a,tính \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+......+\frac{1}{19.21}\)
b,CMR A\(=\frac{1}{1.3}+\frac{1}{3.5}+....+\frac{1}{(2n-1).(2n+1)}\le\frac{1}{2}\)
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
DM
7
H
1 tháng 3 2020
Đặt tên bthuc là A
\(A=\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{19.21}\)
\(2A=\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{19.21}\)
\(2A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{19}-\frac{1}{21}\)
\(2A=1-\frac{1}{21}=\frac{20}{21}\)
=>\(A=\frac{20}{21}:2=\frac{10}{21}\)
1 tháng 3 2020
\(=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{17.19}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{17}-\frac{1}{19}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{19}\right)=\frac{1}{2}.\left(\frac{18}{19}\right)\)
\(=\frac{9}{19}\)
TD
3
XO
3 tháng 4 2020
Ta có:\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{19.21}=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{19.21}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{19}-\frac{1}{21}\right)=\frac{1}{2}\left(1-\frac{1}{21}\right)=\frac{1}{2}.\frac{20}{21}=\frac{10}{21}\)
3 tháng 4 2020
\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}\)\(+...+\frac{1}{19.21}\)
=\(\frac{2}{2}\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{19.21}\right)\)
=\(\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+....+\frac{2}{19.21}\right)\)
=\(\frac{1}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{19}-\frac{1}{21}\right)\)
=\(\frac{1}{2}\left(\frac{1}{1}-\frac{1}{21}\right)\)
=\(\frac{1}{2}.\frac{20}{21}\)
=\(\frac{20}{42}=\frac{10}{21}\)
16 tháng 6 2016
\(I=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{\left(2n+1\right).\left(2n+3\right)}\)
\(\Rightarrow I=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{\left(2n+1\right).\left(2n+3\right)}\right)\)
\(\Rightarrow I=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2n+1}-\frac{1}{2n+3}\right)\)
\(\Rightarrow I=\frac{1}{2}\left(1-\frac{1}{2n+3}\right)\)
\(\Rightarrow I=\frac{1}{2}.\frac{2n+2}{2n+3}\)
\(\Rightarrow I=\frac{n+1}{2n+3}\)
16 tháng 6 2016
\(I=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{\left(2n+1\right)\left(2n+3\right)}=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{\left(2n+1\right)\left(2n+3\right)}\right)\)
\(=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2n+1}-\frac{1}{2n+3}=\frac{1}{1}-\frac{1}{2n+3}\)
\(=\frac{2n+3}{2n+3}-\frac{1}{2n+3}=\frac{2n+2}{2n+3}\)
MN
1
26 tháng 12 2019
gọi biểu thức là A
ta có :
A = \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}...\frac{1}{19.21}\)
=> 2A = \(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}...\frac{2}{19.21}\)
2A = \(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...-\frac{1}{21}\)
2A = 1 - \(\frac{1}{21}\)
2A = \(\frac{20}{21}\)
A = \(\frac{20}{21}:2=\frac{10}{21}\)
17 tháng 3 2018
\(\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+...+\frac{1}{19\cdot21}-\frac{x}{14}=\frac{2}{-7}\)
\(\frac{1}{2}\left(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{19\cdot21}\right)-\frac{x}{14}=\frac{2}{-7}\)
\(\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{19}-\frac{1}{21}\right)-\frac{x}{14}=\frac{2}{-7}\)
\(\frac{1}{2}\left(1-\frac{1}{21}\right)-\frac{x}{14}=\frac{2}{-7}\)
\(\frac{1}{2}\cdot\frac{20}{21}-\frac{x}{14}=\frac{2}{-7}\)
\(\frac{10}{21}-\frac{x}{14}=\frac{2}{-7}\)
\(\frac{x}{14}=\frac{10}{21}-\frac{2}{-7}\)
\(\frac{x}{14}=\frac{16}{21}\)
\(\Rightarrow x\cdot=21=14\cdot16\)
\(\Rightarrow x\cdot21=224\)
\(\Rightarrow x=\frac{224}{21}\)
G6
1
15 tháng 7 2018
\(\frac{4}{1.3}+\frac{4}{3.5}+\frac{4}{5.7}+...+\frac{4}{\left(2n-1\right)\left(2n+1\right)}\)
\(=2.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{\left(2n-1\right)\left(2n+1\right)}\right)\)
\(=2.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2n-1}-\frac{1}{2n+1}\right)\)
\(=2.\left(1-\frac{1}{2n+1}\right)\)
\(=2.\left(\frac{2n}{2n+1}\right)\)
\(=\frac{4n}{2n+1}\)
Tham khảo nhé~
tớ làm câu b thôi, câu a nhân 1/2 lên là đc
\(A=\frac{1}{2}.\left[\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{\left(2n-1\right).\left(2n+1\right)}\right)\right]\)
\(A=\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2.n-1}-\frac{1}{2n+1}\right)\)
\(A=\frac{1}{2}.\left(1-\frac{1}{2n+1}\right)=\frac{1}{2}-\frac{1}{2.\left(2n+1\right)}< \frac{1}{2}\)
p/s: lưu ý không có dấu "=" đâu nhé vì \(\frac{1}{2.\left(2n+1\right)}>0\left(n\text{ thuộc }N\right)\)