Question

So sánh hai số:

\(A=3^{32}-1\)

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

Answer 1

Ta có: \(A=3^{32}-1=\left(3^{16}+1\right)\left(3^{16}-1\right)\)

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

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

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

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

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

Vậy A = 2B