\(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}+\dfrac{1}{110}\)
K
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D
27 tháng 5 2022
`[-1]/2+[-1]/6+[-1]/12+[-1]/20+[-1]/30+[-1]/42+[-1]/56+[-1]/72+[-1]/90`
`=(-1)(1/2+1/6+1/12+1/20+1/30+1/42+1/56+1/72+1/90)`
`=(-1)(1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7+1/7-1/8+1/8-1/9+1/9-1/10)`
`=(-1)(1-1/10)`
`=(-1). 9/10=-9/10`
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
27 tháng 5 2022
A = \(\dfrac{-1}{2}\) + \(\dfrac{-1}{6}\)+ \(\dfrac{-1}{12}\)+ \(\dfrac{-1}{20}\)+ \(\dfrac{-1}{30}\)+ \(\dfrac{-1}{42}\)+ \(\dfrac{-1}{56}\)+ \(\dfrac{-1}{72}\)+ \(\dfrac{-1}{90}\)
A = \(\dfrac{-1}{2}\) + \(\dfrac{-1}{2\times3}\)+ \(\dfrac{-1}{3\times4}\)+ \(\dfrac{-1}{4\times5}\)+ \(\dfrac{-1}{5\times6}\)+ \(\dfrac{-1}{6\times7}\)+ \(\dfrac{-1}{7\times8}\)+ \(\dfrac{-1}{8\times9}\)+ + \(\dfrac{-1}{9\times10}\)
A = - (\(\dfrac{1}{2}\)+ \(\dfrac{1}{2}\)-\(\dfrac{1}{3}\)+ \(\dfrac{1}{3}\)-\(\dfrac{1}{4}\)+ \(\dfrac{1}{4}\)- \(\dfrac{1}{5}\)+ \(\dfrac{1}{5}\)-\(\dfrac{1}{6}\)+ \(\dfrac{1}{6}\)-\(\dfrac{1}{7}\)+\(\dfrac{1}{7}\)-\(\dfrac{1}{8}\)+\(\dfrac{1}{8}\)-\(\dfrac{1}{9}\)+ \(\dfrac{1}{9}\)-\(\dfrac{1}{10}\))
A = -(1-\(\dfrac{1}{10}\))
A = \(\dfrac{-9}{10}\)
2 tháng 8 2023
=9/10-(1/2+1/6+...+1/90)
=9/10-(1-1/2+1/2-1/3+...+1/9-1/10)
=9/10-9/10=0
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
14 tháng 4 2023
A = \(\dfrac{1}{20}\) + \(\dfrac{1}{30}\) + \(\dfrac{1}{42}\) + \(\dfrac{1}{56}\) + \(\dfrac{1}{72}\) + \(\dfrac{1}{90}\) + \(\dfrac{1}{110}\) + \(\dfrac{1}{132}\)
A = \(\dfrac{1}{4\times5}\) + \(\dfrac{1}{5\times6}\) + \(\dfrac{1}{6\times7}\)+ \(\dfrac{1}{7\times8}\)+\(\dfrac{1}{8\times9}\)+ \(\dfrac{1}{9\times10}\) + \(\dfrac{1}{10\times11}\)+\(\dfrac{1}{11\times12}\)
A = \(\dfrac{1}{4}\)-\(\dfrac{1}{5}\) +\(\dfrac{1}{5}\)-\(\dfrac{1}{6}\) +.....+\(\dfrac{1}{11}\) - \(\dfrac{1}{12}\)
A = \(\dfrac{1}{4}\) - \(\dfrac{1}{12}\)
A = \(\dfrac{1}{6}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
10 tháng 7 2023
\(\dfrac{1}{3}-\dfrac{1}{12}-\dfrac{1}{20}-\dfrac{1}{30}-\dfrac{1}{42}-\dfrac{1}{56}-\dfrac{1}{72}-\dfrac{1}{90}-\dfrac{1}{110}=x-\dfrac{5}{13}\)
\(\dfrac{1}{3}\) - \(\dfrac{1}{3.4}\) - \(\dfrac{1}{4.5}\) - \(\dfrac{1}{5.6}\) - \(\dfrac{1}{6.7}\) - \(\dfrac{1}{7.8}\)- \(\dfrac{1}{8.9}\) - \(\dfrac{1}{9.10}\) - \(\dfrac{1}{10.11}\) = \(x\) - \(\dfrac{5}{13}\)
\(\dfrac{1}{3}\) - (\(\dfrac{1}{3.4}\) + \(\dfrac{1}{4.5}\) + \(\dfrac{1}{5.6}\) + \(\dfrac{1}{6.7}\)+ \(\dfrac{1}{7.8}\) + \(\dfrac{1}{8.9}\) + \(\dfrac{1}{9.10}\) + \(\dfrac{1}{10.11}\) =\(x\)-\(\dfrac{5}{13}\)
\(\dfrac{1}{3}\) - (\(\dfrac{1}{3}\) - \(\dfrac{1}{4}\) + \(\dfrac{1}{4}\) - \(\dfrac{1}{5}\) +...+ \(\dfrac{1}{9}\) - \(\dfrac{1}{10}\) + \(\dfrac{1}{10}\) - \(\dfrac{1}{11}\)) = \(x\) - \(\dfrac{5}{13}\)
\(\dfrac{1}{3}\) - (\(\dfrac{1}{3}\) - \(\dfrac{1}{11}\)) = \(x\) - \(\dfrac{5}{13}\)
\(\dfrac{1}{3}\) - \(\dfrac{1}{3}\) + \(\dfrac{1}{11}\) = \(x\) - \(\dfrac{5}{13}\)
\(x-\dfrac{5}{13}=\dfrac{1}{11}\)
\(x\) = \(\dfrac{1}{11}\) + \(\dfrac{5}{13}\)
\(x\) = \(\dfrac{68}{143}\)
27 tháng 5 2022
Quãng sông AB dài là :
8 giờ 24 phú x 10 = 84 (km)
Vận tốc cua ca nô khi xuôi dòng là :
10 + 2 = 12 (km/giờ )
Thời gian ca nô đi xuôi dòng là :
84 : 2 = 7 (giờ )
Đáp số : 7 giờ
O
27 tháng 3 2022
dễ mà
\(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}\)
\(=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}+\dfrac{1}{7\cdot8}+\dfrac{1}{8\cdot9}+\dfrac{1}{9\cdot10}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\)
\(=1-\dfrac{1}{10}\)
\(=\dfrac{10}{10}-\dfrac{1}{10}=\dfrac{9}{10}\)
C
13 tháng 4 2022
1/5x6 + 1/6x7 + 1/7x8 + 1/8 x9 + 1/9x10 + 1 /10x11 + 1/11x12
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
= 1/5 - 1/12
= 7/60
2 tháng 5 2017
\(\dfrac{1}{2}\)+\(\dfrac{1}{6}\)+\(\dfrac{1}{12}\)+\(\dfrac{1}{20}\)+\(\dfrac{1}{30}\)+\(\dfrac{1}{42}\)+\(\dfrac{1}{56}\)+\(\dfrac{1}{72}\)+\(\dfrac{1}{90}\)
=\(\dfrac{1}{1.2}\)+\(\dfrac{1}{2.3}\)+\(\dfrac{1}{3.4}\)+\(\dfrac{1}{4.5}\)+\(\dfrac{1}{5.6}\)+\(\dfrac{1}{6.7}\)+\(\dfrac{1}{7.8}\)+\(\dfrac{1}{8.9}\)+\(\dfrac{1}{9.10}\)
=\(\dfrac{1}{1}\)-\(\dfrac{1}{2}\)+\(\dfrac{1}{2}\)-\(\dfrac{1}{3}+\dfrac{1}{4}-\dfrac{1}{5}\)+\(\dfrac{1}{5}\)-\(\dfrac{1}{6}\)+\(\dfrac{1}{6}\)-\(\dfrac{1}{7}\)+\(\dfrac{1}{7}\)-\(\dfrac{1}{8}\)+\(\dfrac{1}{8}\)-\(\dfrac{1}{9}\)+\(\dfrac{1}{9}\)-\(\dfrac{1}{10}\)
=\(\dfrac{1}{1}\)-\(\dfrac{1}{10}\)=\(\dfrac{10}{10}\)-\(\dfrac{1}{10}\)=\(\dfrac{9}{10}\)
Vậy \(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}=\dfrac{9}{10}\)
TT
2 tháng 5 2017
\(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}\)
=\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}+\dfrac{1}{9.10}\)
= \(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\)
= \(1-\dfrac{1}{10}\) = \(\dfrac{9}{10}\)
31 tháng 3 2023
\(=1-\dfrac{1}{2}+1-\dfrac{1}{6}+...+1-\dfrac{1}{90}\)
\(=10-\left(\dfrac{1}{2}+\dfrac{1}{6}+...+\dfrac{1}{90}\right)\)
\(=10-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{9}-\dfrac{1}{10}\right)\)
=9+1/10
=9,1
Em chú ý mẫu
2 = 1 x 2
6 = 2 x 3
......
110 = 10 x 11
Từ đó em đưa về dạng tổng quát \(\dfrac{1}{n\left(n+1\right)}=\dfrac{1}{n}-\dfrac{1}{n+1}\)
Rút gọn đi ta sẽ có được đáp án.
Chúc em học tốt!
\(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}+\dfrac{1}{110}\)
\(=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{10.11}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{10}-\dfrac{1}{11}\)
\(=1-\dfrac{1}{11}\)
\(=\dfrac{10}{11}\).