Question

Розв'язати систему рівнянь: x2 +2xy+y2=1, x2 —xy = 6​

Answer 1

Объяснение:

$\left \{ {{x^2+2xy+y^2=1} \atop {x^2-xy=6}} \right.\ \ \ \ \left \{ {{(x+y)^2=1} \atop {x^2-xy=6}} \right. .$

1)

x+y=-1.        ⇒

$\left \{ {{y=-x-1} \atop {x^2-x*(-x-1)=6}} \right.\ \ \ \ \left \{ {{y=x-1} \atop {2x^2+x-6=0}} \right. \ \ \ \ \ \left \{ {{y=-x-1} \atop {D=49\ \ \ \ \sqrt{D} =7}} \right. \ \ \ \ \left \{ {{y_1=1\ \ \ \ \ y_2=-2,5} \atop {x_1=-2\ \ \ \ x_2=1,5}} \right. .$

2)

x+y=1.          ⇒

$\left \{ {{y=1-x} \atop {x^2-x*(1-x)=6}} \right.\ \ \ \ \left \{ {{y=1-x} \atop {2x^2-x-6=0}} \right. \ \ \ \ \ \left \{ {{y=1-x} \atop {D=49\ \ \ \ \sqrt{D} =7}} \right. \ \ \ \ \left \{ {{y_1=2,5\ \ \ \ \ y_2=-1} \atop {x_1=-1,5\ \ \ \ x_2=2}} \right. .$

Ответ: (-2;1),  (1,5;-2,5),  (-1,5;2,5),  (2;-1).