Question

rút gọn biểu thức sau:

\((2^2+1).(2^4+1).(2^8+1).(2^{16}+1)\)

Answer 1

\(A=\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

=>  \(3A=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

  \(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

  \(=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

  \(=\left(2^{16}-1\right)\left(2^{16}+1\right)=2^{32}-1\)

=>  \(A=\frac{2^{32}-1}{3}\)