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

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

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

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

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

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

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

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

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

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

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

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

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

=\(\frac{10}{33}\)

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

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

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

=1/3*10/33

=10/99

dễ mà bn                                     

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

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

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

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

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

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

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

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

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

\(=\frac{1}{3\times6}+\frac{1}{6\times9}+...+\)\(\frac{1}{30\times33}\)

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

K hiểu cái quy luật là j luôn

Trần Quang Đài vẫn có quy luật mà

dễ chờ chút để tớ giúp cho

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

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

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

\(M=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{4970}\)

\(M=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{70.71}\)

\(M=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{70}-\frac{1}{71}\)

\(M=1-\frac{1}{71}\)

\(M=\frac{70}{71}\)

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

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

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

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

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

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

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