Ta có:

\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2017.2018}\)

\(\Rightarrow A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2017}-\frac{1}{2018}\)

\(\Rightarrow A=\frac{1}{1}-\frac{1}{2018}=\frac{2017}{2018}\)

\(B=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2015.2017}\)

\(\Rightarrow B=\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2015}-\frac{1}{2017}\right)\)

\(\Rightarrow B=\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{2017}\right)=\frac{1}{2}.\frac{2016}{2017}\)

\(\Rightarrow B=\frac{1008}{2017}\)

Cố gắng lên (tự nhủ) 

\(S=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2017.2019}\)

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

\(2S=1-\frac{1}{2019}=\frac{2018}{2019}\)

\(S=\frac{1009}{2019}\)

Hi

Hình như =98, bạn thử bấm xem đúng không

Nếu đúng thì thanks mình nhé, mình làm violympic vòng 19 rồi

Đề bài cứ sao sao ý bạn, phân số cuối phải là 1/99.101 chứ !

S= 1/1.3+1/3.5+1/5/7+....+1/99.100

S= 1-1/3+1/3-/1/5+1/5-1/7+.....+1/99-1/100

S= 1 - 1/100 

S = 99/100

1-1/100 , ủng hộ mk nha

=>2S=2/1.3+2/3.5+....+2/99.100

ơ bạn nhầm đề bài à

Bài này lớp 6 học rùi! 

S = 312/25

Bạn có cần giải cặn kẽ ko

\(2.S=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{2017.2019}\)

\(=\frac{3-1}{1.3}+\frac{5-3}{3.5}+\frac{7-5}{5.7}+...+\frac{2019-2017}{2017.2019}\)

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

\(=1-\frac{1}{2019}=\frac{2018}{2019}\)

=> \(S=\frac{1009}{2019}\)

Tính: S= 1/1.3 + 1/3.5 +1/5.7 + 1009/2019 .....+ 1/2017.2019

Trả lời:

1009/2019

\(\frac{1}{1\times3}+\frac{1}{3\times5}+\frac{1}{5\times7}+...+\frac{1}{2001\times2003}+\frac{1}{2003\times2005}=\frac{1}{2}\times\left(\frac{2}{1\times3}+\frac{2}{3\times5}+\frac{2}{5\times7}+...+\frac{2}{2001\times2003}+\frac{2}{2003\times2005}\right)\)

\(=\frac{1}{2}\times\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2001}-\frac{1}{2003}+\frac{1}{2003}-\frac{1}{2005}\right)=\frac{1}{2}\times\left(1-\frac{1}{2005}\right)=\frac{1}{2}\times\frac{2004}{2005}=\frac{1002}{2005}\)

Chúc bạn học tốt

\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2017.2019}\)

\(=1-\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+..+\frac{1}{2017}-\frac{1}{2019}\div2\)

\(=\left(1-\frac{1}{2019}\right)\div2\)

\(=\frac{2018}{2019}\div2\)

\(=\frac{1009}{2019}\)

Đặt \(A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2017.2019}\)

\(2A=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{2017.2019}\)

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

\(2A=1-\frac{1}{2017}\)

\(2A=\frac{2016}{2017}\)

\(A=\frac{2016}{2017}:2\)

\(A=\frac{1008}{2017}\)

CÁC CẬU KẾT BẠN VỚI MÌNH NHA