Question

решить систему уравнений">

Answer 1

Объяснение:

$\displaystyle\\\left \{ {{2x^2-5xy-3y^2=0} \atop {x^2-2xy-y^2=2}} \right. ;\left \{ {{2x^2-6xy+xy-3y^2=0} \atop {x^2-2xy-y^2=2}} \right. ;\\\\\\\left \{ {{2x(x-3y)+y(x-3y)=0} \atop {x^2-2xy-y^2=2}} \right. ;\left \{ {{(x-3y)(2x+y)=0} \atop {x^2-2xy-y^2=2}} \right. ;\\\\\\1)\left \{ {{x=3y} \atop {x^2-2xy-y^2=2}} \right. ;\left \{ {{x=3y} \atop {9y^2-6y^2-y^2=2}} \right. ;\\\\\\\left \{ {{x=3y} \atop {y^2=1}} \right. ;\left \{ {{x=3y} \atop {y=\pm1}} \right.$

$y_1=-1;x_1=-3\\y_2=1;x_2=3$

$2)\displaystyle\\\left \{ {{y=-2x} \atop {x^2-2xy-y^2=2}} \right. ;\left \{ {{y=-2x} \atop {x^2+4x^2-4x^2=2}} \right. ;\left \{ {{y=-2x} \atop {x=\pm\sqrt{2} }} \right.$

$x_3=-\sqrt{2} ;y_3=2\sqrt{2} \\x_4=\sqrt{2} ;y_4=-2\sqrt{2}$

ответ: $\bf(-3;-1)~(3;1)~(-\sqrt{2} ;2\sqrt{2} )~(\sqrt{2} ;-2\sqrt{2} )$