PT
1
K
Khách
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PT
1
CM
8 tháng 12 2019
1/72+ 1/90 + 1/ 110 + .....+1/9702+1/9900
= 1/ 8x9 + 1/9x10+1/10x11+.....+1/98x99+1/99x100
.....
= 1/8 - 1/100
=23/200
PP
3
TH
Thầy Hùng Olm
Manager
VIP
4 tháng 6 2023
\(\dfrac{1}{20}=\dfrac{1}{4x5}=\dfrac{1}{4}-\dfrac{1}{5}\)
Tương tự các phân số khác
S= \(\dfrac{1}{4}-\dfrac{1}{12}=\dfrac{1}{6}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
4 tháng 6 2023
\(\dfrac{1}{20}+\dfrac{1}{30}\)+ \(\dfrac{1}{42}\)+\(\dfrac{1}{56}\)+\(\dfrac{1}{72}\)+\(\dfrac{1}{90}\)+\(\dfrac{1}{110}\)+\(\dfrac{1}{132}\)
= \(\dfrac{1}{4\times5}\)+\(\dfrac{1}{5\times6}\)+\(\dfrac{1}{6\times7}\)+\(\dfrac{1}{7\times8}\)+\(\dfrac{1}{8\times9}\)+\(\dfrac{1}{9\times10}\)+\(\dfrac{1}{10\times11}\)+\(\dfrac{1}{11\times12}\)
= \(\dfrac{1}{4}\) - \(\dfrac{1}{5}\)+\(\dfrac{1}{5}\)-\(\dfrac{1}{6}\)+\(\dfrac{1}{6}\)-\(\dfrac{1}{7}\)+\(\dfrac{1}{7}\)-\(\dfrac{1}{8}\)+\(\dfrac{1}{8}\)-\(\dfrac{1}{9}\)+\(\dfrac{1}{9}\)-\(\dfrac{1}{10}\)+\(\dfrac{1}{10}\)-\(\dfrac{1}{11}\)+\(\dfrac{1}{11}\)-\(\dfrac{1}{12}\)
= \(\dfrac{1}{4}\) - \(\dfrac{1}{12}\)
= \(\dfrac{3}{12}\) - \(\dfrac{1}{12}\)
= \(\dfrac{2}{12}\)
=\(\dfrac{1}{6}\)
NT
2
D
29 tháng 5 2022
`=1/[4xx5]+1/[5xx6]+1/[6xx7]+...+1/[11xx12]`
`=1/4-1/5+1/5-1/6+1/6-1/7+...+1/11-1/12`
`=1/4-1/12=3/12-1/12=2/12=1/6`
29 tháng 5 2022
\(\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}+\dfrac{1}{110}+\dfrac{1}{132}\\ =\dfrac{1}{4\times5}+\dfrac{1}{5\times6}+\dfrac{1}{6\times7}+\dfrac{1}{7\times8}+\dfrac{1}{8\times9}+\dfrac{1}{9\times10}+\dfrac{1}{10\times11}+\dfrac{1}{11\times12}\\ =\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{12}\\ =\dfrac{1}{4}-\dfrac{1}{12}\\ =\dfrac{3}{12}-\dfrac{1}{12}=\dfrac{2}{12}=\dfrac{1}{6}\)
NX
0
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
14 tháng 4 2023
A = \(\dfrac{1}{20}\) + \(\dfrac{1}{30}\) + \(\dfrac{1}{42}\) + \(\dfrac{1}{56}\) + \(\dfrac{1}{72}\) + \(\dfrac{1}{90}\) + \(\dfrac{1}{110}\) + \(\dfrac{1}{132}\)
A = \(\dfrac{1}{4\times5}\) + \(\dfrac{1}{5\times6}\) + \(\dfrac{1}{6\times7}\)+ \(\dfrac{1}{7\times8}\)+\(\dfrac{1}{8\times9}\)+ \(\dfrac{1}{9\times10}\) + \(\dfrac{1}{10\times11}\)+\(\dfrac{1}{11\times12}\)
A = \(\dfrac{1}{4}\)-\(\dfrac{1}{5}\) +\(\dfrac{1}{5}\)-\(\dfrac{1}{6}\) +.....+\(\dfrac{1}{11}\) - \(\dfrac{1}{12}\)
A = \(\dfrac{1}{4}\) - \(\dfrac{1}{12}\)
A = \(\dfrac{1}{6}\)
NQ
3
NH
2
29 tháng 10 2021
T = \(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+...+\frac{1}{98\cdot99}+\frac{1}{99\cdot100}=\)
T = \(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}=\)
T = \(\frac{1}{1}-\frac{1}{100}=\)
T = \(\frac{99}{100}\)
LT
1
17 tháng 12 2015
A=1/1*2+...+1/10*11
=1-1/2+...+1/10-1/11
=(1-1/11)+(1/2-1/2)+...+(1/10-1/10)
=(11/11-1/11)+0+0+...+0=10/11
1/72+ 1/90 + 1/ 110 + .....+1/9702+1/9900 = 1/ 8x9 + 1/9x10+1/10x11+.....+1/98x99+1/99x100 ..... = 1/8 - 1/100 =23/200