Ta có:

\(M=\dfrac{100^{100}+1}{100^{99}+1}\)

\(\Rightarrow\dfrac{M}{100}=\dfrac{100^{100}+1}{100\cdot\left(100^{99}+1\right)}\)

\(\Rightarrow\dfrac{M}{100}=\dfrac{100^{100}+1}{100^{100}+100}\)

\(\Rightarrow\dfrac{M}{100}=1-\dfrac{99}{100^{100}+100}\) 

\(N=\dfrac{100^{101}+1}{100^{100}+1}\)

\(\Rightarrow\dfrac{N}{100}=\dfrac{100^{101}+1}{100\cdot\left(100^{100}+1\right)}\)

\(\Rightarrow\dfrac{N}{100}=\dfrac{100^{101}+1}{100^{101}+100}\)

\(\Rightarrow\dfrac{N}{100}=1-\dfrac{99}{100^{101}+100}\)

Mà: \(100^{101}>100^{100}\)

\(\Rightarrow100^{101}+100>100^{100}+100\)

\(\Rightarrow\dfrac{99}{100^{101}+100}< \dfrac{99}{100^{100}+100}\)

\(\Rightarrow1-\dfrac{99}{101^{101}+100}< 1-\dfrac{99}{100^{100}+100}\)

\(\Rightarrow\dfrac{N}{100}< \dfrac{M}{100}\)

\(\Rightarrow N< M\)

M= \(\frac{100^{100}+1}{100^{99}+1}=\frac{100^{100}+100-99}{100^{99}+1}=\frac{100^{100}+100}{100^{99}+1}-\frac{99}{100^{99}+1}=\frac{100.\left(100^{99}+1\right)}{100^{99}+1}-\frac{99}{100^{99}+1}\)

\(=100-\frac{99}{100^{99}+1}\)

N= \(\frac{100^{101}+1}{100^{100}+1}=\frac{100^{101}+100-99}{100^{100}+1}=\frac{100^{101}+100}{100^{100}+1}-\frac{99}{100^{100}+1}\)

\(=\frac{100.\left(100^{100}+1\right)}{100^{100}+1}-\frac{99}{100^{100}+1}=100-\frac{99}{100^{100}+1}\)

Vi 100¹⁰⁰+1>100⁹⁹+1

=> \(\frac{99}{100^{99}+1}>\frac{99}{100^{100}+1}\)

=> \(100-\frac{99}{100^{99}+1}

uk ai cũng có lúc nhầm mà chẳng sao đâu bạn ak

\(100^{99}+1< 100^{100}+1\)

=>A>B

Ta có: Theo cách tính phân số dư , phân số nào có phần dư lớn hơn thì lớn hơn.

\(\frac{100^{^{100^{ }}}+1}{100^{99}+1}\)\(-1\)=\(\frac{100^{100}}{100^{99}+1}-100^{99}\)

\(\frac{100^{101}+1}{100^{100}+1}-1=\frac{100^{101}-100^{100}}{100^{100}+1}\)

Suy ra:A>B

a) Áp dụng \(\frac{a}{b}< 1\Leftrightarrow\frac{a}{b}< \frac{a+m}{b+m}\) (a;b;m \(\in\) N*)

Ta có:

\(A=\frac{2008^{2008}+1}{2008^{2009}+1}< \frac{2008^{2008}+1+2007}{2009^{2009}+1+2007}\)

\(A< \frac{2008^{2008}+2008}{2008^{2009}+2008}\)

\(A< \frac{2008.\left(2008^{2007}+1\right)}{2008.\left(2008^{2008}+1\right)}=\frac{2008^{2007}+1}{2008^{2008}+1}=B\)

=> A < B

b) Áp dụng \(\frac{a}{b}>1\Leftrightarrow\frac{a}{b}>\frac{a+m}{b+m}\) (a;b;m \(\in\) N*)

Ta có: 

\(N=\frac{100^{101}+1}{100^{100}+1}>\frac{100^{101}+1+99}{100^{100}+1+99}\)

\(N>\frac{100^{101}+100}{100^{100}+100}\)

\(N>\frac{100.\left(100^{100}+1\right)}{100.\left(100^{99}+1\right)}=\frac{100^{100}+1}{100^{99}+1}=M\)

=> M > N

1

\(A=\frac{2019^{2019}+1}{2019^{2020}+1}< \frac{2019^{2019}+1+2018}{2019^{2020}+1+2018}=\frac{2019^{2019}+2019}{2019^{2020}+2019}=\frac{2019\left(2019^{2018}+1\right)}{2019\left(2019^{2019}+1\right)}\)

\(=\frac{2019^{2018}+1}{2019^{2019}+1}\)

2

\(M=\frac{100^{101}+1}{100^{100}+1}< \frac{100^{101}+1+99}{100^{100}+1+99}=\frac{100^{101}+100}{100^{100}+100}=\frac{100\left(100^{100}+1\right)}{100\left(100^{99}+1\right)}\)

\(=\frac{100^{100}+1}{100^{99}+1}=N\)

\(\dfrac{1}{2022}\cdot A=\dfrac{2022^{100}+1}{2022^{100}+100}=1-\dfrac{99}{2022^{100}+100}\)

\(\dfrac{1}{2022}B=\dfrac{2022^{101}+1}{2022^{101}+100}=1-\dfrac{9}{2022^{101}+100}\)

2022^100+100<2022^101+100

=>-99/2022^100+100<-99/2022^101+100

=>A<B

=> A/2022 = 2022^100+1/2022^100+2022 = 1- 2021/2022^100+2022

=> B/2022 = 2022^101+1/2022^101+2022 = 1- 2021/2022^101+2022

Nhận thấy 2022^101 + 2022 > 2022^100 + 2022

=> 2021/2022^101 + 2022 < 2021/2022^100 + 2022

=> B/2022 > A/2022 => B>A

Vậy A<B

Ta có: \(A=\frac{2017^{99}+1}{2017^{100}+1}\Rightarrow2017A=\frac{2017^{100}+2017}{2017^{100}+1}=1+\frac{2016}{2017^{100}+1}\)

\(B=\frac{2017^{100}+1}{2017^{101}+1}\Rightarrow2017B=\frac{2017^{101}+2017}{2017^{101}+1}=1+\frac{2016}{2017^{101}+1}\)

\(\frac{2016}{2017^{100}+1}>\frac{2016}{2017^{101}+1}\Rightarrow1+\frac{2016}{2017^{100}+1}>1+\frac{2016}{2017^{101}+1}\)

\(\Rightarrow2017A>2017B\Rightarrow A>B\)

Vậy...

Đặt \(A=\frac{2017^{99}+1}{2017^{100}+1}\)nên \(2017A=\frac{2017^{100}+2017}{2017^{100}+1}=\frac{2017^{100}+1+2016}{2017^{100}+1}=1+\frac{2016}{2017^{100}+1}\)

\(B=\frac{2017^{100}+1}{2017^{101}+1}\)nên \(2017B=\frac{2017^{101}+2017}{2017^{101}+1}=\frac{2017^{101}+1+2016}{2017^{101}+1}=1+\frac{2016}{2017^{101}+1}\)

Vì \(1=1;\frac{2016}{2017^{100}+1}>\frac{2016}{2017^{101}+1}\Rightarrow1+\frac{2016}{2017^{100}+1}>1+\frac{2016}{2017^{101}+1}\)

Hay \(2017A>2017B\)nên \(A>B\)

Vây \(\frac{2017^{99}+1}{2017^{1001}+1}>\frac{2017^{100}+1}{2017^{101}+1}\)

đặt \(A=\frac{2017^{99}+1}{2017^{100}+1}\); \(B=\frac{2017^{100}+1}{2017^{101}+1}\)

Ta có : \(2017A=\frac{2017.\left(2017^{99}+1\right)}{2017^{100}+1}=\frac{2017^{100}+2017}{2017^{100}+1}=\frac{2017^{100}+1+2016}{2017^{100}+1}=1+\frac{2016}{2017^{100}+1}\)

\(2017B=\frac{2017.\left(2017^{100}+1\right)}{2017^{101}+1}=\frac{2017^{101}+2017}{2017^{101}+1}=\frac{2017^{101}+1+2016}{2017^{101}+1}=1+\frac{2016}{2017^{101}+1}\)

Vì \(\frac{2016}{2017^{100}+1}>\frac{2016}{2017^{101}+1}\Rightarrow1+\frac{2016}{2017^{100}+1}>1+\frac{2016}{2017^{101}+1}\Leftrightarrow10A>10B\Rightarrow A>B\)

Ta có:

\(\frac{1}{2}< \frac{2}{3}\)

\(\frac{3}{4}< \frac{4}{5}\)

\(\frac{5}{6}< \frac{6}{7}\)

\(...\)

\(\frac{99}{100}< \frac{100}{101}\)

\(\Rightarrow\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{99}{100}< \frac{2}{3}.\frac{4}{5}.\frac{6}{7}...\frac{100}{101}\)

\(\Rightarrow M< N\)