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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
\(2.A=\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{28.30}\)
\(2.A=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{28}-\frac{1}{30}\)
\(2.A=\frac{1}{2}-\frac{1}{30}\)
\(2.A=\frac{7}{15}\)
\(A=\frac{7}{15}:2=\frac{7}{30}\)
TT
4
EC
11 tháng 8 2016
\(\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{98.100}\)
\(=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{98}-\frac{1}{100}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{100}\right)\)
\(=\frac{1}{2}.\frac{49}{100}\)
\(=\frac{49}{200}\)
NT
1
6 tháng 3 2016
\(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...........+\frac{1}{98.100}\)
\(=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{98}-\frac{1}{100}\)
\(=\frac{1}{2}-\frac{1}{100}=\frac{49}{100}\)
cho mình nha!
NN
Nguyễn Ngọc Anh Minh
CTVHS
VIP
20 tháng 6 2020
Đặt BT trên là A
\(2A=\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{100.102}\)
\(2A=\frac{4-2}{2.4}+\frac{6-4}{4.6}+\frac{8-6}{6.8}+...+\frac{102-100}{100.102}\)
\(2A=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{100}-\frac{1}{102}\)
\(2A=\frac{1}{2}-\frac{1}{102}=\frac{50}{102}\Rightarrow A=\frac{25}{102}\)
20 tháng 6 2020
Đặt A là biểu thức trên ta có :
\(A=\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{100.102}\)
\(=\frac{1}{2}\left(\frac{4-2}{2.4}+\frac{6-4}{4.6}+\frac{8-6}{6.8}+...+\frac{102-100}{100.102}\right)\)
\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{100}-\frac{1}{102}\right)\)
\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{102}\right)=\frac{1}{2}.\frac{50}{102}=\frac{25}{102}\)
L
2
Kudo Shinichi mik sẽ biết ơn ai giải cho mik nhìu nhìu
LV
2 tháng 5 2017
Ta có
=\(\left(1+\frac{1}{1.3}\right).\left(1+\frac{1}{2.4}\right)....\left(1+\frac{1}{8.10}\right)\)
=\(\frac{4}{3}.\frac{9}{8}....\frac{81}{80}\)
=\(\frac{2.2}{1.3}.\frac{3.3}{2.4}....\frac{9.9}{8.10}\)
=\(\frac{2.3....9}{1.2....8}.\frac{2.3....9}{3.4....10}\)
=\(9.\frac{2}{10}\)
=\(\frac{9}{5}\)
N
4
HT
4 tháng 8 2015
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 = \(\frac{503}{2014}\)
4 tháng 8 2015
2A=\(\frac{4-2}{2.4}+\frac{6-4}{4.6}+...+\frac{2014-2012}{2012.2014}\)
\(2A=\frac{4}{2.4}-\frac{2}{2.4}+\frac{6}{4.6}-\frac{4}{4.6}+...+\frac{2014}{2012.2014}-\frac{2012}{2012.2014}\)
\(2A=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{2012}-\frac{1}{2014}\)
\(2A=\frac{1}{2}-\frac{1}{2014}=\frac{503}{1007}\Rightarrow A=\frac{503}{2014}\)
HH
4
3 tháng 9 2017
\(S=\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+\frac{1}{8.10}\)
\(2S=\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+\frac{2}{8.10}\)
\(2S=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{8}-\frac{1}{10}\)
\(2S=\frac{1}{2}-\frac{1}{10}\)
\(2S=\frac{2}{5}\)
\(S=\frac{2}{5}:2\)
\(S=\frac{1}{5}\)
HN
3 tháng 9 2017
S = \(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+\frac{1}{8.10}\)
=> 2S = \(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+\frac{2}{8.10}\)
=> 2S = \(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+\frac{1}{8}-\frac{1}{10}\)
=> 2S = \(\frac{1}{2}-\frac{1}{10}=\frac{5}{10}-\frac{1}{10}=\frac{4}{10}=\frac{2}{5}\)
=> S = \(\frac{2}{5}:2=\frac{2}{5}x\frac{1}{2}=\frac{1}{5}\)
AN
3
19 tháng 3 2017
\(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{38.40}\)
=\(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{38}-\frac{1}{40}\)
=\(\frac{1}{2}-\frac{1}{40}\)
=\(\frac{19}{40}\)
19 tháng 3 2017
= 2 *[1/2 * 1/4 +1/4 *1/6 +1/6*1/8+...+1/38*1/40
=2*[1/2 - 1/40]
=2 * (-19/40)
= -380
\(2.A=\frac{2}{2.4}+\frac{2}{4.6}+...+\frac{2}{26.28}=\frac{4-2}{2.4}+\frac{6-4}{4.6}+...+\frac{28-26}{26.28}\)
\(2.A=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{26}-\frac{1}{28}=\frac{1}{2}+\left(-\frac{1}{4}+\frac{1}{4}\right)+...+\left(-\frac{1}{26}+\frac{1}{26}\right)-\frac{1}{28}\)
\(2.A=\frac{1}{2}-\frac{1}{28}=\frac{26}{56}=\frac{13}{28}\)=> A = \(\frac{13}{56}\)