\(3A=3^2+3^3+...+3^{2020}\)
\(=>3A-A=3^{2020}-3^1\)
\(=>2A=3^{2020}-3\)
\(=>A=\frac{3^{2020}-3}{2}\)
Ta cs :\(2A+3=3^x\)
\(=>3^{2020}-3+3=3^x\)
\(=>3^{2020}=3^x\)
\(=>x=2020\)
Vậy ...
Ta có: A=3+32+33+...+32019
3A=32+32+34+..+32020
\(\Rightarrow\)3A-A=(32+33+34+...+32020)-(3+32+33+...+32019)
2A=32020-3
\(\Rightarrow\)2A=32020-3+3=32020
Mà 2A+3=3x
\(\Rightarrow\)x=2020
Vậy x=2020.
Mà 2A+3=3x