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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
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta phân tích: 1/18=1/3x6;1/54=1/6x9;1/108=1/9x12;.........1/990=1/30x33, ta có Fx3=3/3x6+3/6x9+3/9x12+........+3/30x33=1/3-1/6+1/6-1/9+1/9-1/12+.............+1/30-1/33=1/3-1/33=10/33, suy ra F là: 10/33/3=10/99
F=1/18+1/54+1/108+...+1/990 F=1/3.6 + 1/6.9 + 1/9.12 +...+ 1/30.33 suy ra : 3F= 3/3.6 + 3/6.9 + 3/9.12 +...+3/30.33 3F= 3/3 - 3/6 + 3/6 - 3/9 + 3/9 - 3/12 +...+3/30 - 3/33 3F=1 - 3/33 = 33/33 - 3/33 = 30/33 F= 30/33 : 3 = 30/33 . 1/3 =10/99
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a) \(F=\frac{1}{18}+\frac{1}{54}+\frac{1}{108}+...+\frac{1}{990}\)
\(F=\frac{1}{3.6}+\frac{1}{6.9}+\frac{1}{9.12}+...+\frac{1}{30.33}\)
\(3F=\frac{3}{3.6}+\frac{3}{6.9}+\frac{3}{9.12}+...+\frac{3}{30.33}\)
\(3F=\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+...+\frac{1}{30}-\frac{1}{33}\)
\(3F=\frac{1}{3}-\frac{1}{33}\)
\(F=\frac{1}{3}.\left(\frac{1}{3}-\frac{1}{33}\right)\)
\(F=\frac{1}{3}.\frac{1}{3}-\frac{1}{3}.\frac{1}{33}=\frac{1}{9}-\frac{1}{99}=\frac{11}{99}-\frac{1}{99}=\frac{10}{99}\)
b) \(A=\frac{7}{10.11}+\frac{7}{11.12}+\frac{7}{12.13}+...+\frac{7}{69.70}\)
\(A=7.\left(\frac{1}{10.11}+\frac{1}{11.12}+\frac{1}{12.13}+...+\frac{1}{69.70}\right)\)
\(A=7.\left(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+...+\frac{1}{69}-\frac{1}{70}\right)\)
\(A=7.\left(\frac{1}{10}-\frac{1}{70}\right)=7.\left(\frac{7}{70}-\frac{1}{70}\right)=7.\frac{6}{70}\)
\(A=\frac{7.6}{70}=\frac{1.6}{10}=\frac{1.3}{5}=\frac{3}{5}\)
a, \(F=\frac{1}{18}+\frac{1}{54}+\frac{1}{108}+...+\frac{1}{990}\)
\(F=\frac{1}{3}\cdot\left(\frac{3}{3\cdot6}+\frac{3}{6\cdot9}+\frac{3}{9\cdot12}+...+\frac{3}{30\cdot33}\right)\)
\(F=\frac{1}{3}\cdot\left(\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+\frac{1}{9}-\frac{1}{12}+...+\frac{1}{30}-\frac{1}{33}\right)\)
\(F=\frac{1}{3}\cdot\left(\frac{1}{3}-\frac{1}{33}\right)\)
\(F=\frac{1}{3}-\frac{10}{33}\)
\(F=\frac{10}{99}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(F=\frac{1}{18}+\frac{1}{54}+\frac{1}{108}+...+\frac{1}{990}\)
\(F=\frac{1}{9}\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{110}\right)\)
\(F=\frac{1}{9}\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{10.11}\right)\)
\(F=\frac{1}{9}\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{10}-\frac{1}{11}\right)\)
\(F=\frac{1}{9}\left(1-\frac{1}{11}\right)\)
\(F=\frac{1}{9}.\frac{10}{11}=\frac{10}{99}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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
![](https://rs.olm.vn/images/avt/0.png?1311)
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
![](https://rs.olm.vn/images/avt/0.png?1311)
\(F=\frac{1}{18}+\frac{1}{54}+\frac{1}{108}+...+\frac{1}{990}\)
\(F=\frac{1}{3.6}+\frac{1}{6.9}+\frac{1}{9.12}+...+\frac{1}{30.33}\)
\(F=\frac{1}{3}.\left(\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+\frac{1}{9}-\frac{1}{12}+...+\frac{1}{30}-\frac{1}{33}\right)\)
\(F=\frac{1}{3}.\left(\frac{1}{3}-\frac{1}{33}\right)\)
\(F=\frac{1}{3}.\frac{10}{33}\)
\(F=\frac{10}{99}\)
\(F=\frac{1}{18}+\frac{1}{54}+\frac{1}{108}+...+\frac{1}{990}\)
\(=\frac{1}{3.6}+\frac{1}{6.9}+\frac{1}{9.12}+...+\frac{1}{30.33}\)
\(=\frac{1}{3}\left(\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+\frac{1}{9}-\frac{1}{12}+...+\frac{1}{30}-\frac{1}{33}\right)\)
\(=\frac{1}{3}\left(\frac{1}{3}-\frac{1}{33}\right)=\frac{1}{3}\cdot\frac{10}{33}=\frac{10}{99}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
K:2=2/2.4+2/4.6+2/6.8+...+2/2008.2010
=1/2-1/4+1/4-1/6+1/6-1/8+...+1/2008-1/2010
=1/2-1/2010
=502/1005
K=502/1005.2
=1004/1005
F=1/3.6+1/6.9+1/9.12+...+1/30.33
3F=3/3.6+3/6.9+3/9.12+...+1/30.33
=1/3-1/6+1/6-1/9+1/9-1/12+...+1/30-1/33
=1/3-1-33
=10/33
F=10/33:3
=10/99
Bn sai câu K = 4/2.4 + 4/4.6 + 4/6.8 +....+ 4/2008.2010
![](https://rs.olm.vn/images/avt/0.png?1311)
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