M = 3/1x3 + 3/3x5 + 3/5x7 + ... + 3/45x47 + 3/47x49

M = 3/2 x (2/1x3 + 2/3x5 + 2/5x7 + ... + 2/45x47 + 2/47x49)

M = 3/2 x (1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 + ... + 1/45 - 1/47 + 1/47 - 1/49)

M = 3/2 x (1 - 1/49)

M = 3/2 x 48/49

M = 72/49

N tính tương tự, nhân N với 5/4 

\(\frac{3}{1x3}+\frac{3}{3x5}+...+\frac{3}{49x51}=\frac{3}{2}\left(\frac{2}{1x3}+\frac{2}{3x5}+...+\frac{2}{49x51}\right)=\frac{3}{2}\left(\frac{1}{1}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{51}\right)\)

\(=\frac{3}{2}.\frac{50}{51}=\frac{25}{17}\)

\(\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+\frac{3}{2017.2019}\)

\(=\frac{3}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2017}-\frac{1}{2019}\right)\)

\(=\frac{3}{2}.\left(1-\frac{1}{2019}\right)\)

\(=\frac{3}{2}.\frac{2018}{2019}\)

\(=\frac{1009}{673}\)

\(\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}.....+\frac{3}{2017.2019}\)

\(=\frac{3}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+......+\frac{2}{2017.2019}\right)\)

\(=\frac{3}{2}\left(1-\frac{1}{3}+....+\frac{1}{2017}-\frac{1}{2019}\right)\)

\(=\frac{3}{2}\left(1-\frac{1}{2019}\right)\)

\(=\frac{3}{2}.\frac{2018}{2019}=\frac{1009}{673}\)

\(\frac{3}{1\times3}+\frac{3}{3\times5}+.......+\frac{3}{2017\times2019}\)

\(=\frac{3}{2}\times\left(\frac{2}{1\times3}+\frac{2}{3\times5}+......+\frac{2}{2017\times2019}\right)\)

\(=\frac{3}{2}\times\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+.......+\frac{1}{2017}-\frac{1}{2019}\right)\)

\(=\frac{3}{2}\times\left(1-\frac{1}{2019}\right)\)

\(=\frac{3}{2}\times\frac{2018}{2019}\)

\(=\frac{1009}{673}\)

`2/(1xx3)+2/(3xx5)+2/(5xx7)+...+2/(99xx101)` đề phải ntn chứ mà nhỉ

`=1/1-1/3+1/3-1/5+1/5-1/7+...+1/99-1/101`

`=1/1-1/101`

`=101/101-1/101`

`=100/101`

(Sửa phần 3 / 3 x 5 = 2 / 3 x 5)

\(\dfrac{2}{1\times3}+\dfrac{2}{3\times5}+\dfrac{2}{5\times7}+...+\dfrac{2}{99\times101}\)

Ta có: \(=2\times\left(\dfrac{1}{1\times3}+\dfrac{1}{3\times5}+\dfrac{1}{5\times7}+...+\dfrac{1}{99\times101}\right)\)

\(=2\times\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\)

\(=2\times\left(1-\dfrac{1}{101}\right)\)

\(=2\times\dfrac{100}{101}\)

\(=\dfrac{200}{101}\)

Sửa bài ( dòng 3 đến hết bài )

... = \(2\times\dfrac{1}{2}\times\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\)

\(=1-\dfrac{1}{101}\)

\(=\dfrac{100}{101}\)

=1/14 nhé

\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}\)

\(=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}\right)\)

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

\(=\frac{1}{2}.\left(1-\frac{1}{9}\right)\)

\(=\frac{1}{2}.\frac{8}{9}\)

\(=\frac{4}{9}\)

#)Giải :

\(S=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}\)

\(\Rightarrow2S=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}\)

\(\Rightarrow2S=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}\)

\(\Rightarrow2S=1-\frac{1}{9}=\frac{8}{9}\)

\(S=\frac{8}{9}:2=\frac{4}{9}\)

             #~Will~be~Pens~#

S=\(\frac{1}{1×3}+\frac{1}{3×5}+\frac{1}{5×7}+\frac{1}{7×9}\)

S=\(\frac{1.2}{\left(1.3\right).2}+\frac{1.2}{\left(3.5\right).2}+\frac{1.2}{\left(5.7\right).2}+\frac{1.2}{\left(7.9\right).2}\)

S=\(\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}\right)\)

S= \(\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}\right)\)

S= \(\frac{1}{2}.\left(1-\frac{1}{9}\right)\)

S= 1/2 . 8/9

S = 4/9

Lưu ý : "." có nghĩa là dấu nhân , lên cấp 2 bạn sẽ được học  

\(\dfrac{2}{1\times3}+\dfrac{2}{3\times5}+\dfrac{2}{5\times7}+...+\dfrac{2}{13\times15}+\dfrac{2}{15\times17}\)

\(=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{13}-\dfrac{1}{15}+\dfrac{1}{15}-\dfrac{1}{17}\)

\(=1-\dfrac{1}{17}\)

\(=\dfrac{16}{17}\)

\(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{15\cdot17}\)

\(=2-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{15}-\dfrac{1}{17}\)

\(=2-\dfrac{1}{17}\)

\(=\dfrac{33}{17}\)