KT
1
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)
C
6
1 tháng 5 2016
\(\frac{2}{3.5}+\frac{2}{5.7}+.................+\frac{2}{97.99}\)
=\(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+..................+\frac{1}{97}-\frac{1}{99}\)
=\(\frac{1}{3}-\frac{1}{99}\)
=\(\frac{32}{99}\)
1 tháng 5 2016
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+.....+\frac{1}{97}-\frac{1}{99}\)
\(=\frac{1}{3}-\frac{1}{99}\)
\(=\frac{32}{99}\)
9 tháng 10 2018
Bài 1:
\(\frac{4}{3.5}+\frac{4}{5.7}+...+\frac{4}{97.99}\)
\(=2\left(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{97.99}\right)\)
\(=2\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\right)\)
\(=2\left(\frac{1}{3}-\frac{1}{99}\right)\)
\(=2.\frac{32}{99}=\frac{64}{99}\)
Bài 2:
a) \(2.4^x-18=110\)
\(\Leftrightarrow2.4^x=128\)
\(\Leftrightarrow4^x=64\)
\(\Leftrightarrow4^x=4^3\Leftrightarrow x=3\)
Vậy x = 3
b) \(\left(\frac{3}{2}x-1\right)^5=1\)
\(\Leftrightarrow\frac{3}{2}x-1=1\)
\(\Leftrightarrow\frac{3}{2}x=2\)
\(\Leftrightarrow x=\frac{4}{3}\)
Vậy \(x=\frac{4}{3}\)
SN
9 tháng 10 2018
a) 4/3.5 + 3/5.7 + .... + 4/97.99
= 4( 1/3.5 +1/5.7 + ... + 1/97.99 )
= 4 . 1/2 . 2 ( 1/3.5 +1/5.7 + ... + 1/97.99 )
= 4/2 ( 2/3.5 + 2/5.7 + .... + 2/97.99 )
= 2 ( 5-3/3.5 + 7-5/5.7 + ..... + 99-97/97.99 )
= 2 (5/3.5 - 3/3.5 + 7/5.7 - 5/5.7 + .... + 99/97.99 - 97/97.99 )
= 2 ( 1/3 - 1/5 + 1/5 - 1/7 + ..... + 1/97 - 1/99 )
= 2 ( 1/3 -1/99 )
= 2 (33/99 - 1/99 )
= 2 . 32/99
= 32.2/99
=64/99
AM
17 tháng 6 2015
\(\frac{x-2}{3}+\frac{x-2}{3.5}+\frac{x-2}{5.7}+...+\frac{x-2}{97.99}=\frac{-49}{99}\)
<=>\(\left(x-2\right)\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{97.99}\right)=-\frac{49}{99}\)
<=>\(\left(x-2\right)\cdot\frac{1}{2}\cdot\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{97}-\frac{1}{99}\right)=-\frac{49}{99}\)
<=>\(\left(x-2\right)\cdot\frac{1}{2}\cdot\left(1-\frac{1}{99}\right)=-\frac{49}{99}\)
<=>\(\left(x-2\right)\cdot\frac{49}{99}=-\frac{49}{99}\)
<=>x-2=-1
<=>x=1
TT
5
29 tháng 7 2019
2A=2-1/3+1/3-1/5+...+1/97-1/99
2A=2-1/99
2A=197/99
A=197/198
29 tháng 7 2019
\(\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+...+\frac{1}{97\cdot99}\)
\(=\frac{1}{2}\left(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+...+\frac{2}{97\cdot99}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{97}-\frac{1}{99}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{99}\right)\)
\(=\frac{1}{2}\cdot\frac{98}{99}\)
\(=\frac{49}{99}\)
=))
13 tháng 3 2019
\(A=\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+...+\frac{1}{97\cdot99}-\frac{5}{4}\cdot\frac{13}{99}+\frac{5}{99}\cdot\frac{1}{4}\)
\(A=\frac{1}{2}\left(\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{97\cdot99}\right)-\frac{13}{4}\cdot\frac{5}{99}+\frac{5}{99}\cdot\frac{1}{4}\)
\(A=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\right)-\frac{5}{99}\cdot\left(\frac{13}{4}-\frac{1}{4}\right)\)
\(A=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{99}\right)-\frac{5}{99}\cdot3\)
\(A=\frac{1}{2}\cdot\frac{32}{99}-\frac{5}{33}\)
\(A=\frac{16}{99}-\frac{5}{33}=\frac{1}{99}\)
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
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 \(\frac{1}{2.\left(2n+1\right)}>0\left(n\text{ thuộc }N\right)\)
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_