KT
1
K
Khách
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TG
6
TL
18 tháng 6 2015
Đặt biểu thức đó là A
=> 2A = \(\frac{2}{9.11}+\frac{2}{11.13}+...+\frac{2}{61.63}=\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}+...+\frac{1}{61}-\frac{1}{63}\)
\(=\frac{1}{9}-\frac{1}{63}=\frac{2}{21}\)
=> A = \(\frac{2}{21}.\frac{1}{2}=\frac{1}{21}\)
2 tháng 5 2016
A = \(\frac{5}{1.2}\) + \(\frac{5}{2.3}\) +........+\(\frac{5}{99.100}\)
A = 5.(\(\frac{1}{1.2}\) + \(\frac{1}{2.3}\) +......+\(\frac{1}{99.100}\) )
A = 5. ( \(\frac{1}{1}\) - \(\frac{1}{2}\) +\(\frac{1}{2}-\frac{1}{3}\) +......+\(\frac{1}{99}-\frac{1}{100}\) )
A= 5. (\(1-\frac{1}{100}\))
A= 5.\(\frac{99}{100}\)
A= \(\frac{99}{20}\)
23 tháng 3 2017
B = \(\frac{1}{2.3}\)+ \(\frac{1}{3.4}\)+............+ \(\frac{1}{9.10}\)
= \(\frac{1}{2}\)- \(\frac{1}{3}\)+\(\frac{1}{3}\)- \(\frac{1}{4}\)+ ...................+\(\frac{1}{9}\)- \(\frac{1}{10}\)
= \(\frac{1}{2}\) - \(\frac{1}{10}\)
= \(\frac{2}{5}\)
K
23 tháng 6 2015
\(A=\frac{5}{1.2}+\frac{5}{2.3}+\frac{5}{3.4}+...+\frac{5}{99.100}\)
\(A=5\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\right)\)
\(A=5\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(A=5\left(1-\frac{1}{100}\right)\)
\(A=5.\frac{99}{100}\)
\(A=\frac{99}{20}\)
\(B=\frac{1}{1.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}\)
\(B=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\)
\(B=\frac{1}{2}-\frac{1}{10}\)
\(B=\frac{2}{5}\)
\(C=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}+\frac{2}{13.15}\)
\(C=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}\)
\(C=\frac{1}{3}-\frac{1}{15}\)
\(C=\frac{4}{15}\)
K
23 tháng 6 2015
\(A=\frac{5}{1.2}+\frac{5}{2.3}+\frac{5}{3.4}+...+\frac{5}{99.100}\)
\(A=5\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\right)\)
\(A=5\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(A=5\left(1-\frac{1}{100}\right)\)
\(A=5.\frac{99}{100}\)
\(A=\frac{99}{20}\)
\(B=\frac{1}{1.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}\)
\(B=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\)
\(B=\frac{1}{2}-\frac{1}{10}\)
\(B=\frac{2}{5}\)
\(C=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}+\frac{2}{13.15}\)
\(C=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}\)
\(C=\frac{1}{3}-\frac{1}{15}\)
\(C=\frac{4}{15}\)
NH
20 tháng 4 2019
\(\frac{1}{9.11}+\frac{1}{11.13}+\frac{1}{13.15}+......+\frac{1}{97.99}\)
=\(\frac{1}{2}.\left(\frac{2}{9.11}+\frac{2}{11.13}+\frac{2}{13.15}+......+\frac{2}{97.99}\right)\)
=\(\frac{1}{2}.\left(\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+......+\frac{1}{97}-\frac{1}{99}\right)\)
=\(\frac{1}{2}.\left(\frac{1}{9}-\frac{1}{99}\right)=\frac{1}{2}.\left(\frac{11}{99}-\frac{1}{99}\right)=\frac{1}{2}.\frac{10}{99}=\frac{5}{99}\)
15 tháng 4 2019
Tính :
\(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+\frac{1}{11.13}+\frac{1}{13.15}\Rightarrow\frac{1}{2}.\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}+\frac{2}{13.15}\right)=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{13}-\frac{1}{15}\right)=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{15}\right)=\frac{1}{2}\left(\frac{5}{15}-\frac{1}{15}\right)=\frac{1}{2}.\frac{4}{15}=\frac{2}{15}\)
15 tháng 4 2019
\(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+\frac{1}{11.13}+\frac{1}{13.15}\)
=\(\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}\right)\)
=\(\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{15}\right)=\frac{1}{2}.\frac{4}{15}=\frac{2}{15}\)
5 tháng 5 2018
\(\frac{8^2}{7.9}.\frac{9^2}{8.10}...\frac{14^2}{13.15}\)
\(\frac{8.8}{7.9}.\frac{9.9}{8.10}...\frac{14.14}{13.15}\)
\(\frac{8.9...14}{7.8...13}.\frac{8.9...14}{9.10...15}\)
\(\frac{14}{7}.\frac{8}{15}\)
\(2.\frac{8}{15}\)
\(\frac{16}{15}\)
5 tháng 5 2018
(8.9.10.11.12.13.14)(8.9.10.11.12.13.14)/7.8.9.10.11.12.13).(9.10.11.12.13.14.15)
=14.8/7.15
=16/15
k cho mình nhá
20 tháng 4 2015
\(\left(\frac{2}{11.13}+\frac{2}{13.15}+...+\frac{2}{19.20}\right)-x+\frac{221}{231}=\frac{4}{3}\)
\(=\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+...+\frac{1}{19}-\frac{1}{20}\right)-x=\frac{4}{3}-\frac{221}{231}\)
\(=\left(\frac{1}{11}-\frac{1}{20}\right)-x=\frac{29}{77}\)
\(=\frac{9}{220}-x=\frac{29}{77}\)
\(x=\frac{9}{220}-\frac{29}{77}\)
20 tháng 4 2015
bạn ơi chỗ \(\frac{2}{19.20}\) có phải là \(\frac{2}{19.21}\) không
B
2
6 tháng 4 2017
Đặt \(A=\frac{2}{11.13}+\frac{2}{13.15}+...+\frac{2}{19.21}\)
\(A=\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{14}+...+\frac{1}{19}-\frac{1}{21}\)
\(A=\frac{1}{11}-\frac{1}{21}=\frac{10}{231}\)
thay A vào ta được:
\(\frac{10}{231}-x+\frac{23}{231}=\frac{5}{77}\)
\(x+\frac{23}{231}=\frac{10}{231}-\frac{5}{77}=-\frac{5}{231}\)
\(x=-\frac{5}{231}-\frac{23}{231}=\frac{-4}{33}\)
D
6 tháng 4 2017
\(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+...+\frac{1}{19}-\frac{1}{21}-x+\frac{23}{231}=\frac{5}{77}\)\(\frac{5}{77}\)
=>\(\frac{1}{11}-\frac{1}{21}-x+\frac{23}{231}=\frac{5}{77}\)
=>\(\frac{10}{231}+\frac{23}{231}-x=\frac{5}{77}\)
=>\(\frac{33}{231}-x=\frac{5}{77}\)
=>\(\frac{1}{7}-x=\frac{5}{77}\)
=>\(\frac{11}{77}-x=\frac{5}{77}\)
=>\(x=\frac{11}{77}-\frac{5}{77}=\frac{6}{77}\)
Vậy \(x=\frac{6}{77}\)
= 1/9 - 1/11 + 1/11 - 1/13 + 1/13 - 1/15 + 1/15 - 1/19
= 1/9 - 1/19
= 10/171
Tk mk nha