Tim x biet:
(\(\dfrac{1}{1.4}+\dfrac{1}{4.7}+...+\dfrac{1}{97.100}\))=\(\dfrac{0,33.x}{2009}\)
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26 tháng 12 2015
\(\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+...+\frac{3}{97\cdot100}=\frac{0,33\cdot x}{2009}\cdot3\)
\(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{100}=\frac{0,99\cdot x}{2009}\)
\(\frac{100}{100}-\frac{1}{100}=\frac{0,99x}{2009}\)
\(\frac{99}{100}=\frac{0,99x}{2009}\)
=>0,99x*100=2009*99
99x=2009*99
=>x=2009
Vậy x=2009
26 tháng 12 2015
\(0,33\cdot\frac{x}{2009}\) hay \(\frac{0,33\cdot x}{2009}\)
TK
1
19 tháng 12 2017
\(l=\dfrac{1}{1.4}+\dfrac{1}{4.7}+\dfrac{1}{7.10}+....+\dfrac{1}{97.100}\)
\(=\dfrac{1}{3}\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{97}-\dfrac{1}{100}\right)\)
\(=\dfrac{1}{3}\left(1-\dfrac{1}{100}\right)=\dfrac{1}{3}-\dfrac{1}{300}< \dfrac{1}{3}\left(đpcm\right)\)
26 tháng 9 2021
b: Ta có: \(B=\dfrac{1}{4\cdot9}+\dfrac{1}{9\cdot14}+...+\dfrac{1}{64\cdot69}\)
\(=\dfrac{1}{5}\left(\dfrac{5}{4\cdot9}+\dfrac{5}{9\cdot14}+...+\dfrac{5}{64\cdot69}\right)\)
\(=\dfrac{1}{5}\left(\dfrac{1}{4}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{14}+...+\dfrac{1}{64}-\dfrac{1}{69}\right)\)
\(=\dfrac{1}{5}\cdot\dfrac{65}{4\cdot69}\)
\(=\dfrac{13}{276}\)
26 tháng 9 2021
\(A=\dfrac{2}{1\cdot4}+\dfrac{2}{4\cdot7}+...+\dfrac{2}{97\cdot100}\\ A=\dfrac{2}{3}\left(\dfrac{3}{1\cdot4}+\dfrac{3}{4\cdot7}+...+\dfrac{3}{97\cdot100}\right)\\ A=\dfrac{2}{3}\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{100}\right)\\ A=\dfrac{2}{3}\left(1-\dfrac{1}{100}\right)=\dfrac{2}{3}\cdot\dfrac{99}{100}=\dfrac{33}{50}\\ B=\dfrac{1}{4\cdot9}+\dfrac{1}{9\cdot14}+...+\dfrac{1}{64\cdot69}\\ B=\dfrac{1}{5}\left(\dfrac{5}{4\cdot9}+\dfrac{5}{9\cdot14}+...+\dfrac{5}{64\cdot69}\right)\\ B=\dfrac{1}{5}\left(\dfrac{1}{4}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{14}+...+\dfrac{1}{64}-\dfrac{1}{69}\right)\\ B=\dfrac{1}{5}\left(\dfrac{1}{4}-\dfrac{1}{69}\right)=\dfrac{1}{5}\cdot\dfrac{65}{276}=\dfrac{13}{276}\)
\(C=70\left(\dfrac{13}{56}+\dfrac{13}{72}+\dfrac{13}{90}\right)=70\cdot13\left(\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}\right)\\ C=910\left(\dfrac{1}{7\cdot8}+\dfrac{1}{8\cdot9}+\dfrac{1}{9\cdot10}\right)\\ C=910\left(\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\right)\\ C=910\left(\dfrac{1}{7}-\dfrac{1}{10}\right)=910\cdot\dfrac{3}{70}=39\)
MA
3 tháng 2 2022
1.
`16 + (27 - 7.6 ) - (94 -7 - 27.99)`
`= 16+ 27 - 7.6 - 94 + 7 + 27.99`
`= 16 + 27(99 +1) - 7(6-1) - 94`
`= -78 + 27.100 - 7.5`
`= 2587`
2.
`A = 2/1.4 + 2/4.7 + 2/7.10 +...+ 2/97.100`
`A= 2(1/1.4 + 1/4.7 + 1/7.10 +...+1/97.100)`
`3A = 2 (3/1.4 + 3/4.7 + 3/7.10+...+ 3/97.100)`
`3/2 A = 1 - 1/4 + 1/4 - 1/7 +...+ 1/97 - 1/100`
`3/2A = 1 - 1/100`
`3/2 A= 99/100`
`A= 99/100 : 3/2`
`A=33/50`
Vậy `A= 33/50`
TD
3 tháng 2 2022
1.16+(27-7.6)-(94-7-27.99)=16+27-7.6-94+7+27.99
=(27+27.99)+(27+7-94)+16
=27.100-60+16
=2700-44=2656
2.A=\(\dfrac{2}{1.4}+\dfrac{2}{4.7}+\dfrac{2}{7.10}+...+\dfrac{2}{97.100}\)
=\(\dfrac{1}{1}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{100}\)
=\(1-\dfrac{1}{100}=\dfrac{99}{100}\)
HN
1
KS
11 tháng 4 2022
\(\dfrac{2}{1.4}+\dfrac{2}{4.7}+...+\dfrac{2}{97.100}\)
=> \(\dfrac{2.3}{1.4}+\dfrac{2.3}{4.7}+...+\dfrac{2.3}{97.100}\)
=> \(2.\left(\dfrac{3}{1.4}+\dfrac{3}{4.7}+...+\dfrac{3}{97.100}\right)\)
=> \(2.\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{100}\right)\)
=> \(2.\left(1-\dfrac{1}{100}\right)\)
=>\(2\).\(\dfrac{99}{100}\)
=\(\dfrac{99}{50}\)
H
15 tháng 5 2018
Giải:
\(S=\dfrac{1}{1.4}-\dfrac{1}{4.7}-\dfrac{1}{7.10}-...-\dfrac{1}{97.100}\)
\(\Leftrightarrow S=-\left(-\dfrac{1}{1.4}+\dfrac{1}{4.7}+\dfrac{1}{7.10}+...+\dfrac{1}{97.100}\right)\)
\(\Leftrightarrow S=-\dfrac{1}{3}\left(-\dfrac{1}{1}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{97}-\dfrac{1}{100}\right)\)
\(\Leftrightarrow S=-\dfrac{1}{3}\left(-\dfrac{1}{1}-\dfrac{1}{100}\right)\)
\(\Leftrightarrow S=-\dfrac{1}{3}\left(-\dfrac{101}{100}\right)\)
\(\Leftrightarrow S=\dfrac{101}{300}\)
Vậy ...
15 tháng 5 2018
Bạn ơi cho mình hỏi tại sao phía trước \(-\dfrac{1}{1}-\dfrac{1}{4}+\dfrac{1}{4}-.....+\dfrac{1}{97}-\dfrac{1}{100}\) lại là \(-\dfrac{1}{3}\)
LM
1
\(\Leftrightarrow\dfrac{1}{3}\left(\dfrac{1}{1}-\dfrac{1}{4}+\dfrac{1}{4}-...+\dfrac{1}{97}-\dfrac{1}{100}\right)=\dfrac{0,33x}{2009}\)
\(\Leftrightarrow\dfrac{1}{3}\cdot\dfrac{99}{100}=\dfrac{0,33x}{2009}\)
\(\Leftrightarrow\dfrac{33}{100}=\dfrac{0,33x}{2009}\) <=> x = (tự tính )
⇔13(11−14+14−...+197−1100)=0,33x2009⇔13(11−14+14−...+197−1100)=0,33x2009
⇔13⋅99100=0,33x2009⇔13⋅99100=0,33x2009