1/9.10-1/8.9-......-1/1.2
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
28 tháng 6 2015
1/9.10 - ( 1/1.2 + 1/2.3 +...+ 1/8 .9)
= 1/9 - 1/10 -( 1/1 - 1/2 + 1/2 - 1/3 +. ..+1/8 - 1 9/)
= 1/9 - 1/10 - ( 1 - 1/9 )
= 1/9 - 1/10 - 1 + 1/9
=-79/9
29 tháng 6 2015
thang Tran làm cách giải đúng rồi, nhưng đáp số là -79/90 nha.
Dù sao cũng cám ơn bạn vì giúp mình hiểu bài.
21 tháng 5 2022
\(=\dfrac{1}{9}-\dfrac{1}{10}-\left(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{8}-\dfrac{1}{9}\right)\)
\(=\dfrac{1}{9}-\dfrac{1}{10}-\left(1-\dfrac{1}{9}\right)\)
\(=\dfrac{1}{9}-\dfrac{1}{10}-1+\dfrac{1}{9}\)
\(=-\dfrac{97}{90}\)
LP
1
20 tháng 8 2021
Ta có: \(\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{8\cdot9}+\dfrac{1}{9\cdot10}\right)\cdot100-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\)
\(\Leftrightarrow100\left(\dfrac{1}{1}-\dfrac{1}{10}\right)-\left[\dfrac{5}{2}:\left(x+\dfrac{103}{50}\right)\right]\cdot2=89\)
\(\Leftrightarrow\dfrac{5}{2}:\left(x+\dfrac{103}{50}\right)=\dfrac{1}{2}\)
\(\Leftrightarrow x+\dfrac{103}{50}=5\)
hay \(x=\dfrac{147}{50}\)
MN
3 tháng 6 2016
Ta có: \(M=\frac{1}{9.10}-\frac{1}{8.9}-\frac{1}{7.8}-\frac{1}{6.7}-\frac{1}{5.6}-...-\frac{1}{1.2}\)
\(\Rightarrow-M=-\left(\frac{1}{9.10}-\frac{1}{8.9}-\frac{1}{7.8}-...-\frac{1}{1.2}\right)\)
\(\Rightarrow-M=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{7.8}+\frac{1}{8.9}-\frac{1}{9.10}\)
\(\Rightarrow-M=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}-\frac{1}{90}\)
\(\Rightarrow-M=1-\frac{1}{9}-\frac{1}{90}\)
\(\Rightarrow-M=\frac{8}{9}-\frac{1}{90}=\frac{80}{90}-\frac{1}{90}=\frac{79}{90}\)
\(\Rightarrow M=-\frac{79}{90}\)
BU
2
24 tháng 2 2016
A= 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
=> A = 1/1 - 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 + 1/9 - 1/10
=> A = 1 - 1/10 = 9/10
Vậy A = 9/10
24 tháng 2 2016
A = 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 + 1/9 - 1/10
A = 1 - 1/10 = 9/10
NT
1
DL
22 tháng 6 2018
Đặt \(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{8.9}+\frac{1}{9.10}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\)
\(A=1-\frac{1}{10}=\frac{9}{10}\)
\(\Rightarrow\frac{9}{10}.100-\left[\frac{5}{2}:\left(x+\frac{206}{100}\right)\right]:\frac{1}{2}=89\)
\(\Leftrightarrow90-\left[\frac{5}{2}:\left(x+\frac{206}{100}\right)\right]:\frac{1}{2}=89\)
\(\Rightarrow\frac{5}{2}:\left(x+\frac{206}{100}\right):\frac{1}{2}=90-89=1\)
\(\Leftrightarrow x+\frac{206}{100}=\frac{5}{2}:\frac{1}{2}:1=5\)
\(\Rightarrow x=5-\frac{206}{100}=\frac{147}{50}\)
Vậy \(x=\frac{147}{50}.\)
NT
1
HN
15 tháng 7 2019
\(\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{8.9}+\frac{1}{9.10}\right)\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-........-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\)
\(=1-\frac{1}{10}\)
\(=\frac{10}{10}-\frac{1}{10}=\frac{9}{10}\)
\(\Leftrightarrow\frac{9}{10}.100-\left[\frac{5}{2}:\left(x+\frac{206}{100}\right):\frac{1}{2}\right]=89\)
\(\Leftrightarrow90-\left[\frac{5}{2}:\left(x+\frac{206}{100}\right):\frac{1}{2}\right]=89\)
\(\Leftrightarrow\frac{5}{2}:\left(x+\frac{206}{100}\right):\frac{1}{2}=90-89=1\)
\(\Leftrightarrow\frac{5}{2}:\left(x+\frac{206}{100}\right)=1.\frac{1}{2}=\frac{1}{2}\)
\(\Leftrightarrow x+\frac{206}{100}=\frac{5}{2}:\frac{1}{2}\)
\(\Leftrightarrow x+\frac{103}{50}=\frac{5}{2}.2\)
\(\Leftrightarrow x+\frac{103}{50}=5\)
\(\Leftrightarrow x=5-\frac{103}{50}\)
\(\Leftrightarrow x=\frac{250}{50}-\frac{103}{50}\)
\(\Leftrightarrow x=\frac{147}{50}\)
NT
0
\(\frac{1}{9.10}-\frac{1}{8.9}-...-\frac{1}{1.2}\)
\(=-\left(\frac{1}{9.10}+\frac{1}{8.9}+...+\frac{1}{1.2}\right)\)
\(=-\left(\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{9.10}\right)\)
\(=-\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(=-\left(1-\frac{1}{10}\right)\)
\(=-\frac{9}{10}\)
\(\frac{1}{9.10}-\frac{1}{8.9}-....-\frac{1}{1.2}\)
\(=\frac{1}{9.10}-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{8.9}\right)\)
\(\Rightarrow\frac{1}{9.10}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{8}-\frac{1}{9}\right)\)
\(=\frac{1}{90}-\left(1-\frac{1}{9}\right)=\frac{1}{90}-\frac{80}{90}=\frac{-79}{90}\)