`{(x^2-y^2=3(x-y)(1)),(xy=2(2)):}`
`(1)<=>(x-y)(x+y)-3(x-y)=0`
`<=> (x-y)(x+y-3)=0`
`<=>`\(\left[ \begin{array}{l}x=y\\x+y=3\end{array} \right.\)
Với `x=y` thì `(2)<=>x^2=y^2=2<=>x=y=+-\sqrt{2}`
Với `x+y=3` thì `(2)<=>x=2/y`
`=> 2/y+y=3`
`<=> (2+y^2)/y=3`
`<=> y^2+2=3y`
`<=> y^2-3y+2=0`
`<=> y^2-2y-y+2=0`
`<=> (y-2)(y-1)=0`
`<=>`\(\left[ \begin{array}{l}y=2⇒x=1\\y=1⇒x=2\end{array} \right.\)
Vậy `(x;y)=(\sqrt{2};\sqrt{2});(-\sqrt{2};-\sqrt{2});(1;2);(2;1)`
