}+\frac{1}{108}+...+\frac{1}{990}=\frac{1}{3.6}+\frac{1}{6.9}+\frac{1}{9.12}+...+\frac{1}{30.33}=\frac{1}{3}.\left(\frac{3}{3.6}+\frac{3}{6.9}+\frac{3}{9.12}+...+\frac{3}{30.33}\right)=\frac{1}{3}.\left(\frac{6-3}{3.6}+\frac{9-6}{6.9}+\frac{12-9}{9.12}+...+\frac{33-30}{30.33}\right)=\frac{1}{3}.\left(\frac{6}{3.6}-\frac{3}{3.6}+\frac{9}{6.9}-\frac{6}{6.9}+\frac{12}{9.12}-\frac{9}{9.12}+...+\frac{33}{30.33}-\frac{30}{30.33}\right)=\frac{1}{3}.\left(\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+\frac{1}{9}-\frac{1}{12}+...+\frac{1}{30}-\frac{1}{33}\right)=\frac{1}{3}.\left(\frac{1}{3}-\frac{1}{33}\right)=\frac{1}{3}.\frac{10}{33}=\frac{10}{99}\)

2 tháng 4 2017

Đặt A=1/18+1/54+1/108+...+1/990

=> A=1/3.6+1/6.9+1/9.12+...+1/30.33

=>3A=3/3.6+3/6.9+3/9.12+...+3/30.33

=>3A=1/3-1/6+1/6-1/9+1/9-1/12+...+1/30-1/33

=>3A=1/3-1/33

=>3A=10/33

=>A=10/33:3

=>A=10/99

Vậy 1/18+1/54+1/108+...+1/990=10/99

Các bạn hãy ủng hộ mik nha !!! Mik cảm ơn nhiều .

2 tháng 4 2017

\(\frac{989}{990}\)nha bạn 

tk mk nha ! mk nhanh nhất !

8 tháng 2 2019

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

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

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

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

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

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

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

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

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

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

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

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