`a`

`||x-2024|+|x-2025|+3(x-y)^2|=2(x-y)^2+1`

`|x-2024|+|x-2025|+3(x-y)^2=2(x-y)^2+1`

`|x-2024|+|x-2025|+3(x-y)^2-3(x-y)^2=1`

`|x-2024|+|2025-x|+(x-y)^2=1`

Áp dụng  `|A|+|B| >= |A+B|` ta được 

`|x-2024|+|2025-x| >= |x-2024+2025-x|=1`

Vì `|x-2024|+|2025-x| >=1 AA x`

`(x-y)^2 >=0 AA x;y`

Để `VT=1` thì 

`(x-2024)(2025-x) >=0` và `(x-y)^2=0`

`TH1`

`{(x-2024>=0),(2025-x>=0):} => {(x >=2024),(x<=2025):} => 2024 <=x <=2025`

`=>x in {2024;2025}`

`TH2`

`{(x-2024<=0),(2025-x<=0):} => {(x <=2024),(x>=2025):} (` vô lí )

Ta lại có

`TH1`

`x=2024`

`=>(x-y)^2=0`

`=>(2024-y)^2=0`

`=>2024-y=0`

`=>y=2024`

`TH2`

`x=2025`

`=>(x-y)^2=0`

`=>(2025-y)^2=0`

`=>2025-y=0`

`=>y=2025`

Vậy `(x;y) in {(2024;2024);(2025;2025)}`