A = 5/(3.7) + 5/(7.11) + 5/(11.15) + ... + 5/(2019.2023)

= 5/4 . (1/3 - 1/7 + 1/7 - 1/11 + 1/11 - 1/15 + ... + 1/2019 - 1/2023)

= 5/4 . (1/3 - 1/2023)

= 5/4 . 2020/6069

= 2525/6069

Lời giải:

$A=5(\frac{1}{3.7}+\frac{1}{7.11}+\frac{1}{11.15}+...+\frac{1}{2019.2023})$

$4A=5(\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+...+\frac{4}{2019.2023})$

$=5(\frac{7-3}{3.7}+\frac{11-7}{7.11}+\frac{15-11}{11.15}+...+\frac{2023-2019}{2019.2023})$

$=5(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+....+\frac{1}{2019}-\frac{1}{2023})$

$=5(\frac{1}{3}-\frac{1}{2023})=\frac{2020}{6069}$

$\Rightarrow A=\frac{2020}{6069}:4=\frac{505}{6069}$

A = \(\dfrac{5}{3.7}\) + \(\dfrac{5}{7.11}\) + \(\dfrac{5}{11.15}\)+...+ \(\dfrac{5}{2019.2023}\)

A = 5.(\(\dfrac{1}{3.7}\) + \(\dfrac{1}{7.11}\) + \(\dfrac{1}{11.15}\) + ... + \(\dfrac{1}{2019.2023}\)) 

A = \(\dfrac{5}{4}\).( \(\dfrac{4}{3.7}\) + \(\dfrac{4}{7.11}\) + \(\dfrac{4}{11.15}\) +  ... + \(\dfrac{4}{2019.2023}\))

A = \(\dfrac{5}{4}\).( \(\dfrac{1}{3}\) - \(\dfrac{1}{7}\) +\(\dfrac{1}{7}\) -  \(\dfrac{1}{11}\) + \(\dfrac{1}{11}\) - \(\dfrac{1}{15}\) + ... + \(\dfrac{1}{2019}\) - \(\dfrac{1}{2023}\))

A = \(\dfrac{5}{4}\). ( \(\dfrac{1}{3}\) - \(\dfrac{1}{2023}\))

A = \(\dfrac{5}{4}\) . \(\dfrac{2020}{6069}\)

A = \(\dfrac{2525}{6069}\)

\(\dfrac{4}{3.7}+\dfrac{4}{7.11}+\dfrac{4}{11.15}+...+\dfrac{4}{59.63}\)

\(=\dfrac{1}{3}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{15}+...+\dfrac{1}{59}-\dfrac{1}{63}\)

\(=\dfrac{1}{3}-\dfrac{1}{63}\)

\(=\dfrac{20}{63}\)

\(K=\dfrac{5}{4}\left(\dfrac{1}{3}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+...+\dfrac{1}{85}-\dfrac{1}{89}\right)\)

\(=\dfrac{5}{4}\cdot\left(\dfrac{1}{3}-\dfrac{1}{89}\right)=\dfrac{5}{4}\cdot\dfrac{86}{267}=\dfrac{215}{534}\)

a=4/3.7 +4/7.11+4/11.15 +.....+4/107/111

=1/3-1/7+1/7-1/11+1/11-1/15+......+1/107-1/111

=1/3-1/111

=12/37

12/37nhe

4x(\(\frac{1}{3.7}+...+\frac{1}{107.111}\) )

4(\(\frac{1}{3}-\frac{1}{7}+...+\frac{1}{107}-\frac{1}{111}\))

4(\(\frac{1}{3}-\frac{1}{111}\))

4.\(\frac{12}{37}\)

48/37

Ta thấy \(\frac{1}{3}-\frac{1}{7}=\frac{7-3}{3.7}=\frac{4}{3.7}\) 

             \(\frac{1}{7}-\frac{1}{11}=\frac{11-7}{7.11}=\frac{4}{7.11}\)

             ..........................

             \(\frac{1}{1023}-\frac{1}{1027}=\frac{1027-1023}{1023.1027}=\frac{4}{1023.1027}\)

=> \(\frac{4}{3.7}+\frac{4}{7.11}+....+\frac{4}{1023.1027}=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+....+\frac{1}{1023}-\frac{1}{1027}\)

=>                                                       =\(\frac{1}{3}-\frac{1}{1027}=\frac{1024}{3.1027}\)

Ta có: \(\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+...+\frac{4}{1023.1027}\)

\(=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...+\frac{1}{1023}-\frac{1}{1027}\)

\(=\frac{1}{3}-\frac{1}{1027}=\frac{1024}{3081}\)

\(\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+....+\frac{4}{1023.1027}\)

\(=\frac{1}{3.7}+\frac{1}{7.11}+\frac{1}{11.15}+...+\frac{1}{1023.1027}\)

\(=\frac{1}{3}-\frac{1}{1027}\)

\(=\frac{1024}{3081}\)

ai k mik mik k lại nha