Tính nhanh
1/2+1/6+1/12+1/20+...+1/90+1/110+1/132
K
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Những câu hỏi liên quan
TF
3
XO
6 tháng 1 2020
Ta có : \(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}+\frac{1}{110}+\frac{1}{132}\)
= \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}+\frac{1}{10.11}+\frac{1}{11.12}\)
= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}\)
= \(1-\frac{1}{12}=\frac{11}{12}\)
NK
17 tháng 4 2023
\(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}\)
\(=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}\)
\(=\dfrac{1}{1}\cdot\dfrac{1}{2}+\dfrac{1}{2}\cdot\dfrac{1}{3}+\dfrac{1}{3}\cdot\dfrac{1}{4}+\dfrac{1}{4}\cdot\dfrac{1}{5}+\dfrac{1}{5}\cdot\dfrac{1}{6}+\dfrac{1}{6}\cdot\dfrac{1}{7}\)
\(=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}\)
\(=\dfrac{1}{1}-\dfrac{1}{7}=\dfrac{7}{7}-\dfrac{1}{7}=\dfrac{6}{7}\)
14 tháng 3 2022
1/2 + 1/6 + 1/12 + ... + 1/132
= 1/1.2 + 1/2.3 + 1/3.4 + ... + 1/11.12
= 1/1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/11 - 1/12
= 1 - 1/12
= 11/12
1/90 + 1/110 + 1/132 + ... + 1/10100
= 1/9.10 + 1/10.11 + 1/11.12 + ... + 1/100.101
= ... [như trên]
= 1/9 - 1/100
= 49/450
PH
9
20 tháng 8 2016
1/12 + 1/20 + ... + 1/132
= 1/3×4 + 1/4×5 + ... + 1/11×12
= 1/3 - 1/4 + 1/4 - 1/5 + ... + 1/11 - 1/12
= 1/3 - 1/12
= 4/12 - 1/12
= 3/12 = 1/4
TM
1
VP
2
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
8 tháng 12 2024
A = 1 - \(\dfrac{5}{6}\)+\(\dfrac{7}{12}\)-\(\dfrac{9}{20}\)+\(\dfrac{11}{30}\)-\(\dfrac{13}{42}\)+\(\dfrac{15}{56}\) - \(\dfrac{17}{72}\)+\(\dfrac{19}{90}\)+\(\dfrac{23}{132}\)-\(\dfrac{25}{156}\)
A = 1 - \(\dfrac{5}{2.3}\)+\(\dfrac{7}{3.4}\)-\(\dfrac{9}{4.5}\)+\(\dfrac{11}{5.6}\)-\(\dfrac{13}{6.7}\)+\(\dfrac{15}{7.8}\)-\(\dfrac{17}{8.9}\)+\(\dfrac{19}{9.10}\)+\(\dfrac{23}{11.12}\)-\(\dfrac{25}{12.13}\)
A = 1 - \(\dfrac{1}{2}-\dfrac{1}{3}\)+\(\dfrac{1}{3}+\dfrac{1}{4}\)-\(\dfrac{1}{4}-\dfrac{1}{5}\)+...+\(\dfrac{1}{11}+\dfrac{1}{12}\)- \(\dfrac{1}{12}-\dfrac{1}{13}\)
A = 1 - \(\dfrac{1}{2}\) - \(\dfrac{1}{13}\)
A = \(\dfrac{1}{2}\) - \(\dfrac{1}{13}\)
A = \(\dfrac{11}{26}\)
AM
Tính tổng
\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+.........+\frac{1}{110}+\frac{1}{132}\)
7
2 tháng 4 2017
=1/1*2+1/2*3+1/3*4+...+1*10*11+1/11*12=1-1/2+1/2-1/3+1/3-1/4+...+1/10-1/11+1/11-1/12
=1-1/12=11/12.
DP
2 tháng 4 2017
\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{110}+\frac{1}{132}\)
\(=\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{10\times11}+\frac{1}{11\times12}\)
\(=1-\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{11}+\frac{1}{12}\)
\(=1-\frac{1}{12}\)
\(=\frac{11}{12}\)
\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+..........+\frac{1}{132}\)
\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+..........+\frac{1}{11.12}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...........+\frac{1}{11}-\frac{1}{12}\)
\(=1-\frac{1}{12}\)
\(=\frac{11}{12}\)
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