Tính \(\dfrac{1}{100\times99}-\dfrac{1}{99\times98}-\dfrac{1}{98\times97}-...-\dfrac{1}{3\times2}-\dfrac{1}{2\times1}\)
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7 tháng 7 2017
C= 1/100-(1/1.2+1/2.3+...+1/97.98+1/98.99+1/99.100)
C=1/100-(1-1/2+1/2-1/3+...+1/97-1/98+1/98-1/99+1/99-1/100)
C=1/100-(1-1/100)
C=1/100-99/100
C=-98/100=-49/50
7 tháng 7 2017
\(C=\dfrac{1}{100}-\dfrac{1}{100.99}-\dfrac{1}{99.98}-\dfrac{1}{98.97}-...\dfrac{1}{3.2}-\dfrac{1}{2.1}\)
\(=-\left(\dfrac{1}{100.99}+\dfrac{1}{99.98}+\dfrac{1}{98.97}+...+\dfrac{1}{3.2}+\dfrac{1}{2.1}\right)+\dfrac{1}{100}\)
\(=-\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{97.98}+\dfrac{1}{98.99}+\dfrac{1}{99.100}\right)+\dfrac{1}{100}\)
\(=-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{97}-\dfrac{1}{98}+\dfrac{1}{98}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{100}\right)+\dfrac{1}{100}\)
\(=-\left(1-\dfrac{1}{100}\right)+\dfrac{1}{100}\)
\(=\left(-1\right)+\dfrac{1}{50}=-\dfrac{49}{50}\)
31 tháng 3 2021
Ta có: \(A=\dfrac{1+\dfrac{1}{3}+\dfrac{1}{5}+\dfrac{1}{7}+...+\dfrac{1}{97}+\dfrac{1}{99}}{\dfrac{1}{1\cdot99}+\dfrac{1}{3\cdot97}+\dfrac{1}{5\cdot95}+...+\dfrac{1}{97\cdot3}+\dfrac{1}{99\cdot1}}\)
\(\Leftrightarrow\dfrac{A}{100}=\dfrac{1+\dfrac{1}{3}+\dfrac{1}{5}+\dfrac{1}{7}+...+\dfrac{1}{97}+\dfrac{1}{99}}{\dfrac{100}{1\cdot99}+\dfrac{100}{3\cdot97}+\dfrac{100}{5\cdot95}+...+\dfrac{100}{97\cdot3}+\dfrac{100}{99\cdot1}}\)
\(\Leftrightarrow\dfrac{A}{100}=\dfrac{1+\dfrac{1}{3}+\dfrac{1}{5}+\dfrac{1}{7}+...+\dfrac{1}{97}+\dfrac{1}{99}}{1+\dfrac{1}{99}+\dfrac{1}{3}+\dfrac{1}{97}+\dfrac{1}{5}+\dfrac{1}{95}+...+\dfrac{1}{97}+\dfrac{1}{3}+\dfrac{1}{99}+1}\)
\(\Leftrightarrow\dfrac{A}{100}=\dfrac{1+\dfrac{1}{3}+\dfrac{1}{5}+\dfrac{1}{7}+...+\dfrac{1}{97}+\dfrac{1}{99}}{2\left(1+\dfrac{1}{3}+\dfrac{1}{5}+\dfrac{1}{7}+...+\dfrac{1}{97}+\dfrac{1}{99}\right)}\)
\(\Leftrightarrow\dfrac{A}{100}=\dfrac{1}{2}\)
hay A=50
AD
23 tháng 7 2023
\(1,\\ =\dfrac{2-1}{1\times2}+\dfrac{3-2}{2\times3}+\dfrac{4-3}{3\times4}+\dfrac{5-4}{4\times5}+.....+\dfrac{99-98}{98\times99}+\dfrac{100-99}{99\times100}\\ =1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+....+\dfrac{1}{98}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{100}\\ =1-\dfrac{1}{100}=\dfrac{100-1}{100}=\dfrac{99}{100}\)
\(2,=\dfrac{13-11}{11\times13}+\dfrac{15-13}{13\times15}+....+\dfrac{21-19}{19\times21}+\dfrac{23-21}{21\times23}\\ =\dfrac{1}{11}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{15}+....+\dfrac{1}{19}-\dfrac{1}{21}+\dfrac{1}{21}-\dfrac{1}{23}\\ =\dfrac{1}{11}-\dfrac{1}{23}\\ =\dfrac{23-11}{11\times23}=\dfrac{12}{253}\)
@seven
23 tháng 7 2023
a: 1/1*2+1/2*3+...+1/99*100
=1-1/2+1/2-1/3+...+1/99-1/100
=1-1/100
=99/100
b: 2/11*13+2/13*15+...+2/21*23
=1/11-1/13+1/13-1/15+...+1/21-1/23
=1/11-1/23
=12/253
H
17 tháng 9 2023
`4 1/5 xx 2 1/4`
`= 21/5 xx 9/4`
`= 189/20`
__
`4 1/5 : 2 1/4`
`= 21/5 : 9/4`
`= 21/5 xx 4/9`
`=84/45`
`=28/15`
__
`3 3/5 xx 1 2/3`
`= 18/5 xx 5/3`
`= 90/15`
`=6`
__
`3 3/5 : 1 2/3`
`= 18/5 : 5/3`
`= 18/5 xx 3/5`
`=54/25`
17 tháng 9 2023
\(4\dfrac{1}{5}\times2\dfrac{1}{4}\\ =\dfrac{21}{5}\times\dfrac{9}{4}\\ =\dfrac{21\times9}{5\times4}\\ =\dfrac{189}{20}\)
\(3\dfrac{3}{5}\times1\dfrac{2}{3}\\ =\dfrac{18}{5}\times\dfrac{5}{3}\\ =\dfrac{18\times5}{5\times3}\\ =\dfrac{90}{15}\\ =6\)
\(4\dfrac{1}{5}:2\dfrac{1}{4}\\ =\dfrac{21}{5}:\dfrac{9}{4}\\ =\dfrac{21}{5}\times\dfrac{4}{9}\\ =\dfrac{21\times4}{5\times9}\\ =\dfrac{84}{45}\\ =\dfrac{28}{15}\)
\(3\dfrac{3}{5}:1\dfrac{2}{3}\\ =\dfrac{18}{5}:\dfrac{5}{3}\\ =\dfrac{18}{5}\times\dfrac{3}{5}\\ =\dfrac{18\times3}{5\times5}\\ =\dfrac{54}{25}\)
NT
26 tháng 3 2018
\(\frac{1+\left(1+2\right)+\left(1+2+3\right)+.....+\left(1+2+3+4+......+100\right)}{\left(1.100+2.99+3.98+.......+99.2+100.1\right).2013}\)
\(=\frac{1.100+2.99+3.98+......+99.2+100.1}{\left(1.100+2.99+3.98+.....+99.2+100.1\right).2013}\)
\(=\frac{1}{2013}\)
16 tháng 5 2022
=4/3 x 9/8 x 16/15 x 25/24x.....x100/99 =2x2x3x3x4x4x5x5x.....x10x10/1x3x2x4x3x5x4x6x....x9x11
=(2x3x4x5x....x10 ) x (2x3x4x5x...x10) / (1x2x3x4x....x9 ) x (3x4x5x...x11)
=10x2/11 =20/11
2 tháng 7 2021
Ta có: \(M=\dfrac{\dfrac{1}{99}+\dfrac{2}{98}+\dfrac{3}{97}+\dfrac{4}{96}+...+\dfrac{97}{3}+\dfrac{98}{2}+\dfrac{99}{1}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{100}}\)
\(=\dfrac{\left(1+\dfrac{1}{99}\right)+\left(1+\dfrac{2}{98}\right)+\left(1+\dfrac{3}{97}\right)+\left(1+\dfrac{4}{96}\right)+...+\left(1+\dfrac{98}{2}\right)+1}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{100}}\)
\(=\dfrac{\dfrac{100}{99}+\dfrac{100}{98}+\dfrac{100}{97}+...+\dfrac{100}{1}+\dfrac{100}{2}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{100}}\)
=100
Ta có: \(N=\dfrac{92-\dfrac{1}{9}-\dfrac{2}{10}-\dfrac{3}{11}-...-\dfrac{90}{98}-\dfrac{91}{99}-\dfrac{92}{100}}{\dfrac{1}{45}+\dfrac{1}{50}+\dfrac{1}{55}+...+\dfrac{1}{495}+\dfrac{1}{500}}\)
\(=\dfrac{\left(1-\dfrac{1}{9}\right)+\left(1-\dfrac{2}{10}\right)+\left(1-\dfrac{3}{11}\right)+...+\left(1-\dfrac{90}{98}\right)+\left(1-\dfrac{91}{99}\right)+\left(1-\dfrac{92}{100}\right)}{\dfrac{1}{5}\left(\dfrac{1}{9}+\dfrac{1}{10}+\dfrac{1}{11}+...+\dfrac{1}{99}+\dfrac{1}{100}\right)}\)
\(=\dfrac{\dfrac{8}{9}+\dfrac{8}{10}+\dfrac{8}{11}+...+\dfrac{8}{99}+\dfrac{8}{100}}{\dfrac{1}{5}\left(\dfrac{1}{9}+\dfrac{1}{10}+\dfrac{1}{11}+...+\dfrac{1}{99}+\dfrac{1}{100}\right)}\)
\(=\dfrac{8}{\dfrac{1}{5}}=40\)
\(\Leftrightarrow\dfrac{M}{N}=\dfrac{100}{40}=\dfrac{5}{2}\)
\(\dfrac{1}{100.99}-\dfrac{1}{99.98}-...-\dfrac{1}{2.1}\)
\(=\dfrac{1}{99}-\dfrac{1}{100}-\dfrac{1}{98}+\dfrac{1}{99}-\dfrac{1}{97}+\dfrac{1}{98}-...-\dfrac{1}{2}+\dfrac{1}{3}-1+\dfrac{1}{2}\)
\(=\dfrac{2}{99}-\dfrac{1}{100}-1=-\dfrac{9799}{9900}\)
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