Tham khảo:

a: Đặt \(A=\sqrt{2021}-\sqrt{2020}\) và \(B=\sqrt{2022}-\sqrt{2021}\)

\(A=\sqrt{2021}-\sqrt{2020}=\frac{2021-2020}{\sqrt{2021}+\sqrt{2020}}=\frac{1}{\sqrt{2021}+\sqrt{2020}}\)

\(B=\sqrt{2022}-\sqrt{2021}=\frac{2022-2021}{\sqrt{2022}+\sqrt{2021}}=\frac{1}{\sqrt{2022}+\sqrt{2021}}\)

Ta có: \(\sqrt{2021}+\sqrt{2020}<\sqrt{2022}+\sqrt{2021}\)

=>\(\frac{1}{\sqrt{2021}+\sqrt{2020}}>\frac{1}{\sqrt{2022}+\sqrt{2021}}\)

=>A>B

b: Đặt \(A=\sqrt{2022}-\sqrt{2020}\) và \(B=\sqrt{2020}-\sqrt{2018}\)

\(A=\sqrt{2022}-\sqrt{2020}=\frac{2022-2020}{\sqrt{2022}+\sqrt{2020}}=\frac{2}{\sqrt{2022}+\sqrt{2020}}\)

\(B=\sqrt{2020}-\sqrt{2018}=\frac{2020-2018}{\sqrt{2020}+\sqrt{2018}}=\frac{2}{\sqrt{2020}+\sqrt{2018}}\)

TA có: \(\sqrt{2022}+\sqrt{2020}>\sqrt{2020}+\sqrt{2018}\)

=>\(\frac{2}{\sqrt{2022}+\sqrt{2020}}<\frac{2}{\sqrt{2020}+\sqrt{2018}}\)

=>A<B

$1997/1996$ $>$ $1998/1997$

$=>$ Phân số nào có mẫu số bé hơn thì phân số đó lớn hơn.

Ủa tử có giống nhau đâu bạn

\(\dfrac{2021}{2022}=\dfrac{2020}{2021}\)

\(\dfrac{2021}{2022}\) và \(\dfrac{2020}{2021}\)

\(\dfrac{2021}{2022}=1-\dfrac{1}{2022}\)

\(\dfrac{2020}{2021}=1-\dfrac{1}{2021}\)

\(\text{Vì }\)\(\dfrac{1}{2022}>\dfrac{1}{2021}=>1-\dfrac{1}{2022}>1-\dfrac{1}{2021}=>\dfrac{2021}{2022}>\dfrac{2020}{2021}\)

B = \(\dfrac{1}{2002}\) + \(\dfrac{2}{2021}\) + \(\dfrac{3}{2020}\)+...+ \(\dfrac{2021}{2}\) + \(\dfrac{2022}{1}\)

B = \(\dfrac{1}{2002}\) + \(\dfrac{2}{2021}\) + \(\dfrac{3}{2020}\)+...+ \(\dfrac{2021}{2}\) + 2022

B = 1 + ( 1 + \(\dfrac{1}{2022}\)) + ( 1 + \(\dfrac{2}{2021}\)) + \(\left(1+\dfrac{3}{2020}\right)\)+ ... + \(\left(1+\dfrac{2021}{2}\right)\) 

B = \(\dfrac{2023}{2023}\) + \(\dfrac{2023}{2022}\) + \(\dfrac{2023}{2021}\) + \(\dfrac{2023}{2020}\) + ...+ \(\dfrac{2023}{2}\) 

B = 2023 \(\times\) ( \(\dfrac{1}{2023}\) + \(\dfrac{1}{2022}\) + \(\dfrac{1}{2021}\) + \(\dfrac{1}{2020}\)+ ... + \(\dfrac{1}{2}\))

Vậy B > C 

Ta có: \(B=2020.2021.2022=\left(2021-1\right).\left(2021+1\right).2021=\left(2021-1\right)^2.2021< 2021^2.2021=A\)

1996 < 1997 < 1998 < 1999

-1996x1999=1996x(1998+1)=1996x1998+1996

-1997x1998=(1996+1)x1998=1996x1998+1998

Vì vậy 1996x1999<1997x1998

Có: \(\dfrac{2019}{2021}=1-\dfrac{2}{2021}\)

       \(\dfrac{2020}{2022}=1-\dfrac{2}{2022}\)

Mà \(\dfrac{2}{2021}>\dfrac{2}{2022}\Rightarrow1-\dfrac{2}{2021}< 1-\dfrac{2}{2022}\Rightarrow\dfrac{2019}{2021}< \dfrac{2020}{2022}\)

cảm ơn nhá