\(B=...=\frac{\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)}{2}=\frac{\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)}{2}\)
\(=\frac{\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)}{2}=\frac{\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)}{2}=\frac{\left(3^{16}-1\right)\left(3^{16}+1\right)}{2}=\frac{3^{32}-1}{2}< 3^{32}-1=A\)
Ta có : \(B=\left(3+1\right).\left(3^2+1\right).\left(3^4+1\right).\left(3^8+1\right)+\left(3^{16}+1\right)\)
\(\Rightarrow\) \(2B=2.\left(3+1\right).\left(3^2+1\right).\left(3^4+1\right).\left(3^8+1\right).\left(3^{16}+1\right)\)
\(=\left(3-1\right).\left(3+1\right).\left(3^2+1\right).\left(3^4+1\right).\left(3^8+1\right).\left(3^{16}+1\right)\)
\(=\left(3^2-1\right).\left(3^2+1\right).\left(3^4+1\right).\left(3^8+1\right).\left(3^{16}+1\right)\)
\(=\left(3^4-1\right).\left(3^4+1\right).\left(3^8+1\right).\left(3^{16}+1\right)\)
\(=\left(3^8-1\right).\left(3^8+1\right).\left(3^{16}+1\right)\)
\(=\left(3^{16}-1\right).\left(3^{16}+1\right)\)
\(=3^{32}-1\)
\(\Rightarrow\) \(B=\frac{3^{32}-1}{2}< 3^{32}-1\)
\(\Rightarrow\) \(B< A\)