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

a: \(98^{10}\cdot A=\dfrac{98^{98}+98^{10}}{98^{98}+1}=1+\dfrac{98^{10}-1}{98^{98}+1}\)

\(98^{10}\cdot B=\dfrac{98^{99}+98^{10}}{98^{99}+1}=1+\dfrac{98^{10}-1}{98^{99}+1}\)

98^88+1>98^99+1

=>A<B

b: \(\dfrac{1}{2022^2}\cdot C=\dfrac{2022^{2023}+1}{2022^{2023}+2022^2}=1+\dfrac{1-2022^2}{2022^{2023}+2022^2}\)

\(\dfrac{1}{2022^2}\cdot D=\dfrac{2022^{2021}+1}{2022^{2021}+2022^2}=1+\dfrac{1-2022^2}{2022^{2021}+2022^2}\)

2022^2023>2022^2021

=>2022^2023+2022^2>2022^2021+2022^2

=>\(\dfrac{2022^2-1}{2022^{2023}+2022^2}< \dfrac{2022^2-1}{2022^{2021}+2022^2}\)

=>\(\dfrac{1-2022^2}{2022^{2023}+2022^2}>\dfrac{1-2022^2}{2022^{2021}+2022^2}\)

=>C>D

a: 21^15=3^15*7^15

27^5*49^8=3^15*7^14

mà 15>14

nên 21^15>27^5*49^8

b: \(2020^{2020}-2020^{2019}=2020^{2019}\left(2020-1\right)=2020^{2019}\cdot2019\)

\(2020^{2019}-2020^{2018}=2020^{2018}\cdot2019\)

mà 2019>2018

nên 2020^2020-2020^2019>2020^2019-2020^2018

2020+2022/2022+2024 lớn hơn

lm sao hở c ?

2020/2022 > 2020/2022+2024     (1)

2022/2024 > 2022/2022+2024      (2)

từ (1) và (2) cộng vế theo vế ta có :

2020/2022 + 2022/2024 > 2020/2022+2024 + 2022/2022+2024

=> 2020/2022 + 2022/2024 > 2020+2022/2022+2024

a/

2020.2021=(2019+1)(2022-1)=

=2019.2022-2019+2022-1=2019.2022+2>2019.2022

b/

\(4^7=\left(2^2\right)^7=2^{14}< 2^{15}\)

c/

\(199^{20}< 200^{20}=\left(8.25\right)^{20}=\left(2^3.5^2\right)^{20}=2^{60}.5^{40}\)

\(2000^{15}=\left(16.125\right)^{15}=\left(2^4.5^3\right)^{15}=2^{60}.5^{45}\)

\(\Rightarrow2000^{15}=2^{60}.5^{45}>2^{60}.5^{40}>199^{20}\)

d/

\(31^{31}< 32^{31}=\left(2^5\right)^{31}=2^{155}\)

\(17^{39}>16^{39}=\left(2^4\right)^{39}=2^{156}\)

\(\Rightarrow17^{39}=2^{156}>2^{155}>31^{31}\)

A<B

B nha bn

Lý thuyết : Những số nào có chữ số tạn cùng là 0 hoặc 5 thì chia hết cho 5.

⇒ Đáp án B. 2020 + 2000 + 2030

Tham khảo: