Ta có : \(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{x\left(x+1\right)}=\frac{2018}{2019}\)
\(\Rightarrow\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+...+\left(\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2018}{2019}\)
\(\Rightarrow1-\frac{1}{x+1}=\frac{2018}{2019}\)
\(\Rightarrow\frac{1}{x+1}=1-\frac{2018}{2019}=\frac{1}{2019}\)
\(\Rightarrow x+1=2019\)
\(\Rightarrow x=2018\)
Vậy x = 2018
Nhớ t.i.c.k cho mình nha!
Chỗ \(x(x+1)\Rightarrow\frac{1}{x(x+1)}\) nhé
\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x(x+1)}=\frac{2018}{2019}\)
\(\Rightarrow\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{x(x+1)}=\frac{2018}{2019}\)
\(\Rightarrow1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2018}{2019}\)
\(\Rightarrow1-\frac{1}{x+1}=\frac{2018}{2019}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{2019}\Leftrightarrow x+1=2019\Leftrightarrow x=2018\)
\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+.....+\frac{1}{x.\left(x+1\right)}=\frac{2018}{2019}\)
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+.......+\frac{1}{x.\left(x+1\right)}=\frac{2018}{2019}\)
\(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+......+\frac{1}{x}-\frac{1}{x+1}=\frac{2018}{2019}\)
\(\frac{1}{1}-\frac{1}{x+1}=\frac{2018}{2019}\)
\(\frac{1}{x+1}=\frac{2018}{2019}\)
=>x+1 =2019
=> x = 2018
