\(N=\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(=\left(2^{16}-1\right)\left(2^{16}+1\right)=2^{32}-1\)

=>N<M

Vì N<1

=> N= 20^31+2/20^32+2

<20^31+2+38/ 20^32+2+38

=20^31+40/ 20^32+40

=20.(20^30+2) / 20.(20^31+2)

=20^30+2 / 20^32+2 = M

Vậy N<M

\(N=\frac{20^{31}+2}{20^{32}+2}=\frac{20^{31}+2+18}{20^{32}+2+18}=\frac{20^{31}+20}{20^{32}+20}=\frac{10.\left(20^{30}+2\right)}{10.\left(20^{31}+2\right)}\)\(=M\)

\(\Rightarrow M=N\)
 

\(M=\frac{10^{30}+1}{10^{31}+1}\)

\(\Rightarrow10M=\frac{10\cdot(10^{30}+1)}{10^{31}+1}\)

\(\Rightarrow10M=\frac{10^{31}+10}{10^{31}+1}\)

\(\Rightarrow10M=\frac{10^{31}+1+9}{10^{31}+1}\)

\(\Rightarrow10M=1+\frac{9}{10^{31}+1}\)

\(N=\frac{10^{31}+1}{10^{32}+1}\)

\(\Rightarrow10N=\frac{10\cdot(10^{31}+1)}{10^{32}+1}\)

\(\Rightarrow10N=\frac{10^{32}+10}{10^{32}+1}\)

\(\Rightarrow10N=\frac{10^{32}+1+9}{10^{32}+1}\)

\(\Rightarrow10N=1+\frac{9}{10^{32}+1}\)

Mà\(1+\frac{9}{10^{31}+1}>1+\frac{9}{10^{32}+1}\)

Nên \(10M>10N\)

Hay \(M>N\)

a)>

b)<

Bài 1: a)  \(M=1+5+5^2+...+5^{100}\)

\(5M=5+5^2+5^3+...+5^{101}\)

\(5M-M=\left(5+5^2+5^3+...+5^{101}\right)-\left(1+5+5^2+...+5^{100}\right)\)

\(4M=5^{101}-1\)

\(M=\frac{5^{101}-1}{4}\)

b) \(N=2+2^2+...+2^{100}\)

\(2N=2^2+2^3+...+2^{101}\)

\(2N-N=\left(2^2+2^3+...+2^{101}\right)-\left(2+2^2+...+2^{100}\right)\)

\(N=2^{101}-2\)

Bài 2:

a) \(16^{32}=\left(2^4\right)^{32}=2^{128}\) 

\(32^{16}=\left(2^5\right)^{16}=2^{80}\)

Vì \(2^{128}>2^{80}\Rightarrow16^{32}>32^{16}\)

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