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/

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

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

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

ta có : 

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

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

\(\Leftrightarrow\) 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

18:

a: \(S=3\left(\dfrac{2}{2\cdot4}+\dfrac{2}{4\cdot6}+...+\dfrac{2}{98\cdot100}\right)\)

=3*(1/2-1/4+1/4-1/6+...+1/98-1/100)

=3*49/100=147/100

b: Để A là số nguyên thì n-1 thuộc Ư(2)

=>n-1 thuộc {1;-1;2;-2}

=>n thuộc {2;0;3;-1}

Ta có: \(A=2019.2021=\left(2020-1\right)\left(2020+1\right)=2020^2-1\)

\(B=2020^2\)

=> A <  B

A=2019.2021

A=(2020-1)(2020+1)

A=2020²-1

Vậy: A<B

TA CÓ \(A=2019.2020\)

\(=\left(2020-1\right)\left(2020+1\right)\)

\(=2020^2-1\)

\(B=2020^2\)

\(\Leftrightarrow A\le B\)

HOK TỐT

Ta có \(\left(\sqrt{2018}+\sqrt{2020}\right)^2=4038+2\sqrt{4076360}\) và \(\left(2\sqrt{2019}\right)^2=8076=4038+4038\)

Mà \(\left(2\sqrt{4076360}\right)^2=16305440\) và \(4038^2=16305444\)

\(\Rightarrow2\sqrt{4076360}< 4038\)

\(\Rightarrow\sqrt{2018}+\sqrt{2020}< 2\sqrt{2019}\)

\(\left(\sqrt{2018}+\sqrt{2020}\right)^2=4038+2\cdot\sqrt{2018\cdot2020}\)

\(\left(2\sqrt{2019}\right)^2=8076=4038+4038\)

mà \(2\cdot\sqrt{2018\cdot2020}< 4038\)

nên \(\sqrt{2018}+\sqrt{2020}< 2\sqrt{2019}\)

\(A=2019\times2021=\left(2021-1\right)\times\left(2021+1\right)=2021^2-1< 2021^2=B.\)

sai mất rồi nhưng dù sao cũng cảm ơn bn nhé

Ta có : A=2019 x 2021

A=(2020-1) x 2021

A=2020 x 2021-2021

 Ta có :B=2020²

=>B=2020 x 2020

B=(2021-1) x 2020

B=2020 x 2021-2020

 Vậy ta có 2020 x 2021- 2021 < 2020 x 2021-2020 (vì hai bên đều có 2020 x 2021 chung nhưng bên A bị trừ nhiều hơn nên nhỏ hơn)nên A<B

bài 1:

ssh của A là:

(151-3):2+1=75

A=(151+3)x75:2=5775

đáp số: 5775