ĐKXĐ: x∉{2019;2020;2021}

Ta có: \(\frac{1}{2019-x}+\frac{1}{2020-x}+\frac{1}{2021-x}=\frac{13}{12}\)

=>\(\frac{1}{2019-x}-\frac12+\frac{1}{2020-x}-\frac13+\frac{1}{2021-x}-\frac14=0\)

=>\(\frac{2-2019+x}{2\left(2019-x\right)}+\frac{3-2020+x}{3\left(2020-x\right)}+\frac{4-2021+x}{4\left(2021-x\right)}=0\)

=>\(\left(x-2017\right)\left(\frac{1}{2\left(2019-x\right)}+\frac{1}{3\left(2020-x\right)}+\frac{1}{4\left(2021-x\right)}\right)=0\)

=>x-2017=0
=>x=2017(nhận)

câu hỏi 4 năm ms có ng trả lời anh ạ ;))))

Ta có: \(\dfrac{x+1}{2018}+\dfrac{x+1}{2019}+\dfrac{x+1}{2020}+\dfrac{x+1}{2021}=0\)

\(\Leftrightarrow x+1=0\)

hay x=-1

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

khó hiểu vcl

đúng lun ko hiểu một chút nào
 

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