tính:\(A=\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{2012.2014}\)
K
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Những câu hỏi liên quan
20 tháng 4 2019
\(=2.\left(\frac{2}{2.4}+\frac{2}{4.6}+...+\frac{2}{2014.2016}\right)\)
\(=2.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{2014}-\frac{1}{2016}\right)\)
\(=2.\left(\frac{1}{2}-\frac{1}{2016}\right)\)
\(=2.\frac{1007}{2016}\)
\(=\frac{2007}{1008}\)
KT
20 tháng 4 2019
giải:
4/2.4+4/4.6+4/6.8+...+4/2012.2014+4/2014.2016
=2.(2/2.4+2/4.6+2/6.8+...+2/2012.2014+2/2014.2016
=2.(1/2-1/4+1,4-1/6+1/6-1/8+...+1/2012-1/2014+1/2014-1/2016)
=2.(1/2-1/2016)
=2.1007/2016
=1007/1008
xong rùi đó
TD
1
13 tháng 4 2016
\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.......+\frac{1}{2014}-\frac{1}{2016}\)\(=1-\frac{1}{2016}=\frac{2015}{2016}\)
3 tháng 6 2020
\(b,\frac{10}{99}\)+\(\frac{11}{199}\)+\(\frac{12}{299}\).\(\frac{1}{2}\)-\(\frac{1}{3}\)+\(\frac{-1}{6}\)
S
9 tháng 5 2019
=1/1x2+1/2x3+1/3x4+...+1/1006x1007+1/1007x1008
=1/1-1/2+1/2-1/3+1/3-1/4+...+1/1006-1/1007+1/1007-1/1008
=1/1-1/1008
=1007/1008
~-~:33
L
2
Kudo Shinichi mik sẽ biết ơn ai giải cho mik nhìu nhìu
NN
2
MA
9 tháng 9 2016
\(A=\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{28.30}\)
\(A=\frac{2}{2}\left(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{28.30}\right)\)
\(A=\frac{1}{2}\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{28.30}\right)\)
\(A=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{28}-\frac{1}{30}\right)\)
\(A=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{30}\right)\)
\(A=\frac{1}{2}.\frac{7}{15}\)
\(A=\frac{7}{30}\)
9 tháng 9 2016
\(2.A=\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{28.30}\)
\(2.A=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{28}-\frac{1}{30}\)
\(2.A=\frac{1}{2}-\frac{1}{30}\)
\(2.A=\frac{7}{15}\)
\(A=\frac{7}{15}:2=\frac{7}{30}\)
PA
3 tháng 12 2016
\(A=\frac{1}{1\times3}+\frac{1}{2\times4}+\frac{1}{3\times5}+\frac{1}{4\times6}+\frac{1}{5\times7}+\frac{1}{6\times8}+\frac{1}{7\times9}+\frac{1}{8\times10}\)
\(2A=\frac{2}{1\times3}+\frac{2}{2\times4}+\frac{2}{3\times5}+\frac{2}{4\times6}+\frac{2}{5\times7}+\frac{2}{6\times8}+\frac{2}{7\times9}+\frac{2}{8\times10}\)
\(2A=1-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-\frac{1}{5}+\frac{1}{4}-\frac{1}{6}+\frac{1}{5}-\frac{1}{7}+\frac{1}{6}-\frac{1}{8}+\frac{1}{7}-\frac{1}{9}+\frac{1}{8}-\frac{1}{10}\)
\(2A=1+\frac{1}{2}-\frac{1}{9}-\frac{1}{10}\)
\(2A=\frac{58}{45}\)
\(A=\frac{58}{45}\div2\)
\(A=\frac{29}{45}\)
LT
3 tháng 12 2016
\(2A=\frac{2}{1.3}+\frac{2}{2.4}+...+\frac{2}{8.10}=1-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-....+\frac{1}{8}-\frac{1}{10}\)
\(=1+\frac{1}{2}-\frac{1}{9}-\frac{1}{10}=\frac{58}{45}\)
\(A=\frac{29}{45}\)
FZ
1
1 tháng 10 2015
\(2.A=\frac{2}{2.4}+\frac{2}{4.6}+...+\frac{2}{26.28}=\frac{4-2}{2.4}+\frac{6-4}{4.6}+...+\frac{28-26}{26.28}\)
\(2.A=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{26}-\frac{1}{28}=\frac{1}{2}+\left(-\frac{1}{4}+\frac{1}{4}\right)+...+\left(-\frac{1}{26}+\frac{1}{26}\right)-\frac{1}{28}\)
\(2.A=\frac{1}{2}-\frac{1}{28}=\frac{26}{56}=\frac{13}{28}\)=> A = \(\frac{13}{56}\)
A = \(\frac{1}{2.4}+\frac{1}{4.6}+....+\frac{1}{2012.2014}\)
A = \(\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+....+\frac{1}{2012}-\frac{1}{2014}\right)\)
A = \(\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{2014}\right)\)
A = \(\frac{1}{2}.\frac{503}{1007}\)
A = \(\frac{503}{2014}\)
2A=\(\frac{4-2}{2.4}+\frac{6-4}{4.6}+...+\frac{2014-2012}{2012.2014}\)
\(2A=\frac{4}{2.4}-\frac{2}{2.4}+\frac{6}{4.6}-\frac{4}{4.6}+...+\frac{2014}{2012.2014}-\frac{2012}{2012.2014}\)
\(2A=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{2012}-\frac{1}{2014}\)
\(2A=\frac{1}{2}-\frac{1}{2014}=\frac{503}{1007}\Rightarrow A=\frac{503}{2014}\)