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![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:
\(\frac{2021}{1}+\frac{2020}{2}+\frac{2019}{3}+...+\frac{2}{2020}+\frac{1}{2021}\)
\(=1+\frac{2020}{2}+1+\frac{2019}{3}+...+1+\frac{2}{2020}+1+\frac{1}{2021}+1\)
\(=\frac{2022}{2}+\frac{2022}{3}+...+\frac{2022}{2020}+\frac{2022}{2021}+\frac{2022}{2022}\)
\(=2022\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2020}+\frac{1}{2021}+\frac{1}{2022}\right)\)
Do đó giá trị của \(M\)là:
\(M=\frac{2022\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2020}+\frac{1}{2021}+\frac{1}{2022}\right)}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2020}+\frac{1}{2021}+\frac{1}{2022}}=2022\)
2021/1 + 2/2020 + 2019/3 + ... + 2/2020 + 1/2021
M = _______________________________________
1/2 + 1/3 + ... + 1/2020 + 1/2021 + 1/2022
=\(\frac{2022+\frac{2022}{2}+...+\frac{2022}{2021}-2021}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2022}}=\frac{1+\frac{2022}{2}+...+\frac{2022}{2021}}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2022}}\)
=2022
![](https://rs.olm.vn/images/avt/0.png?1311)
( 3 + 6 + 9 + 12 + 15 + 18 + 21 ) - x = 13
84-x =13
x =84-13
x =74
k nha
![](https://rs.olm.vn/images/avt/0.png?1311)
c)\(1-\left(x-\frac{1}{5}\right)^2=\frac{21}{25}\)
\(\Leftrightarrow\left(x-\frac{1}{5}\right)^2=\frac{4}{25}\)
\(\Leftrightarrow x-\frac{1}{5}=\frac{2}{5}\)
\(\Leftrightarrow x=\frac{3}{5}\)
d)\(\left|x+\frac{1}{3}\right|.3\frac{2}{5}=\frac{2}{5}\)
\(\Leftrightarrow\left|x+\frac{1}{3}\right|.\frac{17}{5}=\frac{2}{5}\)
\(\Leftrightarrow\left|x+\frac{1}{3}\right|=\frac{2}{17}\)
\(\Leftrightarrow\orbr{\begin{cases}x+\frac{1}{3}=\frac{2}{17}\\x+\frac{1}{3}=-\frac{2}{17}\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-\frac{11}{51}\\x=-\frac{23}{51}\end{cases}}}\)
#H
1-(x-1/5)2=21/25
(x-1/5)2=1-21/25
(x-1/5)2=4/25
(x-1/5)2=(2/5)2
x-1/5=2/5
x=2/5+1/5
x=3/5
k nha
![](https://rs.olm.vn/images/avt/0.png?1311)
123.123= 15129
ko chắc lắm nha
ai chên 10 k mình nha
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
=0 nha