Nghiệm của phương trình \(\left|x-\dfrac{1}{2020}\right|+\left|x-\dfrac{2}{2020}\right|+\left|x-\dfrac{3}{2020}\right|+....+\left|x-\dfrac{2019}{2020}\right|=2020x-2020\) là
\(x=\dfrac{2021}{2}\).\(x=2020\).\(x=2019\).\(x=\dfrac{2019}{2}\).Hướng dẫn giải:Nhận xét: \(VT\ge0\) nên \(VP=2020x-2020\ge0\Leftrightarrow x\ge1\)
Khi đó: \(x-\dfrac{1}{2020}>0\) ; \(x-\dfrac{2}{2020}>0\) ; ..... ; \(x-\dfrac{2019}{2020}>0\)
Nên phương trình đã cho trở thành:
\(x-\dfrac{1}{2020}+x-\dfrac{2}{2020}+x-\dfrac{3}{2020}+....+x-\dfrac{2019}{2020}=2020x-2020\)
\(\Leftrightarrow2019x-\left(\dfrac{1}{2020}+\dfrac{2}{2020}+\dfrac{3}{2020}+...+\dfrac{2019}{2020}\right)=2020x-2020\)
\(\Leftrightarrow2019x-\dfrac{1+2+3+...+2019}{2020}=2020x-2020\)
\(\Leftrightarrow2019x-\dfrac{\left(1+2019\right).2019}{2.2020}=2020x-2020\)
\(\Leftrightarrow2019x-\dfrac{2019}{2}=2020x-2020\)
\(\Leftrightarrow2020x-2019x=2020-\dfrac{2019}{2}\)
\(\Leftrightarrow x=\dfrac{2021}{2}\).
Vậy phương trình có nghiệm \(x=\dfrac{2021}{2}\).