\(\dfrac{5}{3\cdot7}+\dfrac{5}{7\cdot11}+\dfrac{5}{11\cdot15}+...+\dfrac{5}{\left(4n-1\right)\left(4n+3\right)}\\ =\dfrac{5}{4}\cdot\left(\dfrac{4}{3\cdot7}+\dfrac{4}{7\cdot11}+\dfrac{4}{11\cdot15}+...+\dfrac{4}{\left(4n-1\right)\left(4n+3\right)}\right)\\ =\dfrac{5}{4}\cdot\left(\dfrac{1}{3}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{15}+...+\dfrac{1}{4n-1}-\dfrac{1}{4n+3}\right)\\ =\dfrac{5}{4}\cdot\left(\dfrac{1}{3}-\dfrac{1}{4n+3}\right)\\ =\dfrac{5}{4}\cdot\dfrac{4n}{12n+9}\\ =\dfrac{5n}{12n+9}\)

Mk thực sự nghĩ đề hình như bị sai hay sao ấy!

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

\(=\frac{5}{4}\left(\frac{7-3}{7.3}+\frac{11-7}{7.11}+........+\frac{\left(4n+3\right)-\left(4n-1\right)}{\left(4n-1\right)\left(4n+3\right)}\right)\)

\(=\frac{5}{4}\left(\frac{7}{7.3}-\frac{3}{7.3}+\frac{11}{7.11}-\frac{7}{7.11}+......+\frac{4n+3}{\left(4n-1\right)\left(4n+3\right)}-\frac{4n-1}{\left(4n-1\right)\left(4n+3\right)}\right)\)

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

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

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

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

\(=\frac{5}{4}.\frac{4n}{3\left(4n+3\right)}=\frac{4.n.5}{3\left(4n+3\right).4}=\frac{5n}{3\left(4n+3\right)}\)

ban nen xem lai dau bai di minh giai dung 100% do

ma neu dau bai ra nhu ket qua cua to thi tick cho minh nha

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Tôi sắp tử vong rồi

2+12345678-5=

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