Giải thích các bước giải:
a,
Ta có:
\[\begin{array}{l}
A = 2 + {2^2} + {2^3} + ... + {2^{2019}}\\
\Leftrightarrow 2A = {2^2} + {2^3} + {2^4} + .... + {2^{2020}}\\
\Leftrightarrow 2A - A = \left( {{2^2} + {2^3} + {2^4} + ... + {2^{2020}}} \right) - \left( {2 + {2^2} + {2^3} + ... + {2^{2019}}} \right)\\
\Leftrightarrow A = {2^{2020}} - 2
\end{array}\]
\[\begin{array}{l}
A + 2 = {2^x}\\
\Leftrightarrow {2^{2020}} - 2 + 2 = {2^x}\\
\Leftrightarrow {2^{2020}} = {2^x}\\
\Rightarrow x = 2020
\end{array}\]
b,
\[\begin{array}{l}
{\left( {2x - 1} \right)^{2018}} = 2x - 1\\
\Rightarrow 2x - 1 = 0\\
\Rightarrow x = \frac{1}{2}
\end{array}\]
