\(\frac{112}{305}\)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: \(A=\frac{3^2}{2\cdot5}+\frac{3^2}{5\cdot8}+\frac{3^2}{8\cdot11}\)

\(=3\left(\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+\frac{3}{8\cdot11}\right)\)

\(=3\left(\frac12-\frac15+\frac15-\frac18+\frac18-\frac{1}{11}\right)\)

\(=3\left(\frac12-\frac{1}{11}\right)=3\cdot\frac{9}{22}=\frac{27}{22}\) >1

Ta có: \(B=\frac{4}{5\cdot7}+\frac{4}{7\cdot9}+\cdots+\frac{4}{59\cdot61}\)

\(=2\left(\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+\cdots+\frac{2}{59\cdot61}\right)\)

\(=2\left(\frac15-\frac17+\frac17-\frac19+\cdots+\frac{1}{59}-\frac{1}{61}\right)\)

\(=2\left(\frac15-\frac{1}{61}\right)=2\cdot\frac{61-5}{305}=2\cdot\frac{56}{305}=\frac{112}{305}<1\)

Ta có: A>1

B<1

Do đó: A>B

Ta có: \(A=\frac{3^2}{2\cdot5}+\frac{3^2}{5\cdot8}+\frac{3^2}{8\cdot11}\)

\(=3\left(\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+\frac{3}{8\cdot11}\right)\)

\(=3\left(\frac12-\frac15+\frac15-\frac18+\frac18-\frac{1}{11}\right)=3\left(\frac12-\frac{1}{11}\right)=3\cdot\frac{9}{22}=\frac{27}{22}>1\)

TA có: \(B=\frac{4}{5\cdot7}+\frac{4}{7\cdot9}+\cdots+\frac{4}{59\cdot61}\)

\(=2\left(\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+\cdots+\frac{2}{59\cdot61}\right)\)

\(=2\left(\frac15-\frac17+\frac17-\frac19+\cdots+\frac{1}{59}-\frac{1}{61}\right)\)

\(=2\left(\frac15-\frac{1}{61}\right)=2\cdot\frac{56}{305}=\frac{112}{305}<1\)

Ta có: B<1

1<A

Do đó: B<A

Ta có:

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

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

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

\(\dfrac{A}{2}=\dfrac{1}{6}-\dfrac{1}{61}\)

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

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

Em nhớ nhân 1/2 trong tất cả dấu bằng thì biểu thức này mới không thay đổi kết quả nhé.

`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`

\(A=\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{59.61}\)

\(A=\dfrac{5-3}{3.5}+\dfrac{7-5}{5.7}+\dfrac{9-7}{7.9}+...+\dfrac{61-59}{59.61}\)

\(A=\dfrac{5}{3.5}-\dfrac{3}{3.5}+\dfrac{7}{5.7}-\dfrac{5}{5.7}+\dfrac{9}{7.9}-\dfrac{7}{7.9}+...+\dfrac{61}{59.61}-\dfrac{59}{59.61}\)

\(A=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{59}-\dfrac{1}{61}\)

\(A=\dfrac{1}{3}-\dfrac{1}{61}=\dfrac{61}{183}-\dfrac{3}{183}=\dfrac{58}{183}\)

\(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{59.61}\)

= \(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{59}-\dfrac{1}{61}\)

= \(\dfrac{1}{3}-\dfrac{1}{61}\)

= \(\dfrac{58}{183}\)

b,=1/5-1/7+1/7-1/9+...+1/59-1/61

=1/5-1/61

=54/115

Theo quy luật thì mình nghĩ đáng lẽ \(\dfrac{4}{5.9}\)phải là\(\dfrac{4}{7.9}\)Bạn có chép sai đề ko?

A=1-\(\dfrac{4}{5.7}-\dfrac{4}{7.9}-\dfrac{4}{9.11}...-\dfrac{4}{59.61}\)

A=\(1-\left(\dfrac{4}{5.7}+\dfrac{4}{7.9}+\dfrac{4}{9.11}+...+\dfrac{4}{59.61}\right)\)

Đặt B=​\(\dfrac{4}{5.7}+\dfrac{4}{7.9}+\dfrac{4}{9.11}+...+\dfrac{4}{59.61}\)​