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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
![](https://rs.olm.vn/images/avt/0.png?1311)
\(3F=\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+...+\frac{1}{10}-\frac{1}{33}\)
\(3F=\frac{1}{3}-\frac{1}{33}=\frac{30}{99}\)
\(F=\frac{10}{99}\)
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
F =10/99
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{1}{18}\)+\(\frac{1}{54}\)+\(\frac{1}{108}\)+...+\(\frac{1}{990}\)
=\(\frac{1}{3.6}\)+\(\frac{1}{6.9}\)+\(\frac{1}{9.12}\)+...+\(\frac{1}{30.33}\)
=\(\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+\frac{1}{9}-\frac{1}{12}+...+\)\(\frac{1}{30}-\frac{1}{33}\)
=\(\frac{1}{3}-\frac{1}{33}\)
=\(\frac{10}{33}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{1}{18}+\frac{1}{54}+\frac{1}{108}+...+\frac{1}{990}\)
=\(\frac{1}{3}.\left(\frac{3}{3.6}+\frac{3}{6.9}+\frac{3}{9.12}+...+\frac{1}{30.33}\right)\)
=\(\frac{1}{3}.\left(\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+\frac{1}{9}-\frac{1}{12}+...+\frac{1}{30}-\frac{1}{33}\right)\)
=\(\frac{1}{3}.\left(\frac{1}{3}-\frac{1}{33}\right)\)
=\(\frac{1}{3}.\frac{10}{33}\)
=\(\frac{10}{99}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
đặt A=1/18+1/54+1/108+...+1/990
\(A=\frac{1}{3.6}+\frac{1}{6.9}+\frac{1}{9.12}+...+\frac{1}{30.33}\)
\(=\frac{1}{3}\times\left(\frac{3}{3.6}+\frac{3}{6.9}+...+\frac{3}{30.33}\right)\)
\(=\frac{1}{3}\times\left(\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+...+\frac{1}{30}-\frac{1}{33}\right)\)
\(=\frac{1}{3}\times\left(\frac{1}{3}-\frac{1}{33}\right)\)
\(=\frac{1}{3}\times\frac{10}{33}\)
\(=\frac{10}{99}\)
1/18 + 1/54 + ..... + 1/990
= 1/3.6 +1/6.9 +...... + 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
![](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
\(D=\frac{1}{18}+\frac{1}{54}+\frac{1}{108}+...+\frac{1}{990}\)
\(\Rightarrow D=\frac{1}{3\cdot6}+\frac{1}{6\cdot9}+\frac{1}{9\cdot12}+...+\frac{1}{30\cdot33}\)
\(\Rightarrow D=\frac{1}{3}\left(\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+\frac{1}{9}-\frac{1}{12}+...+\frac{1}{30}-\frac{1}{33}\right)\)
\(\Rightarrow D=\frac{1}{3}\left(\frac{1}{3}-\frac{1}{33}\right)=\frac{1}{3}\cdot\frac{10}{33}=\frac{10}{99}\)
k ủng hộ nha!!!!!!!!