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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 : \(\dfrac{2019}{2020}>\dfrac{2019}{2020+2021}\)
\(\dfrac{2020}{2021}>\dfrac{2020}{2020+2021}\)
\(\Rightarrow\dfrac{2019}{2020}+\dfrac{2020}{2021}>\dfrac{2019+2020}{2020+2021}\)
\(\Rightarrow M>N\)
B/A
\(=\dfrac{1+\dfrac{2020}{2}+1+\dfrac{2019}{3}+...+1+\dfrac{1}{2021}+1}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2021}+\dfrac{1}{2022}}\)
\(=\dfrac{2022\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2021}+\dfrac{1}{2022}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2021}+\dfrac{1}{2022}}=2022\)
\(2a^{2020}+2b^{2020}+2c^{2020}-2\left(ab\right)^{1010}-2\left(bc\right)^{1010}-2\left(ca\right)^{1010}=0\)
\(\Leftrightarrow\left(a^{1010}-b^{1010}\right)^2+\left(b^{1010}-c^{1010}\right)^2+\left(c^{1010}-a^{1010}\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}a^{1010}-b^{1010}=0\\b^{1010}-c^{1010}=0\\c^{1010}-a^{1010}=0\end{matrix}\right.\)
\(\Rightarrow\left|a\right|=\left|b\right|=\left|c\right|\)
Nếu đề không cho a;b;c dương thì không tính được cụ thể giá trị A
Nếu a;b;c dương thì \(a=b=c\Rightarrow A=0\)
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: \(2020^2-1< 2020^2\)
\(\Leftrightarrow\sqrt{2020^2-1}< \sqrt{2020^2}\)
\(\Leftrightarrow2\cdot\sqrt{2020^2-1}< 2\cdot\sqrt{2020^2}\)
\(\Leftrightarrow4040+2\cdot\sqrt{2020^2-1}< 4040+2\cdot\sqrt{2020^2}\)
\(\Leftrightarrow\left(\sqrt{2019}+\sqrt{2021}\right)^2< \left(2\sqrt{2020}\right)^2\)
\(\Leftrightarrow\sqrt{2019}+\sqrt{2021}< 2\sqrt{2020}\)
