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10 tháng 7 2018

a, M=2011.2013=2011.(2012+1)=2011.2012+2011

N=2012^2=2012.(2011+1)=2012.2011+2012

=>M<N

b, M=2015^2015+2015^2016=2015^2015.(1+2015)=2015^2015.2016

N=2016^2016=2016^2015.2016

=>M<N

k cho k nha

9 tháng 5 2021
5 tháng 9 2021

\(N=\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)...\left(2^{1008}+1\right)=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)...\left(2^{1008}+1\right)=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)...\left(2^{1008}+1\right)=2^{2016}-1< 2^{2016}=M\)

Ta có:

\(N=\left(1+2\right)\left(2-1\right)\left(2^2+1\right)\left(2^4+1\right)...\left(2^{2008}+1\right)\)

\(\Leftrightarrow N=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)...\left(2^{2008}+1\right)\)

\(\Leftrightarrow N=\left(2^4-1\right)\left(2^4+1\right)...\left(2^{2008}+1\right)\)

\(\Leftrightarrow N=\left(2^8-1\right)...\left(2^{2008}+1\right)\)

\(\Leftrightarrow N=2^{4016}-1>2^{2016}=M\)

 

 

2 tháng 9 2021

Ta có:

N=(1+2)(2−1)(22+1)(24+1)...(22008+1)N=(1+2)(2−1)(22+1)(24+1)...(22008+1)

⇔N=(22−1)(22+1)(24+1)...(22008+1)⇔N=(22−1)(22+1)(24+1)...(22008+1)

⇔N=(24−1)(24+1)...(22008+1)⇔N=(24−1)(24+1)...(22008+1)

⇔N=(28−1)...(22008+1)⇔N=(28−1)...(22008+1)

⇔N=24016−1>22016=M