Hãy so sánh : \(\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+...+\dfrac{1}{9900}\)với \(\dfrac{1}{2}\)
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Những câu hỏi liên quan
O
27 tháng 3 2022
dễ mà
\(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}\)
\(=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}+\dfrac{1}{7\cdot8}+\dfrac{1}{8\cdot9}+\dfrac{1}{9\cdot10}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\)
\(=1-\dfrac{1}{10}\)
\(=\dfrac{10}{10}-\dfrac{1}{10}=\dfrac{9}{10}\)
LB
15 tháng 11 2017
\(A=\dfrac{1}{2}+\dfrac{3-2}{3.2}+\dfrac{4-3}{3.4}+...+\dfrac{100-99}{100.99}\)
\(A=\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+....+\dfrac{1}{99}-\dfrac{1}{100}\)
\(A=1-\dfrac{1}{100}\)
\(A=\dfrac{99}{100}\)
LB
15 tháng 11 2017
\(2B=\dfrac{2}{1.3}+\dfrac{2}{3.5}+....+\dfrac{2}{2007.2009}+\dfrac{2}{2009..2011}\)
\(2B=\dfrac{3-1}{1.3}+\dfrac{5-3}{3,5}+...+\dfrac{2009-2007}{2009.2007}+\dfrac{2011-2009}{2011.2009}\)
\(2B=\dfrac{3}{3}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2007}-\dfrac{1}{2009}+\dfrac{1}{2009}-\dfrac{1}{2011}\)
\(2B=1-\dfrac{1}{2011}\)
\(2B=\dfrac{2010}{2011}\)
\(B=\dfrac{2010}{4022}\)
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 = \(\dfrac{-9}{10}\)
21 tháng 11 2021
\(\Rightarrow x+\dfrac{1}{1\times2}+\dfrac{1}{2\times3}+\dfrac{1}{3\times4}+\dfrac{1}{4\times5}+\dfrac{1}{5\times6}=\dfrac{47}{42}\\ \Rightarrow x+1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}=\dfrac{47}{42}\\ \Rightarrow x+1-\dfrac{1}{6}=\dfrac{47}{42}\\ \Rightarrow x=\dfrac{47}{42}-\dfrac{5}{6}=\dfrac{2}{7}\)
DA
1
23 tháng 10 2023
ta nhận thấy
1/2=1-1/2
1/6=1/2-1/3
1/12=1/3-1/4
1/20=1/4-1/5
1/30=1/5-1/6
1/42=1/6-1/7
ta có:
1/2+1/6+1/12+1/20+1/30+1/42
=1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7
=1-1/7
=6/7
bn tự hiểu nha
TT
8 tháng 12 2021
\(A=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}\)
\(A=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{7}-\dfrac{1}{8}\)
\(A=1-\dfrac{1}{8}=\dfrac{7}{8}\)
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
LS
6
O
25 tháng 3 2022
\(B=\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}\)
\(B=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}\)
\(B=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}\)
\(B=1-\dfrac{1}{7}\)
\(B=\dfrac{6}{7}\)
\(\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+...+\dfrac{1}{9900}\)
\(=\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{99.100}\)
\(=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(=\dfrac{1}{2}-\dfrac{1}{100}< \dfrac{1}{2}\)
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