NT
0
K
Khách
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NV
2
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
14 tháng 3
\(\dfrac{3}{2.6}\) + \(\dfrac{3}{6.10}\) + \(\dfrac{3}{10.14}\)
= \(\dfrac{3}{4}\).(\(\dfrac{4}{2.6}\) + \(\dfrac{4}{6.10}\) + \(\dfrac{4}{10.14}\))
= \(\dfrac{3}{4}\).(\(\dfrac{1}{2}-\dfrac{1}{6}\) + \(\dfrac{1}{6}\) - \(\dfrac{1}{10}\) + \(\dfrac{1}{10}\) - \(\dfrac{1}{14}\))
= \(\dfrac{3}{4}\).(\(\dfrac{1}{2}\) - \(\dfrac{1}{14}\))
= \(\dfrac{3}{4}\). \(\dfrac{3}{7}\)
= \(\dfrac{9}{28}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
14 tháng 3
B = \(\dfrac{4}{1.3.5}\) + \(\dfrac{4}{3.5.7}\) + \(\dfrac{4}{5.7.9}\)
B = \(\dfrac{1}{1.3}\) - \(\dfrac{1}{3.5}\) + \(\dfrac{1}{3.5}\) - \(\dfrac{1}{5.7}\) + \(\dfrac{1}{5.7}\) - \(\dfrac{1}{7.9}\)
B = \(\dfrac{1}{1.3}\) - \(\dfrac{1}{7.9}\)
B = \(\dfrac{1}{3}\) - \(\dfrac{1}{63}\)
B = \(\dfrac{20}{63}\)
NP
0
HN
3
NL
2
8 tháng 5 2018
\(A=\frac{2}{2.4.6}+\frac{2}{4.6.8}+\frac{2}{6.8.10}+\frac{2}{8.10.12}\)
\(A=\frac{2}{48}+\frac{2}{192}+\frac{2}{480}+\frac{2}{960}\)
\(A=\frac{1}{24}+\frac{1}{96}+\frac{1}{240}+\frac{1}{480}\)
\(A=\frac{20}{480}+\frac{5}{480}+\frac{2}{480}+\frac{1}{480}\)
\(A=\frac{7}{120}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
14 tháng 3
A = \(\dfrac{2}{2.4.6}\) + \(\dfrac{2}{4.6.8}\) + \(\dfrac{2}{6.8.10}\) + \(\dfrac{2}{8.10.12}\)
A = \(\dfrac{2}{2}\).(\(\dfrac{2}{2.4.6}\) + \(\dfrac{2}{4.6.8}\) + \(\dfrac{2}{6.8.10}\) + \(\dfrac{2}{8.10.12}\))
A = \(\dfrac{1}{2}\).(\(\dfrac{2.2}{2.4.6}\) + \(\dfrac{2.2}{4.6.8}\) + \(\dfrac{2.2}{6.8.10}+\dfrac{2.2}{8.10.12}\))
A = \(\dfrac{1}{2}\).( \(\dfrac{4}{2.4.6}+\dfrac{4}{4.6.8}+\dfrac{4}{6.8.10}+\dfrac{4}{8.10.12}\))
A = \(\dfrac{1}{2}\).(\(\dfrac{1}{2.4}\) - \(\dfrac{1}{4.6}\) +\(\dfrac{1}{4.6}\) - \(\dfrac{1}{6.8}\) + \(\dfrac{1}{6.8}\) - \(\dfrac{1}{8.10}\) + \(\dfrac{1}{8.10}\) - \(\dfrac{1}{10.12}\))
A = \(\dfrac{1}{2}\).(\(\dfrac{1}{2.4}\) - \(\dfrac{1}{10.12}\))
A = \(\dfrac{1}{2}\).(\(\dfrac{1}{8}-\dfrac{1}{120}\))
A = \(\dfrac{1}{2}\).\(\dfrac{7}{60}\)
A = \(\dfrac{7}{120}\)
CM
1
MV
5 tháng 5 2017
Ta nhận thấy
\(\dfrac{1}{n\cdot\left(n+2\right)}-\dfrac{1}{\left(n+2\right)\cdot\left(n+4\right)}\\ =\dfrac{n+4}{n\cdot\left(n+2\right)\cdot\left(n+4\right)}-\dfrac{n}{n\cdot\left(n+2\right)\cdot\left(n+4\right)}\\ =\dfrac{n+4-n}{n\cdot\left(n+2\right)\cdot\left(n+4\right)}\\ =\dfrac{4}{n\cdot\left(n+2\right)\cdot\left(n+4\right)}\)
\(A=\dfrac{4}{2\cdot4\cdot6}+\dfrac{4}{4\cdot6\cdot8}+\dfrac{4}{6\cdot8\cdot10}+...+\dfrac{4}{46\cdot48\cdot50}\\ =\dfrac{1}{2\cdot4}-\dfrac{1}{4\cdot6}+\dfrac{1}{4\cdot6}-\dfrac{1}{6\cdot8}+\dfrac{1}{6\cdot8}-\dfrac{1}{8\cdot10}+...+\dfrac{1}{46\cdot48}-\dfrac{1}{48\cdot50}\\ =\dfrac{1}{2\cdot4}-\dfrac{1}{48\cdot50}\\ =\dfrac{1}{8}-\dfrac{1}{2400}\\ =\dfrac{300}{2400}-\dfrac{1}{2400}\\ =\dfrac{299}{2400}\)
Số nghịch đảo của \(A\) là \(\dfrac{2400}{299}\)