Câu hỏi của Anh GoBi

Question

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

Hiiiii~ · Accepted Answer

Giải:

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

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

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

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

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

$=2^{32}-1$

Vậy ...Câu trả lời: Giải:

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

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

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

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

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

$=2^{32}-1$

Vậy ...

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