`\color{#7FFFD4}{g}\color{#54FF9F}{b}`

`{(x^2+xy+y^2=7  (1)),(x^2-xy+y^2=3  (2)):}`

Lấy `(1) ` trừ `(2)=>2xy=4 =>xy=2=>x=2/y`

Thay `x=2/y` vào `(1) ` ta có:

`(2/y)^2 +2/y*y +y^2=7`

`4/(y^2) +2+y^2=7`

`4/(y^2) +y^2=5`

`4/(y^2) + (y^4)/(y^2) =5`

`(y^4+4)/(y^2) =5`

`y^4+4=5y^2`

`y^4-5y^2+4=0`

`y^4-y^2-4y^4+4=0`

`y^2(y^2-1) -4(y^2-1) =0`

`(y^2-4)(y^2-1)=0`

`(y-2)(y+2)(y-1)(y+1) =0`

`TH1:y-2=0=>y=2`

`x=2/y =2/2=1`

`TH2:y+2=0=>y=-2`

`x=2/y =2/(-2) =-1`

`TH3:y-1=0=>y=1`

`x=2/y=2/1=2`

`TH4:y+1=0=>y=-1`

`x=2/y=2/(-1) =-2`

Vậy `(x;y) in {(2;1);(-2;-1);(1;2);(-1;-2)}`