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

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

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

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

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

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

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

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

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

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

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

=1/5-1/61

=56/305

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

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

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

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

đồ con gà ko biết làm

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!

`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`

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

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

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

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

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

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

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

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

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

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

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

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

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

a) Sửa tí: \(1+\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2006}}\)

Đặt \(A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2006}}\)

\(\Rightarrow2A=2.\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2006}}\right)\)

\(\Rightarrow2A=2+1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2005}}\)

\(\Rightarrow2A-A=2+1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2005}}-\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2006}}\right)\)

\(\Rightarrow A=2+1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2005}}-1-\dfrac{1}{2}-\dfrac{1}{2^2}-\dfrac{1}{2^3}-...-\dfrac{1}{2^{2006}}\)

\(\Rightarrow A=2-\dfrac{1}{2^{2006}}\)

b) Đặt \(A=\dfrac{2}{5.7}+\dfrac{2}{7.9}+\dfrac{2}{9.11}+...+\dfrac{1}{50.61}\)

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

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

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

c) Đặt \(A=\dfrac{7}{3}+\dfrac{7}{15}+\dfrac{7}{35}+...+\dfrac{7}{9999}\)

\(A=\dfrac{7}{2}.2.\left(\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{9999}\right)\)

\(A=\dfrac{7}{2}.\left(1-\dfrac{1}{101}\right)\)

\(A=\dfrac{7}{2}.\dfrac{100}{101}\)

\(A=\dfrac{256}{101}\)