\(A=4\left(3^2+1\right)\left(3^4+1\right)....\left(3^{64}+1\right)\)
\(2A=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)....\left(3^{64}+1\right)\)
\(2A=\left(3^4-1\right)\left(3^4+1\right)....\left(3^{64}+1\right)\)
\(2A=\left(3^{16}-1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\left(3^{64}+1\right)\)
\(2A=\left(3^{32}-1\right)\left(3^{32}+1\right)\left(3^{64}+1\right)\)
\(2A=\left(3^{64}-1\right)\left(3^{64}+1\right)\)
\(2A=3^{128}-1\Rightarrow A=\dfrac{3^{128}-1}{2}< 3^{128}-1=B\)
Vậy \(A< B\)