Ta có A = \(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2019}}\)(1)
=> 3A = \(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2018}}\)(2)
Lấy (2) trừ (1) theo vế ta có :
3A - A = \(\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2018}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2019}}\right)\)
2A = \(1-\frac{1}{3^{2019}}\)
Khi đó : \(\left(2A+\frac{1}{3^{2019}}\right).x=2\)
\(\Leftrightarrow\left(1-\frac{1}{3^{2019}}+\frac{1}{3^{2019}}\right).x=2\)
\(\Rightarrow x=2\)
