Sửa đề: \(\left(\frac12+\frac13+\cdots+\frac{1}{2021}\right)\cdot x=\frac{2020}{1}+\frac{2019}{2}+\cdots+\frac{1}{2020}\)

Ta có: \(\frac{2020}{1}+\frac{2019}{2}+\cdots+\frac{1}{2020}\)

\(=\left(1+\frac{2019}{2}\right)+\left(1+\frac{2018}{3}\right)+\cdots+\left(1+\frac{1}{2020}\right)+1\)

\(=\frac{2021}{2}+\frac{2021}{3}+\cdots+\frac{2021}{2021}=2021\left(\frac12+\frac13+\cdots+\frac{1}{2021}\right)\)

Ta có: \(\left(\frac12+\frac13+\cdots+\frac{1}{2021}\right)\cdot x=\frac{2020}{1}+\frac{2019}{2}+\cdots+\frac{1}{2020}\)

=>\(x\left(\frac12+\frac13+\cdots+\frac{1}{2021}\right)=2021\left(\frac12+\frac13+\cdots+\frac{1}{2021}\right)\)

=>x=2021

Dễ thấy A chia hết cho 10 nên A có tận cùng là 0

còn 1x 3 x 5 x... x 2021 là một số lẻ và chia hết cho 5 nên có tận cùng là 5

help me !!!!!!!

\(=2021\cdot2\cdot\left(1+\dfrac{1}{2}:\dfrac{3}{2}-\dfrac{4}{3}\right)=4042\cdot\left(1+\dfrac{1}{3}-\dfrac{4}{3}\right)=0\)

Ta có: \(\left|x+\frac{1}{2021}\right|\ge0\) ; \(\left|x+\frac{2}{2021}\right|\ge0\) ; ... ; \(\left|x+\frac{2020}{2021}\right|\ge0\) \(\left(\forall x\right)\)

\(\Rightarrow\left|x+\frac{1}{2021}\right|+\left|x+\frac{2}{2021}\right|+...+\left|x+\frac{2020}{2021}\right|\ge0\left(\forall x\right)\)

\(\Rightarrow2021x\ge0\Rightarrow x\ge0\)

Từ đó ta được: \(x+\frac{1}{2021}+x+\frac{2}{2021}+...+x+\frac{2020}{2021}=2021x\)

\(\Leftrightarrow2020x+\frac{1+2+...+2020}{2021}=2021x\)

\(\Leftrightarrow x=\frac{\left(2020+1\right)\left[\left(2020-1\right)\div1+1\right]}{2021}\)

\(\Leftrightarrow x=\frac{2021\cdot2020}{2021}=2020\)

Vậy x = 2020

\(\left|\frac{x+1}{2021}\right|+\left|\frac{x+2}{2021}\right|+...+\left|\frac{x+2020}{2021}\right|=2021x\)

Ta có:\(\left|\frac{x+1}{2021}\right|\ge0;\left|\frac{x+2}{2021}\right|\ge0;....;\left|\frac{x+2020}{2021}\right|\ge0\forall x\)

\(\Rightarrow\left|\frac{x+1}{2021}\right|+\left|\frac{x+2}{2021}\right|+...+\left|\frac{x+2020}{2021}\right|\ge0\forall x\)

\(\Rightarrow2021x\ge0\Rightarrow x\ge0\)

\(\Rightarrow\frac{x+1}{2021}+\frac{x+2}{2021}+...+\frac{x+2020}{2021}=2021x\)

\(\Rightarrow x+\frac{1}{2021}+x+\frac{2}{2021}+...+x+\frac{2020}{2021}=2021x\)

\(\Rightarrow2020x+\frac{1+2+...+2020}{2021}=2021x\)

\(\Rightarrow x=2020\)

A

\(\dfrac{x+1}{2023}+\dfrac{x+2}{2022}=\dfrac{x+3}{2021}+\dfrac{x+4}{2020}\\ \Leftrightarrow\dfrac{x+1}{2023}+1+\dfrac{x+2}{2022}+1=\dfrac{x+3}{2021}+1+\dfrac{x+4}{2020}+1\\ \Leftrightarrow\dfrac{x+1+2023}{2023}+\dfrac{x+2+2022}{2022}-\dfrac{x+3+2021}{2021}-\dfrac{x+4+2020}{2020}=0\\ \Leftrightarrow\left(x+2024\right)\times\left(\dfrac{1}{2023}+\dfrac{1}{2022}-\dfrac{1}{2021}-\dfrac{1}{2020}\right)=0\\ \Rightarrow x+2024=0:\left(\dfrac{1}{2023}+\dfrac{1}{2022}-\dfrac{1}{2021}-\dfrac{1}{2020}\right)\\ \Rightarrow x+2024=0\\ \Rightarrow x=-2024\)

Tham khảo câu trả lời:

Nếu |x+1|>1 thì \(\left|x+1\right|^{2020}>1\) , \(\left|x+2\right|^{2021}>1\)

=>Phương trình sẽ vô lý

=>\(\begin{cases}\left|x+1\right|\le1\\ \left|x+2\right|\le1\end{cases}\Rightarrow\begin{cases}-1\le x+1\le1\\ -1\le x+2\le1\end{cases}\)

=>-2<=x<=-1

TH1: x=-2

Phương trình sẽ trở thành: \(\left|-2+1\right|^{2020}+\left|-2+2\right|^{2021}=1\)

=>1=1(đúng)

=>NHận

TH2: x=-1

Phương trình sẽ trở thành: \(\left|-1+1\right|^{2020}+\left|-1+2\right|^{2021}=1\)

=>1=1(đúng)

=>Nhận

TH3: -2<x<-1

Đặt t=x+2

=>t-1=x+1

Phương trình sẽ trở thành: \((1-t)^{2020} + t^{2021} = 1\)

Vì t∈(0;1) nên 0<t<1 và 0<1-t<1

nên ta có: \((1-t)^{2020}<1-t\quad;\quad t^{2021}

=>\((1-t)^{2020} + t^{2021} < (1-t) + t = 1\)

=>Loại

(\(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2023}\)). x = (\(\dfrac{2021}{2}+1\))+(\(\dfrac{2020}{3}+1\))+....+(\(\dfrac{1}{2022}+1\))

(\(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2023}\)). x = \(\dfrac{2023}{2}\)+\(\dfrac{2023}{3}\)+....+ \(\dfrac{2023}{2022}\)

(\(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2023}\)). x = 2023.( \(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2023}\))

vậy x= 2023