\(\frac{1}{1.2}.x+\frac{1}{2.3}.x.........+\frac{1}{9.10}=\frac{18}{5}\)
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NM
3
28 tháng 3 2016
A=\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\)
A=1-\(\frac{1}{10}\)
A=\(\frac{9}{10}\)
28 tháng 3 2016
\(\frac{1}{1.2}.\frac{1}{2.3}....\frac{1}{9.10}=\frac{1.1.1.1.1.1}{1.2.2.3.3....9.9.10}=\frac{1}{1.4.9.16.25.36....100}=\frac{1}{13168189440000}\)
MH
2 tháng 2 2016
\(\text{Đề }\Leftrightarrow\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right).\left(x-1\right)=x-\frac{1}{3}\)
=> \(\left(1-\frac{1}{10}\right).\left(x-1\right)=x-\frac{1}{3}\)
=> \(\frac{9}{10}.\left(x-1\right)=x-\frac{1}{3}\)
=> \(\frac{9x}{10}-\frac{9}{10}=\frac{3x-1}{3}\)
=> \(\frac{27x}{30}-\frac{27}{30}=\frac{10.\left(3x-1\right)}{30}\)
=> 27x - 27 = 30x - 10
=> 27x - 30x = -10 + 27
=> -3x = 17
=> x = -17/3.
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
ND
2
6 tháng 3 2017
\(\frac{1}{1.2}-\frac{1}{2.3}-\frac{1}{3.4}-......-\frac{1}{9.10}\)
\(=\frac{1}{2}-\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+.....+\frac{1}{9.10}\right)\)
\(=\frac{1}{2}-\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{9}-\frac{1}{10}\right)\)
\(=\frac{1}{2}-\left(\frac{1}{2}-\frac{1}{10}\right)\)
\(=\frac{1}{2}-\frac{1}{2}+\frac{1}{10}=\frac{1}{10}\)
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 \(1+\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{9.10}\)\(=\frac{19}{10}\)
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
15 tháng 7 2018
\(\frac{1}{9.10}-\frac{1}{8.9}-\frac{1}{7.8}-\frac{1}{6.7}-\frac{1}{5.6}-\frac{1}{4.5}-\frac{1}{3.4}-\frac{1}{2.3}-\frac{1}{1.2}\)
\(=-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}\right)+\frac{1}{9.10}\)
\(=-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{9}\right)+\frac{1}{90}\)
\(=-\left(1-\frac{1}{10}\right)+\frac{1}{90}\)
\(=-\frac{9}{10}+\frac{1}{90}\)
= ...
bn tự tính nha!
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
\(\frac{1}{9.10}\)có x k bn?
k có x như trên đề thì hơi vl đấy.
phiền bn xem lại.
\(\frac{1}{1\cdot2}\times+\cdot\cdot\cdot+\frac{1}{9\cdot10}\times=\frac{18}{5}\)
\(\Rightarrow\left(\frac{1}{1\cdot2}+\cdot\cdot\cdot+\frac{1}{9\cdot10}\right)\times=\frac{18}{5}\)
\(\Rightarrow\left(1-\frac{1}{2}+\cdot\cdot\cdot+\frac{1}{9}-\frac{1}{10}\right)\times=\frac{18}{5}\)
\(\Rightarrow\left(1-\frac{1}{10}\right)\times=\frac{18}{5}\)
\(\Rightarrow\frac{9}{10}\times=\frac{18}{5}\)
\(\Rightarrow\times=\frac{18}{5}:\frac{9}{10}\)
\(\Rightarrow\times=4\)