= 1-1/x+1 = 17/18

=> 1/x+1 = 1-17/18= 1/18

=> x+1 = 18 => x=17

ta có 1/1.2+1/2.3+1/3.4+1/4.5+...+1/x.(x+1)=17/18

1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+...+1/x-1/x+1=17/18

1-1/x+1=17/18

1/x+1=1-17/18

1/x+1=1/18

suy ra: x+1=18

x=18-1

x=17

=> 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ........ + 1 /x - 1/x + 1 = 17 /18

=> 1 - 1/x+1 = 17/18

=> 1/x+1 = 1/18

=> x + 1 = 18

=> x = 17 (tm)

Vậy x = 17 nha!

Ai mk mk lại !!

1/1.2 + 1/2.3 + 1/3.4 +......+ 1/x(x+1) = 17/18

=> 1- 1/x+1 = 17/18

=>  1/x +1 = 1-17/18

=> 1/x+1 = 1/18

=> x= 17 

ta có: 1/1.2+1/2.3+1/3.4+1/4.5+...+1/x(x+1)=17/18

 1-1/2+1/2-1/3+1/3+1/4+1/4-1/5+...+1/x-1/x+1=17/18

 1-1/x+1=17/18

 1/x+1=1-17/18

 1/x+1=1/18

suy ra: x+1=18

x=18-1

x=17

Số x cần tìm là : 99

Ta có : \(\frac{5}{1.2}+\frac{5}{2.3}+\frac{5}{3.4}+.....+\frac{5}{x\left(x+1\right)}=\frac{99}{20}\)

\(\Rightarrow5\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+......+\frac{1}{x\left(x+1\right)}\right)=\frac{99}{20}\)

\(\Rightarrow1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{x}-\frac{1}{x+1}=\frac{99}{20}.\frac{1}{5}\)

\(\Rightarrow1-\frac{1}{x+1}=\frac{99}{100}\)

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

=> x + 1 = 100

=> x = 99

Ta có  \(\frac{5}{1.2}+\frac{5}{2.3}+...+\frac{5}{x\left(x+1\right)}=\frac{99}{20}\)

   =>  \(5\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{99}{20}\)

   =>    \(5\left(1-\frac{1}{x+1}\right)=\frac{99}{20}\)

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

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

 =>   x+1 = 100

=>  x = 99

\(\text{Đề }\Leftrightarrow\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right).\left(x-1\right)=x-\frac{1}{3}\)

=> \(\left(1-\frac{1}{10}\right).\left(x-1\right)=x-\frac{1}{3}\)

=> \(\frac{9}{10}.\left(x-1\right)=x-\frac{1}{3}\)

=> \(\frac{9x}{10}-\frac{9}{10}=\frac{3x-1}{3}\)

=> \(\frac{27x}{30}-\frac{27}{30}=\frac{10.\left(3x-1\right)}{30}\)

=> 27x - 27 = 30x - 10

=> 27x - 30x = -10 + 27

=> -3x = 17

=> x = -17/3.

\(\frac{2}{2\cdot3}+\frac{2}{3\cdot4}+\frac{2}{4\cdot5}+...+\frac{2}{x\left(x+1\right)}=\frac{2019}{2020}\)

\(\Rightarrow2\left(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{x\cdot\left(x+1\right)}\right)=\frac{2019}{2020}\)

\(\Rightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2019}{4040}\)

\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2019}{4040}\)

...

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

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

\(=2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2019}{2020}\)

\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2019}{4040}\)

Rồi bn tự tìm x nha!hok tot

\(\frac{2}{2.3}+\frac{2}{3.4}+......+\frac{2}{x\left(x+1\right)}=\frac{2019}{2020}\)

\(\Leftrightarrow2\left[\frac{1}{2.3}+\frac{1}{3.4}+........+\frac{1}{x\left(x+1\right)}\right]=\frac{2019}{2020}\)

\(\Leftrightarrow\frac{1}{2.3}+\frac{1}{3.4}+........+\frac{1}{x\left(x+1\right)}=\frac{2019}{4040}\)

\(\Leftrightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.......+\frac{1}{x}-\frac{1}{x+1}=\frac{2019}{2020}\)

\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2019}{4040}\)\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{2019}{4040}\)

\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{4040}\)\(\Leftrightarrow x+1=4040\)\(\Leftrightarrow x=4039\)

Vậy \(x=4039\)

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

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

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

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

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

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

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

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

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

\(\Leftrightarrow x=99\)

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

=> 3x + 3/2 = 1

=> 3x = 1 - 3/2

=> 3x = -1/2

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

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

\(\Leftrightarrow3\left(x+\frac{1}{2}\right)=1\)

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

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

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

\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{\left(x-1\right)\times x}=\frac{15}{16}\)

\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x-1}-\frac{1}{x}=\frac{15}{16}\)

\(1-\frac{1}{x}=\frac{15}{16}\)

\(\frac{1}{x}=\frac{1}{16}\)

\(\Rightarrow x=16\)

đề chưa đầy đủ

à đề thiếu tổng các giá trị tuyệt đối ở trên =100x

What wrong with you ?

\(\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+....+\frac{1}{2013.2014}-\frac{1}{2014.2015}\right)x=\frac{1}{3}\left(2014.2015.2016-2013.2014.2015........+2.3.4-1.2.3+1.2.3-0.1.2\right)\)

\(\left(\frac{1}{2}-\frac{1}{2014.2015}\right)x=\frac{1}{3}.2014.2015.2016\)

\(x=\frac{1}{3.2029104}.2014^2.2015^2.2016=\)

\(\left(\frac{1}{2}-\frac{1}{2014.2015}\right)x=\frac{1}{3}.2014.2015.2016\)

vào câu hỏi tương tự nha bạn