\(\frac{2}{2.3}\) + \(\frac{2}{3.4}\) + \(\frac{2}{4.5}\) + .......+ \(\frac{2}{x.\left(x+1\right)}\) = \(\frac{2017}{2019}\)
2 . ( \(\frac{1}{2}\) - \(\frac{1}{3}\) + \(\frac{1}{3}\) - \(\frac{1}{4}\) + .......+ \(\frac{1}{x+1}\) ) = \(\frac{2017}{2019}\)
2 . ( \(\frac{1}{2}\) - \(\frac{1}{x+1}\) ) = \(\frac{2017}{2019}\)
\(\frac{1}{2}\) - \(\frac{1}{x+1}\) = \(\frac{2017}{2019}\) : 2
\(\frac{1}{2}\) - \(\frac{1}{x+1}\) = \(\frac{2017}{4038}\)
\(\frac{1}{x+1}\) = \(\frac{1}{2}\) - \(\frac{2017}{4038}\)
\(\frac{1}{x+1}\) = \(\frac{1}{2019}\)
<=> x + 1 = 2019 => x = 2018
vậy x = 2018
\(\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{x\left(x+1\right)}=\frac{2017}{2019}\)
\(\Leftrightarrow2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2017}{2019}\)
\(\Leftrightarrow2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2017}{2019}\)
\(\Leftrightarrow2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2017}{2019}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2017}{4038}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{2019}\)
\(\Rightarrow x+1=2019\)
\(\Leftrightarrow x=2018\)
Vậy \(x=2018\)
\(\Leftrightarrow2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+....+\frac{1}{x\left(x+1\right)}\right)=\frac{2017}{2019}\)
\(\Leftrightarrow\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+....+\frac{1}{x\left(x+1\right)}=\frac{2017}{2019}:2\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{x}-\frac{1}{x+1}=\frac{2017}{4038}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2017}{4038}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{2017}{4038}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{2019}\)
=> x + 1 = 2019
=> x = 2018
\(\Rightarrow2\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2017}{2019}\)
\(\Rightarrow2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2017}{2019}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2017}{2019}:2\)
\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2017}{4038}\Rightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{2017}{4038}\Rightarrow\frac{1}{x+1}=\frac{1}{2019}\)
=>x+1=2019=>x=2018
\(\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+.......+\frac{2}{x\left(x+1\right)}=\frac{2017}{2019}\)
\(\Rightarrow2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+.........+\frac{1}{x\left(x+1\right)}\right)=\frac{2017}{2019}\)
\(\Rightarrow2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-...........-\frac{1}{\left(x+1\right)}\right)=\frac{2017}{2019}\)
\(\Rightarrow2\left(\frac{1}{2}-\frac{1}{\left(x+1\right)}\right)=\frac{2017}{2019}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{\left(x+1\right)}=\frac{2017}{2019}:2\)
\(\Rightarrow\frac{1}{2}-\frac{1}{\left(x+1\right)}=\frac{2017}{4038}\)
\(\Rightarrow\frac{1}{\left(x+1\right)}=\frac{1}{2}-\frac{2017}{4038}\)
\(\Rightarrow\frac{1}{\left(x+1\right)}=\frac{2019}{4038}-\frac{2017}{4038}\)
\(\Rightarrow\frac{1}{\left(x+1\right)}=\frac{2}{4038}=\frac{1}{2019}\)
\(\Rightarrow x=2018\)
Vậy x = 2018
\(\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{x.\left(x+1\right)}=\frac{2017}{2019}\)
\(2.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2017}{2019}\)
\(2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2017}{2019}\)
\(2.\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2017}{2019}\)
\(1-\frac{2}{x+1}=\frac{2017}{2019}\)
\(\frac{2}{x+1}=1-\frac{2017}{2019}\)
\(\frac{2}{x+1}=\frac{2}{2019}\)
\(\Rightarrow x+1=2019\)
\(x=2019-1\)
\(x=2018\)
Vậy \(x=2018\)
\(\frac{2}{2\cdot3}+\frac{2}{3\cdot4}+\frac{2}{4\cdot5}+...+\frac{2}{x\left(x+1\right)}=\frac{2017}{2019}\)
\(\Rightarrow2\left(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2017}{2019}\)
\(\Rightarrow2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2017}{2019}\)
\(\Rightarrow2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2017}{2019}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2017}{2019}:2\)
\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2017}{4038}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{2017}{4038}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{2019}\)
\(\Rightarrow x+1=2019\)
\(\Rightarrow x=2019-1\)
\(\Rightarrow x=2018\)
Vậy x=2018
