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NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
14 tháng 3 2024
\(\dfrac{3}{2.6}\) + \(\dfrac{3}{6.10}\) + \(\dfrac{3}{10.14}\)
= \(\dfrac{3}{4}\).(\(\dfrac{4}{2.6}\) + \(\dfrac{4}{6.10}\) + \(\dfrac{4}{10.14}\))
= \(\dfrac{3}{4}\).(\(\dfrac{1}{2}-\dfrac{1}{6}\) + \(\dfrac{1}{6}\) - \(\dfrac{1}{10}\) + \(\dfrac{1}{10}\) - \(\dfrac{1}{14}\))
= \(\dfrac{3}{4}\).(\(\dfrac{1}{2}\) - \(\dfrac{1}{14}\))
= \(\dfrac{3}{4}\). \(\dfrac{3}{7}\)
= \(\dfrac{9}{28}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
14 tháng 3 2024
B = \(\dfrac{4}{1.3.5}\) + \(\dfrac{4}{3.5.7}\) + \(\dfrac{4}{5.7.9}\)
B = \(\dfrac{1}{1.3}\) - \(\dfrac{1}{3.5}\) + \(\dfrac{1}{3.5}\) - \(\dfrac{1}{5.7}\) + \(\dfrac{1}{5.7}\) - \(\dfrac{1}{7.9}\)
B = \(\dfrac{1}{1.3}\) - \(\dfrac{1}{7.9}\)
B = \(\dfrac{1}{3}\) - \(\dfrac{1}{63}\)
B = \(\dfrac{20}{63}\)
16 tháng 3 2016
\(A=\frac{1}{6.10}+\frac{1}{10.14}+\frac{1}{14.18}+...+\frac{1}{402.406}\)
4\(A=\frac{1}{6}-\frac{1}{10}+\frac{1}{10}-\frac{1}{14}+\frac{1}{14}-\frac{1}{18}+...+\frac{1}{402}-\frac{1}{406}\)
4\(A=\frac{1}{6}-\frac{1}{406}\)
4\(A=\frac{100}{609}\)
\(\Rightarrow A=\frac{100}{609}:4\)\(=\frac{25}{609}\)
16 tháng 3 2016
=1/6-1/10+1/10-1/14+1/14-1/18+...........+1/402-1/406
=1/6-1/406
AT
1
22 tháng 4 2018
\(S=\dfrac{2}{2\cdot6}+\dfrac{2}{6\cdot10}+...+\dfrac{2}{96\cdot100}\\ =\dfrac{1}{2}\left(\dfrac{4}{2\cdot6}+\dfrac{4}{6\cdot10}+...+\dfrac{4}{96\cdot100}\right)\\ =\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{10}+...+\dfrac{1}{96}-\dfrac{1}{100}\right)\\ =\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{100}\right)=\dfrac{1}{2}\cdot\dfrac{49}{100}\\ =\dfrac{49}{100}\)
YK
17 tháng 1 2015
Nếu ai có giải dùm mình thì giải từng phần nhưng đừng chỉ ghi kết quả nhé~
CN
20 tháng 1 2016
a,\(\frac{2004}{10045}\)
b,\(\frac{25}{609}\)
c,\(\frac{1000}{3549}\)
d,\(\frac{25}{258}\)
11 tháng 5 2017
\(A=\dfrac{3}{1.4}+\dfrac{3}{2.6}+\dfrac{3}{3.8}+...............+\dfrac{1}{2012.1342}\)
\(A=\dfrac{3}{1.4}+\dfrac{3}{2.6}+\dfrac{3}{3.8}+...........................+\dfrac{3}{2012.4026}\)
\(A=\dfrac{6}{2.4}+\dfrac{6}{4.6}+\dfrac{6}{6.8}+..........................+\dfrac{6}{4024.4026}\)
\(A=3\left(\dfrac{2}{2.4}+\dfrac{2}{4.6}+\dfrac{2}{6.8}+...................+\dfrac{2}{4024.4026}\right)\)
\(A=3\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+....................+\dfrac{1}{4024}-\dfrac{1}{4026}\right)\)
\(A=3\left(\dfrac{1}{2}-\dfrac{1}{4026}\right)\)
\(A=3.\dfrac{1}{2}-3.\dfrac{1}{4026}\)
\(A=1,5-\dfrac{3}{4026}< 1,5\)
DT
Đỗ Thanh Hải
CTVVIP
11 tháng 5 2017
Ta có
A = \(\dfrac{3}{1.4}\) + \(\dfrac{3}{2.6}\) + \(\dfrac{3}{3.8}\) + ... + \(\dfrac{1}{2012.1342}\)
A = \(\dfrac{3}{1.4}\) + \(\dfrac{3}{2.6}\) + \(\dfrac{3}{3.8}\) + ... + \(\dfrac{3}{2012.4026}\)
A = \(\dfrac{6}{2.4}\) + \(\dfrac{6}{4.6}\) + \(\dfrac{6}{6.8}\) + ... + \(\dfrac{6}{4024.4026}\)
A = \(3\left(\dfrac{2}{2.4}+\dfrac{2}{4.6}+\dfrac{2}{6.8}+...+\dfrac{2}{4024.4026}\right)\)
A = \(3\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{4024}-\dfrac{1}{4026}\right)\)
A = \(3\left(\dfrac{1}{2}-\dfrac{1}{4026}\right)\)
A = 3.\(\dfrac{1}{2}\) - 3.\(\dfrac{1}{4026}\)
A = 1,5 - \(3.\dfrac{1}{4026}\) < 1,5
=> A < 1,5
=> đpcm
11 tháng 5 2017
\(A=\)\(\frac{3}{1.4}\)\(+\)\(\frac{3}{2.6}\)\(+\)\(\frac{3}{2.8}\)\(+\).........\(+\)\(\frac{1}{2012.1342}\)\(< 1,5\)
\(=\)\(\frac{3}{1.4}\)\(+\)\(\frac{3}{2.6}\)\(+\)\(\frac{3}{3.8}\)\(+\)............\(+\)\(\frac{3}{2012.4026}\)
\(=\)\(\frac{6}{2.4}\)\(+\)\(\frac{6}{4.6}\)\(+\)\(\frac{6}{6.8}\)\(+\)..............\(+\)\(\frac{6}{4024.4026}\)
\(=\)\(3.\)\(\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...........+\frac{2}{4024.4026}\right)\)
\(=\)\(3.\)\(\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+....+\frac{1}{4024}-\frac{1}{4026}\right)\)
\(=\)\(3.\)\(\left(\frac{1}{2}-\frac{1}{4026}\right)\)
\(=\)\(3.\)\(\frac{1}{2}\)\(-\)\(3.\)\(\frac{1}{4026}\)
\(=\)\(1,5\)\(-\)\(\frac{3}{4026}\)\(< \)\(1,5\)
Vậy \(A< 1,5\)
8 tháng 5 2018
\(\frac{3}{1.4}+\frac{3}{2.6}+\frac{3}{3.8}+...+\frac{1}{2012.1342}\)
\(=\frac{3}{1.4}+\frac{3}{2.6}+\frac{3}{3.8}+...+\frac{3}{2012.4026}\)
\(=\frac{6}{2.4}+\frac{6}{4.6}+\frac{6}{4.8}+...+\frac{6}{4024.4026}\)
\(=3\cdot\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{4024.4026}\right)\)
\(=3\cdot\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{4024}-\frac{1}{4026}\right)\)
\(=3\cdot\left(\frac{1}{2}-\frac{1}{4026}\right)\)
\(=3\cdot\frac{1}{2}-3\cdot\frac{1}{4026}\)
\(=1,5-\frac{3}{4026}< 1,5\)
=3/12+3/60+3/140
=>1/4+1/20+3/140
=>4+20+3/140
=>3363/140