TT
5
K
Khách
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3
30 tháng 10 2018
\(S=\frac{5}{1.3}+\frac{5}{3.5}+\frac{5}{5.7}+..+\frac{5}{97.99}\)
\(=\frac{5}{2}.\left(5+\frac{5}{3}+\frac{5}{5}+\frac{5}{7}+...+\frac{5}{97}+\frac{5}{99}\right)\)
\(=\frac{5}{2}.\left(5+\frac{5}{99}\right)\)
\(=\frac{5}{2}.\frac{500}{99}\)
\(=\frac{1250}{99}\)(có gì sai sót xin bỏ qua cho T^T)
S
20 tháng 6 2018
\(=\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{97.99}\right)-\left(\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{98.100}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{97}-\frac{1}{99}\right)-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{98}-\frac{1}{100}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{99}\right)-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{100}\right)\)
\(=\frac{1}{2}\cdot\frac{98}{99}-\frac{1}{2}\cdot\frac{49}{100}\)
\(=\frac{1}{2}\left(\frac{98}{99}-\frac{49}{100}\right)=\frac{1}{2}\cdot\frac{4949}{9900}=\frac{4949}{19800}\)
KT
1
2 tháng 1 2022
A = \(\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{97.99}\)
2A = 2 . \(\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{97.99}\)
2A = \(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{97.99}\)
2A = \(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\)
2A = \(\frac{1}{3}-\frac{1}{99}\)
2A = \(\frac{32}{99}\)
A = \(\frac{32}{99}\div2\)
A =\(\frac{16}{99}\)
_HT_
N
11 tháng 12 2018
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 \(\frac{1}{2.\left(2n+1\right)}>0\left(n\text{ thuộc }N\right)\)
HT
8 tháng 8 2015
\(\frac{1}{3.1}-\frac{1}{2.4}+\frac{1}{3.5}-\frac{1}{4.6}+...+\frac{1}{97.99}-\frac{1}{98.100}\)
= \(\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{97.99}\right)-\left(\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{98.100}\right)\)
= \(\frac{1}{2}\left(1-\frac{1}{3}+...+\frac{1}{97}-\frac{1}{99}\right)-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{100}\right)\)
= \(\frac{1}{2}\left(1-\frac{1}{99}\right)-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{100}\right)=\frac{1}{2}.\frac{98}{99}-\frac{1}{2}.\frac{49}{100}\)
= \(\frac{49}{99}-\frac{49}{200}\)
= \(\frac{4949}{19800}\)
8 tháng 8 2015
bn zô xem nha, ko hiểu thì cứ hỏi bn ấy nhá
http://olm.vn/hoi-dap/question/154321.html
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
\(=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{17.19}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{17}-\frac{1}{19}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{19}\right)=\frac{1}{2}.\left(\frac{18}{19}\right)\)
\(=\frac{9}{19}\)
CR
2
S
14 tháng 1 2018
\(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{47.49}=\frac{1}{x}\)
\(\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{47.49}\right)=\frac{1}{x}\)
\(\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{37}-\frac{1}{49}\right)=\frac{1}{x}\)
\(\frac{1}{2}\left(1-\frac{1}{49}\right)=\frac{1}{x}\)
\(\frac{1}{2}\cdot\frac{48}{49}=\frac{1}{x}\)
\(\frac{1}{x}=\frac{24}{49}\)
=>x=49/24
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 = \(\frac{20}{21}:2=\frac{10}{21}\)
2A=2-1/3+1/3-1/5+...+1/97-1/99
2A=2-1/99
2A=197/99
A=197/198
\(\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+...+\frac{1}{97\cdot99}\)
\(=\frac{1}{2}\left(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+...+\frac{2}{97\cdot99}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{97}-\frac{1}{99}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{99}\right)\)
\(=\frac{1}{2}\cdot\frac{98}{99}\)
\(=\frac{49}{99}\)
=))