Question

пожалуйста подарите срочно пожалуйста​

Answer 1

Объяснение:

1.

$\left \{ {{x^2+xy^2+y^2=13} \atop {x+y=4}} \right.\ \ \ \ \left \{ {{x^2+y*(x+y)=13} \atop {x+y=4} \right.\ \ \ \ \left \{ {{x^2+4y=13} \atop {x+y=4}} \right. \ \ \ \ \left \{ {{(4-y)^2+4y=13} \atop {x=4-y}} \right. \ \ \ \ \left \{ {16-8y+y^2+4y=13} \atop {x=4-x}} \right. \\\left \{ {{y^2-4y+3=0} \atop {x=4-y}} \right.\ \ \ \ \ \left \{ {{D=4\ \ \ \ \sqrt{D}=2 } \atop {x=4-y}} \right. \ \ \ \ \left \{ {{y_1=1\ \ \ \ y_2=3} \atop {x_1=3\ \ \ \ x_2=1}} \right..$

Ответ: (3;1), (1;3).

$\left \{ {{x-3y=0} \atop {x^2+y^2+2y=9}} \right. \ \ \ \ \left \{ {{x=3y} \atop {(3y)^2+y^2+2y=9}} \right. \ \ \ \ \left \{ {{x=3y} \atop {9y^2+y^2+2y-9=0}} \right. \ \ \ \ \left \{ {{x=3y} \atop {10y^2+2y-9=0}} \right. \\\left \{ {{x=3y} \atop {D=364\ \ \ \ \sqrt{D}=2\sqrt{91} }} \right. \ \ \ \ \left \{ {{x_1=-0,3-0,3\sqrt{91}\ \ \ \ x_2=-0,3+0,3\sqrt{91} } \atop {y_1=-0,1-0,1\sqrt{91}\ \ \ \ y_2=-0,1+0,1\sqrt{91} }} \right. .$

Ответ: (-0,3-0,3√91;-0,1-0,1√91), (-0,3+0,3√91;-0,1+0,1√91).