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25 tháng 6 2017

\(\frac{5}{1\cdot3}+\frac{5}{3\cdot5}+...+\frac{5}{97\cdot99}=\frac{5}{2}\left[\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+...+\frac{1}{97\cdot99}\right]\)

\(=\frac{5}{2}\left[\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{97}-\frac{1}{99}\right]=\frac{5}{2}\left[1-\frac{1}{99}\right]\)

\(=\frac{5}{2}\cdot\frac{98}{99}=\frac{245}{99}\)

25 tháng 6 2017

\(=\frac{5}{2}\left(\frac{2}{1\times3}+\frac{2}{3\times5}+\frac{2}{5\times7}+...+\frac{2}{97\times99}\right)\)

\(=\frac{5}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\right)\)

\(=\frac{5}{2}\left(1-\frac{1}{99}\right)\)

\(=\frac{5}{2}\times\frac{98}{99}\)

\(=\frac{245}{99}\)

19 tháng 8 2023

a)\(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{99.101}\)

\(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\)

\(1-\dfrac{1}{101}\)

=\(\dfrac{100}{101}\) 

 

 

19 tháng 8 2023

\(\dfrac{5}{1.3}+\dfrac{5}{3.5}+\dfrac{5}{5.7}+...+\dfrac{5}{99.101}\)

=\(\dfrac{5}{2}.\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{99+101}\right)\)

=\(\dfrac{5}{2}.\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\) 

=\(\dfrac{5}{2}.\left(1-\dfrac{1}{101}\right)\)

\(\dfrac{5}{2}-\dfrac{100}{101}\)

\(\dfrac{305}{202}\)

21 tháng 8 2015

\(a,\frac{5}{1.3}+\frac{5}{3.5}+\frac{5}{5.7}+...+\frac{5}{43.45}=\frac{5}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{43.45}\right)=\frac{5}{3}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{43}-\frac{1}{45}\right)=\frac{5}{3}.\frac{44}{45}=\frac{44}{27}\)

24 tháng 6 2017

\(M=\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{99.100}\)

\(M=2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\right)\)

\(M=2\left(1-\frac{1}{100}\right)\)

\(M=2.\frac{99}{100}\)

\(M=\frac{99}{50}\)

\(N=\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+...+\frac{3}{97.99}\)

\(N=\frac{3}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\right)\)

\(N=\frac{3}{2}\left(1-\frac{1}{99}\right)\)

\(N=\frac{3}{2}.\frac{98}{99}\)

\(N=\frac{49}{33}\)

19 tháng 2 2017

\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\)

\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right)\)

\(=\frac{1}{2}\left(1-\frac{1}{11}\right)=\frac{1}{2}.\frac{10}{11}=\frac{5}{11}\)

20 tháng 6 2016

\(\frac{5}{1.3}+\frac{5}{3.5}+\frac{5}{5.7}=5\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}\right)\)

\(=\frac{5}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}\right)\)

\(=\frac{5}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}\right)\)

\(=\frac{5}{2}\left(1-\frac{1}{7}\right)=\frac{5}{2}\left(\frac{7}{7}-\frac{1}{7}\right)=\frac{5}{2}.\frac{6}{7}=\frac{15}{7}\)

15 tháng 8 2016

M = 3/1x3 + 3/3x5 + 3/5x7 + ... + 3/45x47 + 3/47x49

M = 3/2 x (2/1x3 + 2/3x5 + 2/5x7 + ... + 2/45x47 + 2/47x49)

M = 3/2 x (1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 + ... + 1/45 - 1/47 + 1/47 - 1/49)

M = 3/2 x (1 - 1/49)

M = 3/2 x 48/49

M = 72/49

N tính tương tự, nhân N với 5/4