=1/5-1/7+1/7-1/12+1/12-1/20+3/140

=3/20+3/140

=21/140+3/140=24/140=6/35

\(\frac{2}{5\cdot7}+\frac{5}{7\cdot12}+\frac{8}{12\cdot20}+\frac{3}{140}\)

Ta có công thức:\(\frac{a}{n\cdot\left(n+a\right)}=\frac{1}{n}-\frac{1}{n+a}\)Lên mạng gõ dãy phân số viết theo quy luật là thấy

=>ta có \(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{12}+\frac{1}{12}-\frac{1}{20}+\frac{3}{140}\)

=>\(\frac{1}{5}-\frac{1}{20}+\frac{3}{140}\)

=>\(\frac{6}{35}\)

gõ rồi mà ko thấy

a) 1/5.6 + 1/6.7 + 1/7.8 + ... + 1/24.25

= 1/5 - 1/6 + 1/6 - 1/7 + 1/7 - 1/8 + ... + 1/24 - 1/25

= 1/5 - 1/25

= 4/25

b) 2/1.3 + 2/3.5 + 2/5.7 + ... + 2/99.101

= 1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 + ... + 1/99 -1/101

= 1 - 1/101

= 100/101

c) 3/1.4 + 3/4.7 + ... + 3/2002.2005

= 1 - 1/4 + 1/4 - 1/7 + ... + 1/2002 - 1/2005

= 1 - 1/2005

= 2004/2005

d) 5/2.7 + 5/7.12 + ... + 5/1997.2002

= 1/2 - 1/7 + 1/7 - 1/12 + ... + 1/1997 - 1/2002

= 1/2 - 1/2002

= 500/1001

a,A =  \(\frac{1}{5\times6}+\frac{1}{6\times7}+\frac{1}{7\times8}+...+\frac{1}{24\times25}\)

A\(=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{24}-\frac{1}{25}\)

A\(=\frac{1}{5}-\frac{1}{25}=\frac{5}{25}-\frac{1}{25}=\frac{4}{25}\)

b, B=\(\frac{2}{1\times3}+\frac{2}{3\times5}+\frac{2}{5\times7}+...+\frac{2}{99\times101}\)

B= \(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\)

B=\(1-\frac{1}{101}=\frac{100}{101}\)

c, \(C=\frac{3}{1\times4}+\frac{3}{4\times7}+...+\frac{3}{2002\times2005}\)

C= \(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{2002}-\frac{1}{2005}\)

C= \(1-\frac{1}{2005}=\frac{2004}{2005}\)

d, D= \(\frac{5}{2\times7}+\frac{5}{7\times12}+...+\frac{5}{1997\times2002}\)

D= \(\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{12}+...+\frac{1}{1997}-\frac{1}{2002}\)

D= \(\frac{1}{2}-\frac{1}{2002}=\frac{1001}{2002}-\frac{1}{2002}=\frac{1000}{2002}=\frac{500}{1001}\)

cacwj con

bằng 2270/5049 đó bạn

\(\frac{5}{2.7}+\frac{5}{7.12}+\frac{5}{12.17}+\frac{5}{22.27}\)

\(=\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{12}+\frac{1}{12}-\frac{1}{17}+\frac{5}{22.27}\)

\(=\frac{1}{2}-\frac{1}{17}+\frac{5}{22.27}\)

\(=\frac{2270}{5049}\)

\(A=\frac{2}{5.7}+\frac{5}{7.12}+\frac{7}{12.19}+\frac{9}{19.28}+\frac{11}{28.39}+\frac{1}{39.40}\)

\(=\frac{7-5}{5.7}+\frac{12-7}{7.12}+\frac{19-12}{12.19}+\frac{28-19}{19.28}+\frac{39-28}{28.39}+\frac{40-39}{39.40}\)

\(=\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{12}+\frac{1}{12}-\frac{1}{19}+\frac{1}{19}-\frac{1}{28}+\frac{1}{28}-\frac{1}{39}+\frac{1}{39}-\frac{1}{40}\)

\(=\frac{1}{5}-\frac{1}{40}=\frac{7}{40}\)

\(B=\frac{1}{20}+\frac{1}{44}+\frac{1}{77}+\frac{1}{119}+\frac{1}{170}\)

\(=\frac{2}{40}+\frac{2}{88}+\frac{2}{154}+\frac{2}{238}+\frac{2}{340}\)

\(=\frac{2}{5.8}+\frac{2}{8.11}+\frac{2}{11.14}+\frac{2}{14.17}+\frac{2}{17.20}\)

\(=\frac{2}{3}\left(\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+\frac{3}{14.17}+\frac{3}{17.20}\right)\)

\(=\frac{2}{3}\left(\frac{8-5}{5.8}+\frac{11-8}{8.11}+\frac{14-11}{11.14}+\frac{17-14}{14.17}+\frac{20-17}{17.20}\right)\)

\(=\frac{2}{3}\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}+\frac{1}{17}-\frac{1}{20}\right)\)

\(=\frac{2}{3}\left(\frac{1}{5}-\frac{1}{20}\right)=\frac{1}{10}\)

\(\frac{A}{B}=\frac{\frac{7}{40}}{\frac{1}{10}}=\frac{7}{4}\)

A = \(\dfrac{2}{5.7}\) + \(\dfrac{5}{7.12}\) + \(\dfrac{7}{12.19}\) + \(\dfrac{9}{19.28}\) + \(\dfrac{11}{28.39}\) + \(\dfrac{1}{30.40}\)

A = \(\dfrac{1}{5}\) - \(\dfrac{1}{7}\) + \(\dfrac{1}{7}\) - \(\dfrac{1}{12}\) + \(\dfrac{1}{12}\) - \(\dfrac{1}{19}\)  + \(\dfrac{1}{19}\) - \(\dfrac{1}{28}\) + \(\dfrac{1}{28}\) - \(\dfrac{1}{39}\) + \(\dfrac{1}{1200}\)

A = \(\dfrac{1}{5}\) - \(\dfrac{1}{39}\) + \(\dfrac{1}{1200}\)

A = \(\dfrac{34}{195}\) + \(\dfrac{1}{1200}\) 

B = \(\dfrac{1}{20}\) + \(\dfrac{1}{44}\) + \(\dfrac{1}{77}\) + \(\dfrac{1}{119}\) + \(\dfrac{1}{170}\)

B = 2 \(\times\) ( \(\dfrac{1}{2.20}\) + \(\dfrac{1}{2.44}\) + \(\dfrac{1}{2.77}\) + \(\dfrac{1}{2.119}\) + \(\dfrac{1}{2.170}\))

B = 2 \(\times\) ( \(\dfrac{1}{40}\) + \(\dfrac{1}{88}\) + \(\dfrac{1}{154}\) + \(\dfrac{1}{238}\) + \(\dfrac{1}{340}\))

B = 2 \(\times\) ( \(\dfrac{1}{5.8}\) + \(\dfrac{1}{8.11}\) + \(\dfrac{1}{11.14}\) + \(\dfrac{1}{14.17}\) + \(\dfrac{1}{17.20}\))

B = \(\dfrac{2}{3}\) \(\times\) ( \(\dfrac{3}{5.8}\) + \(\dfrac{3}{8.11}\)+ \(\dfrac{3}{11.14}\) + \(\dfrac{3}{14.17}\) + \(\dfrac{3}{17.20}\))

B = \(\dfrac{2}{3}\) \(\times\) ( \(\dfrac{1}{5}\) - \(\dfrac{1}{8}\) + \(\dfrac{1}{8}\) - \(\dfrac{1}{11}\) + \(\dfrac{1}{11}\) - \(\dfrac{1}{14}\) + \(\dfrac{1}{14}\) - \(\dfrac{1}{17}\) + \(\dfrac{1}{17}\) - \(\dfrac{1}{20}\))

B = \(\dfrac{2}{3}\) \(\times\) ( \(\dfrac{1}{5}\) - \(\dfrac{1}{20}\))

B = \(\dfrac{2}{3}\) \(\times\) \(\dfrac{3}{20}\)

B = \(\dfrac{1}{10}\)  = \(\dfrac{34}{340}\) < \(\dfrac{34}{195}\) + \(\dfrac{1}{1200}\)

Vậy A > B 

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