Rút gọn biểu thức: 

- Hướng dẫn giải

$$\begin{gathered} 3\left(2^2+1\right)\left(2^2+1\right)\left(2^8+1\right)\left(2^{16}+1\right) \\ =\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right) \\ (vì 2^2-1=4-1=3) \\ =\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right) \\ =\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right) \\ =\left(2^{16}-1\right)\left(2^{16}+1\right) \\ =2^{32}-1 \end{gathered}$$