4/5.7+4/7.9+...+4/59.61

=2.(2/5.7+2/7.9+...+2/59.61)

=2.(1/5-1/7+1/7-1/9+...+1/59-1/61)

=2.(1/5-1/61)

=2.56/305

=112/203

A=( 2/5.7+2/7.9+.........+2/59.61).2

 A = (1/5-1/7+1/7-1/9+.......+1/59-1/61).2

A= (  1/5-1/61)2

B=4/5.7+4/7.9+4/9.11+...+4/59.61

Đặt : A = \(\frac{4}{5.7}+\frac{4}{7.9}+...+\frac{4}{59.61}\)

      A = \(2.\)\(\left(\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{59.61}\right)\)

      A = 2 . ( \(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{59}-\frac{1}{61}\)) 

      A = 2 . \(\left(\frac{1}{5}-\frac{1}{61}\right)\)

      A = 2 . \(\frac{56}{305}\)= \(\frac{112}{305}\)  

Đặt A=như đã cho.

=>1/2A=2/5*7+2/7*9+2/9*11+...+2/59*61.

=>1/2A=1/5-1/7+1/7-1/9+1/9-1/11+...+1/59-1/61.

=>1/2A=1/5-1/61=56/305.

=>A=56/305*2=112/305.

k nha đúng đó.Có j kb nha.

pn kia nhầm zùi

Bài này dễ thôi mà bạn

\(\frac{112}{305}\)nha 

Đặt A=\(\frac{4}{5.7}\)+\(\frac{4}{7.9}\)+...+\(\frac{4}{59.61}\)

      A=2( \(\frac{2}{5.7}\)+\(\frac{2}{7.9}\)+...+\(\frac{2}{59.61}\))

       A=2( \(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\)\(\frac{1}{59}-\frac{1}{61}\))

         =2( \(\frac{1}{5}-\frac{1}{61}\))=2.\(\frac{56}{305}\)=\(\frac{112}{305}\)

Ta có:

\(S=\frac{4}{5.7}+\frac{4}{7.9}+\frac{4}{9.11}+...+\frac{4}{59.61}\)

\(=2.\left(\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{59.61}\right)\)

\(=2.\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{59}-\frac{1}{61}\right)\)

\(=2.\left(\frac{1}{5}-\frac{1}{61}\right)\)

\(=2.\left(\frac{61}{305}-\frac{5}{305}\right)\)

\(=2.\frac{56}{305}\)

\(=\frac{112}{305}\)

Vậy \(S=\frac{112}{305}\)

Ta có :

\(A=\dfrac{4}{5.7}+\dfrac{4}{7.9}+............+\dfrac{4}{59.61}\)

\(\dfrac{A}{2}=\dfrac{2}{5.7}+\dfrac{2}{7.9}+..............+\dfrac{2}{59.61}\)

\(\dfrac{A}{2}=\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+.......+\dfrac{1}{59}-\dfrac{1}{61}\)

\(\dfrac{A}{2}=\dfrac{1}{5}-\dfrac{1}{61}\)

\(\dfrac{A}{2}=\dfrac{56}{305}\)

\(\Rightarrow A=\dfrac{112}{305}\)

Chúc bn học tốt!!

\(A=\dfrac{4}{5.7}+\dfrac{4}{7.9}+...+\dfrac{4}{59.61}\)

\(A=2\left(\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{59.61}\right)\)

\(A=2\left(\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{59}-\dfrac{1}{61}\right)\)

\(A=2\left(\dfrac{1}{5}-\dfrac{1}{61}\right)\)

\(A=2.\dfrac{56}{305}\)

\(A=\dfrac{112}{305}\)

\(A=\dfrac{4}{5.7}+\dfrac{4}{7.9}..........+\dfrac{4}{59.61}\)

\(\dfrac{1}{2}A=\dfrac{2}{5.7}+\dfrac{2}{7.9}+.........+\dfrac{2}{59.61}\)

\(\dfrac{1}{2}A=\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+.....+\dfrac{1}{59}-\dfrac{1}{61}\)

\(\dfrac{1}{2}A=\dfrac{1}{5}-\dfrac{1}{61}\)

\(\dfrac{1}{2}A=\dfrac{56}{305}\)

\(A=\dfrac{112}{305}\)

4/5.7+4/7.9+...+4/59.61

= 2. (2/5.7+2/7.9+...+2/59.61)

= 2. (1/5-1/7+1/7-1/9+...+1/59-1/61)

= 2. (1/5-1/61)

= 2. 56/305

= 112/305

Ta có 1/2B=2/5.7+2/7.9+...+2/59.61

1/2B=1/5-1/7+1/7-1/9+1/9-...+1/59-1/61

1/2B=1/5-1/61

1/2B=56/305

B=56/305:1/2

B=112/305