Ta có: \(\frac{2}{3\times5}+\frac{2}{5\times7}+\cdots+\frac{2}{49\times51}\)

\(=\frac13-\frac15+\frac15-\frac17+\cdots+\frac{1}{49}-\frac{1}{51}\)

\(=\frac13-\frac{1}{51}=\frac{17}{51}-\frac{1}{51}=\frac{16}{51}\)

\(S.2=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\)

\(S.2=\frac{1}{1}-\frac{1}{11}\)

\(S.2=\frac{10}{11}\)

\(S=\frac{10}{11}:2\)

\(S=\frac{5}{11}\)

S = 5/11

Tính thuận tiện

Ta có : 

\(\frac{2}{1\times3}+\frac{2}{3\times5}+\frac{2}{5\times7}+\frac{2}{7\times9}+\frac{2}{9\times11}\)

= \(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\)

= \(\frac{1}{1}-\frac{1}{11}\)

= \(\frac{10}{11}\)

\(=\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{9.11}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{11}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{11}\right)\)

\(=\frac{1}{2}.\frac{10}{11}\)

\(=\frac{5}{11}\)

\(=\frac{1}{2}\times\left(\frac{2}{1\times3}+\frac{2}{3\times5}+....+\frac{2}{9\times11}\right)\)

\(=\frac{1}{2}\times\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{9}-\frac{1}{11}\right)\)

\(=\frac{1}{2}\times\left(1-\frac{1}{11}\right)\)

\(=\frac{1}{2}\times\frac{10}{11}\)

\(=\frac{5}{11}\)

           p=1/(3*5)+1/(5*7)+.....+1/(2015*2017)+1/(2017*2019)

<=> p = 1/3-1/5+1/5-1/7+1/7-......+1/2017-1/2019

<=> p = 1/3 - 1/2019

<=> p = 224/673

\(P=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{2015.2017}+\frac{1}{2017.2019}\)

\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{2017}-\frac{1}{2019}\right)\)

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

\(=\frac{112}{673}\)

\(\left(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{19.21}\right)x=\frac{9}{7}\)

\(\left[\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{19}-\frac{1}{21}\right)\right]x=\frac{9}{7}\)

\(\left[\frac{1}{2}\left(\frac{1}{3}-\frac{1}{21}\right)\right]x=\frac{9}{7}\)

\(\left(\frac{1}{2}.\frac{2}{7}\right)x=\frac{9}{7}\)

\(\frac{1}{7}.x=\frac{9}{7}\)

\(x=\frac{9}{7}\div\frac{1}{7}\)

\(x=9\)

Vậy ...

\(A=\frac{1}{3}-\frac{1}{17}=\frac{14}{51}\)

cách làm thì tự biết

trên mạng đầy

kết quả đúng phải là 7/51 chứ bn 

mk cần cách trình bày thôi 

 câu trả lời của bn hơi lạnh nhạt tí ^.^

Ta có : \(\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+......+\frac{3}{49.51}\)

\(=\frac{3}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+......+\frac{2}{49.51}\right)\)

\(=\frac{3}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+......+\frac{1}{49}-\frac{1}{50}\right)\)

\(=\frac{3}{2}\left(1-\frac{1}{50}\right)\)

\(=\frac{3}{2}.\frac{49}{50}=\frac{147}{100}\)

Fast,fast,fast.