\(a^{2019}+b^{2019}=a^{2020}+b^{2020}\\ \Leftrightarrow a^{2020}-a^{2019}=b^{2019}-b^{2020}=0\\ \Leftrightarrow a^{2019}\left(a-1\right)=b^{2019}\left(1-b\right)\\ \Leftrightarrow\dfrac{a^{2019}}{b^{2019}}=\dfrac{1-b}{a-1}\left(1\right)\\ a^{2020}+b^{2020}=a^{2021}+b^{2021}\\ \Leftrightarrow a^{2021}-a^{2020}=b^{2020}-b^{2021}\\ \Leftrightarrow a^{2020}\left(a-1\right)=b^{2020}\left(1-b\right)\\ \Leftrightarrow\dfrac{a^{2020}}{b^{2020}}=\dfrac{1-b}{a-1}\left(2\right)\\ \left(1\right)\left(2\right)\Leftrightarrow\dfrac{a^{2019}}{b^{2019}}=\dfrac{a^{2020}}{b^{2020}}\Leftrightarrow\dfrac{a}{b}=1\Leftrightarrow a=b\\ \Leftrightarrow2a^{2019}=2a^{2020}\\ \Leftrightarrow a=1=b\\ \Leftrightarrow P=2022-\left(1+1-1\right)^{2022}=2021\)

ghê wa b ưi, nhma mình hông hỉu j hết

bài 1:

ssh của A là:

(151-3):2+1=75

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

đáp số: 5775

B/A

\(=\dfrac{1+\dfrac{2020}{2}+1+\dfrac{2019}{3}+...+1+\dfrac{1}{2021}+1}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2021}+\dfrac{1}{2022}}\)

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

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: \(2020^2-1< 2020^2\)

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

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

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

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

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

Ta có:

\(A=\dfrac{7\left(4-7^{2020}\right)}{7^{2021}}+\dfrac{5+7^{2021}}{7^{2021}}\)

\(A=\dfrac{28-7^{2021}+5+7^{2021}}{7^{2021}}=\dfrac{33}{7^{2021}}\)

Ta có: \(B=\dfrac{7^2}{7^{2021}}=\dfrac{49}{7^{2021}}\)

=> B>A

Thank you☺

D

D nha

a) Ta có:

Suy ra: 

Do đó .

Lại có .

Vậy A < B.

Ta có:2019>4
=>2019/2020+2020/2021+2021/2022+2019>4
=>a>4(dpcm)

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

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

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