A = B

a) \(M=2020+2020^2+...+2020^{10}\)

\(M=\left(2020+2020^2\right)+\left(2020^3+2020^4\right)+...+\left(2020^9+2020^{10}\right)\)

\(M=2020\left(1+2020\right)+2020^3\left(1+2020\right)+...+2020^9\left(1+2020\right)\)

\(M=2021\left(2020+2020^3+...+2020^9\right)⋮2021\).

b) Bạn làm tương tự câu a). 

b, \(A=2021+2021^2+...+2021^{2020}\)

\(=2021\left(1+2021\right)+...+2021^{2019}\left(1+2021\right)\)

\(=2022\left(2021+...+2021^{2019}\right)⋮2022\)

Vậy ta có đpcm 

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

id của mik 88461550

còn câu hỏi thì mik ko bt nhé mik mới lớp4

nick là ri nhé

Ta thấy mẫu số ở PS A = mẫu số PS B nên ta xét ở tử số của 2 số.

2020+201<2020+2019 nên PS B lớn hơn

Mik ko chơi mini world nha k mik vs

ok các bạn

Ta có : A = \(\frac{10^{2020}+1}{10^{2021}+1}\)

=> 10A = \(\frac{10^{2021}+10}{10^{2021}+1}=1+\frac{9}{10^{2021}+1}\)

Lại có : \(B=\frac{10^{2021}+1}{10^{2022}+1}\)

=> \(10B=\frac{10^{2022}+10}{10^{2022}+1}=1+\frac{9}{10^{2022}+1}\)

Vì \(\frac{9}{10^{2022}+1}< \frac{9}{10^{2021}+1}\)

=> \(1+\frac{9}{10^{2022}+1}< 1+\frac{9}{10^{2022}+1}\)

=> 10B < 10A

=> B < A

b) Ta có : \(\frac{2019}{2020+2021}< \frac{2019}{2020}\)

Lại có : \(\frac{2020}{2020+2021}< \frac{2020}{2021}\)

=> \(\frac{2019}{2020+2021}+\frac{2020}{2020+2021}< \frac{2019}{2020}+\frac{2020}{2021}\)

=> \(\frac{2019+2020}{2020+2021}< \frac{2019}{2020}+\frac{2020}{2021}\)

=> B < A

a) A=10^2020+1/10^2021+1 < 10^2020+1+9/10^2022+1+9 =         

10.(10^2021+1)/10.(10^2022+1) = 10^2021+1/10^2022+1 = B

Vậy A < B.

tk mik nha

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

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

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

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

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

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

B/A

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

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

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

=> \(\frac{2019}{2020}+\frac{2020}{2021}>\frac{2019}{2020+2021}+\frac{2020}{2020+2021}=\frac{2019+2020}{2020+2021}\)

=> A > B.