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 \(minA=2\Leftrightarrow x=2019\)
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