Bài 5.
a) Ta có:
\(A=\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(=\dfrac{1}{2}.2\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(=\dfrac{1}{2}\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(=\dfrac{1}{2}\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(=\dfrac{1}{2}\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(=\dfrac{1}{2}\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(=\dfrac{1}{2}\left(3^{16}-1\right)\left(3^{16}+1\right)\)
\(=\dfrac{1}{2}\left(3^{32}-1\right)\)
Do \(1< 2\Rightarrow\dfrac{1}{2}< \dfrac{2}{2}\Rightarrow\dfrac{1}{2}< 1\Rightarrow\dfrac{1}{2}\left(3^{32}-1\right)< 3^{32}-1\Rightarrow A< B\)
b) Ta có:
\(A=2011.2013=\left(2012-1\right)\left(2012+1\right)=2012^2-1\)
Do \(-1< 0\Rightarrow2012^2-1< 2012^2\Rightarrow A< B\)
