B=1/1.3+1/3.5+...+1/2017.2019
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11 tháng 4 2023
=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
11 tháng 4 2023
=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
TT
2
1 tháng 6 2020
\(2.S=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{2017.2019}\)
\(=\frac{3-1}{1.3}+\frac{5-3}{3.5}+\frac{7-5}{5.7}+...+\frac{2019-2017}{2017.2019}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2017}-\frac{1}{2019}\)
\(=1-\frac{1}{2019}=\frac{2018}{2019}\)
=> \(S=\frac{1009}{2019}\)
GE
1 tháng 6 2020
Tính: S= 1/1.3 + 1/3.5 +1/5.7 + 1009/2019 .....+ 1/2017.2019
Trả lời:
1009/2019
8 tháng 6 2020
Cố gắng lên (tự nhủ)
\(S=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2017.2019}\)
\(2S=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2017}-\frac{1}{2019}\)
\(2S=1-\frac{1}{2019}=\frac{2018}{2019}\)
\(S=\frac{1009}{2019}\)
DP
22 tháng 3 2017
\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2017.2019}\)
\(=1-\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+..+\frac{1}{2017}-\frac{1}{2019}\div2\)
\(=\left(1-\frac{1}{2019}\right)\div2\)
\(=\frac{2018}{2019}\div2\)
\(=\frac{1009}{2019}\)
S
22 tháng 3 2017
Đặt \(A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2017.2019}\)
\(2A=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{2017.2019}\)
\(2A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2017}-\frac{1}{2019}\)
\(2A=1-\frac{1}{2017}\)
\(2A=\frac{2016}{2017}\)
\(A=\frac{2016}{2017}:2\)
\(A=\frac{1008}{2017}\)
S
4
8 tháng 8 2021
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
8 tháng 8 2021
\(A=\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+...+\frac{1}{2017\cdot2019}\)
\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2017}-\frac{1}{2019}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{2019}\right)=\frac{1}{2}\cdot\frac{2018}{2019}\)
\(=\frac{1009}{2019}\)
1
8
16 tháng 4 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
18 tháng 12 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 )
HN
18 tháng 12 2019
\(M=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+.....+\frac{1}{2017.2019}\)
\(\Rightarrow M=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-......-\frac{1}{2017}+\frac{1}{2017}-\frac{1}{2019}\)
\(\Rightarrow M=1-\frac{1}{2019}\)
\(\Rightarrow M=\frac{2019}{2019}-\frac{1}{2019}\)
\(\Rightarrow M=\frac{2018}{2019}\)
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 < \(\frac{1}{2}\)
\(B=\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{2017.2019}\)
\(2B=\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{2017.2019}\)
\(2B=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2017}-\frac{1}{2019}\)
\(2B=\frac{1}{1}-\frac{1}{2019}\)
\(2B=\frac{2018}{2019}\)
\(\Rightarrow B=\frac{2018}{2019}:2\Rightarrow B=\frac{1009}{2019}\)
\(B=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+.....+\frac{2}{2017.2019}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+........+\frac{1}{2017}-\frac{1}{2019}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{2019}\right)=\frac{1009}{2019}\)