Giải:
a) \(A=\left(3^8+1\right)\left(3^4+1\right)\left(3^2+1\right)\left(3+1\right)\)
\(\Leftrightarrow2A=2\left(3^8+1\right)\left(3^4+1\right)\left(3^2+1\right)\left(3+1\right)\)
\(\Leftrightarrow2A=\left(3^8+1\right)\left(3^4+1\right)\left(3^2+1\right)\left(3+1\right)\left(3-1\right)\)
\(\Leftrightarrow2A=\left(3^8+1\right)\left(3^4+1\right)\left(3^2+1\right)\left(3^2-1\right)\)
\(\Leftrightarrow2A=\left(3^8+1\right)\left(3^4+1\right)\left(3^4-1\right)\)
\(\Leftrightarrow2A=\left(3^8+1\right)\left(3^8-1\right)\)
\(\Leftrightarrow2A=3^{16}-1\)
\(\Leftrightarrow A=\dfrac{3^{16}-1}{2}\)
Vậy ...
b) \(B=12\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)...\left(5^{32}+1\right)\)
\(\Leftrightarrow2B=2.12\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)...\left(5^{32}+1\right)\)
\(\Leftrightarrow2B=24\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)...\left(5^{32}+1\right)\)
\(\Leftrightarrow2B=\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)...\left(5^{32}+1\right)\)
\(\Leftrightarrow2B=\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)...\left(5^{32}+1\right)\)
...
\(\Leftrightarrow2B=\left(5^{32}-1\right)\left(5^{32}+1\right)\)
\(\Leftrightarrow2B=5^{64}-1\)
\(\Leftrightarrow B=\dfrac{5^{64}-1}{2}\)
Vậy ...
