`3.`

`{(x^2(x-y)+(y-1)^2=0 (1)),(4x^3-9x^2+7x+3y^2-10y+5=0 (2)):}`

Từ `(1)`: `x^2(y-x)=(y-1)^2`

Đặt: `y-x=t^2` `(đk: t>=0)`

`y=x+t^2`

`=>x^2t^2=(x+t^2-1)^2`

`=>xt=x+t^2-1` hoặc `xt=-x-t^2+1`

`TH1: xt=x+t^2-1`

`x(t-1)=t^2-1`

`x=t+1`

`y=t^2+t+1`

Thay vào `(2)`:

`t(t+3)(3t^2+t-1)=0`

`t=0` hoặc `t=(\sqrt13-1)/6`

`t=0=>(x;y)=(1;1)`

`t=(\sqrt13-1)/6=>(x;y)=((5+\sqrt13)/6;(11+\sqrt13)/9)`

`TH2: xt=-x-t^2+1`

`x(t+1)=1-t^2`

`x=1-t`

`y=t^2-t+1`

Thay vào `(2)`:

`t(t-3)(3t^2-t-1)=0`

`t=0; t=3; t=(1+\sqrt13)/6`

`t=0=>(x;y)=(1;1)`

`t=3=>(x;y)=(-2;7)`

`t=(1+\sqrt13)/6=>(x;y)=((5-\sqrt13)/6;(11-\sqrt13)/9)`

Vậy `S={(1;1); (-2;7); ((5+\sqrt13)/6;(11+\sqrt13)/9); ((5-\sqrt13)/6;(11-\sqrt13)/9)}`