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16 tháng 4 2017
Ta có:
\(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-...+\dfrac{1}{98}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{100}\)
=1-\(\left(\dfrac{1}{2}+\dfrac{1}{2}\right)-\left(\dfrac{1}{3}+\dfrac{1}{3}\right)-...-\left(\dfrac{1}{99}+\dfrac{1}{99}\right)-\dfrac{1}{100}\)
=\(1-\dfrac{1}{100}=\dfrac{100}{100}-\dfrac{1}{100}=\dfrac{99}{100}\)
JJ
1
3 tháng 4 2017
A = \(\dfrac{2}{1\times2}+\dfrac{2}{2\times3}+...+\dfrac{2}{98\times99}+\dfrac{2}{99\times100}\)
A = \(2\left(\dfrac{1}{1\times2}+\dfrac{1}{2\times3}+...+\dfrac{1}{98\times99}+\dfrac{1}{99\times100}\right)\)
= \(2\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{98}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{100}\right)\)
= \(2\left(1-\dfrac{1}{100}\right)\)
= \(2\times\dfrac{99}{100}=\dfrac{99}{50}\)
17 tháng 4 2017
A = \(\dfrac{9}{1.2}\)+ \(\dfrac{9}{2.3}\)+\(\dfrac{9}{3.4}\)+......+\(\dfrac{99}{99.100}\)
A = 9( \(\dfrac{1}{1.2}\)+\(\dfrac{1}{2.3}\)+\(\dfrac{1}{3.4}\)+.......+\(\dfrac{1}{99.100}\))
A = 9( 1-\(\dfrac{1}{2}\)+\(\dfrac{1}{2}\)-\(\dfrac{1}{3}\)+........+\(\dfrac{1}{99}\)-\(\dfrac{1}{100}\))
A = 9 ( 1 - \(\dfrac{1}{100}\))
A = 9 . \(\dfrac{99}{100}\)
A = \(\dfrac{891}{100}\)
18 tháng 4 2017
\(A=\dfrac{9}{1\cdot2}+\dfrac{9}{2\cdot3}+\dfrac{9}{3\cdot4}+...+\dfrac{9}{98\cdot99}+\dfrac{9}{99\cdot100}\)
\(=9\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{98\cdot99}+\dfrac{1}{99\cdot100}\right)\)
\(=9\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{98}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{100}\right)\)
\(=9\left(1-\dfrac{1}{100}\right)\)
\(=9\left(\dfrac{100}{100}-\dfrac{1}{100}\right)\)
\(=9\cdot\dfrac{99}{100}\)
\(=\dfrac{891}{100}\)
LT
1
16 tháng 10 2017
\(a=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}=1-\dfrac{1}{100}=\dfrac{99}{100}\)
C
2
23 tháng 4 2017
Đặt :
\(A=\dfrac{1}{1.2}+\dfrac{1}{2.3}+................+\dfrac{1}{99.100}\)
\(A=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+............+\dfrac{1}{99}-\dfrac{1}{100}\)
\(A=1-\dfrac{1}{100}\)
\(A=\dfrac{99}{100}< 1\)
\(\Rightarrow A< 1\)
\(\Rightarrow\dfrac{1}{1.2}+\dfrac{1}{2.3}+.........+\dfrac{1}{99.100}< 1\rightarrowđpcm\)
NH
10 tháng 3 2023
a)
`1/1-1/2`
`=2/2-1/2`
`=1/2`
b)
`1/(1*2)+1/(2*3)`
`=1/1-1/2+1/2-1/3`
`=1/1-1/3`
`=3/3-1/3`
`=2/3`
c)
\(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{99\cdot100}\\ =\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\\ =\dfrac{1}{1}-\dfrac{1}{100}\\ =\dfrac{99}{100}\)
d)
\(\dfrac{3}{1\cdot2}+\dfrac{3}{2\cdot3}+...+\dfrac{3}{99\cdot100}\) đề phải như thế này chứ nhỉ?
\(=\dfrac{1\cdot3}{1\cdot2}+\dfrac{1\cdot3}{2\cdot3}+...+\dfrac{1\cdot3}{99\cdot100}\\ =3\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{99\cdot100}\right)\\ =3\left(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\\ =3\left(\dfrac{1}{1}-\dfrac{1}{100}\right)\\ =3\cdot\dfrac{99}{100}\\ =\dfrac{297}{100}\)
25 tháng 3 2017
\(\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}\right)-2x=\dfrac{1}{2}\\ \left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)-2x=\dfrac{1}{2}\\ \left(1-\dfrac{1}{100}\right)-2x=\dfrac{1}{2}\\ \dfrac{99}{100}-2x=\dfrac{1}{2}\\ 2x=\dfrac{99}{100}-\dfrac{1}{2}\\ 2x=\dfrac{49}{100}\\ x=\dfrac{49}{100}:2\\ x=\dfrac{49}{200}\)
25 tháng 3 2017
\(\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}\right)-2x=\dfrac{1}{2}\)
\([\left(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)]-2x=\dfrac{1}{2}\)
\(\left(\dfrac{1}{1}-\dfrac{1}{100}\right)-2x=\dfrac{1}{2}\)
\(\dfrac{99}{100}-2x=\dfrac{1}{2}\)
\(2x=\dfrac{99}{100}-\dfrac{1}{2}\)
\(2x=\dfrac{49}{100}\)
\(x=\dfrac{49}{100}:2\)
\(x=\dfrac{49}{200}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-...+\dfrac{1}{98}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{100}\)
\(=1-\dfrac{1}{100}=\dfrac{99}{100}\)