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

`{(3(x^2+2xy-y^2)=6(1)),(2(x^2+xy+y^2)=6(2)):}`

Lấy `(1)` trừ `(2)` vế theo vế:

`3(x^2+2xy-y^2)-2(x^2+xy+y^2)=6-6`

`3x^2+6xy-3y^2-2x^2-2xy-2y^2=0`

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

`x^2-xy+5xy-5y^2=0`

`x(x-y)+5y(x-y)=0`

`(x-y)(x+5y)=0`

`text{TH1}: x=y`

Thay vào `(**)`:

`y^2+y*y+y^2=3`

`3y^2=3`

`y^2=1`

`[(y=1->x=1),(y=-1->x=-1):}`

`text{TH2}: x=-5y`

Thay vào `(**)`:

`(-5y)^2+(-5y)*y+y^2=3`

`25y^2-5y^2+y^2=3`

`21y^2=3`

`y^2=1/7`

`[(y=1/sqrt(7)=sqrt(7)/7->x=-(5sqrt(7))/7),(y=-1/sqrt(7)=-sqrt(7)/7->x=(5sqrt(7))/7):}`

Vậy hệ có nghiệm:

`(x;y) = (1;1),(-1;-1),(-(5sqrt(7))/7;sqrt(7)/7),((5sqrt(7))/7;-sqrt(7)/7)`