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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
![](https://rs.olm.vn/images/avt/0.png?1311)
=> 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
![](https://rs.olm.vn/images/avt/0.png?1311)
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
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\text{Đề }\Leftrightarrow\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right).\left(x-1\right)=x-\frac{1}{3}\)
=> \(\left(1-\frac{1}{10}\right).\left(x-1\right)=x-\frac{1}{3}\)
=> \(\frac{9}{10}.\left(x-1\right)=x-\frac{1}{3}\)
=> \(\frac{9x}{10}-\frac{9}{10}=\frac{3x-1}{3}\)
=> \(\frac{27x}{30}-\frac{27}{30}=\frac{10.\left(3x-1\right)}{30}\)
=> 27x - 27 = 30x - 10
=> 27x - 30x = -10 + 27
=> -3x = 17
=> x = -17/3.
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{2}{2\cdot3}+\frac{2}{3\cdot4}+\frac{2}{4\cdot5}+...+\frac{2}{x\left(x+1\right)}=\frac{2019}{2020}\)
\(\Rightarrow2\left(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{x\cdot\left(x+1\right)}\right)=\frac{2019}{2020}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2019}{4040}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2019}{4040}\)
...
\(=2\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x\left(x+1\right)}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2019}{2020}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2019}{4040}\)
Rồi bn tự tìm x nha!hok tot
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(x+\frac{1}{2}\right)+\left(x+\frac{1}{2}\right)+\left(x+\frac{1}{2}\right)=1\)
\(\Leftrightarrow3x+\left(\frac{1}{2}+\frac{1}{2}+\frac{1}{2}\right)=1\)
\(\Leftrightarrow3x+\frac{3}{2}=1\)
\(\Leftrightarrow3x=-\frac{1}{2}\)
\(\Leftrightarrow x=-\frac{1}{2}\div3=-\frac{1}{6}\)
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
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{1}{1.2}+\frac{1}{2.3}+......+\frac{1}{2019.2020}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+..-\frac{1}{2020}=1-\frac{1}{2020}=\frac{2019}{2020}\)
\(\Rightarrow a=\frac{2020}{2019}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x-\frac{1}{1.2}-\frac{1}{2.3}-\frac{1}{3.4}=\frac{1}{4.5}\)
\(x-\frac{1}{1.2}-\frac{1}{2.3}-\frac{1}{3.4}-\frac{1}{4.5}=0\)
\(x-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}\right)=0\)
\(x-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}\right)=0\)
\(x-\left(1-\frac{1}{5}\right)=0\)
\(x-\frac{4}{5}=0\)
\(x=\frac{4}{5}\)
\(x-\frac{1}{1.2}-\frac{1}{2.3}-\frac{1}{3.4}=\frac{1}{4.5}\)
\(x-\left(1-\frac{1}{2}\right)-\left(\frac{1}{2}-\frac{1}{3}\right)-\left(\frac{1}{3}-\frac{1}{4}\right)=\frac{1}{4}-\frac{1}{5}\)
\(x-1+\frac{1}{2}-\frac{1}{2}+\frac{1}{3}-\frac{1}{3}+\frac{1}{4}-\frac{1}{4}=\frac{-1}{5}\)
\(x-1=-\frac{1}{5}\)
\(x=\frac{4}{5}\)
\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{x(x+1)}=\frac{2019}{2020}\)
\(\Rightarrow1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2019}{2020}\)
\(\Rightarrow1-\frac{1}{x+1}=\frac{2019}{2020}\)
\(\Rightarrow\frac{1}{x+1}=1-\frac{2019}{2020}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{2020}\)
\(\Rightarrow x+1=2020\Leftrightarrow x=2019\)
Vậy x = 2019