Question

Решить систему уравнений(желательно с подробным решением)
x+y-xy=1
x^3-y^3=13(x-y)

Answer 1

Объяснение:

$\left \{ {{x+y-xy=1} \atop {x^3-y^3=13*(x-y)}} \right.\ \ \ \ \left \{ {{x+y-xy=1} \atop {(x-y)*(x^2+xy+y^2)-13*(x-y)=0}} \right. \ \ \ \ \left \{ {{x+y-xy=1} \atop {(x-y)*(x^2+xy+y^2-13)=0}} \right. .$

1.

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

2.

$\left \{ {{x+y-xy=1} \atop {x^2+xy+y^2-13=0}} \right.\ \ \ \ \left \{ {{x+y=xy+1} \atop {x^2+2xy+y^2=xy+13}} \right. \ \ \ \ \left \{ {{x+y=xy+1} \atop {(x+y)^2=xy+1+12}} \right.\ \ \ \ \left \{ {{x+y=xy+1} \atop {(x+y)^2=(x+y)+12}} \right.\\\left \{ {{x+y=xy+1} \atop {(x+y)^2-(x+y)-12=0}} \right. .$

Пусть х+у=t       ⇒

$t^2-t-12=0 \\D=49\ \ \ \ \sqrt{D}=7\\t_1=x+y=4. \\4=xy+1\\xy=3.\\\left \{ {{x+y=4} \atop {xy=3}} \right. \ \ \ \ \Rightarrow\ \ \ \ \left \{ {{x_2=3\ \ \ \ x_3=1} \atop {y_2=1\ \ \ \ y_3=3}} \right. .$

$t_2=x+y=-3.\\-3=xy+1\\xy=-4.\\\left \{ {{x+y=-3} \atop {xy=-4}} \right. \ \ \ \ \Rightarrow\ \ \ \ \left \{ {{x_4=1\ \ \ \ x_5=-4} \atop {y_4=-4\ \ \ \ y_5=1}} \right. .$

Ответ: (1;1), (3;1), (1;3), (1;-4), (-4;1).