`A=|2018-x|+|2019-x|`
`A=|x-2018|+|2019-x|`
Áp dụng bđt `|a|+|b|>=|a+b|` ta có:
`|x-2018|+|2019-x|>=|x-2018+2019-x|=1`
Dấu = xảy ra khi `(x-2018)(2019-x)>=0`
⇔\(\left[ \begin{array}{l}\begin{cases}x-2018≥ 0\\2019-x ≥0\end{cases}\\\begin{cases}x-2018 ≤0\\2019-x≤ 0\end{cases}\end{array} \right.\)
⇔\(\left[ \begin{array}{l}\begin{cases}x≥2018\\x≤2019\end{cases}\\\begin{cases}x≤2018\\x≥2019 \end{cases}\end{array} \right.\)
`<=> 2018<=x<=2019`
Vậy `A_min=1<=>2018<=x<=2019`
