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

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

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

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

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

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

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

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

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

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

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

=1/5-1/61

=56/305

Đặt 2/5 ra ngoài rồi tách từng cặp phân số ra  sau đó bn tự làm nhé!

A=\(\frac{5}{5.7}+\frac{5}{7.9}+.........+\frac{5}{59.61}\)

=\(\frac{5}{2}.\left(\frac{2}{5.7}+\frac{2}{7.9}+.........+\frac{2}{59.61}\right)\)

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

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

=\(\frac{5}{2}.\frac{56}{305}\)

=\(\frac{28}{61}\)

Nhầm nhé! Đặt 5/2 ra ngoài chứ ko phải 2/5.

`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`

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`

= 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!

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

gọi biểu thức trên là A

gọi biểu thức trên là A. ta có: 

3A = 1/5.7+1/7.9+......+ 1/59.61

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

3A = 1/5 - 1/61

3A = 56/305

A = 56/305 : 3 = 56/915

dell btt

\(M=\frac{3}{5\cdot7}+\frac{3}{7\cdot9}+...+\frac{3}{59\cdot61}\)

\(M=\frac{3}{2}\left[\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+...+\frac{2}{59\cdot61}\right]\)

\(M=\frac{3}{2}\left[\frac{1}{5}-\frac{1}{7}+...+\frac{1}{59}-\frac{1}{61}\right]\)

\(M=\frac{3}{2}\left[\frac{1}{5}-\frac{1}{61}\right]\)

\(M=\frac{3}{2}\cdot\frac{56}{305}=\frac{84}{305}\)

M=\(\frac{3}{5.7}\)+\(\frac{3}{7.9}\)+\(\frac{3}{9.11}\)+......................+\(\frac{3}{59.61}\)

M= 2.\((\frac{3}{5.7}+\frac{3}{7.9}+\frac{3}{9.11}+...........+\frac{3}{59.61})\):2

M=3.\((\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+...........+\frac{2}{59.61})\):2

M=3.\((\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+........+\frac{1}{59}-\frac{1}{61})\):2

M=3,\((\frac{1}{5}-\frac{1}{61})\):2

M=3.\(\frac{56}{305}\):2

M=\(\frac{168}{305}\):2

M=\(\frac{84}{305}\)