Bài 1:
Ta có: \(F=\left(2^2+1\right)\left(2^4+1\right)\cdot\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{16}+1\right)\)
\(=\dfrac{\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{16}+1\right)}{3}\)
\(=\dfrac{\left(2^4+1\right)\left(2^4-1\right)\left(2^8+1\right)\left(2^{16}+1\right)}{3}\)
\(=\dfrac{\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)}{3}\)
\(=\dfrac{\left(2^{16}-1\right)\left(2^{16}+1\right)}{3}\)
\(=\dfrac{2^{32}-1}{3}\)
Đúng 2
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