Question

A= 4 x ( 3^2+1 ) x ( 3^4+1 ) x ( 3^8+1 ) x ( 3^16+1 ) và B= 3 ^32 -1

Answer 1

Có: \(A=4\cdot\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)

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

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

          \(=...........................\)

           \(=\frac{3^{32}-1}{2}\)

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

=> \(A< B\)