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

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

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

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

=> \(S=\frac{1009}{2019}\)

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

Trả lời:

1009/2019

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

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

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

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

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

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

=1/2(2/1*3+2/3*5+...+2/2017*2019)

=1/2(1-1/3+1/3-1/5+...+1/2017-1/2019)

=1/2*2018/2019

=1009/2019

=1/2(2/1x3+2/3x5+...+2/2017x2019)

=1/2(1-1/3+1/3-1/5+...+1/2017-1/2019)

=1/2x2018/2019

=1008/2019

A = 1/1.3 + 1/3.5 + 1/5.7 + ... + 1/2017.2019

A = 1/2 (1 - 1/3 + 1/3 - 1/5 + 1/5 - ... - 1/2019)

A = 1/2 (1 - 1/2019)

A = 1/2 . 2018/2019

A = 1009/2019

@Cỏ

\(A=\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+...+\frac{1}{2017\cdot2019}\)

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

\(=\frac{1}{2}\left(1-\frac{1}{2019}\right)=\frac{1}{2}\cdot\frac{2018}{2019}\)

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

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

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

\(=\frac{1}{2}\left(1-\frac{1}{2019}\right)=\frac{1}{2}.\frac{2018}{2019}=\frac{1009}{2019}\)

A=1/1*3+1/3*5+...+1/2017*2019

2A=2/1*3+2/3*5+...+2/2017*2019

2A=1-1/3+1/3-1/5+..+1/2017-1/2019

2A=1-1/2019

2A=2018/2019

A=(2018/2019):2

A=1009/2019

A=1009/2019

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

\(A=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{2017.2019}\right)\)

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

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

\(A=\frac{1}{2}.\frac{2018}{2019}\)

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

co ban nao ra chua de minh do ke qua coi dung ko?

\(S=\frac{1}{1.2}+\frac{1}{3.4}+.........+\frac{1}{199.200}\)

Bài 2:

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

\(=\frac{1}{2}\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{2017.2019}\right)\) 

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

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

\(=\frac{1}{2}\cdot\frac{2018}{2019}=\frac{1009}{2019}\)

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

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

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

\(=\frac{1}{2}.\frac{2018}{2019}\)

\(=\frac{2018}{4038}\)

\(\Rightarrow\frac{2018}{4038}< \frac{1}{2}\)( lấy máy tính ) 

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

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

\(\Rightarrow M=1-\frac{1}{2019}\)

\(\Rightarrow M=\frac{2019}{2019}-\frac{1}{2019}\)

\(\Rightarrow M=\frac{2018}{2019}\)

Có \(\frac{2018}{2019}=\frac{2018.2}{2019.2}=\frac{4036}{4038}\)

\(\frac{1}{2}=\frac{1.2019}{2.2019}=\frac{2019}{4038}\)

Mà \(\frac{4036}{4038}< \frac{2019}{4038}\Rightarrow M< \frac{1}{2}\)

Vậy M < \(\frac{1}{2}\)

a) + \(\frac{2}{n\left(n+2\right)}=\frac{\left(n+2\right)}{n\left(n+2\right)}-\frac{n}{n\left(n+2\right)}=\frac{1}{n}-\frac{1}{n+2}\)

Do đó :

+ \(A=\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+...+\frac{2}{2017\cdot2019}\)

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

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

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