C = 1/5.10 + 1/10.15 + 1/15.20 +...+ 1/95.100
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DL
2 tháng 9 2019
\(1-\frac{1}{5\cdot10}-\frac{1}{10\cdot15}-\frac{1}{15\cdot20}-...-\frac{1}{95\cdot100}\)
\(=1-\left(\frac{1}{5\cdot10}+\frac{1}{10\cdot15}+...+\frac{1}{95\cdot100}\right)\)
\(=1-\frac{1}{5}\left(\frac{1}{5}-\frac{1}{10}+\frac{1}{10}-\frac{1}{15}+\frac{1}{15}-...+\frac{1}{95}-\frac{1}{100}\right)\)
\(=1-\frac{1}{5}\left(\frac{1}{5}-\frac{1}{100}\right)=1-\frac{19}{500}=\frac{481}{500}\)
MM
4
27 tháng 6 2018
\(1-\frac{1}{5.10}-\frac{1}{10.15}-\frac{1}{15.20}-...-\frac{1}{95.100}\)
\(=1-\left(\frac{1}{5.10}+\frac{1}{10.15}+\frac{1}{15.20}+...+\frac{1}{95.100}\right)\)
\(=1-\frac{1}{5}.\left(\frac{1}{5}-\frac{1}{10}+\frac{1}{10}-\frac{1}{15}+\frac{1}{15}-\frac{1}{20}+...+\frac{1}{95}-\frac{1}{100}\right)\)
\(=1-\frac{1}{5}.\left(\frac{1}{5}-\frac{1}{100}\right)\)
\(=1-\frac{1}{5}.\frac{19}{100}\)
\(=1-\frac{19}{500}\)
\(=\frac{481}{500}\)
27 tháng 6 2018
\(1-\frac{1}{5.10}-\frac{1}{10.15}-\frac{1}{15.20}-.....-\frac{1}{95.100}\)
\(=1-\left(\frac{1}{5.10}+\frac{1}{10.15}+\frac{1}{15.20}+...+\frac{1}{95.100}\right)\)
Đặt \(C=\frac{1}{5.10}+\frac{1}{10.15}+\frac{1}{15.20}+....+\frac{1}{95.100}\)
\(\Rightarrow C=\frac{1}{5}.\left(\frac{5}{5.10}+\frac{5}{10.15}+\frac{5}{15.20}+....+\frac{5}{95.100}\right)\)
\(=\frac{1}{5}.\left(\frac{1}{5}-\frac{1}{10}+\frac{1}{10}-\frac{1}{15}+....+\frac{1}{95}-\frac{1}{100}\right)\)
\(=\frac{1}{5}.\left(\frac{1}{5}-\frac{1}{100}\right)=\frac{1}{5}.\frac{19}{100}=\frac{19}{500}\)
\(\Rightarrow1-C=1-\frac{19}{500}=\frac{481}{500}\)
Chúc bạn học tốt
LT
1
YN
22 tháng 7 2018
ta có B = 1- 1/5.10 - 1/10.15 -.......- 1/95 .100
=> 5B = 5 -( 5/5.10+5/10.15 +....+ 5/95.100
= > 5B = 5 - ( 1/5 -1/100 )
=> 5B= 481/100
=> B = 481/500
JL
4
8 tháng 8 2018
\(B=\frac{5}{5\cdot10}+\frac{5}{10\cdot15}+...+\frac{5}{95\cdot100}\)
\(B=\frac{1}{5}-\frac{1}{10}+\frac{1}{10}-\frac{1}{15}+...+\frac{1}{95}-\frac{1}{100}\)
\(B=\frac{1}{5}-\frac{1}{100}\)
\(B=\frac{19}{100}\)
9 tháng 7 2017
B=1+\(\dfrac{1}{5.10}\)+\(\dfrac{1}{10.15}\)+\(\dfrac{1}{15.20}\)+......+\(\dfrac{1}{95.100}\)
5B = 5 +\(\dfrac{5}{5.10}+\dfrac{5}{10.15}+\dfrac{5}{15.20}+........+\dfrac{5}{95.100}\)
5B=5+\(\dfrac{1}{5}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{15}+\dfrac{1}{15}-\dfrac{1}{20}+.........+\dfrac{1}{95}-\dfrac{1}{100}\)
5B=5+\(\dfrac{1}{5}-\dfrac{1}{100}\)
5B=\(\dfrac{519}{100}\)
=>B= \(\dfrac{519}{100}:5=\dfrac{519}{500}\)
9 tháng 7 2017
A= \(\dfrac{1}{3.7}\) +\(\dfrac{1}{7.11}\)+\(\dfrac{1}{11.15}\)+\(\dfrac{1}{15.19}\)+\(\dfrac{1}{19.23}\)+\(\dfrac{1}{23.27}\)
A= 4.(\(\dfrac{1}{3.7}+\dfrac{1}{7.11}+\dfrac{1}{11.15}+\dfrac{1}{15.19}+\dfrac{1}{19.23}\)+\(\dfrac{1}{23.27}\)
A=4.\(\dfrac{1}{3.7}+4.\dfrac{1}{7.11}+4.\dfrac{1}{11.15}+4.\dfrac{1}{15.19}+4.\dfrac{1}{19.23}+4.\dfrac{1}{23.27}\)
A=\(4.(\dfrac{1}{3}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{15}+\dfrac{1}{15}-\dfrac{1}{19}+\dfrac{1}{19}-\dfrac{1}{23}+\dfrac{1}{23}-\dfrac{1}{25}+\dfrac{1}{25}-\dfrac{1}{29}+\dfrac{1}{29})\)
A= 4 (.\(\dfrac{1}{3}-\dfrac{1}{29}\))
A=\(\dfrac{104}{87}\)
WW
9 tháng 7 2017
a) Mình ko ghi lại đề nhé!
= \(\frac{1}{2}\) - ( \(\frac{1}{3.7}\) + \(\frac{1}{7.11}\) + ... + \(\frac{1}{23.27}\) )
= \(\frac{1}{2}\) - \(\frac{1}{4}\) . ( \(\frac{1}{3}\) - \(\frac{1}{7}\) + \(\frac{1}{7}\) - .... - \(\frac{1}{27}\) )
= \(\frac{1}{2}\) - \(\frac{1}{4}\) . ( \(\frac{1}{3}\) - \(\frac{1}{27}\) )
= \(\frac{1}{2}\) - \(\frac{1}{4}\) . \(\frac{8}{27}\)
= \(\frac{1}{2}\) - \(\frac{2}{27}\) = \(\frac{23}{54}\)
b) ..............................................................................
= \(\frac{1}{5}\) . ( \(\frac{5}{5.10}\) - \(\frac{5}{10.15}\) - ... - \(\frac{5}{95.100}\) )
= \(\frac{1}{5}\) . ( \(\frac{1}{5}\) - \(\frac{1}{10}\) + \(\frac{1}{10}\) - ... - \(\frac{1}{100}\) )
= \(\frac{1}{5}\) . ( \(\frac{1}{5}\) - \(\frac{1}{100}\) )
= \(\frac{1}{5}\) . \(\frac{19}{100}\)
= \(\frac{19}{500}\)
k mình nha! Chúc bạn học tốt và được nhiều k!
C=1/5.10+1/10.15+...+1/95.100
= 5/5.10+5/10.15+...+5/95.100
= 1/5-1/10+1/10-1/15+...+1/95-1/100
= 1/5-1/100
= 19/100
\(C=5\times\left(1+\frac{1}{5}-\frac{1}{10}+\frac{1}{10}-\frac{1}{15}+..+\frac{1}{95}-\frac{1}{100}\right)\)
\(C=5\times\left(1-\frac{1}{100}\right)\)
\(C=5\times\frac{99}{100}\)
\(C=\frac{99}{20}\)