Question

помогит пожалуйста!!!​

Answer 1

Объяснение:

                                               Способ №1.

$\displaystyle\\\left \{ {{x-y=3} \atop {x^2+y^2=17}} \right.\ \ \ \ \left \{ {{x-y=3} \atop {x^2-2xy+y^2=17-2xy}} \right. \ \ \ \ \left \{ {{x-y=3} \atop {(x-y)^2=17-2xy}} \right. \\\\\\\left \{ {{x-y=3} \atop {3^2=17-2xy}} \right.\ \ \ \ \left \{ {{x-y=3} \atop {9=17-2xy}} \right. \ \ \ \ \left \{ {{x-y=3} \atop {-8=-2xy\ |:(-2)}} \right. \ \ \ \ \left \{ {{x=y+3} \atop {xy=4}}\right.$

$\displaystyle\\\left \{ {{x=y+3} \atop {(y+3)*y=4}} \right.\ \ \ \ \left \{ {{x=y+3} \atop {y^2+3y-4=0}} \right. \ \ \ \ \left \{ {{x=y+3} \atop {y^2+4y-y-4=0}} \right. \\\\\\\left \{ {{x=y+3} \atop {y*(y+4)-(y+4)=0}} \right. \ \ \ \ \left \{ {{x=y+3} \atop {(y+4)(y-1)=0}} \ \ \ \ \left \{ {{x_1=-1\ \ \ \ x_2=4} \atop {y_1=-4\ \ \ \ y_2=1}} \right. \right..$

Ответ: (-1;-4),  (4;1).

                                                   Способ №2.

$\displaystyle\\\left \{ {{x-y=3} \atop {x^2+y^2=17}} \right. \ \ \ \ \left \{ {{x=y+3} \atop {(y+3)^2+y^2=17}} \right. \ \ \ \ \left \{ {{x=y+3} \atop {y^2+6y+9+y^2-17=0}} \right. \\\\\\\left \{ {{x=y+3} \atop {2y^2+6y-8=0\ |:2}} \right. \ \ \ \ \left \{ {x=y+3} \atop {y^2+3y-4=0}} \right. \ \ \ \ \left \{ {{x=y+3} \atop {y^2+4y-y-4=0}} \right.$

$\displaystyle\\\left \{ {{x=y+3} \atop {(y+3)*y=4}} \right.\ \ \ \ \left \{ {{x=y+3} \atop {y^2+3y-4=0}} \right. \ \ \ \ \left \{ {{x=y+3} \atop {y^2+4y-y-4=0}} \right. \\\\\\\left \{ {{x=y+3} \atop {y*(y+4)-(y+4)=0}} \right. \ \ \ \ \left \{ {{x=y+3} \atop {(y+4)(y-1)=0}} \ \ \ \ \left \{ {{x_1=-1\ \ \ \ x_2=4} \atop {y_1=-4\ \ \ \ y_2=1}} \right. \right..$

Ответ: (-1;-4),  (4;1).