Đáp án: x=3
Giải thích các bước giải:
$\begin{array}{l}
{2^x} + {2^{x + 1}} + {2^{x + 2}} + {2^{x + 3}} + ... + {2^{x + 2015}} = {2^{2019}} - 8\\
\Rightarrow {2^x}\left( {1 + {2^1} + {2^2} + ... + {2^{2015}}} \right) = {2^{2019}} - 8\\
ĐẶt\,A = 1 + {2^1} + {2^2} + ... + {2^{2015}}\\
\Rightarrow {2^x}.A = {2^{2019}} - 8\\
\Rightarrow 2A = 2\left( {1 + {2^1} + {2^2} + ... + {2^{2015}}} \right)\\
\Rightarrow 2A = 2 + {2^2} + {2^3} + ... + {2^{2015}} + {2^{2016}}\\
\Rightarrow 2A - A = \left( {2 + {2^2} + {2^3} + ... + {2^{2015}} + {2^{2016}}} \right)\\
- \left( {1 + {2^1} + {2^2} + ... + {2^{2015}}} \right)\\
\Rightarrow A = {2^{2016}} - 1\\
\Rightarrow {2^x}.\left( {{2^{2016}} - 1} \right) = {2^{2019}} - 8\\
\Rightarrow {2^{x + 2016}} - {2^x} = {2^{2019}} - {2^3}\\
\Rightarrow \left\{ \begin{array}{l}
x + 2016 = 2019\\
x = 3
\end{array} \right. \Rightarrow x = 3\\
Vậy\,x = 3
\end{array}$
