Question

Cho\(A=2^{32}+1;B=\left(2+1\right)\left(2^2+1\right)\left(2^8+1\right)\left(2^{16+1}\right)\)

so sánh A Và B

Answer 1

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

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

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

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

\(=\dfrac{1}{17}\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(=\dfrac{1}{17}\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(=\dfrac{1}{17}\left(2^{16}-1\right)\left(2^{16}+1\right)\)

\(=\dfrac{1}{17}\left(2^{32}-1\right)< A\)

Vậy \(B< A\)