\(A=\dfrac{10^{99}+1}{10^{100}+1}\)

\(\Leftrightarrow10A=\dfrac{10\left(10^{99}+1\right)}{10^{100}+1}\)

\(\Leftrightarrow10A=\dfrac{10^{100}+10}{10^{100}+1}=\dfrac{10^{100}+1+9}{10^{100}+1}=1+\dfrac{9}{10^{100}+1}\)

\(B=\dfrac{10^{100}+1}{10^{101}+1}\)

\(\Leftrightarrow10B=\dfrac{10\left(10^{100}+1\right)}{10^{101}+1}\)

\(\Leftrightarrow10B=\dfrac{10^{101}+10}{10^{101}+1}=\dfrac{10^{101}+1+9}{10^{101}+1}=1+\dfrac{9}{10^{101}+1}\)

Do \(\dfrac{9}{10^{100}+1}>\dfrac{9}{10^{101}+1}\) nên \(10A>10B\)

\(\Rightarrow A>B\)

Áp dụng tính chất:

\(\dfrac{a}{b}< 1\Rightarrow\dfrac{a+m}{b+m}< 1\left(m\in N\right)\)

\(B=\dfrac{10^{100}+1}{10^{101}+1}< 1\)

\(B< \dfrac{10^{100}+1+9}{10^{101}+1+9}\)

\(B< \dfrac{10^{100}+10}{10^{101}+10}\)

\(B< \dfrac{10\left(10^{99}+1\right)}{10\left(10^{100}+1\right)}\)

\(B< \dfrac{10^{99}+1}{10^{100}+1}=A\)

\(B< A\)

Ta có : \(\dfrac{a}{b}< 1\Rightarrow\dfrac{a+m}{b+m}< 1\left(m\in N\right)\)

\(B=\dfrac{10^{100}+1}{10^{101}+1}< 1\)

\(B< \dfrac{10^{100}+1+9}{10^{101}+1+9}\)

\(B< \dfrac{10^{100}+10}{10^{101}+10}\)

\(B< \dfrac{10.\left(10^{99}+1\right)}{10.\left(10^{100}+1\right)}\)

\(B< \dfrac{10^{99}+1}{10^{100}+1}=A\)

Vậy \(B< A\)

c: \(100C=\dfrac{100^{100}+100}{100^{100}+1}=1+\dfrac{99}{100^{100}+1}\)

\(100D=\dfrac{100^{101}+100}{100^{101}+1}=1+\dfrac{99}{100^{101}+1}\)

100^100+1<100^101+1

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

=>100C>100D

=>C>D

b: \(2020E=\dfrac{2020^{2022}+2020}{2020^{2022}+1}=1+\dfrac{2019}{2020^{2022}+1}\)

\(2020F=\dfrac{2020^{2021}+2020}{2020^{2021}+1}=1+\dfrac{2019}{2020^{2021}+1}\)

2020^2022+1>2020^2021+1(Do 2022>2021)

=>\(\dfrac{2019}{2020^{2022}+1}< \dfrac{2019}{2020^{2021}+1}\)

=>2020E<2020F

=>E<F

\(10A=\dfrac{10^{2023}+10}{10^{2023}+1}=1+\dfrac{9}{10^{2023}+1}\)

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

2023>2022

=>10^2023+1>10^2022+1

=>10A<10B

=>A<B

Ta có:

10A=10^102-10/10^102-1

10A=1-9/10^102-1

10B=10^101+10/10^101+1

10B=1+9/10^101+1

suy ra 10B>10A

Vậy B>A

Áp dụng bất đẳng thức :

\(\dfrac{a}{b}< \dfrac{a+m}{b+m}\)

Ta có :

\(A=\dfrac{10^{101}-1}{10^{102}-1}< \dfrac{10^{101}-1+11}{10^{102}-1+11}=\dfrac{10^{101}+10}{10^{102}+10}=\dfrac{10\left(10^{100}+1\right)}{10\left(10^{101}+1\right)}=\dfrac{10^{100}+1}{10^{101}+1}=B\)

\(\Leftrightarrow A< B\)

Ta có:

\(1-A=1-\dfrac{10^{101}-1}{10^{102}-1}=\dfrac{10^{102}-1\left(10^{101}-1\right)}{10^{102}-1}\) \(=\dfrac{10^{102}-1-10^{101}+1}{10^{102}-2}=\dfrac{10^{102}-10^{101}}{10^{102}-1}\)

\(=\dfrac{10^{101}\left(10-1\right)}{10^{101}\left(10-\dfrac{1}{10^{101}}\right)}=\dfrac{10-1}{10-\dfrac{1}{10^{101}}}=\dfrac{9}{10-\dfrac{1}{10^{101}}}\)\(\left(1\right)\)

\(1-B=1-\dfrac{10^{100}+1}{10^{101}+1}=\dfrac{10^{101}+1-\left(10^{100}+1\right)}{10^{101}+1}\)

\(=\dfrac{10^{101}+1-10^{100}-1}{10^{101}+1}\) \(=\dfrac{10^{101}-10^{100}}{10^{101}+1}=\dfrac{10^{100}\left(10-1\right)}{10^{100}\left(10+\dfrac{1}{10^{100}}\right)}\)

\(=\dfrac{10-1}{10+\dfrac{1}{10^{100}}}=\dfrac{9}{10+\dfrac{1}{100}}\)\(\left(2\right)\)

\(Từ\left(1\right);\left(2\right)\) \(=>A< B\)\(\left(đpcm\right)\)

CHÚC BẠN HỌC TỐT

M=\(\dfrac{10^{100^{ }}+1}{10^{101}+1}\)

M=\(\dfrac{10^{99+1}+1}{10^{100+1}+1}\)

M=\(\dfrac{10^{99}.10+1}{10^{100}.10+1}\)

N=\(\dfrac{10^{99^{ }}+1}{10^{100}+1}\)

=>M lớn hơn N

M>N,vì:\(\dfrac{10^{100}+1}{10^{101}+1}=\dfrac{10^{100}}{10^{101}}\)

\(\dfrac{10^{99}+1}{10^{100}+1}=\dfrac{10^{99}}{10^{100}}\)

\(\dfrac{10^{100}}{10^{101}}>\dfrac{10^{99}}{10^{100}}\)

B>A?

Tham khảo:

https://lazi.vn/edu/exercise/so-sanh-a-1-2-3-4-5-6-99-100-va-b-1-10

\(B>A\)