Question

Решите систему уравнений ​

Answer 1

Объяснение:

$\left \{ {{x^3y+xy^3=10} \atop {x^2+y^2=5}} \right. \left \{ {{xy*(x^2+y^2)=10} \atop {x^2+y^2=5}} \right. \ \ \ \ \Rightarrow\ \ \ \ xy*5=10\ |:5\ \ \ \ \\xy=2\ |*2\\2xy=4\ \ \ \ \Rightarrow\\x^2+2xy+y^2=5+2xy\\(x+y)^2=5+4\\(x+y)^2=9.$

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

$2)\ x+y=-3.\\\left \{ {{x+y=-3} \atop {xy=2}} \right.\ \ \ \ \left \{ {{y=-3-x} \atop {x*(-3-x)=2}} \right.\ \ \ \ \left \{ {{y=-3-x} \atop {-3x-x^2=2}} \right.\ \ \ \ \left \{ {{y=-3-x} \atop {x^2+3x+2=0}} \right.\ \ \ \ \left \{ {{y=-3-x} \atop {D=\sqrt{D}=1 }} \right.\ \ \ \ \left \{ {{y_3=-1\ \ \ y_4=-2} \atop {x_3=-2\ \ \ x_4=-1}} \right. .$

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