Question

Chứng minh:

(2^1+1)(2^2+1)(2^4+1)(2^6+1)(2^8+1)=28^2-1

Answer 1

Biến đổi vế trái ta có:

\(\left(2^1+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^6+1\right)\left(2^8+1\right)\)

= \(\left(2-1\right)\left(2+1\right)\left(2^4+1\right)\left(2^6+1\right)\left(2^8+1\right)\)

= \(\left(2^4-1\right)\left(2^4+1\right)\left(2^6+1\right)\left(2^8+1\right)\)

= \(\left(2^8-1\right)\left(2^6+1\right)\left(2^8+1\right)\)

= \(\left(2^{16}-1\right)\left(2^6+1\right)\)

=> Sai đề

Answer 2

\(VT=\left(2^1+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^6+1\right)\left(2^8+1\right)\)

\(=\left(2^1-1\right)\left(2^1+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^6+1\right)\left(2^8+1\right)\)

\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^6+1\right)\left(2^8+1\right)\)

\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^6+1\right)\left(2^8+1\right)\)

\(=\left(2^{16}-1\right)\left(2^6+1\right)\left(2^8+1\right)\)

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