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\(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+.......+\frac{2}{97\cdot99}\)
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Q
2
28 tháng 6 2016
\(E=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{97.99}\)
\(E=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\)
\(E=\frac{1}{1}-\frac{1}{99}\)
\(E=\frac{98}{99}\)
28 tháng 6 2016
E= \(\frac{2}{1.3}.\frac{2}{3.5}+...+\frac{2}{97.99}\)
E = 1 - \(\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{97}-\frac{1}{99}\)
E = 1 - 1/99
E = 98 / 99
Chúc bạn học tốt
DL
13 tháng 7 2017
\(2S=\frac{2}{1}-\frac{2}{3}+\frac{2}{3}-\frac{2}{5}+...+\frac{2}{97}-\frac{2}{99}\)
\(2S=2-\frac{2}{99}\)
\(2S=\frac{196}{99}\)
\(S=\frac{196}{99}\cdot\frac{1}{2}=\frac{98}{99}\)
13 tháng 7 2017
Ta có: S=2/1.3+2/3.5+...+2/97.99
S= 2/2.(1-1/3+1/3-1/5+...+1/97-1/99)
S= 1-1/99=98/99
TM
20 tháng 6 2016
Đặt \(A=\frac{1}{1.3}+\frac{1}{2.4}+\frac{1}{3.5}+...+\frac{1}{97.99}+\frac{1}{99.100}\)
\(\Rightarrow2A=2\left(\frac{1}{1.3}+\frac{1}{2.4}+\frac{1}{3.5}+...+\frac{1}{97.99}+\frac{1}{98.100}\right)\)
\(\Rightarrow2A=\frac{2}{1.3}+\frac{2}{2.4}+\frac{2}{3.5}+...+\frac{2}{97.99}+\frac{2}{98.100}\)
\(\Rightarrow2A=\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{97.99}\right)+\left(\frac{2}{2.4}+\frac{1}{4.6}+...+\frac{1}{98.100}\right)\)
\(\Rightarrow2A=\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{97}-\frac{1}{99}\right)+\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{98}-\frac{1}{100}\right)\)
\(\Rightarrow2A=\left(1-\frac{1}{99}\right)+\left(\frac{1}{2}-\frac{1}{100}\right)\)
\(\Rightarrow2A=\left(\frac{99}{99}-\frac{1}{99}\right)+\left(\frac{50}{100}-\frac{1}{100}\right)\)
\(\Rightarrow2A=\frac{98}{99}+\frac{49}{100}=\frac{9800}{9900}+\frac{4851}{9900}=\frac{14651}{9900}\)
\(\Rightarrow A=\frac{14651}{9900}:2=\frac{14651}{9900}.\frac{1}{2}=\frac{14651}{19800}\)
bạn nhớ thử lại nhé :)
6 tháng 5 2018
\(\left(1-\frac{1}{3}+...+\frac{1}{97}-\frac{1}{99}\right)-x\)\(=\frac{-100}{99}\)
\(\left(1-\frac{1}{99}\right)-x=\frac{-100}{99}\)
\(\frac{98}{99}-x=\frac{-100}{99}\)
\(x=\frac{98}{99}-\left(-\frac{100}{99}\right)\)
\(x=\frac{198}{99}=2\)
CHÚC BN HOK TỐT!
ĐÚNG THÌ K CHO MK NHA!
4 tháng 4 2018
\(A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{97.99}\)
\(A=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{97.99}\right)\)
\(A=\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{97}-\frac{1}{99}\right)\)
\(A=\frac{1}{2}.\left(1-\frac{1}{99}\right)\)
\(A=\frac{1}{2}.\frac{98}{99}\)
\(A=\frac{49}{99}\)
4 tháng 4 2018
=\(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{97}-\frac{1}{99}\)
=1-\(\frac{1}{99}\)
=\(\frac{98}{99}\)
QD
19 tháng 2 2017
\(\frac{2.6.10+6.10.14+10.14.18+...+194.198.202}{1.3.5+3.5.7+...+97.99.101}\)
\(=\frac{2^3.1.3.5+2^3.3.5.7+2^3.97.99.101}{1.3.5+3.5.7+...+97.99.101}\)
\(=\frac{2^3\left(1.3.5+3.5.7+...+97.99.101\right)}{1.3.5+3.5.7+...+97.99.101}\)
\(=\frac{2^3}{1}=8\)
Vậy A = 8
\(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{97.99}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{97}-\frac{1}{99}\)
\(=1-\frac{1}{99}\)
\(=\frac{98}{99}\)
\(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{97.99}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{97}-\frac{1}{99}\)
\(=1-\frac{1}{99}\)
\(=\frac{98}{99}\)