S=\(\frac{2}{1X3}\)+\(\frac{2}{3X5}\)+\(\frac{2}{5X7}\)+.....+\(\frac{2}{2015X2017}\)
K
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NN
2
HN
4 tháng 7 2018
\(B=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+........+\frac{2}{57.59}\)
\(B=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-.........-\frac{1}{59}\)
\(B=1-\frac{1}{59}\)
\(B=\frac{59}{59}-\frac{1}{59}=\frac{58}{59}\)
Vậy B = \(\frac{58}{59}\)
Lưu ý: Dấu "." là dấu nhân
4 tháng 7 2018
\(B=\frac{2}{1.3}+\frac{1}{3.5}+\frac{2}{5.7}+...+\frac{1}{57.59}\)
\(B=1.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{57}-\frac{5}{59}\right)\)
\(B=1.\left(1-\frac{1}{59}\right)\)
\(B=1.\frac{58}{59}\)
\(B=\frac{58}{59}\)
8 tháng 10 2017
\(\left(\frac{2}{1x3}+\frac{2}{3x5}+\frac{2}{5x7}+\frac{2}{7x9}+\frac{2}{9x11}\right).y=\frac{2}{3}\)
\(\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right)y=\frac{2}{3}\)
\(\left(1-\frac{1}{11}\right).y=\frac{2}{3}\)
\(\frac{10}{11}.y=\frac{2}{3}\)
\(y=\frac{2}{3}.\frac{11}{10}\)
\(y=\frac{22}{30}\)
31 tháng 8 2016
\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{11.13}+\frac{2}{13.15}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}\)
\(=1-\frac{1}{15}\)
\(=\frac{14}{15}\)
AH
3
11 tháng 5 2017
\(S=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2015.2017}\)
\(S=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2015}-\frac{1}{2017}\right)\)
\(S=\frac{1}{2}.\left(1-\frac{1}{2017}\right)\)
\(S=\frac{1}{2}.\frac{2016}{2017}\)
\(S=\frac{1008}{2017}< \frac{1}{2}\)
S
11 tháng 5 2017
\(S=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2015.2017}\)
\(2S=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{2015.2017}\)
\(2S=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2015}-\frac{1}{2017}\)
\(2S=1-\frac{1}{2017}< 1\)
=> 2S < 1
=> S < \(\frac{1}{2}\)(đpcm)
M
12 tháng 8 2018
p=1/(3*5)+1/(5*7)+.....+1/(2015*2017)+1/(2017*2019)
<=> p = 1/3-1/5+1/5-1/7+1/7-......+1/2017-1/2019
<=> p = 1/3 - 1/2019
<=> p = 224/673
KT
12 tháng 8 2018
\(P=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{2015.2017}+\frac{1}{2017.2019}\)
\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{2017}-\frac{1}{2019}\right)\)
\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{2019}\right)\)
\(=\frac{112}{673}\)
16 tháng 5 2019
\(\frac{2}{1\times3}+\frac{2}{3\times5}+\frac{2}{5\times7}+\frac{2}{7\times9}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}\)
\(=1-\frac{1}{9}=\frac{8}{9}\)
16 tháng 5 2019
\(\frac{2}{1\times3}+\frac{2}{3\times5}+\frac{2}{5\times7}+\frac{2}{7\times9}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}\)
\(=1+\left(\frac{1}{3}-\frac{1}{3}\right)+\left(\frac{1}{5}-\frac{1}{5}\right)+\left(\frac{1}{7}-\frac{1}{7}\right)-\frac{1}{9}\)
\(=1-\frac{1}{9}=\frac{8}{9}\)
~ Hok tốt ~
HT
1
24 tháng 3 2017
\(\frac{2^2}{1x3}\)x \(\frac{4^2}{3x5}\)x \(\frac{6^2}{5x7}\) x \(\frac{8^2}{7x9}\)
= \(\frac{4}{3}\)x \(\frac{16}{15}\)x \(\frac{36}{35}\)x \(\frac{64}{63}\)
= \(1.486077098\)
2016/2017 nhé
k cho mình nha
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