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

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

gợi ý :1/18 +1/54 + ... +1/990

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

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

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

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

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

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

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

Đúng không Bạch Dương ? 

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

\(=\frac{1}{3x6}+\frac{1}{6x9}+\frac{1}{9x12}+...+\frac{1}{30x33}\)

\(=\frac{1}{3}x\left(\frac{3}{3x6}+\frac{3}{6x9}+....+\frac{3}{30x33}\right)\)

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

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

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

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

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

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

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

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

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

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

sorry nó dài giải ko nổi

K hiểu cái quy luật là j luôn

Trần Quang Đài vẫn có quy luật mà

dễ chờ chút để tớ giúp cho

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

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

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

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

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

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

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

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

= 2(2/2.4 + 2/4.6 +.....+ 2/2008.2016)

= 2(1/2 - 1/4 + 1/4 - 1/6 +....+ 1/2008 - 1/2016)

= 2(1/2 - 1/2016)

=2 . 1007/2016

=1007/1008

=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

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

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 .

= \(\frac{989}{990}\)nha bạn 

tk mk nha ! mk nhanh nhất !

10/99 mk nghĩ  thế bởi hôm nọ cô giáo chữa cho mk rùi

hello

6/u

trrtu

\(B=\frac{10}{2.7}+\frac{10}{7.12}+...+\frac{10}{502.507}\)

\(B=2.\left(\frac{5}{2.7}+\frac{5}{7.12}+...+\frac{5}{502.507}\right)\)

\(B=2.\left(\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{12}+...+\frac{1}{502}-\frac{1}{507}\right)\)

\(B=2.\left(\frac{1}{2}-\frac{1}{507}\right)\)

\(B=2.\frac{505}{1014}\)
\(B=\frac{505}{507}\)

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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