`\color{#08192D}{|}\color{#102A43}{=}\color{#1F3B73}{****}\color{#2A4D8F}{+}\color{#3A63B8}{**}\color{#4A7EDC}{+}\color{#5AA6FF}{*}\color{#2E6BD3}{o}\color{#3A7FE8}{l}\color{#4693FF}{a}\color{#5AA6FF}{l}\color{#6FB9FF}{a}\color{#82CCFF}{9}\color{#95DFFF}{9}\color{#A8F2FF}{9}\color{#5AA6FF}{*}\color{#4A7EDC}{+}\color{#3A63B8}{**}\color{#2A4D8F}{+}\color{#1F3B73}{***}\color{#102A43}{=}\color{#08192D}{|}`

Ta có:

`|3x-4y-2|^2021≥0` và `(x-2y)≥0`

Vì `|3x - 4y - 2|^2021 + (x - 2y)^2020 = 0`

`=>|3x - 4y - 2|^2021=0`

`=>(x-2y)^2020=0`

`=>x-2y=0`

`x=2y`

Thay `x=2y` vào `|3x - 4y - 2|^2021` ta có:

`|3x - 4y - 2|^2021=0`

`=>3(2y)-4y=2`

`=>6y-4y=2`

`=>2y=2`

`=>y=1`

`=>x=2`

Vậy `x=1;y=2`