Ответ:

(2;-1), (-2;1)

Объяснение:

     [tex]\left \{ {{x^2-xy=6} \atop {y^2-xy=3}} \right.\Leftrightarrow \left \{ {{(x^2-xy)+(y^2-xy)=6+3} \atop {(x^2-xy)-(y^2-xy)=6-3}} \right.\Leftrightarrow \left \{ {{(x-y)^2=9} \atop {x^2-y^2=3}} \right.\Leftrightarrow\left \{ {{x-y=\pm 3} \atop {(x-y)(x+y)=3}} \right.[/tex]

1-й случай. [tex]\left \{ {{x-y=3} \atop {(x-y)(x+y)=3}} \right. \Leftrightarrow \left \{ {{x-y=3} \atop {x+y=1}} \right.\Leftrightarrow \left \{ {{(x-y)+(x+y)=3+1} \atop {(x+y)-(x-y)=1-3}} \right.\Leftrightarrow \left \{ {{x=2} \atop {y=-1}} \right. .[/tex]

2-й случай. [tex]\left \{ {{x-y=-3} \atop {(x-y)(x+y)=3}} \right. \Leftrightarrow \left \{ {{x-y=-3} \atop {x+y=-1}} \right.\Leftrightarrow \left \{ {{(x-y)+(x+y)=-3-1} \atop {(x+y)-(x-y)=-1+3}} \right.\Leftrightarrow \left \{ {{x=-2} \atop {y=1}} \right. .[/tex]