Câu 6:
1: Ta có: \(B=6\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(=\dfrac{\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)}{4}\)
\(=\dfrac{\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)}{4}\)
\(=\dfrac{5^{32}-1}{4}< 5^{32}-1\)
hay A>B
Câu 61
Ta có: \(B=6\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
⇒ \(4B=24\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(\Rightarrow4B=\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
⇒ \(4B=\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
⇒ \(4B=\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
⇒ \(4B=\left(5^{16}-1\right)\left(5^{16}+1\right)\)
⇒ \(4B=5^{32}-1\)
⇒ \(B=\dfrac{5^{32}-1}{4}\)
Vì \(5^{32}-1>\dfrac{5^{32}-1}{4}\) nên A>B