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22 tháng 7 2017

\(\left(x+\frac{1}{2}\right)+\left(x+\frac{1}{2}\right)+\left(x+\frac{1}{2}\right)=1\)

\(\Leftrightarrow3x+\left(\frac{1}{2}+\frac{1}{2}+\frac{1}{2}\right)=1\)

\(\Leftrightarrow3x+\frac{3}{2}=1\)

\(\Leftrightarrow3x=-\frac{1}{2}\)

\(\Leftrightarrow x=-\frac{1}{2}\div3=-\frac{1}{6}\)

Sửa đề \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{x.\left(x+1\right)}=\frac{99}{100}\)

\(\Leftrightarrow1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2}-\frac{1}{x+1}=\frac{99}{100}\)

\(\Leftrightarrow1-\frac{1}{x+1}=\frac{99}{100}\)

\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{100}\)

\(\Leftrightarrow x=99\)

22 tháng 7 2017

a) => ( x + 1/2 ) . 3 = 1

=> 3x + 3/2 = 1

=> 3x = 1 - 3/2

=> 3x = -1/2

=> x = -1/2 : 3 = -1/6

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

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.

22 tháng 4 2017

\(1.\)\(M=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{42}\)

\(M=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{6.7}\)

\(M=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-...+\frac{1}{6}-\frac{1}{7}\)

\(M=1-\frac{1}{7}=\frac{6}{7}\)

Mình làm câu 1 thoi nha!

22 tháng 4 2017

1.

\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\)

=\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\)

=\(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{6}-\frac{1}{7}\)

=\(1-\frac{1}{7}\)

=\(\frac{6}{7}\)

2 tháng 4 2023

1+1=3 :)))

24 tháng 8 2019

\(\frac{1}{1.2}+\frac{1}{2.3}+......+\frac{1}{2019.2020}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+..-\frac{1}{2020}=1-\frac{1}{2020}=\frac{2019}{2020}\) 

\(\Rightarrow a=\frac{2020}{2019}\)

24 tháng 8 2019

=.> 1-1/2+1/2-1/3+.......+1/2019-1/2020=1/x

=>1-1/2020=1/x

=>2019/2020=1/x

=>2019x=2020

=>x=2020/2019

    k nha

 giúp mk lên 300sp

27 tháng 2 2018

Ta có : 

\(\frac{1^2}{1.2}.\frac{2^2}{2.3}.\frac{3^2}{3.4}.....\frac{99^2}{99.100}\)

\(=\)\(\frac{1^2.2^2.3^2.....99^2}{1.2.2.3.3.4.....99.100}\)

\(=\)\(\frac{1^2.2^2.3^2.....99^2}{1^2.2^2.3^2.4^2.....99^2}.\frac{1}{100}\)

\(=\)\(\frac{1}{100}\)

13 tháng 7 2017

\(1+\frac{1}{3}+\frac{1}{6}+....+\frac{2}{x\left(x+1\right)}=4\)

\(\Leftrightarrow1+\frac{2}{6}+\frac{2}{12}+....+\frac{2}{x\left(x+1\right)}=4\)

\(\Leftrightarrow1+\frac{2}{2.3}+\frac{2}{3.4}+....+\frac{2}{x\left(x+1\right)}=4\)

\(\Leftrightarrow1+\left[2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{x}-\frac{1}{\left(x+1\right)}\right)\right]=4\)

\(\Leftrightarrow1+2\left(\frac{1}{2}-\frac{1}{\left(x+1\right)}\right)=4\)

\(\Leftrightarrow\frac{1}{2}-\frac{1}{\left(x+1\right)}=\frac{4-1}{2}=\frac{3}{2}\)

\(\Leftrightarrow\frac{1}{\left(x+1\right)}=\frac{1}{2}-\frac{3}{2}=-1\)

\(\Leftrightarrow x=-1+1=-2\)

Vậy x = -2 

16 tháng 7 2017

\(\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{2.6}+\frac{2}{2.10}+....+\frac{2}{x\left(x+1\right)}=1\frac{1991}{1993}\)

\(\Leftrightarrow\frac{2}{2}+\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+....+\frac{2}{x\left(x+1\right)}=1\frac{1991}{1993}\)

\(\Leftrightarrow\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+....+\frac{2}{x\left(x+1\right)}=1\frac{1991}{1993}\)

\(\Leftrightarrow2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{x}-\frac{1}{\left(x+1\right)}\right)=1\frac{1991}{1993}\)

\(\Leftrightarrow2\left(1-\frac{1}{\left(x+1\right)}\right)=1\frac{1991}{1993}\)

\(\Leftrightarrow1-\frac{1}{\left(x+1\right)}=1\frac{1991}{1993}\div2\)

\(\Leftrightarrow1-\frac{1}{\left(x+1\right)}=\frac{1992}{1993}\)

\(\Leftrightarrow\frac{1}{\left(x+1\right)}=1-\frac{1992}{1993}=\frac{1}{1993}\)

\(\Leftrightarrow x+1=1993\)

\(\Leftrightarrow x=1992\)