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LG
0
TG
4
6 tháng 5 2016
1/20 + 1/30 + 1/42 + ... + 1/156
= 1/4.5 + 1/5.6 + 1/6.7 + .... + 1/12.13
= 1/4 - 1/5 + 1/5 - 1/6 + 1/6 - 1/7 + ... + 1/12 - 1/13
= 1/4 - 1/13
= 9/52
6 tháng 5 2016
1/20 + 1/30 + 1/42 + ... + 1/156
= 1/4.5 + 1/5.6 + 1/6.7 + .... + 1/12.13
= 1/4 - 1/5 + 1/5 - 1/6 + 1/6 - 1/7 + ... + 1/12 - 1/13
= 1/4 - 1/13
= 9/52
TT
2
30 tháng 3 2016
\(\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{110}+\frac{1}{132}\)
=\(\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+\frac{1}{8\cdot9}+\frac{1}{9\cdot10}+\frac{1}{10\cdot11}+\frac{1}{11\cdot12}\)
=\(\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}\)
=\(\frac{1}{5}-\frac{1}{12}=\frac{12}{60}-\frac{5}{60}=\frac{7}{60}\)
30 tháng 3 2016
\(\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{110}+\frac{1}{132}\)
= \(\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+\frac{1}{8\cdot9}+\frac{1}{9\cdot10}+\frac{1}{10\cdot11}+\frac{1}{11\cdot12}\)
= \(\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}\)
=\(\frac{1}{5}-\frac{1}{12}\)
=\(\frac{7}{60}\)
NH
1
2 tháng 5 2023
Ta viết lại biểu thức A như sau:
\(A=-\left(\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+...+\dfrac{1}{11.12}\right)\)
\(A=-\left(\dfrac{5-4}{4.5}+\dfrac{6-5}{5.6}+\dfrac{7-6}{6.7}+...+\dfrac{12-11}{11.12}\right)\)
\(A=-\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{11}-\dfrac{1}{12}\right)\)
\(A=-\left(\dfrac{1}{4}-\dfrac{1}{12}\right)\)
\(A=-\dfrac{1}{6}\)
V
7
4 tháng 8 2016
Đặt tổng trên = A
\(A=\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{110}+\frac{1}{132}\)
\(A=\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}+\frac{1}{10.11}+\frac{1}{11.12}\)
\(A=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}\)
\(A=\frac{1}{5}-\frac{1}{12}\)
\(A=\frac{7}{60}\)
DD
4 tháng 8 2016
\(=\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}+\frac{1}{10.11}+\frac{1}{11.12}\)
\(=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}\)
\(=\frac{1}{5}-\frac{1}{12}=\frac{7}{60}\)
HP
9 tháng 5 2016
\(A=\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{110}+\frac{1}{132}\)
\(A=\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}+\frac{1}{10.11}+\frac{1}{11.12}\)
Ta có: \(\frac{1}{n\left(n+1\right)}=\frac{1}{n}-\frac{1}{n+1}\) với mọi số tự nhiên n
\(\Rightarrow A=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}\)
\(A=\frac{1}{5}-\frac{1}{12}=\frac{7}{60}\)
Vậy A=7/60
5 tháng 6 2015
\(A=\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+...+\frac{1}{11.12}\)
\(;A=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{11}-\frac{1}{12}\)
\(;A=\frac{1}{5}-\frac{1}{12}=\frac{7}{60}\)
5 tháng 6 2015
=1/5.6+1/6.7+1/7.8+`1/8.9+1/9.10+1/10.11+1/11.12
=1/5-1/12
=7/60
C
12 tháng 4 2016
\(A=\frac{1}{5.6}+\frac{1}{6.7}+......+\frac{1}{10.11}+\frac{1}{11.12}=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+......+\frac{1}{11}-\frac{1}{12}=\frac{1}{5}-\frac{1}{12}=\frac{7}{60}\)
12 tháng 4 2016
A = 1/30 + 1/42 + 1/56 + 1/72 + 1/90 + 1/110 + 1/132
A = 1/5.6 + 1/6.7 + 1/7.8 + 1/8.9 + 1/9.10 + 1/10.11 + 1/11.12
A = 1/5 - 1/6 + 1/6 - 1/7 + 1/7 - 1/8 + 1/8 - 1/9 + 1/9 - 1/10 + 1/10 - 1/11 + 1/11 - 1/12
A = 1/5 - 1/12
A = 7/12
\(A=\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{110}+\frac{1}{132}\)
\(=\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+\frac{1}{8\cdot9}+\frac{1}{9\cdot10}+\frac{1}{10\cdot11}+\frac{1}{11\cdot12}\)
\(=\frac{1}{5}-\frac{1}{6}+...+\frac{1}{11}-\frac{1}{12}\)
\(=\frac{1}{5}-\frac{1}{12}\)
\(=\frac{12}{60}-\frac{5}{60}=\frac{7}{60}\)
\(A=\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{110}+\frac{1}{132}\)
=\(\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}+\frac{1}{10.11}+\frac{1}{11.12}\)
\(=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{11}-\frac{1}{12}\)
\(=\frac{1}{5}-\frac{1}{12}=\frac{7}{60}\)