\(A=\left(\left|x-1\right|+\left|2020-x\right|\right)+\left(\left|x-2\right|+\left|2019-x\right|\right)+...+\left(\left|x-1009\right|+\left|1010-x\right|\right)\\ A\ge\left|x-1+2020-x\right|+\left|x-2+2019-x\right|+...+\left|x-1009+1010-x\right|\\ A\ge2019+2017+...+1=\dfrac{2020\left[\left(2019-1\right):2+1\right]}{2}=1020100\)

Dấu \("="\Leftrightarrow\left\{{}\begin{matrix}\left(x-1\right)\left(2020-x\right)\ge0\\...\\\left(x-1009\right)\left(1010-x\right)\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}1\le x\le2020\\...\\1009\le x\le1010\end{matrix}\right.\)

\(\Leftrightarrow1009\le x\le1010\)

Theo bđt cosi 

\(P=\left|x-2019\right|+\dfrac{2020}{\left|x-2019\right|}+2021\ge2\sqrt{\dfrac{\left|x-2019\right|.2020}{\left|x-2019\right|}}+2021=4\sqrt{505}+2021\)

Dấu ''='' xảy ra khi \(x-2019=2020\Leftrightarrow x=4039\)

mình ko giúp đc rồi

Ta có: \(A=\left|x-2018\right|+\left|2019-x\right|+\left|x-2020\right|\)

\(A=\left(\left|x-2018\right|+\left|2020-x\right|\right)+\left|2019-x\right|\)

\(\Rightarrow A\ge\left|x-2018+2020-x\right|+\left|2019-x\right|=2+\left|2019-x\right|\)

Dấu "=" xảy ra <=> \(\left(x-2018\right)\left(2020-x\right)\ge0\)

\(\Rightarrow\left(x-2018\right)\left(x-2020\right)\le0\)

\(\Rightarrow\hept{\begin{cases}x-2018\ge0\\x-2020\le0\end{cases}\Rightarrow\hept{\begin{cases}x\ge2018\\x\le2020\end{cases}\Rightarrow}2018\le x\le2020}\)

Và \(\left|2019-x\right|\ge0\), Min (A) = 2 <=> |2019-x| = 0 <=> x= 2019

\(A=\left|x-2018\right|+\left|2019-x\right|+\left|x-2020\right|\)

\(=\left|x-2018\right|+\left|2020-x\right|+\left|2019-x\right|\)

\(\Rightarrow A\ge\left|x-2018+2020-x\right|+\left|2019-x\right|\)

\(=\left|2\right|+\left|2019-x\right|=2+\left|2019-x\right|\ge2\)

Dấu " = " xảy ra \(\Leftrightarrow\hept{\begin{cases}\left(x-2018\right)\left(2020-x\right)\ge0\\2019-x=0\end{cases}}\)(1)

Xét \(\left(x-2018\right)\left(2020-x\right)\ge0\)ta có:

TH1: \(\hept{\begin{cases}x-2018< 0\\2020-x< 0\end{cases}}\Leftrightarrow\hept{\begin{cases}x< 2018\\2020< x\end{cases}}\Leftrightarrow\hept{\begin{cases}x< 2018\\x>2020\end{cases}}\)( vô lý )

TH2: \(\hept{\begin{cases}x-2018\ge0\\2020-x\ge0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ge2018\\2020\ge x\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ge2018\\x\le2020\end{cases}}\Leftrightarrow2018\le x\le2020\)(2)

Từ (1) và (2) \(\Rightarrow\hept{\begin{cases}2018\le x\le2020\\2019-x=0\end{cases}}\Leftrightarrow\hept{\begin{cases}2018\le x\le2020\\x=2019\end{cases}}\Leftrightarrow x=2019\)

Vậy \(minA=2\Leftrightarrow x=2019\)

\(A=\left|2018-x\right|+\left|2019-x\right|+\left|2020-x\right|\)

\(=\left|2018-x\right|+\left|2019-x\right|+\left|x-2020\right|\)

Áp dụng BĐT \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\) :

\(A\ge\left|2018-x+x-2020\right|+\left|2019-x\right|=2+\left|2019-x\right|\ge2\)

Dấu "=" xảy ra \(\Leftrightarrow\left(2018-x\right)\left(x-2020\right)\ge0;2019-x=0\Leftrightarrow x=2019\left(tm\right)\)

Vậy GTNN của A là 2 tại x=2019

\(A=\left(|2018-x|+|2020-x\right)+|2019-x|\)

Đặt \(B=|2018-x|+|2020-x|\)

\(=|2018-x|+|x-2020|\ge|2018-x+x-2020|\)

Hay \(B\ge2\left(1\right)\)

Dấu "=" xảy ra\(\Leftrightarrow\left(2018-x\right)\left(x-2020\right)\ge0\)

\(\Leftrightarrow\hept{\begin{cases}2018-x\ge0\\x-2020\ge0\end{cases}}\)hoặc \(\hept{\begin{cases}2018-x< 0\\x-2020< 0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x\le2018\\x\ge2020\end{cases}\left(loai\right)}\)hoặc \(\hept{\begin{cases}x>2018\\x< 2020\end{cases}}\)

\(\Leftrightarrow2018< x< 2020\)

Đặt \(C=|2019-x|\)

Vì \(|2019-x|\ge0;\forall x\)

Hay \(C\ge0;\forall x\left(2\right)\)

Dấu "=" xảy ra\(\Leftrightarrow2019-x=0\)

                       \(\Leftrightarrow x=2019\)

Từ (1) và (2) \(\Rightarrow B+C\ge2+0\)

Hay \(A\ge2\)

Dấu "=" xảy ra\(\Leftrightarrow\hept{\begin{cases}2018< x< 2020\\x=2019\end{cases}\Leftrightarrow}x=2019\)

Vậy MIN A=2 \(\Leftrightarrow x=2019\)

Duc Dinh

Làm tắt vậy đi thi sao được điểm ? 

Áp dụng BĐT trị tuyệt đối:

\(M=\left|x-2019\right|+\left|2021-x\right|+2020\left|x-2020\right|\)

\(M\ge\left|x-2019+2021-x\right|+2020\left|x-2020\right|=2+2020\left|x-2020\right|\ge2\)

\(\Rightarrow M_{min}=2\) khi \(\left\{{}\begin{matrix}\left(x-2019\right)\left(2021-x\right)\ge0\\\left|x-2020\right|=0\end{matrix}\right.\) \(\Rightarrow x=2020\)