Question

So Sánh A và B biết:\(\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{10}+1\right)\)và B=\(2^{32}-1\)

Answer 1

Mk sửa đề chút

\(A=\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(=\left(2^{16}-1\right)\left(2^{16}+1\right)\)

\(=2^{32}-1\) = B

\(=\)