\(\frac{3}{2.6}+\frac{3}{6.10}+\frac{3}{10.14}\)
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Những câu hỏi liên quan
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
14 tháng 3
\(\dfrac{3}{2.6}\) + \(\dfrac{3}{6.10}\) + \(\dfrac{3}{10.14}\)
= \(\dfrac{3}{4}\).(\(\dfrac{4}{2.6}\) + \(\dfrac{4}{6.10}\) + \(\dfrac{4}{10.14}\))
= \(\dfrac{3}{4}\).(\(\dfrac{1}{2}-\dfrac{1}{6}\) + \(\dfrac{1}{6}\) - \(\dfrac{1}{10}\) + \(\dfrac{1}{10}\) - \(\dfrac{1}{14}\))
= \(\dfrac{3}{4}\).(\(\dfrac{1}{2}\) - \(\dfrac{1}{14}\))
= \(\dfrac{3}{4}\). \(\dfrac{3}{7}\)
= \(\dfrac{9}{28}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
14 tháng 3
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 = \(\dfrac{20}{63}\)
1 tháng 5 2021
Sửa đề: \(D=\dfrac{3}{2\cdot6}+\dfrac{3}{6\cdot10}+\dfrac{3}{10\cdot14}+...+\dfrac{3}{26\cdot30}\)
Ta có: \(D=\dfrac{3}{2\cdot6}+\dfrac{3}{6\cdot10}+\dfrac{3}{10\cdot14}+...+\dfrac{3}{26\cdot30}\)
\(=\dfrac{3}{4}\left(\dfrac{4}{2\cdot6}+\dfrac{4}{6\cdot10}+\dfrac{4}{10\cdot14}+...+\dfrac{4}{26\cdot30}\right)\)
\(=\dfrac{3}{4}\left(\dfrac{1}{2}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{14}+...+\dfrac{1}{26}-\dfrac{1}{30}\right)\)
\(=\dfrac{3}{4}\left(\dfrac{1}{2}-\dfrac{1}{30}\right)\)
\(=\dfrac{3}{4}\cdot\dfrac{28}{60}\)
\(=\dfrac{21}{60}=\dfrac{7}{20}\)
NP
4
4 tháng 8 2020
\(A=\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{2015.2018}\)
\(=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{2015}-\frac{1}{2018}\)
\(=\frac{1}{2}-\frac{1}{2018}=\frac{504}{1009}\)
Vậy \(A=\frac{504}{1009}.\)
\(B=\frac{4}{2.6}+\frac{4}{6.10}+\frac{4}{10.14}+...+\frac{4}{102.106}\)
\(=\frac{1}{2}-\frac{1}{6}+\frac{1}{6}-\frac{1}{10}+\frac{1}{10}-\frac{1}{14}+...+\frac{1}{102}-\frac{1}{106}\)
\(=\frac{1}{2}-\frac{1}{106}=\frac{26}{53}\)
Vậy \(B=\frac{26}{53}.\)
4 tháng 8 2020
Bài làm:
a) \(A=\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{2015.2018}\)
\(A=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{2015}-\frac{1}{2018}\)
\(A=\frac{1}{2}-\frac{1}{2018}\)
\(A=\frac{504}{1009}\)
b) \(B=\frac{4}{2.6}+\frac{4}{6.10}+\frac{4}{10.14}+...+\frac{4}{102.106}\)
\(B=\frac{1}{2}-\frac{1}{6}+\frac{1}{6}-\frac{1}{10}+\frac{1}{10}-\frac{1}{14}+...+\frac{1}{102}-\frac{1}{106}\)
\(B=\frac{1}{2}-\frac{1}{106}\)
\(B=\frac{26}{53}\)
DC
4
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
NN
Nguyễn Ngọc Anh Minh
CTVHS
VIP
11 tháng 9 2023
12.T=2.6.12+6.10.12+10.14.12+...+102.106.12=
=2.6.(10+2)+6.10.(14-2)+10.14.(18-6)+...+102.106.(110-98)=
=2.2.6+2.6.10-2.6.10+6.10.14-6.10.14+10.14.18-...-98.102.106+102.106.110=
=2.2.6+102.106.110
\(\Rightarrow T=\dfrac{2.2.6+102.106.110}{12}=99112\)
LT
7
S
17 tháng 6 2019
Đặt A = 1/2.6 + 1/6.10 + 1/10.14 + ..... + 1/102.106
=> 4A = 4/2.6 + 4/6.10 + 4/10.14 + ..... + 4/102.106
=> 4A = 1/2 - 1/6 + 1/6 - 1/10 + 1/10 - 1/14 + ... + 1/102 - 1/106
=> 4A = 1/2 - 1/106
=> 4A = 26/53
=> A = 13/106
~Study well~
#QASJ
17 tháng 6 2019
\(\frac{1}{2.6}+\frac{1}{6.10}+...+\frac{1}{102.106}\)
\(=\frac{1}{4}.\left(\frac{4}{2.6}+\frac{4}{6.10}+...+\frac{4}{102.106}\right)\)
\(=\frac{1}{4}.\left(\frac{1}{2}-\frac{1}{6}+\frac{1}{6}-\frac{1}{10}+...+\frac{1}{102}-\frac{1}{106}\right)\)
\(=\frac{1}{4}.\left(\frac{1}{2}-\frac{1}{106}\right)\)
\(=\frac{1}{4}.\frac{26}{53}\)
\(=\frac{13}{106}\)
24 tháng 2 2018
\(\frac{1}{6.10}\)+ \(\frac{1}{10.14}\)+ ... + \(\frac{1}{402.406}\)
= \(\frac{1}{4}\). \(\left(\frac{4}{6.10}+\frac{4}{10.14}+...+\frac{4}{402.406}\right)\)
= \(\frac{1}{4}\). ( \(\frac{10-6}{6.10}\)+ \(\frac{14-10}{10.14}\)+ ... + \(\frac{406-402}{402.406}\))
= \(\frac{1}{4}\). ( \(\frac{10}{6.10}\)- \(\frac{6}{6.10}\)+ ... + \(\frac{406}{402.406}\)- \(\frac{402}{402.406}\))
= \(\frac{1}{4}\). ( \(\frac{1}{6}\)- \(\frac{1}{406}\))
= \(\frac{1}{4}\). \(\frac{100}{609}\)
= \(\frac{25}{609}\)
=3/12+3/60+3/140
=>1/4+1/20+3/140
=>4+20+3/140
=>3363/140