Đáp án: $A>B$
Giải thích các bước giải:
$B=(3+1)(3^2+1)(3^4+1)(3^8+1)(3^{16}+1)$
$\rightarrow B.(3-1)=(3-1)(3+1)(3^2+1)(3^4+1)(3^8+1)(3^{16}+1)$
$\rightarrow 2B=(3^2-1)(3^2+1)(3^4+1)(3^8+1)(3^{16}+1)$
$\rightarrow 2B=(3^4-1)(3^4+1)(3^8+1)(3^{16}+1)$
$\rightarrow 2B=(3^8-1)(3^8+1)(3^{16}+1)$
$\rightarrow 2B=(3^{16}-1)(3^{16}+1)$
$\rightarrow 2B=3^{32}-1$
$\rightarrow 2B=A$
$\rightarrow \dfrac{A}{B}=2>1$
$\rightarrow A>B$
