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a) \(A=\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)=\dfrac{1}{2}\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)=\dfrac{1}{2}\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)=\dfrac{1}{2}\left(3^{32}-1\right)< 3^{32}-1=B\)
b) \(A=2011.2013=\left(2012-1\right)\left(2012+1\right)=2012^2-1< 2012^2=B\)
a) Ta có : 2005.2007 = (2006 - 1)(2006 + 1) = 20062 - 12 = 20062 - 1 ( cái khúc này sửa : 2005.2001 thành 2005.2007)
Mà B = 20062
=> 20062 - 1 < 20062
=> A < B
b) Ta có : B = (2 + 1)(22 + 1)(24 + 1)(28 + 1)(216 + 1)
B = (2 - 1)(2 + 1)(22 + 1)(24 + 1)(28 + 1)(216 + 1)
B = (22 - 1)(22 + 1)(24 + 1)(28 + 1)(216 + 1)
B = (24 - 1)(24 + 1)(28 + 1)(216 + 1)
B = (28 - 1)(28 + 1)(216 + 1) = (216 - 1)(216 + 1) = 232 - 1
Mà C = 232
=> B < C
c) Tương tự như câu b
\(\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)=\dfrac{1}{2}\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)=\dfrac{1}{2}.\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)=\dfrac{1}{2}\left(3^{32}-1\right)=\dfrac{3^{32}}{2}-\dfrac{1}{2}\)
\(\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(=\dfrac{\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)}{2}\)
\(=\dfrac{3^{32}-1}{2}\)
Ta có:
\(A=\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(\Leftrightarrow2A=\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(=\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(=\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(=\left(3^{16}-1\right)\left(3^{16}+1\right)=3^{32}-1\)
\(\Leftrightarrow A=\frac{2A}{2}=\frac{3^{32}-1}{2}< B=3^{32}-1\)
Vậy A<B
\(\text{ (a+1)^2 - (a-1)^2 - 3(a+1)(a-1)}\)
\(=a^2+2a+1-a^2+2a-1-3a^2+3\)
\(=-3a^2+4a+3\)