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FZ
2
13 tháng 8 2015
1/3+1/6+1/10+1/15+......+1/4950
=2x（1/6+1/12+1/20+1/30+……+1/9900）
=2x（1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+……+1/99-1/100）
=2x（1/2-1/100）
=1-1/50
=49/50
**** nhé
NP
1
13 tháng 8 2015
D=1/3+1/6+1/10+1/15+......+1/4950
=2x（1/6+1/12+1/20+1/30+……+1/9900）
=2x（1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+……+1/99-1/100）
=2x（1/2-1/100）
=1-1/50
=49/50
**** nhé
VA
22 tháng 9 2017
Bài 1:
a) [ (1/6 + 1/10 + 1/15) : (1/6 + 1/10 - 1/15) phần 1/2 - 1/3 + 1/4 - 1/5 ] : (1/4 - 1/6)
= [ (1/6 : 1/6) + (1/10 : 1/10) - (1/15 : 1/15) phần 30/60 - 20/60 + 15/60 - 12/60 ] : (3/12 - 2/12)
= [ 1 + 1 - 1 phần 13/60 ] : 1/12
= [ 1 : 13/60 ] x 12
= 60/13 x 12
=720/ 13
b) (3/20 + 1/2 - 1/15) x 12/49 phần 3 và 1/3 + 2/9
= (9/60 + 30/60 - 4/60) x 12/49 phần 10/3 + 2/9
= 7/12 x 12/49 phần 30/9 + 2/9
= 1/7 : 32/9
= 1/7 x 9/32
= 9/224
HS
1
VT
0
LL
1
IA
8 tháng 3 2020
\(\frac{1\cdot3\cdot9+2\cdot6\cdot18+3\cdot9\cdot27}{1\cdot5\cdot18+2\cdot10\cdot36+3\cdot15\cdot54}\)
\(=\frac{1\cdot3\cdot9+2\left(1\cdot3\cdot9\right)+3\left(1\cdot3\cdot9\right)}{1\cdot5\cdot18+2\left(1\cdot5\cdot18\right)+3\left(1\cdot5\cdot18\right)}\)
\(=\frac{\left(1\cdot3\cdot9\right)\left(1+2+3\right)}{\left(1\cdot5\cdot18\right)\left(1+2+3\right)}\)
\(=\frac{3}{10}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
25 tháng 3 2023
A = 1 + \(\dfrac{1}{3}\) + \(\dfrac{1}{6}\) + \(\dfrac{1}{10}\) + .....+ \(\dfrac{1}{4950}\)
A = \(\dfrac{2}{2}\) \(\times\) ( 1 + \(\dfrac{1}{3}\) + \(\dfrac{1}{6}\) + \(\dfrac{1}{10}\)+.......+ \(\dfrac{1}{4950}\))
A = 2 \(\times\) ( \(\dfrac{1}{2}\) + \(\dfrac{1}{6}\) + \(\dfrac{1}{12}\) + \(\dfrac{1}{20}\)+......+ \(\dfrac{1}{9900}\))
A = 2 \(\times\) ( \(\dfrac{1}{1.2}\) + \(\dfrac{1}{2.3}\)+ \(\dfrac{1}{3.4}\) + \(\dfrac{1}{4.5}\)+....+ \(\dfrac{1}{99.100}\))
A = 2 \(\times\) ( \(\dfrac{1}{1}\) - \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) - \(\dfrac{1}{3}\)+ \(\dfrac{1}{3}\) - \(\dfrac{1}{4}\)+ \(\dfrac{1}{4}\) - \(\dfrac{1}{5}\) +....+ \(\dfrac{1}{99}\) - \(\dfrac{1}{100}\))
A = 2 \(\times\) ( 1 - \(\dfrac{1}{100}\))
A = 2 \(\times\) \(\dfrac{99}{100}\)
A = \(\dfrac{99}{50}\)
\(A=\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+...+\dfrac{1}{4950}\)
\(\dfrac{1}{2}A=\dfrac{1}{2}\left(\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+...+\dfrac{1}{4950}\right)\)
\(\dfrac{1}{2}A=\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+...+\dfrac{1}{9900}\)
\(\dfrac{1}{2}A=\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{99.100}\)
\(\dfrac{1}{2}A=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(\dfrac{1}{2}A=\dfrac{1}{2}-\dfrac{1}{100}\)
\(\dfrac{1}{2}A=\dfrac{49}{100}\)
\(A=\dfrac{49}{50}\)