Đáp án:
A<$2^{32}$
Giải thích các bước giải:
\(
\begin{array}{l}
A=(2 + 1)(2^2 + 1)(2^4 + 1)(2^8 + 1)(2^{16} + 1) \\
= 1(2 + 1)(2^2 + 1)(2^4 + 1)(2^8 + 1)(2^{16} + 1) \\
= (2 - 1)(2 + 1)(2^2 + 1)(2^4 + 1)(2^8 + 1)(2^{16} + 1) \\
= (2^2 - 1)(2^2 + 1)(2^4 + 1)(2^8 + 1)(2^{16} + 1) \\
= (2^4 - 1)(2^4 + 1)(2^8 + 1)(2^{16} + 1) \\
= (2^8 - 1)(2^8 + 1)(2^{16} + 1) \\
= (2^{16} - 1)(2^{16} + 1) \\
= 2^{32} - 1 < 2^{32} \\
\end{array}
\)
