Câu hỏi của Nguyễn Mạnh Hùng

Question

Rút gọn biểu thức:

$B=(2^2+1).(2^4+1).(2^8+1).(2^{16}+1)$

Không Tên · Accepted Answer

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

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

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

$=\frac{\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)}{3}$

$=\frac{\left(2^{16}-1\right)\left(2^{16}+1\right)}{3}$

$=\frac{2^{32}-1}{3}$

Câu trả lời:

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

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

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

$=\frac{\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)}{3}$

$=\frac{\left(2^{16}-1\right)\left(2^{16}+1\right)}{3}$

$=\frac{2^{32}-1}{3}$