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TT
1
29 tháng 3 2018
1/1x3 + 1/3x5 + 1/5x7 + ...............................+ 1/97x99
=1-1/3 + 1/3 - 1/5 + 1/5 - 1/7 +.............................+ 1/97-1/99
=1-1/99
=98/99
TT
1
3 tháng 8 2015
=1-1/3-1/2+1/4+1/3-1/5-1/4+1/6+...+1/97-1/99-1/98+1/100
=1-1/2-1/99-1/98=2327/4851
NT
0
MR
4
1 tháng 8 2016
\(\frac{1}{1.3}-\frac{1}{2.4}+\frac{1}{3.5}-\frac{1}{4.6}+...+\frac{1}{97.99}-\frac{1}{98.100}\)
\(=1-\frac{1}{3}-\frac{1}{2}+\frac{1}{4}+\frac{1}{3}-\frac{1}{5}-\frac{1}{4}+\frac{1}{6}+...+\frac{1}{97}-\frac{1}{99}-\frac{1}{98}+\frac{1}{100}\)
\(=1-\frac{1}{2}-\frac{1}{99}-\frac{1}{98}\)
\(=\frac{2327}{4851}\)
1 tháng 8 2016
Đặt A=1/1.3 - 1/2.4 +1/3.5 -1/4.6 +.....+1/97.99 -1/98.100
4A= 4/1.3 -4/2.4 +4/3.5 -4/4.6 +.....+4/97.99 -4/98.100
=(4/1.3 +4/3.5 +...+4/97.99) - (4/2.4 +4/4.6 +...+4/98.100)
=(1/1 -1/3+1/3-1/5+...+1/97-1/99)-(1/2 -1/4 -....1/98-1/100)
=(1/1-1/99)-(1/2-1/100)
4A=98/99 - 99/100
A= (98/99-99/100) :4
NB
0
ML
1
11 tháng 10 2023
\(S=\left(\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+...+\dfrac{1}{97\cdot99}\right)+\left(\dfrac{1}{2\cdot4}+\dfrac{1}{4\cdot6}+...+\dfrac{1}{98\cdot100}\right)-\dfrac{49}{99}\)
\(=\dfrac{1}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{97\cdot99}\right)+\dfrac{1}{2}\left(\dfrac{2}{2\cdot4}+\dfrac{2}{4\cdot6}+...+\dfrac{2}{98\cdot100}\right)-\dfrac{49}{99}\)
\(=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{97}-\dfrac{1}{99}\right)+\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{98}-\dfrac{1}{100}\right)-\dfrac{49}{99}\)
\(=\dfrac{1}{2}\cdot\dfrac{98}{99}+\dfrac{1}{2}\cdot\dfrac{49}{100}-\dfrac{49}{99}\)
\(=\dfrac{49}{200}\)
VV
1
5 tháng 8 2016
=>S<1/1.2+1/2.3+1/3.4+...+1/98.99+1/99.100=1-1/2+1/2-1/3+1/3-1/4+...+1/99-1/100
=1-(1/2-1/2)-(1/3-1/3)-...-(1/99-1/99)-1/100
=1-1/100 <1
=>S<1
Vậy S<1
NK
19 tháng 3 2023
\(B=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{97.99}\)
\(B=\dfrac{1}{1}\cdot\dfrac{1}{3}+\dfrac{1}{3}\cdot\dfrac{1}{5}+\dfrac{1}{5}\cdot\dfrac{1}{7}+...+\dfrac{1}{97}\cdot\dfrac{1}{99}\)
\(B=\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{99}\)
\(B=\dfrac{1}{1}-\dfrac{1}{99}\)
\(B=\dfrac{99}{99}-\dfrac{1}{99}\)
\(B=\dfrac{98}{99}\)
#YVA
H1
22 tháng 3 2023
B=\(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{97.99}\)
B=\(\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{97.99}\right):2\)
B=\(\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{97}-\dfrac{1}{99}\right):2\)
B=\(\left(\dfrac{1}{1}-\dfrac{1}{99}\right):2\)
B=\(\dfrac{98}{99}:2\)
B=\(\dfrac{49}{99}\)
\(A=\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{97.99}\)
\(2A=\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{97.99}\)
\(2A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{97}-\frac{1}{99}\)
\(2A=1-\frac{1}{99}=\frac{98}{99}\)
vậy \(A=\frac{98}{99}:2=\frac{49}{99}\)
chúc bạn học tốt
Ta có : \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+......+\frac{1}{97.99}\)
\(=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+.....+\frac{2}{97.99}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+......+\frac{1}{97}-\frac{1}{99}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{99}\right)\)
\(=\frac{1}{2}.\frac{98}{99}=\frac{49}{99}\)