Ta có : A = \(\frac{10^{2020}+1}{10^{2021}+1}\)

=> 10A = \(\frac{10^{2021}+10}{10^{2021}+1}=1+\frac{9}{10^{2021}+1}\)

Lại có : \(B=\frac{10^{2021}+1}{10^{2022}+1}\)

=> \(10B=\frac{10^{2022}+10}{10^{2022}+1}=1+\frac{9}{10^{2022}+1}\)

Vì \(\frac{9}{10^{2022}+1}< \frac{9}{10^{2021}+1}\)

=> \(1+\frac{9}{10^{2022}+1}< 1+\frac{9}{10^{2022}+1}\)

=> 10B < 10A

=> B < A

b) Ta có : \(\frac{2019}{2020+2021}< \frac{2019}{2020}\)

Lại có : \(\frac{2020}{2020+2021}< \frac{2020}{2021}\)

=> \(\frac{2019}{2020+2021}+\frac{2020}{2020+2021}< \frac{2019}{2020}+\frac{2020}{2021}\)

=> \(\frac{2019+2020}{2020+2021}< \frac{2019}{2020}+\frac{2020}{2021}\)

=> B < A

a) A=10^2020+1/10^2021+1 < 10^2020+1+9/10^2022+1+9 =         

10.(10^2021+1)/10.(10^2022+1) = 10^2021+1/10^2022+1 = B

Vậy A < B.

Ta có \(b-a=9.10^{2019}-\dfrac{9}{10^{2021}}>0\Rightarrow b>a\).

A = \(\dfrac{5^{2020}+1}{5^{2021}+1}\) ⇒ A \(\times\) 10 = 2 \(\times\)5 \(\times\) \(\dfrac{5^{2020}+1}{5^{2021}+1}\) =2\(\times\) \(\dfrac{5^{2021}+5}{5^{2021}+1}\)

10A =2 \(\times\) \(\dfrac{5^{2021}+5}{5^{2021}+1}\) = 2 \(\times\)(1 + \(\dfrac{4}{5^{2021}+1}\) )= 2 + \(\dfrac{8}{5^{2021}+1}\) >2

B = \(\dfrac{10^{2019}+1}{10^{2020}+1}\) ⇒ B \(\times\) 10 = 10 \(\times\) \(\dfrac{10^{2019}+1}{10^{2020}+1}\)= \(\dfrac{10^{2020}+10}{10^{2020}+1}\)

10B = \(\dfrac{10^{2020}+10}{10^{2020}+1}\) = 1 + \(\dfrac{9}{10^{2020}+1}\) < 2

10A > 2 > 10B ⇒ 10A>10B ⇒ A>B

ta có : 

A = \(\dfrac{5^{2020}+1}{5^{2020}+1}\)

B = \(\dfrac{5^{2019}+1}{5^{2020}+1}\)

\(\Leftrightarrow\) B < A

a) Ta có : \(\frac{-60}{12}=-5=-\frac{25}{5}\)

\(-0,8=-\frac{8}{10}=-\frac{4}{5}\)

Mà -25 < -4 nên \(\frac{-25}{5}< \frac{-4}{5}\)=> \(\frac{-60}{12}< -0,8\)

b) Ta có : \(\frac{2020}{2019}=1+\frac{1}{2019}\)

\(\frac{2021}{2020}=1+\frac{1}{2020}\)

Vì \(\frac{1}{2019}>\frac{1}{2020}\)nên \(\frac{2020}{2019}>\frac{2021}{2020}\)

c) \(\frac{10^{2018}+1}{10^{2019}+1}=\frac{10\left(10^{2018}+1\right)}{10^{2019}+1}=\frac{10^{2019}+10}{10^{2019}+1}=\frac{10^{2019}+1+9}{10^{2019}+1}=1+\frac{9}{10^{2019}+1}\)(1)

\(\frac{10^{2019}+1}{10^{2020}+1}=\frac{10\left(10^{2019}+1\right)}{10^{2020}+1}=\frac{10^{2020}+10}{10^{2020}+1}=\frac{10^{2020}+1+9}{10^{2020}+1}=1+\frac{9}{10^{2020}+1}\)(2)

Đến đây tự so sánh rồi nhé

 ta có: M=10^2020 +1 / 10^2019 +1

=> M/10= 10^2020 +1 / 10( 10^2019 +1 )

= 10^2020+1/ 10^2020 +10

=>  10/A=  10^2020 +10/10^2020 +1

=(10^2020 +1) +9/ 10^2020+1

=10^2020+1 /10^2020+1 + 9/10^2020+1

=1+ 9/10^2020+1

ta lại có: N=10^2021 +1/10^2020 +1

=> N/10= 10^2021+1/ 10(10^2020+1)

= 10^2021+1 / 10^2021+10

=> 10/N=10^2021+10 / 10^2021+1

=(10^2021+1) +9/10^2021+1

=10^2021+1/10^2021+1 +9/10^2021+1

=1+ 9/10^2021+1

ta thấy: 10/M>10N

=>M<N

\(M=\dfrac{10^{2020}+1}{10^{2019}+1}=1-\dfrac{9}{10^{2019}+1}\)

\(N=\dfrac{10^{2021}+1}{10^{2020}+1}=1-\dfrac{9}{10^{2020}+1}\)

Ta có: \(10^{2019}+1< 10^{2020}+1\)

\(\Leftrightarrow\dfrac{9}{10^{2019}+1}>\dfrac{9}{10^{2020}+1}\)

\(\Leftrightarrow-\dfrac{9}{10^{2019}+1}< -\dfrac{9}{10^{2020}+1}\)

\(\Leftrightarrow M< N\)

Lời giải:
$10A=\frac{10^{2021}-10}{10^{2021}-1}=\frac{10^{2021}-1-9}{10^{2021}-1}$

$=1-\frac{9}{10^{2021}-1}>1$

$10B=\frac{10^{2022}+10}{10^{2022}+1}=\frac{10^{2022}+1+9}{10^{2022}+1}$

$=1+\frac{9}{10^{2022}+1}<1$

$\Rightarrow 10A> 1> 10B$

Suy ra $A> B$

a) Ta có A = \(\frac{2^{2018}+1}{2^{2019}+1}\)

=> 2A = \(\frac{2^{2019}+2}{2^{2019}+1}=1+\frac{1}{2^{2019}+1}\)

Lại có B = \(\frac{2^{2017}+1}{2^{2018}+1}\)

=> 2B = \(\frac{2^{2018}+2}{2^{2018}+1}=\frac{2^{2018}+1+1}{2^{2018}+1}=1+\frac{1}{2^{2018}+1}\)

Vì \(\frac{1}{2^{2018}+1}>\frac{1}{2^{2019}+1}\Rightarrow1+\frac{1}{2^{2018}+1}>1+\frac{1}{2^{2019}+1}\Rightarrow2B>2A\Rightarrow B>A\)