Tính
\(\frac{6}{18}\)+\(\frac{6}{54}\)+\(\frac{6}{108}\)+...+\(\frac{6}{990}\)
Giúp mình với nhé!
K
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Những câu hỏi liên quan
9 tháng 4 2019
\(N=\frac{1}{3.6}+\frac{1}{6.9}+...+\frac{1}{30.33}\)
=\(\frac{1}{3}\left(\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+...+\frac{1}{30}-\frac{1}{33}\right)\)
=\(\frac{1}{3}\left(\frac{1}{3}-\frac{1}{33}\right)=\frac{10}{33}\)
9 tháng 4 2019
\(M=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{4970}\)
\(M=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{70.71}\)
\(M=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{70}-\frac{1}{71}\)
\(M=1-\frac{1}{71}\)
\(M=\frac{70}{71}\)
\(N=\frac{1}{18}+\frac{1}{54}+\frac{1}{108}+...+\frac{1}{990}\)
\(N=\frac{1}{3.6}+\frac{1}{6.9}+\frac{1}{9.12}+...+\frac{1}{30.33}\)
\(N=\frac{1}{3}.\left(\frac{3}{3.6}+\frac{3}{6.9}+\frac{3}{9.12}+...+\frac{3}{30.33}\right)\)
\(N=\frac{1}{3}.\left(\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+\frac{1}{9}-\frac{1}{12}+...+\frac{1}{30}-\frac{1}{33}\right)\)
\(N=\frac{1}{3}.\left(\frac{1}{3}-\frac{1}{33}\right)\)
\(N=\frac{1}{3}.\frac{10}{33}\)
\(N=\frac{10}{99}\)
8 tháng 2 2019
\(\frac{1}{18}+\frac{1}{54}+\frac{1}{108}+...+\frac{1}{990}\)
\(=\frac{1}{3\cdot6}+\frac{1}{6\cdot9}+\frac{1}{9\cdot12}+...+\frac{1}{30\cdot33}\)
\(=\frac{1}{3}\left(\frac{3}{3\cdot6}+\frac{3}{6\cdot9}+\frac{3}{9\cdot12}+...+\frac{3}{30\cdot33}\right)\)
\(=\frac{1}{3}\left(\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+\frac{1}{9}-\frac{1}{12}+...+\frac{1}{30}-\frac{1}{33}\right)\)
\(=\frac{1}{3}\left(\frac{1}{3}-\frac{1}{33}\right)\)
\(=\frac{1}{3}\cdot\frac{10}{33}=\frac{10}{99}\)
5 tháng 7 2019
\(\frac{1}{18}+\frac{1}{54}+\frac{1}{108}+...+\frac{1}{990}\)
\(=\frac{1}{3.6}+\frac{1}{6.9}+\frac{1}{9.12}+...+\frac{1}{30.33}\)
\(=\frac{1}{3}\left(\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+\frac{1}{9}-\frac{1}{12}+...+\frac{1}{30}-\frac{1}{33}\right)\)
\(=\frac{1}{3}\left(\frac{1}{3}-\frac{1}{33}\right)\)
\(=\frac{1}{3}.\frac{10}{33}\)
\(=\frac{10}{99}\)
17 tháng 4 2017
A=.....
=\(7.\left(\frac{1}{10.11}+\frac{1}{11.12}+\frac{1}{12.13}+....+\frac{1}{69.70}\right)=7.\left(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+.....+\frac{1}{69}-\frac{1}{70}\right)\)
=\(7.\left(\frac{1}{10}-\frac{1}{70}\right)=7.\frac{3}{35}=\frac{3}{5}\)
MẤY PHẦN SAU CX TÁCH MẪU RA RÙI LÀM NHƯ VẬY
TỰ LÀM NHE
9 tháng 8 2018
\(B=\frac{1}{3\cdot6}+\frac{1}{6\cdot9}+...+\frac{1}{30\cdot33}\)
\(B=\frac{1}{3}\cdot\left(\frac{3}{3\cdot6}+\frac{3}{6\cdot9}+...+\frac{3}{30\cdot33}\right)\)
\(B=\frac{1}{3}\cdot\left(\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+...+\frac{1}{30}-\frac{1}{33}\right)\)
\(B=\frac{1}{3}\cdot\left(\frac{1}{3}-\frac{1}{33}\right)\)
\(B=\frac{1}{3}\cdot\frac{10}{33}=\frac{10}{99}\)
\(C=\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{6}\right)+...+\left(1-\frac{1}{90}\right)\)
\(C=\left(1-\frac{1}{1\cdot2}\right)+\left(1-\frac{1}{2\cdot3}\right)+...+\left(1-\frac{1}{9\cdot10}\right)\)
\(C=9-\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{9\cdot10}\right)\)
\(C=9-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(C=9-\left(1-\frac{1}{10}\right)\)
\(C=9-\frac{9}{10}=\frac{81}{10}\)
BD
6
ML
1 tháng 4 2017
\(\frac{1}{18}+\frac{1}{54}+\frac{1}{108}+...+\frac{1}{990}\)
\(=\frac{1}{3}.\left(\frac{3}{3.6}+\frac{3}{6.9}+\frac{3}{9.12}+...+\frac{3}{30.33}\right)\)
\(=\frac{1}{3}.\left(\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+\frac{1}{9}-\frac{1}{12}+...+\frac{1}{30}-\frac{1}{33}\right)\)
\(=\frac{1}{3}.\left(\frac{1}{3}-\frac{1}{33}\right)\)
\(=\frac{1}{3}.\frac{10}{33}\)
\(=\frac{10}{99}\)
Đúng không Bạch Dương ?
HB
1 tháng 4 2017
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 \(\frac{1}{18}+\frac{1}{54}+\frac{1}{108}+...+\frac{1}{990}=\frac{10}{99}\)
TQ
6
25 tháng 4 2017
A=\(\frac{1}{18}+\frac{1}{54}+\frac{1}{108}+....+\frac{1}{990}\) =\(\frac{1}{3.6}+\frac{1}{6.9}+\frac{1}{9.12}+...+\frac{1}{30.33}\) =\(\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{33}\right)=\frac{1}{2}.\frac{10}{33}=\frac{5}{33}\)
M
25 tháng 4 2017
\(A=\frac{1}{18}+\frac{1}{54}+\frac{1}{108}+...+\frac{1}{990}\)
\(A=\frac{1}{9}\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{110}\right)\)
\(A=\frac{1}{9}\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{10.11}\right)\)
\(A=\frac{1}{9}\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...-\frac{1}{11}\right)\)
\(A=\frac{1}{9}.\frac{10}{11}=\frac{10}{99}\)
NV
2
XN
10 tháng 4 2017
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-\(\frac{1}{33}\)
=\(\frac{32}{33}\)
10 tháng 4 2017
gợi ý :1/18 +1/54 + ... +1/990
= 1/3*6 + 1/6*9 + 1/9*13 + ... +1/30*33
GS
17 tháng 4 2016
=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
17 tháng 4 2016
\(\frac{1}{18}+\frac{1}{54}+\frac{1}{108}+...+\frac{1}{990}=\frac{1}{3.6}+\frac{1}{6.9}+\frac{1}{9.12}+...+\frac{1}{30.33}=\frac{1}{3}.\left(\frac{3}{3.6}+\frac{3}{6.9}+\frac{3}{9.12}+...+\frac{3}{30.33}\right)=\frac{1}{3}.\left(\frac{6-3}{3.6}+\frac{9-6}{6.9}+\frac{12-9}{9.12}+...+\frac{33-30}{30.33}\right)=\frac{1}{3}.\left(\frac{6}{3.6}-\frac{3}{3.6}+\frac{9}{6.9}-\frac{6}{6.9}+\frac{12}{9.12}-\frac{9}{9.12}+...+\frac{33}{30.33}-\frac{30}{30.33}\right)=\frac{1}{3}.\left(\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+\frac{1}{9}-\frac{1}{12}+...+\frac{1}{30}-\frac{1}{33}\right)=\frac{1}{3}.\left(\frac{1}{3}-\frac{1}{33}\right)=\frac{1}{3}.\frac{10}{33}=\frac{10}{99}\)
TK
3
PT
8 tháng 4 2016
\(\frac{1}{18}\)+ \(\frac{1}{54}\)+ \(\frac{1}{108}\)+ ... + \(\frac{1}{990}\)
=\(\frac{1}{3}\).(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!!!
8 tháng 4 2016
\(\frac{1}{18}+\frac{1}{54}+\frac{1}{108}+....+\frac{1}{990}\)
\(=\frac{1}{3.6}+\frac{1}{6.9}+\frac{1}{9.12}+.....+\frac{1}{30.33}\)
\(=\frac{1}{3}.\left(\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+\frac{1}{9}-\frac{1}{12}+.....+\frac{1}{30}-\frac{1}{33}\right)\)
\(=\frac{1}{3}.\left(\frac{1}{3}-\frac{1}{33}\right)\)
\(=\frac{1}{3}.\left(\frac{11}{33}-\frac{1}{33}\right)\)
\(=\frac{1}{3}.\frac{10}{33}\)
\(=\frac{10}{99}\)
M
3
2 tháng 4 2017
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 .
2 tháng 4 2017
= \(\frac{989}{990}\)nha bạn
tk mk nha ! mk nhanh nhất !
Ta làm như sau:
\(\frac{6}{18}\)+\(\frac{6}{54}\)+\(\frac{6}{108}\)+...+\(\frac{6}{990}\)
=\(\frac{6}{3.6}\)+\(\frac{6}{6.9}\)+\(\frac{6}{9.12}\)+...\(\frac{6}{30.33}\)
=2 (\(\frac{3}{3.6}\)+\(\frac{3}{6.9}\)+\(\frac{3}{9.12}\)+...+\(\frac{3}{30.33}\)
=2 (\(\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+\frac{1}{9}-\frac{1}{12}+...+\frac{1}{30}-\frac{1}{33}\))
=2 ( \(\frac{1}{3}-\frac{1}{33}\))
=2.\(\frac{10}{33}\)=\(\frac{2.10}{33}\)=\(\frac{20}{33}\)
\(\frac{6}{18}+\frac{6}{54}+\frac{6}{108}+...+\frac{6}{990}\)
=\(\frac{6}{3.6}+\frac{6}{6.9}+\frac{6}{9.12}+...+\frac{6}{30.33}\)
= 2.(\(\frac{3}{3.6}+\frac{3}{6.9}+\frac{3}{9.12}+...+\frac{3}{30.33}\))
=2.(\(\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+\frac{1}{9}-\frac{1}{12}+...+\frac{1}{30}-\frac{1}{33}\))
=2.[\(\frac{1}{3}+\left(\frac{-1}{6}+\frac{1}{6}\right)+\left(\frac{-1}{9}+\frac{1}{9}\right)+...+\left(\frac{-1}{30}+\frac{1}{30}\right)+\frac{-1}{33}\)]
=2.\(\left[\frac{1}{3}+\frac{-1}{33}\right]\)
=2.\(\left[\frac{11}{33}+\frac{-1}{33}\right]\)
=2.\(\frac{10}{33}\)
=\(\frac{20}{33}\)