=5/4(4/3*7+4/7*11+...+4/81*85)

=5/4(1/3-1/7+1/7-1/11+...+1/81-1/85)

=5/4*82/255=41/102

Viết vậy đúng đó em

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

= 5/4 . [4/(3.7) + 4/(7.11) + 4/(11.15) + ... + 4/(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

\(A=\frac{5}{3\cdot7}+\frac{5}{7\cdot11}+\frac{5}{11\cdot15}+...+\frac{5}{81\cdot85}+\frac{5}{85\cdot89}\\ A=\frac{5}{4}\left(\frac{4}{3\cdot7}+\frac{4}{7\cdot11}+\frac{4}{11\cdot15}+...+\frac{4}{81\cdot85}+\frac{4}{85\cdot89}\right)\\ A=\frac{5}{4}\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...+\frac{1}{81}-\frac{1}{85}+\frac{1}{85}-\frac{1}{89}\right)\\ A=\frac{5}{4}\left(\frac{1}{3}-\frac{1}{89}\right)\\ A=\frac{5}{4}\left(\frac{89}{267}-\frac{3}{267}\right)\\ A=\frac{5}{4}\cdot\frac{86}{267}=\frac{215}{534}\)

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

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

\(A=\frac{5}{3.7}+\frac{5}{7.11}+...+\frac{5}{\left(4n-1\right).\left(4n+3\right)}\)

\(\frac{4}{5}.A=\frac{4}{3.7}+\frac{4}{7.11}+...+\frac{4}{\left(4n-1\right).\left(4n+3\right)}\)

\(\frac{4}{5}.A=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{4n-1}-\frac{1}{4n+3}\)

\(\frac{4}{5}.A=\frac{1}{3}-\frac{1}{4n+3}\)

\(\frac{4}{5}.A=\frac{4n+3}{12n+9}-\frac{3}{12n+9}\)

\(\frac{4}{5}.A=\frac{4n}{12n+9}\)

\(A=\frac{4n}{12n+9}:\frac{4}{5}\)

\(A=\frac{4n}{12n+9}.\frac{5}{4}\)

\(A=\frac{5n}{12n+9}\)

Đề bài sai nha bn

Ủng hộ mk nha ^_^

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 = \(\dfrac{2525}{6069}\)

nếu lm sai thì thôi na

đặt vế trái lak A

4A=5(5/3.7+5/7.11+.....+5/(4n-1)(4n+3)

4A=(1/3-1/4n+3)

4A=4n+3-3/3(4n+3)

4A=4/12n+9

A=5/4n+3

bn ở đâu

\(\frac{5}{3.7}+\frac{5}{7.11}+...+\frac{5}{\left(4n-1\right).\left(4n+3\right)}=\frac{5}{4n+3}\)

\(\Leftrightarrow\frac{5}{4}\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{4n-1}-\frac{1}{4n+3}\right)=\frac{5}{4n+3}\)

\(\Leftrightarrow\frac{1}{3}-\frac{1}{4n+3}=\frac{5}{4n+3}\cdot\frac{4}{5}\)

\(\Leftrightarrow\frac{1}{3}-\frac{1}{4n+3}=\frac{5\cdot4}{5\left(4n+3\right)}\)

\(\Leftrightarrow\frac{1}{3}-\frac{1}{4n+3}=\frac{4}{4n+3}\)

\(\Leftrightarrow\frac{1}{4n+3}=\frac{1}{3}-\frac{4}{4n+3}\)

\(\Leftrightarrow\frac{1}{4n+3}=\frac{4n+3-12}{3\left(4n+3\right)}\)

\(\Leftrightarrow\frac{1}{4n+3}=\frac{4n-9}{12n+9}\)

\(\Leftrightarrow\frac{3}{12n+9}=\frac{4n-9}{12n+9}\)

\(\Leftrightarrow4n-9=3\)

\(\Leftrightarrow4n=12\Leftrightarrow n=3\)

Vậy n = 3