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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) \(A=2011.2013=\left(2012-1\right)\left(2012+1\right)=2012^2-1< 2012^2=B\)
\(A=\left(2012-1\right)\left(2012+1\right)-2012^2=2012^2-1-2012^2=-1\)
\(A=\left(2012-1\right)\left(2012+1\right)-2012^2=2012^2-1-2012^2=-1\)
a) \(A=1999\cdot2001=\left(2000-1\right)\left(2000+1\right)=2000^2-1\)
=> \(A< B\)
b) \(A=12^6\)
\(B=\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)
\(=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)
\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\)
\(=\left(2^8-1\right)\left(2^8+1\right)=2^{16}-1\)
=> \(A>B\)
c) \(A=2011\cdot2013=\left(2012-1\right)\left(2012+1\right)=2012^2-1\)
\(B=2012^2\)
=> \(A< B\)
d) \(A=4\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)
\(=\frac{\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)}{2}\)
\(=\frac{\left(3^4-1\right)\left(3^4+1\right)..\left(3^{64}+1\right)}{2}\)
\(=\frac{\left(3^8-1\right).....\left(3^{64}+1\right)}{2}\)
\(=\frac{3^{128}-1}{2}\)
\(B=3^{128}-1\)
=> \(A< B\)
2011.2013+2012.2014
=(2013-2).2013+2012.(2012+2)
=20132-4026+20122+4024
=20132+20122+(-4026+4024)
=20132+20122-2
Ta có:\(2011.2013+2012.2014\)
\(=\left(2013-2\right).2013+\left(2012+2\right).2012\)
\(=2013^2-4026+2012^2+4024\)
\(=2012^2+2013^2-2\)
nên hai phép tính trên bằng nhau.
a) \(A=1999.2001=\left(2000-1\right)\left(2000+1\right)=2000^2-1< 2000^2=B\)
b) \(B=\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)
\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\)
\(=\left(2^8-1\right)\left(2^8+1\right)\)
\(=2^{16}-1< 2^{16}=A\)
c) Tương tự a).
d) Tương tự b).
Ta có : \(\hept{\begin{cases}A=1999.2001\\B=2000^2\end{cases}}\)
\(< =>\hept{\begin{cases}A=1999.2000+1999\\B=2000\cdot2000\end{cases}}\)
\(< =>\hept{\begin{cases}A=1999.2000+2000+1\\B=1999.2000+2000\end{cases}}\)
\(< =>\hept{\begin{cases}A=2000.2000+1\\B=2000.2000\end{cases}}\)
\(< =>A>B\)
a. Ta có : \(A=1999.2021=\left(2000-1\right)\left(2000+1\right)=2020^2-1< 2020\)
\(\Rightarrow A< B\)
b. Ta có : \(B=\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)
\(=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)
...
\(=\left(2^8-1\right)\left(2^8+1\right)=2^{16}-1< 2^{16}\)
\(\Rightarrow A>B\)
c,d tương tự
A = 2011.2013
A = 2011.(2012+1)
A = 2011.2012+2011
B = 20122
B = 2012.2012
B = (2011+1).2012
B = 2011.2012+2012
Vì 2011 < 2012
=> 2011.2012 + 2011 < 2011.2012 + 2012
=> A < B
Vậy a<b nha bạn