Do \(\left|a\right|=\left|-a\right|\) nên:
\( A=\left|x-2008\right|+\left|x-2020\right|\)
\(=\left|x-2008\right|+\left|2020-x\right|\)
\(\ge\left|x-2008+2020-x\right|=12\)
Dấu "=" xảy ra \(\Leftrightarrow\left(x-2008\right)\left(2020-x\right)\ge0\)
hay \(\orbr{\begin{cases}x-2008\ge0\\2020-x\ge0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x\ge2008\\x\le2020\end{cases}\Leftrightarrow2008\le}x\le2020\)
Thêm xíu:
Vậy \(A_{min}=12\Leftrightarrow2008\le x\le2020\)
\(A=\left|x-2008\right|+\left|x-2020\right|\)
\(\Rightarrow A=\left|x-2008\right|+\left|-x+2020\right|\ge\left|x-2008-x+2020\right|=12\)
dấu = xảy ra khi \(\left(x-2008\right).\left(-x+2020\right)\ge0\)
\(\Rightarrow2018\le x\le2020\)
vậy min A=12 khi và chỉ khi \(2018\le x\le2020\)
