1/6+1/12+1/20+1/30+1/42+....+1/90+1/11
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
NH
0
HM
8 tháng 9 2016
em lớp 6 nha
B= 1/2 + 1/6 + 1/12 +1/20 + 1/30 + 1/42 + 1/56 + 1/72
B= 1/1*2 + 1/2*3 + 1/3*4 + 1/4*5 + 1/5*6 + 1/6*7 + 1/7*8 + 1/8*9
B=1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7+1/7-1/8+1/8-1/9
B=1+0-0-0-0-0-0-0-1/9
B=1-1/9
B=8/9
k em nha
NT
7
LM
9 tháng 8 2016
\(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+...+\frac{1}{90}+\frac{1}{110}=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{9.10}+\frac{1}{10.11}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}\)\(=\frac{1}{2}-\frac{1}{11}=\frac{9}{22}\)
9 tháng 8 2016
\(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}+\frac{1}{110}\)
\(=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{9.10}+\frac{1}{10.11}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-...+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}\)
\(=\frac{1}{2}-\frac{1}{11}=\frac{9}{22}\)
15 tháng 5 2016
Đặt A = \(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+...+\frac{1}{90}+\frac{1}{110}\)
A=\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{9.10}+\frac{1}{10.11}\)
A=\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}\)
A=\(\frac{1}{2}-\frac{1}{11}\)
A=\(\frac{9}{22}\)
15 tháng 5 2016
\(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+.......+\frac{1}{90}+\frac{1}{110}\)
\(=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...........+\frac{1}{9.10}+\frac{1}{10.11}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...........+\frac{1}{10}-\frac{1}{11}\)
\(=\frac{1}{2}-\frac{1}{11}=\frac{9}{22}\)
HP
8 tháng 5 2016
Tổng quát: \(\frac{1}{n\left(n+1\right)}=\frac{1}{n}-\frac{1}{n+1}\)
\(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+..+\frac{1}{110}=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+....+\frac{1}{10.11}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+....+\frac{1}{10}-\frac{1}{11}\)
\(=\frac{1}{2}-\frac{1}{11}=\frac{9}{22}\)
NM
2
28 tháng 6 2016
Ta có: \(\frac{1}{6}+\frac{1}{12}+....+\frac{1}{110}\)
\(\Rightarrow\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+.....+\frac{1}{10.11}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{10}-\frac{1}{11}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{11}=\frac{9}{22}\)
28 tháng 6 2016
\(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+........+\frac{1}{110}\)
\(=\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+.......+\frac{1}{10\times11}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+.....+\frac{1}{10}-\frac{1}{11}\)
\(=\frac{1}{2}-\frac{1}{11}\)
\(=\frac{11}{22}-\frac{2}{22}\)
\(=\frac{9}{22}\)
7 tháng 2 2016
\(\frac{1}{2}+\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+...+\frac{55}{56}+\frac{71}{72}+\frac{89}{90}\)
\(=1-\frac{1}{2}+1-\frac{1}{6}+1-\frac{1}{12}+....+1-\frac{1}{56}+1-\frac{1}{72}+1-\frac{1}{90}\)
\(=9-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+....+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}\right)\)
\(=9-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}\right)\)
\(=9-\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\right)\)
\(=9-\left(1-\frac{1}{10}\right)=9-\frac{9}{10}=\frac{81}{10}\)
7 tháng 2 2016
b.=3/2.4/3....2012/2011
=3.4....2012/2.3....2011=2012/2=1006
NT
2
20 tháng 6 2016
\(B=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+...+\frac{1}{90}+\frac{1}{110}\)
\(B=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{9.10}+\frac{1}{10.11}\)
\(B=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{9}-\frac{1}{10}\)\(+\frac{1}{10}-\frac{1}{11}\)
\(B=\frac{1}{2}-\frac{1}{11}\)
\(B=\frac{9}{22}\)
20 tháng 6 2016
Thôi chết! Ở hàng thứ 3 của bài Mình có ghi 1/2 - 1/3 + 1/3 - 1/4 +1/4 - 1/5 .... nha bạn! Không phải bằng đâu ^.^
18 tháng 6 2021
`A=1/2+5/6+11/12+19/20+29/30+41/42+55/56+71/72+89/90`
`=1-1/2+1-1/6+1-1/12+1-1/20+1-1/30+1-1/42+1-1/56+1-1/72+1-1/90`
`=9-(1/2+1/6+1/12+1/20+1/30+1/42+1/56+1/72+1/90)`
`=9-(1/(1.2)+1/(2.3)+1/(3.4)+1/(4.5)+1/(5.6)+1/(6.7)+1/(7.8)+1/(8.9)+1/(9.10))`
`=9-(1-1/2+1/2-1/3+.....+1/9-1/10)`
`=9-(1-1/10)`
`=9-1+1/10=8+1/10=81/10`
18 tháng 6 2021
A = \(\left(1-\dfrac{1}{2}\right)+\left(1-\dfrac{1}{6}\right)+...+\left(1-\dfrac{1}{90}\right)\)
= \(9-\left(\dfrac{1}{2}+\dfrac{1}{6}+...+\dfrac{1}{90}\right)\)
=\(9-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{9}-\dfrac{1}{10}\right)\)
= \(9-\left(1-\dfrac{1}{10}\right)\)
= \(9-\dfrac{9}{10}=\dfrac{81}{10}\)
\(=\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{9.10}+\dfrac{1}{11}\)
\(=\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}\right)+\dfrac{1}{11}\)
\(=\left(\dfrac{1}{2}-\dfrac{1}{10}\right)+\dfrac{1}{11}\)
\(=\dfrac{2}{5}+\dfrac{1}{11}\)
\(=\dfrac{27}{55}\)
Kết quả = 27/55