Tham khảo:

Giải:

Ta có: N=2019+2020/2020+2021

=>N=2019/2020+2021 + 2020/2020+2021

Vì 2019/2020 > 2019/2020+2021 ; 2020/2021 > 2020/2020+2021

=>M>N

Vậy ...

Chúc bạn học tốt!

Ta có : \(\dfrac{2019}{2020}>\dfrac{2019}{2020+2021}\)

            \(\dfrac{2020}{2021}>\dfrac{2020}{2020+2021}\)

\(\Rightarrow\dfrac{2019}{2020}+\dfrac{2020}{2021}>\dfrac{2019+2020}{2020+2021}\)

\(\Rightarrow M>N\)

Lời giải:
$A=1-\frac{1}{2019}+1-\frac{1}{2020}+1-\frac{1}{2021}+1+\frac{3}{2018}$

$=4+(\frac{1}{2018}-\frac{1}{2019}+\frac{1}{2018}-\frac{1}{2020}+\frac{1}{2018}-\frac{1}{2021})$

$> 4+0+0+0+0=4$

a) 29^57 < 29^75

b) 1011^22 < 1101^22

c) ( 2021 + 2018 )^2019 < ( 2020 + 2019 )^2020

d) 2^5000 > 7^2000

Chúc bạn học tốt @!!!

Nếu có thể thì t.i.c.k cho mình nha ! Thank

Đại ca toàn học giỏi nhất hành tinh đây

a) 29⁵⁷và 29⁷⁵

 ^=> 29⁵⁷<29⁷⁵

2019 / 2020 và 2020 / 2021

2019 / 2020  <  2020 / 2021

\(\frac{2019}{2020}=1-\frac{1}{2020}\)

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

Vì \(\frac{1}{2020}>\frac{1}{2021}\Rightarrow1-\frac{1}{2020}< 1-\frac{1}{2021}\Rightarrow\frac{2019}{2020}< \frac{2020}{2021}\)

Chúc bạn học tốt ^^!!!

2019 / 2020 = 1 - 1 / 2020 

2020 / 2021 = 1 - 1 / 2021 

Vì 1 / 2020 > 1 / 2021 

Nên 1 - 1 / 2020 < 1 - 1 / 2021 

Vậy 2019 / 2020 < 2020 / 2021 

\(\dfrac{-2019}{2019}=-1\)

\(\dfrac{-2021}{2020}=-1,004\)

\(\Rightarrow\dfrac{-2019}{2019}>\dfrac{-2021}{2020}\)

\(-\dfrac{2019}{2019}=-1\)

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

Do đó: \(-\dfrac{2019}{2019}>\dfrac{-2021}{2020}\)