a ) \(A=2015.2017=\left(2016-1\right)\left(2016+1\right)=2016^2-1\)
Do \(2016^2>2016^2-1\)
\(\Rightarrow B>A\)
b ) \(C=\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^{16}-1\right)\left(2^{16}+1\right)\)
\(=2^{32}-1< 2^{32}=D\)
Vậy \(C< D\)
so sánh :a)A=2015.2017 va B=20162
Ta có: A = 2015.2017 = (2016-1)(2016+1)
= 20162-1<20162
=> A < B
