Đáp án:

`|3x-2y| + |5z-7x| + (xy+yz+xz-500)^2018 = 0`

Ta có: `|3x-2y| >= 0 AA x,y`

`|5z-7x| >= 0 AA x,z`

`(xy+yz+xz-500)^2018 >= 0 AA x,y,z`

`=>` `|3x-2y| + |5z-7x| + (xy+yz+xz-500)^2018 >= 0 AA x,y,z`

Mà `|3x-2y| + |5z-7x| + (xy+yz+xz-500)^2018 = 0` (Theo đề bài)

`=>` `{(|3x-2y| = 0),(|5z-7x|= 0),((xy+yz+xz-500)^2018=0):}`

`=>` `{(3x=2y),(5z=7x),((xy+yz+xz-500)^2018=0):}`

`=>` `{(3x=2y),(5z=7x),(xy+yz+xz-500=0):}`

Ta có: `3x=2y`

`=>` `x/2 = y/3`

`x/10 = y/15`

Ta có: `5z = 7x`

`=>` `z/7 = x/5`

`z/14 = x/10`

`=> x/10 = y/15 = z/14`

Đặt `x/10 = y/15 = z/14 = k`

`=>` `{(x=10k),(y=15k),(z=14k):}` `(1)`

Thay `(1)` vào `xy+yz+xz-500=0`

`10k*15k+15k*14k+10k*14k = 500`

`150k^2 + 210k^2 + 140k^2 = 500`

`500*k^2 = 500`

`k^2 = 1`

`k = 1` hoặc `k=-1`

`TH1: k=1`

`=>` `{(x=10*1),(y=15*1),(z=14*1):}`

`{(x=10),(y=15),(z=14):}`

`=> A = (3*10-15-14)^2019 = 1^2019 = 1`

`TH2: k=-1`

`=>` `{(x=10*(-1)),(y=15*(-1)),(z=14*(-1)):}`

`{(x=-10),(y=-15),(z=-14):}`

`=> A = [3*(-10)-(-15)-(-14)]^2019 = (-1)^2019 = -1`

Vậy `...`