Chào bạn, bạn hãy theo dõi bài giải của mình nhé!

Ta có : 

\(\frac{4}{5.7}+\frac{4}{7.9}+\frac{4}{9.11}+...+\frac{4}{53.55}\)

\(=\frac{4}{2}\left(\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+...+\frac{2}{53.55}\right)\)

\(=2.\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{53}-\frac{1}{55}\right)\)

\(=2.\left(\frac{1}{5}-\frac{1}{55}\right)=2.\left(\frac{11}{55}-\frac{1}{55}\right)=2.\frac{10}{55}=2.\frac{2}{11}=\frac{4}{11}\)

Có gì không hiểu bạn hỏi lại mình nhé! Chúc bạn học tốt!

Ta có: \(\frac{4}{5.7}+\frac{4}{7.9}+.....+\frac{4}{53.55}\)

Đặt C = \(\frac{4}{5.7}+\frac{4}{7.9}+...+\frac{4}{53.55}\)

     \(\frac{1}{2}C=\left(\frac{1}{5}-\frac{1}{7}\right)+\left(\frac{1}{7}-\frac{1}{9}\right)+....+\left(\frac{1}{53}-\frac{1}{55}\right)\)

      \(\frac{1}{2}C=\frac{1}{5}-\frac{1}{55}\)

      \(\frac{1}{2}C=\frac{2}{11}\)

\(C=\frac{2}{11}:\frac{1}{2}\)

Vậy C = \(\frac{4}{11}\)
Có gì sai thì mong bạn thông cảm

Ta có : 

\(A=2.\left(\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+............+\frac{2}{53.55}\right)\)

\(\Rightarrow A=2.\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+..............+\frac{1}{53}-\frac{1}{55}\right)\)

\(\Rightarrow A=2.\left(\frac{1}{5}-\frac{1}{55}\right)=2.\frac{2}{11}=\frac{4}{11}\)

k nha bạn !!!

\(\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)\left(1+\frac{1}{4}\right).....\left(1+\frac{1}{99}\right)\)

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

\(=\frac{3}{2}.\frac{4}{3}.\frac{5}{4}.....\frac{100}{99}\)

\(=\frac{3.4.5....100}{2.3.4....99}=\frac{100}{2}=50\)

`A=4/[5.7]+4/[7.9]+4/[9.11]+...+4/[21.23]+4/[23.25]`

`A=2.(2/[5.7]+2/[7.9]+....+2/[23.25])`

`A=2.(1/5-1/7+1/7-1/9+....+1/23-1/25)`

`A=2.(1/5-1/25)`

`A=2. 4/25`

`A=8/25`

\(\frac{4}{1.3}+\frac{4}{3.5}+...+\frac{4}{2013.2015}=2.\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{2013.2015}\right)=2.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2013}-\frac{1}{2015}\right)\)

\(=2.\left(\frac{2015}{2015}-\frac{1}{2015}\right)\)

\(=2.\frac{2014}{2015}\)

\(=\frac{4028}{2015}\)

\(j,\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{53.55}=\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{53}-\dfrac{1}{55}=\dfrac{1}{5}-\dfrac{1}{55}=\dfrac{11}{55}-\dfrac{1}{55}=\dfrac{10}{55}=\dfrac{2}{11}\\ k,\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}.\dfrac{4}{5}...\dfrac{99}{100}=\dfrac{1}{100}.\dfrac{2}{2}.\dfrac{3}{3}...\dfrac{99}{99}=\dfrac{1}{100}.1.1...1=\dfrac{1}{100}\)

4/3.5+4/5.7+4/7.9+4/9.11

=4.(1/3.5+1/5.7+1/7.9+1/9.11)

=4.1/2.(2/3.5+2/5.7+2/7.9+2/9.11)

=2.(1/3-1/5+1/5-1/7+1/7-1/9+1/9-1/11)

=2.(1/3-1/11)

=2.8/33

=16/33

  4/3.5+4/5.7+4/7.9+4/9.11

=4.2/2.3.5+4.2/2.5.7+4.2/2.7.9+4.2/2.9.11

=4/2.2/3.5+4/2.2/5.7+4/2.2/7.9+4/2.2/9.11

=4/2.(2/3.5+2/5.7+2/7.9+2/9.11)

=4/2.(1/3-1/5+1/5-1/7+1/7-1/9+1/9-1/11)

=2.(1/3-1/11)

=2.8/33

=16/33

16/33 nha