Đáp án:
`1, 2|x+4|+2021`
Do `2|x+4| >= 0 AAx`
`-> 2|x+4| + 2021 >=0 AAx`
Dấu ''='' xảy ra `<=> x+4 = 0`
`-> x = 0-4 = -4`
Vậy `MinA = 2021 <=> x = -4`
`2, |x+2021|+|y+2022|-10`
Do `|x+2021| >= 0 AAx`
`|y+2022| >= 0 AAy`
`-> |x+2021|+|y+2022| >=0 AAx,y`
`-> |x+2021|+|y+2022| - 10 >= -10 AAx,y`
Dấu ''='' xảy ra `<=>` $\begin{cases} x+2021 = 0\\y+2022 = 0 \end{cases}$
`->` $\begin{cases} x = -2021\\y = -2022 \end{cases}$
Vậy `MinA = -10 <=> = -2021; y = -2022`