So sánh M =

- Hướng dẫn giải

Ta có

$$\begin{gathered} N = (2 + 1)(2^2 + 1)(2^4 + 1)(2^8 + 1)(2^{16} + 1) (2^{16} + 1) \\ = 3(2^2 + 1)(2^4 + 1)(2^8 + 1) (2^{16} + 1) \\ = \left[\right(2^2 – 1\left)\right(2^2 + 1\left)\right](2^4 + 1)(2^8 + 1)(2^{16} + 1) \\ = (2^4 – 1)(2^4 + 1)(2^8 + 1)(2^{16} + 1) \\ = (2^8 – 1)(2^8 + 1)(2^{16} + 1) \\ = (2^{16} - 1)(2^{16} + 1) \\ =\left({\left(2^{16}\right)}^2-1\right)=2^{32}-1 \\ Mà 2^{32}-1>2^{32}\Rightarrow N<M \end{gathered}$$

Đáp án cần chọn là: A