F = 1/3.6 + 1/6.9 + 1/9.12 + ... + 1/30.33

F = 1/3.(1.3-1/6+1/6-1/9+1/9-1/12+...+1/30-1/33)

F = 1/3.(1/3-1/33)

F = 1/3.10/33

F = 10/99

F = 1/ 3.6 + 1/ 6.9 + 1/ 9.12 +...+1/ 30.33

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

F = 1/3 . ( 1/3 - 1/33 )

F = 1/3 . 10/ 33

F = 10 /99

1/18+1/54+1/108+……+1/810+1/990

=（1/3-1/6+1/6-1/9+1/9-1/12+……+1/27-1/30+1/30-1/33）÷3

=（1/3-1/33）÷3

=10/33÷3

=10/99

Ta có; F=1/3.6 +1 /6.9 + 1/9.12+......+1/30.33

         F=1.3/3.6.3 + 1.3/6.9.3+......+1.3/30.33.3

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

         F=1/3.(1/3-1/33)

         F=1/3.10/33

        F=10/99

                                                 giải
1/18 + 1/54 +1/108 + ......+ 1/990
ta tách mẫu số ra thành 1 tích của 2 số :
1/3x6 + 1/6x9 + 1/9x12 +........ + 1/30x33
theo quy tắc ta có : nếu tử nhân với 3 thì mẩu cũng sẽ nhân với 3 :
1x3/3x6x3 +1x3/6x9x3 + 1x3/9x11x3 + .........+ 1x3/30x33x3
= 1/3 x  (  3/3x6 + 3/6x9 + 3/9x11 +.....+3/30x33
= 1/3 x ( 1/3 - 1/33 )
= 1/3 x 10/33
=10/99

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

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

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

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

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

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

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

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

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

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

1/18+1/54+1/108+...+1/990

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

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

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

=1/3-1/33

=10/33

                                                 giải

1/18 + 1/54 +1/108 + ......+ 1/990

ta tách mẫu số ra thành 1 tích của 2 số :

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

theo quy tắc ta có : nếu tử nhân với 3 thì mẩu cũng sẽ nhân với 3 :

1x3/3x6x3 +1x3/6x9x3 + 1x3/9x11x3 + .........+ 1x3/30x33x3

= 1/3 x  (  3/3x6 + 3/6x9 + 3/9x11 +.....+3/30x33

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

= 1/3 x 10/33

=10/99

Mình giải hơi khó hiểu 1 chút nha

1/18+1/54+1/108+...+1/990

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

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

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

=1/3-1/33

=10/33

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

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

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

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

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

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

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

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

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

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

\(F=\frac{11}{33}-\frac{1}{33}\)

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

Chúc bạn học tốt !!!! 

 f =1/18+1/54+1/108+...+1/990

F=1/3.6+1/6.9+1/9.12+...+1/30.33

F=3/3.6+3/6.9+3/9.12+...+3/30.33

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

F=1/3-1/33

F=11/33-1/33

F=10/33

Vậy F=10/33.

dấu chấm là dấu nhân nha bạn

chúc bạn học tốt ^^. cho mình nha