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![](https://rs.olm.vn/images/avt/0.png?1311)
1/1-1/2+1.2-1/3+1/3-1/4+..+1/x-1/x+1=2018/2019
1-1/x+1=2018/2019
1-2018/2019=1/x+1
1/2019=1/x+1
=>x+1=2019
=>x=2018
vậy...
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x\left(x+1\right)}=\frac{2018}{2019}.\)
\(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2018}{2019}.\)
\(\frac{1}{1}-\frac{1}{x+1}=\frac{2018}{2019}\)
\(\frac{1}{1}-\frac{2018}{2019}=\frac{1}{x+1}\)
\(\frac{1}{2019}=\frac{1}{x+1}\)
=> \(2019=x+1\)
\(x+1=2019\)
\(x=2019-1\)
\(x=2018\)
Vậy x = 2018
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{1313}{1212}:x=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\)\(\frac{1}{5.6}\)
\(\Leftrightarrow\frac{13}{12}:x=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}\)
\(\Leftrightarrow\frac{13}{12}:x=1-\frac{1}{6}\)
\(\Leftrightarrow\frac{13}{12}:x=\frac{5}{6}\)
\(\Leftrightarrow x=\frac{13}{12}:\frac{5}{6}\)
\(\Leftrightarrow x=\frac{13}{10}\)
Vậy \(x=\frac{13}{10}\)
~~~~~Hok tốt ~~~~~
a,\(\frac{1313}{1212}\div x=\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+\frac{1}{5\times6}\)
\(\frac{13}{12}\div x=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}\)
\(\frac{13}{12}\div x=1-\frac{1}{6}\)
\(\frac{13}{12}\div x=\frac{5}{6}\)
\(x=\frac{13}{12}\div\frac{5}{6}\)
\(x=\frac{13}{12}\times\frac{6}{5}\)
\(x=\frac{13}{10}\)
Chúc bạn hok tốt !
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{1313}{1212}:x=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\)\(\frac{1}{5.6}\)
\(\Leftrightarrow\frac{13}{12}:x=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}\)
\(\Leftrightarrow\frac{13}{12}:x=1-\frac{1}{6}\)
\(\Leftrightarrow\frac{13}{12}:x=\frac{5}{6}\)
\(\Leftrightarrow x=\frac{13}{12}:\frac{5}{6}\)
\(\Leftrightarrow x=\frac{13}{10}\)
Hok tốt
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{8.9}+\dfrac{1}{9.10}\right).100-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\)
\(\Rightarrow\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\right).100-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\)
\(\Rightarrow\left(1-\dfrac{1}{10}\right).100-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\)
\(\Rightarrow\left(100-10\right)-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\)
\(\Rightarrow90-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\)
\(\Rightarrow\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=1\)
\(\Rightarrow\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)=1.2=2\)
\(\Rightarrow\left(x+\dfrac{206}{100}\right)=\dfrac{5}{2}:2=\dfrac{5}{2}.\dfrac{1}{2}=\dfrac{5}{4}\)
\(\Rightarrow x=\dfrac{5}{4}-\dfrac{206}{100}=\dfrac{125}{100}-\dfrac{206}{100}\)
\(\Rightarrow x=-\dfrac{81}{100}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(C=\frac{9}{10}-\frac{1}{10.9}-\frac{1}{9.8}-...-\frac{1}{3.2}-\frac{1}{2.1}\)
\(C=\frac{9}{10}-\left(\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{9.8}+\frac{1}{9.10}\right)\)
\(C=\frac{9}{10}-\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\right)\)
\(C=\frac{9}{10}-\left(\frac{1}{1}-\frac{1}{10}\right)\)
\(C=\frac{9}{10}-\frac{9}{10}=0\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{2020\cdot2021}+\dfrac{1}{2021\cdot2022}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2020}-\dfrac{1}{2021}+\dfrac{1}{2021}-\dfrac{1}{2022}\)
\(=1-\dfrac{1}{2022}=\dfrac{2021}{2022}\)
1/1x2+1/2x3+1/3x4+...+1/2020x2021+1/2021x2022
=1/1-1/2+1/2-1/3+1/3-1/4+...+1/2020-1/2021+1/2021-1/2022.
=1/1-1/2022
=2021/2022
\(\dfrac{1}{2\times3}+\dfrac{1}{3\times4}+\dfrac{1}{4\times5}+...+\dfrac{1}{x\left(x+1\right)}=\dfrac{9}{20}\)
\(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...-\dfrac{1}{x}+\dfrac{1}{x}-\dfrac{1}{x+1}=\dfrac{9}{20}\)
\(\dfrac{1}{2}+0+0+0+...+0-\dfrac{1}{x+1}=\dfrac{9}{20}\)
\(\dfrac{1}{2}-\dfrac{1}{x+1}=\dfrac{9}{20}\)
\(\dfrac{1}{x+1}=\dfrac{1}{20}\)
\(x+1=20\)
\(x=20-1\)
\(x=19\)