A=\(\frac{1}{1.2}\)+\(\frac{5}{2.3}\)+\(\frac{11}{3.4}\)+.....+\(\frac{89}{9.10}\)
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18 tháng 8 2019
\(\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{9\cdot10}\right)\cdot100-\left[\frac{5}{2}:\left(X+\frac{206}{100}\right)\right]:\frac{1}{2}=89\\ \left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\right)\cdot100-\left[\frac{5}{2}:\left(X+\frac{206}{100}\right)\right]:\frac{1}{2}=89\\ \left(1-\frac{1}{10}\right)\cdot100-\left[\frac{5}{2}:\left(X+\frac{206}{100}\right)\right]:\frac{1}{2}=89\\ \frac{9}{10}\cdot100-\left[\frac{5}{2}:\left(X+\frac{206}{100}\right)\right]:\frac{1}{2}=89\\ 90-\left[\frac{5}{2}:\left(X+\frac{206}{100}\right)\right]:\frac{1}{2}=89\\ \left[\frac{5}{2}:\left(X+\frac{206}{100}\right)\right]:\frac{1}{2}=1\\ \frac{5}{2}:\left(X+\frac{206}{100}\right)=\frac{1}{2}\\ X+\frac{206}{100}=5\\ X=\frac{500}{100}-\frac{206}{100}\\ X=\frac{294}{100}=\frac{147}{50}\)
Vậy \(X=\frac{147}{50}\)
18 tháng 8 2019
( 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ......+ 1/9 - 1/10) . 100 - [ 5/2 : ( x + 103/50 ) ] = 89 . 1/2
( 1 - 1/10) . 100 - [ 5/2 : ( x + 103/50 ) ] = 89/2
90 - 5/2 : ( x + 103/50 ) = 89/2
5/2 : ( x + 103/50 ) = 90 - 89/2
5/2 : ( x + 103/50 ) = 91/2
x + 103/50 = 5/2 : 91/2
x + 103/50 = 5/91
x = 5/91 - 103/50
x = -9,123/4550
DG
1
5 tháng 2 2018
\(\frac{5}{1.2}+\frac{13}{2.3}+\frac{25}{3.4}+...+\frac{181}{9.10}\)
=\(\frac{4+1}{2}+\frac{12+1}{6}+\frac{24+1}{12}+...+\frac{180+1}{90}\)
=\(2+\frac{1}{1.2}+2+\frac{1}{2.3}+2+\frac{1}{3.4}+...+2+\frac{1}{9.10}\)
=\(18+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)
=\(9-\frac{1}{10}\)
=\(\frac{189}{10}\)
ND
2
6 tháng 3 2017
\(\frac{1}{1.2}-\frac{1}{2.3}-\frac{1}{3.4}-......-\frac{1}{9.10}\)
\(=\frac{1}{2}-\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+.....+\frac{1}{9.10}\right)\)
\(=\frac{1}{2}-\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{9}-\frac{1}{10}\right)\)
\(=\frac{1}{2}-\left(\frac{1}{2}-\frac{1}{10}\right)\)
\(=\frac{1}{2}-\frac{1}{2}+\frac{1}{10}=\frac{1}{10}\)
EL
4
A
9 tháng 3 2017
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\)
\(A=\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+...+\left(\frac{1}{9}-\frac{1}{10}\right)\)
\(A=1-\frac{1}{10}\)
\(A=\frac{9}{10}\)
9 tháng 3 2017
dế mà em, giải thế này nè
A=1-1/2 +1/2-1/3 +1/3-1/4 +......+1/9-1/10
A=1-1/10+9/10
A
2
TD
1
IM
1 tháng 10 2016
Ta có
\(1+\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{9.10}\)
\(=1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{9}-\frac{1}{10}\)
\(=2-\frac{1}{10}\)
\(=\frac{19}{10}\)
Vậy \(1+\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{9.10}\)\(=\frac{19}{10}\)
CT
4
28 tháng 3 2019
Có: A=\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\)
A=\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)
A=\(1-\frac{1}{10}\)
A=\(\frac{9}{10}\)
Vậy A=\(\frac{9}{10}\)
DM
2
19 tháng 4 2017
\(=\frac{1}{1.2}-\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\right)\)
\(=\frac{1}{2}-\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(=\frac{1}{2}-\left(\frac{1}{2}-\frac{1}{10}\right)\)
\(=\frac{1}{2}-\frac{1}{2}+\frac{1}{10}\)
\(=\frac{1}{10}\)
LT
19 tháng 4 2017
(1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-......+1/9-1/10)
1-1/10=9/10
nhớ cho mk
