Ta có:

2019.2021=2019.(2020+1)=2019.2020+2019 (1)

Lại có:

2020.2020=(2019+1).2020=2019.2020+2020 (2)

Vì 2019.2020=2019.2020 mà 2019<2020

=>(1)<(2)

=>..... 

                                                                          Bài giải

Ta có : \(2019\text{ x }2021=2019\text{ x }2020+2019\)

          \(2020\text{ x }2020=2019\text{ x }2020+2020\)

\(\text{Vì }2019\text{ x }2020+2019< 2019\text{ x }2020+2020\text{ }\Rightarrow\text{ }2019\text{ x }2021< 2020\text{ x }2020\)

                                    Bài giải

Ta có : 

\(2019\text{ x }2021=2019\text{ x }2020+2019\)

 \(2020\text{ x }2020=2019\text{ x }2020+2020\)

\(\text{Vì }2019\text{ x }2020+2019< 2019\text{ x }2020+2020\text{ }\Rightarrow\text{ }2019\text{ x }2021< 2020\text{ x }2020\)

\(\frac{2019}{2021}+\frac{2021}{2019}< 2\)

Ta có: \(\frac{a}{b}+\frac{b}{a}\le2\)

Dấu bằng xảy ra khi : a=b

=>\(\frac{2021}{2019}+\frac{2019}{2021}< 2\)

có 2=2019/2021+2/2021+1  (1)

2019/2021+2021/2019=2019/2021+2/2019+1  (2)

có 2/2021 < 2/2019 (3)

từ (1)(2)(3) => 2<2019/2021+2021/2019

k đúng cho mk vs

Tham khảo:

a 

oki

Ta có: \(\sqrt{2020}-\sqrt{2019}=\frac{\left(\sqrt{2020}-\sqrt{2019}\right)\left(\sqrt{2020}+\sqrt{2019}\right)}{\sqrt{2020}+\sqrt{2019}}\)

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

\(\sqrt{2021}-\sqrt{2020}=\frac{\left(\sqrt{2021}-\sqrt{2020}\right)\left(\sqrt{2021}+\sqrt{2020}\right)}{\sqrt{2021}+\sqrt{2020}}\)

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

Vì \(\sqrt{2020}+\sqrt{2019}< \sqrt{2021}+\sqrt{2020}\)

\(\Rightarrow\) \(\frac{1}{\sqrt{2020}+\sqrt{2019}}>\frac{1}{\sqrt{2021}+\sqrt{2020}}\)

Hay \(\sqrt{2020}-\sqrt{2019}>\sqrt{2021}-\sqrt{2020}\)

Chúc bn học tốt!

a) Ta có: \(\sqrt{2021}-\sqrt{2020}\)

\(=\frac{\left(\sqrt{2021}-\sqrt{2020}\right)\left(\sqrt{2021}+\sqrt{2020}\right)}{\sqrt{2021}+\sqrt{2020}}\)

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

Ta có: \(\sqrt{2020}-\sqrt{2019}\)

\(=\frac{\left(\sqrt{2020}-\sqrt{2019}\right)\left(\sqrt{2020}+\sqrt{2019}\right)}{\sqrt{2020}+\sqrt{2019}}\)

\(=\frac{1}{\sqrt{2019}+\sqrt{2020}}\)

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

\(\Leftrightarrow\frac{1}{\sqrt{2020}+\sqrt{2021}}< \frac{1}{\sqrt{2019}+\sqrt{2020}}\)

hay \(\sqrt{2021}-\sqrt{2020}< \sqrt{2020}-\sqrt{2019}\)

b) Ta có: \(\sqrt{2019\cdot2021}\)

\(=\sqrt{\left(2020-1\right)\left(2020+1\right)}\)

\(=\sqrt{2020^2-1}\)

Ta có: \(2020=\sqrt{2020^2}\)

Ta có: \(2020^2-1< 2020^2\)

nên \(\sqrt{2020^2-1}< \sqrt{2020^2}\)

\(\Leftrightarrow\sqrt{2019\cdot2021}< 2020\)

c) Ta có: \(\left(\sqrt{2019}+\sqrt{2021}\right)^2\)

\(=2019+2021+2\cdot\sqrt{2019\cdot2021}\)

\(=4040+2\sqrt{2019\cdot2021}\)

\(=4040+2\cdot\sqrt{2020^2-1}\)

Ta có: \(\left(2\sqrt{2020}\right)^2\)

\(=4\cdot2020\)

\(=4040+2\cdot2020\)

\(=4040+2\cdot\sqrt{2020^2}\)

Ta có: \(2020^2-1< 2020^2\)

\(\Leftrightarrow\sqrt{2020^2-1}< \sqrt{2020^2}\)

\(\Leftrightarrow2\cdot\sqrt{2020^2-1}< 2\cdot\sqrt{2020^2}\)

\(\Leftrightarrow4040+2\cdot\sqrt{2020^2-1}< 4040+2\cdot\sqrt{2020^2}\)

\(\Leftrightarrow\left(\sqrt{2019}+\sqrt{2021}\right)^2< \left(2\sqrt{2020}\right)^2\)

\(\Leftrightarrow\sqrt{2019}+\sqrt{2021}< 2\sqrt{2020}\)

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

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

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