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20 tháng 3 2017

\(\dfrac{3}{5\cdot7}+\dfrac{3}{7\cdot9}+...+\dfrac{3}{59\cdot61}\)

\(=3\left(\dfrac{1}{5\cdot7}+\dfrac{1}{7\cdot9}+...+\dfrac{1}{59\cdot61}\right)\)

\(=\dfrac{3}{2}\left(\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{59}-\dfrac{1}{61}\right)\) \(=\dfrac{3}{2}\left(\dfrac{1}{5}-\dfrac{1}{61}\right)=\dfrac{3}{2}\cdot\dfrac{56}{305}=\dfrac{84}{305}\)

20 tháng 3 2017

A=\(\dfrac{3}{5.7}\)+\(\dfrac{3}{7.9}\)+...+\(\dfrac{3}{59.61}\)

A=\(\dfrac{3}{2}\)(\(\dfrac{2}{5.7}\)+\(\dfrac{2}{7.9}\)+...+\(\dfrac{2}{59.61}\))

A=\(\dfrac{3}{2}\)(\(\dfrac{1}{5}\)-\(\dfrac{1}{7}\)+\(\dfrac{1}{7}\)-\(\dfrac{1}{9}\)+..+\(\dfrac{1}{59}\)-\(\dfrac{1}{61}\))

A=\(\dfrac{3}{2}\)(\(\dfrac{1}{5}\)-\(\dfrac{1}{61}\))

A=\(\dfrac{3}{2}\)(\(\dfrac{61-5}{5.61}\))

A=\(\dfrac{3}{2}\).\(\dfrac{56}{305}\)

A=\(\dfrac{84}{305}\)

leuleuok

14 tháng 3 2017

\(T=\dfrac{3}{5\cdot7}+\dfrac{3}{7\cdot9}+\dfrac{3}{9\cdot11}+...+\dfrac{3}{59\cdot61}\)

\(=\dfrac{3}{2}\cdot\left(\dfrac{2}{5\cdot7}+\dfrac{2}{7\cdot9}+\dfrac{2}{9\cdot11}+...+\dfrac{2}{59\cdot61}\right)\)

\(=\dfrac{3}{2}\cdot\left(\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{59}-\dfrac{1}{61}\right)\)

\(=\dfrac{3}{2}\cdot\left(\dfrac{1}{5}-\dfrac{1}{61}\right)=\dfrac{3}{2}\cdot\dfrac{56}{305}=\dfrac{84}{305}\)

14 tháng 3 2017

\(\dfrac{3}{5.7}+\dfrac{3}{7.9}+\dfrac{3}{9.11}+...+\dfrac{3}{59.61}\)

\(=3.\left(\dfrac{1}{5.7}+\dfrac{1}{7.9}+\dfrac{1}{9.11}+...+\dfrac{1}{59.61}\right)\)

\(=3.\dfrac{1}{2}.\left(\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+....+\dfrac{1}{59}-\dfrac{1}{61}\right)\)

\(=\dfrac{3}{2}.\left(\dfrac{1}{5}-\dfrac{1}{61}\right)\)

\(=\dfrac{3}{2}.\dfrac{56}{305}\)

\(=\dfrac{84}{305}\)

10 tháng 5 2018

A=3/4.(1/5.7+1/7.9+....+1/59.61)

A=3/4.(1/5-1/7+1/7-1/9+...+1/59-1/61)

A=3/4.(1/5-1/61)

A=3/4.56/305

A=42/305

mình làm cho bạn phần A thôi nhé còn phần B mình chưa nghĩ ra cách làm ahihi!

HQ
Hà Quang Minh
Giáo viên
2 tháng 8 2023

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

2 tháng 8 2023

`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 tháng 3 2017

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!!

11 tháng 3 2017

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

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

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

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

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

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

2 tháng 5 2017

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

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

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

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

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

2 tháng 5 2017

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

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

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

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

17 tháng 4 2017

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

=1/5-1/61

=54/115