1/1.3+1/3.5+1/5.7+...+1/97.99
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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
NT
1
V
2
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=\(\dfrac{49}{99}\)
NP
1
VQ
30 tháng 3 2023
A= \(\dfrac{1}{1.3}\)+\(\dfrac{1}{3.5}\)+\(\dfrac{1}{5.7}\)+\(\dfrac{1}{7.9}\)+...+\(\dfrac{1}{97.99}\)
2A= 1 - \(\dfrac{1}{3}\)+\(\dfrac{1}{3}\) - \(\dfrac{1}{5}\)+\(\dfrac{1}{5}\) - \(\dfrac{1}{7}\)+\(\dfrac{1}{7}\) - \(\dfrac{1}{9}\)+...+\(\dfrac{1}{97}\)-\(\dfrac{1}{99}\)
2A= 1-\(\dfrac{1}{99}\)
2A= \(\dfrac{98}{99}\)
A= \(\dfrac{98}{99}\) : 2
A=\(\dfrac{49}{99}\)
NP
1
S
30 tháng 3 2023
\(A=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+...+\dfrac{1}{97.99}\)
\(=\dfrac{1}{2}\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{97.99}\right)\)
\(=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{97}-\dfrac{1}{99}\right)\)
\(=\dfrac{1}{2}\left(1-\dfrac{1}{99}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{98}{99}\)
\(=\dfrac{49}{99}\)
DT
1
K
27 tháng 6 2015
\(\frac{5}{1.2}+\frac{5}{2.3}+...+\frac{5}{99.100}-2x=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{97.99}\)
\(5\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\right)-2x=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{97.99}\right)\)
\(5\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)-2x=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\right)\)\(5\left(1-\frac{1}{100}\right)-2x=\frac{1}{2}\left(1-\frac{1}{99}\right)\)
\(5.\frac{99}{100}-2x=\frac{1}{2}.\frac{98}{99}\)
\(\frac{99}{20}-2x=\frac{49}{99}\)
\(2x=\frac{99}{20}-\frac{49}{99}\)
\(2x=\frac{8821}{1980}\)
\(x=\frac{8821}{1980}:2\)
\(x=\frac{8821}{3960}\)
TN
6
23 tháng 4 2019
\(A=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{97.99}\right)\)
\(A=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{97}-\frac{1}{99}\right)\)
\(A=\frac{1}{2}.\left(1-\frac{1}{99}\right)\)
\(A=\frac{1}{2}.\frac{98}{99}\)
\(A=\frac{49}{99}\)
L
23 tháng 4 2019
\(A=\frac{1}{1\cdot3} +\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+...+\frac{1}{95\cdot97}+\frac{1}{97\cdot99}\)
\(2A=\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{95\cdot97}+\frac{2}{97\cdot99}\)
\(2A=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{95}-\frac{1}{97}+\frac{1}{97}-\frac{1}{99}\)
\(2A=1-\frac{1}{99}\)
\(2A=\frac{98}{99}\)
\(A=\frac{98}{99}\text{ : }2\)
\(A=\frac{98}{99}\cdot\frac{1}{2}\)
\(A=\frac{49}{99}\)
D
6
EC
8 tháng 8 2016
\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{97.99}\)
\(=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\right)\)
\(=\frac{1}{3}.\left(1-\frac{1}{99}\right)\)
\(=\frac{1}{3}.\frac{98}{99}\)
\(=\frac{98}{297}\)
Chuc bn học tốt
8 tháng 8 2016
\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{97.99}\)
\(=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{99}\)
\(=1-\frac{1}{99}\)
\(=\frac{98}{99}\)
\(\frac{1}{1.3}+\frac{1}{3.5}+......+\frac{1}{97.99}\)
\(=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+.........+\frac{2}{97.99}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+.........+\frac{1}{97}-\frac{1}{99}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{99}\right)\)
\(=\frac{1}{2}.\frac{98}{99}=\frac{49}{99}\)
Đặt \(A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{97.99}\)
\(2A=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{97.99}\)
\(2A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\)
\(2A=1-\frac{1}{99}\)
\(A=\frac{98}{99}:2\)
\(A=\frac{49}{99}\)
Ủng hộ mk nha !!! ^_^