1/1.3 + 1/3.5+...+ 1/97.99
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NT
2
10 tháng 4 2016
B=1.3+2.4+3.5+...+97.99+98.100B=1.3+2.4+3.5+...+97.99+98.100
B=1(2+1)+2(3+1)+....+97(98+1)+98(99+1)B=1(2+1)+2(3+1)+....+97(98+1)+98(99+1)
B=1.2+1+2.3+2+....+97.98+97+98.99+98B=1.2+1+2.3+2+....+97.98+97+98.99+98
B=(1.2+2.3+3.4+....+97.98+98.99)+(1+2+3+...+98)B=(1.2+2.3+3.4+....+97.98+98.99)+(1+2+3+...+98)
B=98.99.1003+98.992B=98.99.1003+98.992
B=323400+4851=328251B=323400+4851=328251
10 tháng 4 2016
Số đó=1.3 + 2.4 + 3.5 +....+ 98.100
= 1(2+1) + 2.(3+1) + 3.(4+1) +...+ 98(99+1)
= 1.2 + 1 + 2.3 + 2 + 3.4 + 3+....+ 98.99 +98
= (1.2 + 2.3 + 3.4+....98.99) + (1+2+3+....+98)
=323400 + 4851=328251
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
10 tháng 10 2023
Ta viết lại tổng này thành:
\(P=\left(\dfrac{1}{1.3}+\dfrac{1}{3.5}+...+\dfrac{1}{97.99}\right)+\left(\dfrac{1}{2.4}+\dfrac{1}{4.6}+...+\dfrac{1}{98.100}\right)-\dfrac{49}{99}\)
\(P=\dfrac{1}{2}.\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{97.99}\right)+\dfrac{1}{2}\left(\dfrac{2}{2.4}+\dfrac{2}{4.6}+...+\dfrac{2}{98.100}-\dfrac{49}{99}\right)\)
\(P=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{97}-\dfrac{1}{99}\right)+\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{98}-\dfrac{1}{100}\right)-\dfrac{49}{99}\)
\(P=\dfrac{1}{2}\left(1-\dfrac{1}{99}\right)+\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{100}\right)-\dfrac{49}{99}\)
\(P=\dfrac{1}{2}-\dfrac{1}{198}+\dfrac{1}{4}-\dfrac{1}{200}-\dfrac{49}{99}\)
\(P=\dfrac{49}{200}\)
11 tháng 10 2023
\(S=\left(\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+...+\dfrac{1}{97\cdot99}\right)+\left(\dfrac{1}{2\cdot4}+\dfrac{1}{4\cdot6}+...+\dfrac{1}{98\cdot100}\right)-\dfrac{49}{99}\)
\(=\dfrac{1}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{97\cdot99}\right)+\dfrac{1}{2}\left(\dfrac{2}{2\cdot4}+\dfrac{2}{4\cdot6}+...+\dfrac{2}{98\cdot100}\right)-\dfrac{49}{99}\)
\(=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{97}-\dfrac{1}{99}\right)+\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{98}-\dfrac{1}{100}\right)-\dfrac{49}{99}\)
\(=\dfrac{1}{2}\cdot\dfrac{98}{99}+\dfrac{1}{2}\cdot\dfrac{49}{100}-\dfrac{49}{99}\)
\(=\dfrac{49}{200}\)
20 tháng 7 2016
\(\frac{1}{1.3}+\frac{1}{3.5}+......+\frac{1}{97.99}\)
\(=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+.........+\frac{2}{97.99}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+.........+\frac{1}{97}-\frac{1}{99}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{99}\right)\)
\(=\frac{1}{2}.\frac{98}{99}=\frac{49}{99}\)
VA
20 tháng 7 2016
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 !!! ^_^
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
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=\(\dfrac{49}{99}\)
\(A=\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{97.99}\)
\(2A=\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{97.99}\)
\(2A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{97}-\frac{1}{99}\)
\(2A=1-\frac{1}{99}=\frac{98}{99}\)
vậy \(A=\frac{98}{99}:2=\frac{49}{99}\)
chúc bạn học tốt
Ta có : \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+......+\frac{1}{97.99}\)
\(=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+.....+\frac{2}{97.99}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+......+\frac{1}{97}-\frac{1}{99}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{99}\right)\)
\(=\frac{1}{2}.\frac{98}{99}=\frac{49}{99}\)