A=4(3^2+1)(3^4+1)(3^8+1)...(3^64+1)
2A=8(3^2+1)(3^4+1)(3^8+1)...(3^64+1)
2A=(3^2-1)(3^2+1)(3^4+1)(3^8+1)...(3^64+1)
2A=(3^4-1)(3^4+1)(3^8+1)...(3^64+1)
2A=(3^8-1)(3^8+1)....(3^64+1)
2A=(3^16-1)...(3^64+1)
......
2A=(3^64-1)(3^64+1)
2A=3^128-1
A=(3^128-1)/2
=> A>B
\(A=4\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)
\(\Leftrightarrow4A=\left(3^2-1\right)\left(3^2+1\right)...\left(3^{64}+1\right)\)
\(\Leftrightarrow4A=\left(3^4-1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)
\(\Leftrightarrow4A=\left(3^8-1\right)\left(3^8+1\right)...\left(3^{64}+1\right)\)
\(\Leftrightarrow4A=\left(3^{16}-1\right)\left(3^{16}+1\right)...\left(3^{64}+1\right)\)
\(\Leftrightarrow4A=\left(3^{32}-1\right)\left(3^{32}+1\right)\left(3^{64}+1\right)\)
\(\Leftrightarrow4A=\left(3^{64}-1\right)\left(3^{64}+1\right)\Leftrightarrow4A=3^{128}-1\Leftrightarrow A=\frac{3^{128}-1}{4}\)
Ta có \(\frac{3^{128}-1}{4}< 3^{128}-1\Rightarrow A< B\)
Lâm Huyền:Bạn sai đề rồi B phải là 3128-1 chứ !
Sorry,mình tính sai.Bạn thay 4A thành 2A va các số 4 thành số 2 nhé
