`11/(5.7) + 11/(7.9) + 11/(9.11) + ... + 11/(59.61)`

`= 2.(11/(5.7) + 11/(7.9) + ... + 11/(59.61))`

`= 11.(2/(5.7) + 2/(7.9) + ... + 2/(59.61))`

`= 11.(1/5 - 1/7 + 1/7 - 1/9 + ... +1/59 - 1/61)`

`= 11.(1/5 - 1/61)`

`= 11.56/305`

`= 616/305`

`11/(5.7) + 11/(7.9) + 11/(9.11) + ... + 11/(59.61)`

`= 2.(11/(5.7) + 11/(7.9) + ... + 11/(59.61))`

`= 11.(2/(5.7) + 2/(7.9) + ... + 2/(59.61))`

`= 11.(1/5 - 1/7 + 1/7 - 1/9 + ... +1/59 - 1/61)`

`= 11.(1/5 - 1/61)`

`= 11.56/305`

`= 616/305`

`= 616/305 : 2`

`= 308/305`

=> \(\Rightarrow\left(\frac{11}{5}-\frac{11}{7}+\frac{11}{7}-\frac{11}{9}+...+\frac{11}{59}-\frac{11}{61}\right):2=\left(\frac{11}{5}-\frac{11}{61}\right):2=\frac{616}{305}:2=\frac{308}{305}\)

Đặt \(A=\frac{11}{5.7}+\frac{11}{7.9}+...+\frac{11}{59.61}\)

\(\Rightarrow2A:11=\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{59.61}\)

\(\Rightarrow2A:11=\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{59}-\frac{1}{61}\)

\(\Rightarrow2A:11=\frac{1}{5}-\frac{1}{61}\)

\(\Rightarrow2A:11=\frac{56}{305}\)

\(\Rightarrow2A=\frac{56}{305}.11=\frac{616}{305}\)

\(\Rightarrow A=\frac{616}{305}:2=\frac{308}{305}\)

Vậy kết quả của phép tính trên là \(\frac{308}{305}\)

11/2x(1/5-1/7+1/7+1/9+.....+1/59-1/61)

=11/2x(1/5-1/61)

=11/2x56/305

=308/305

a,Ta có : \(\frac{1}{a}+\frac{-1}{a+1}=\frac{1}{a}-\frac{1}{a+1}\)

=\(\frac{a+1-a}{a\left(a+1\right)}=\frac{1}{a\left(a+1\right)}\)(Đpcm)

b,\(\frac{11}{5.7}+\frac{11}{7.9}+\frac{11}{9.11}+.....+\frac{11}{59.61}\)

=\(\frac{11}{2}.\left(\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+.....+\frac{2}{59.61}\right)\)

=\(\frac{11}{2}.\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+......+\frac{1}{59}-\frac{1}{61}\right)\)

=\(\frac{11}{2}.\left(\frac{1}{5}-\frac{1}{61}\right)=\frac{308}{305}\)

a,Gọi tổng trên là A.

Xét \(\frac{4}{5}-\frac{4}{7}=\frac{8}{35};...;\frac{4}{59}-\frac{4}{61}=\frac{8}{3599}\)=>\(A=\frac{1}{2}.\left(\frac{4}{5}-\frac{4}{7}+\frac{4}{7}-\frac{4}{9}+...+\frac{4}{59}-\frac{4}{61}\right)\)\(=\frac{1}{2}.\left(\frac{4}{5}-\frac{4}{61}\right)=\frac{1}{2}.\frac{224}{305}=\frac{112}{305}\)

b,Gọi tổng trên là B

Theo đề bài ta có:\(B=\frac{24.47-23}{24+47.23}.\frac{3+\frac{3}{7}-\frac{3}{11}+\frac{3}{1001}-\frac{3}{13}}{\frac{9}{1001}-\frac{9}{13}+\frac{9}{7}-\frac{9}{11}+9}\)=\(\frac{\left(23+1\right).47-23}{24+47.23}.\frac{3+\frac{3}{7}-\frac{3}{11}+\frac{3}{1001}-\frac{3}{13}}{\frac{9}{1001}-\frac{9}{13}+\frac{9}{7}-\frac{9}{11}+9}=\frac{47.23+24}{24+47.23}.\frac{3.\left(1+\frac{1}{7}-\frac{1}{11}+\frac{1}{1001}-\frac{1}{13}\right)}{3.\left(3+\frac{3}{1001}-\frac{3}{13}+\frac{3}{7}-\frac{3}{11}\right)}\)\(=\frac{1+\frac{1}{1001}-\frac{1}{13}+\frac{1}{7}-\frac{1}{11}}{3+\frac{3}{1001}-\frac{3}{13}+\frac{3}{7}-\frac{3}{11}}=\frac{1+\frac{1}{1001}-\frac{1}{13}+\frac{1}{7}-\frac{1}{11}}{3.\left(1+\frac{1}{1001}-\frac{1}{13}+\frac{1}{7}-\frac{1}{11}\right)}=\frac{1}{3}\)

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

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

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

Đặt A=B*C

B=\(\frac{24\cdot47-23}{24+47-23}=\frac{1128-23}{71-23}=\frac{1105}{48}\)

C=\(\frac{3\cdot\left(1+\frac{1}{7}-\frac{1}{11}+\frac{1}{1001}-\frac{1}{13}\right)}{9\cdot\left(\frac{1}{1001}-\frac{1}{13}+\frac{1}{7}-\frac{1}{11}+1\right)}=\frac{1}{3}\)

Suy ra A =\(\frac{1105}{144}\)

Giải:

S=3/5.7+3/7.9+...+3/59.61

S=3/2.(2/5.7+2/5.7+...+2/59.61)

S=3/2.(1/5-1/7+1/7-1/9+...+1/59-1/61)

S=3/2.(1/5-1/61)

S=3/2.56/305

S=84/305

Chúc bạn học tốt!

`S=3/(5.7)+3/(7.9)+....+3/(59.61)`

`=>2S=3(2/(5.7)+2/(7.9)+....+2/(59.61))`

`=>2S=3(1/5-1/7+.....+1/59-1/61)`

`=>2S=3(1/5-1/61)=168/305`

`=>S=84/305`

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

= 3(1/5.7+1/7.9+...+1/59.61)

= 3/2(2/5.7+2/7.9+...+2/59.61)

= 3/2(1-1/5+1/5-1/7+1/7-1/9+...+1/59-1/61)

= 3/2(1-1/61)=3/2.60/61=90/61

Chẳng biết mk làm đúng ko nữa!

\(\frac{3}{5.7}+\frac{3}{7.9}+...+\frac{3}{59.61}=\frac{3}{2}\cdot\left(\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{59.61}\right)=\frac{3}{2}\cdot\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{59}-\frac{1}{61}\right)=\frac{3}{2}\cdot\left(\frac{1}{5}-\frac{1}{61}\right)=\frac{84}{305}\)