Question

x³+y³=9
xy=2
решите неравенство пж​

Answer 1

Ответ:

Объяснение:

!!!!!!!!!!!!!!!!!!!!!!!!

Answer 2

Объяснение:

$\left \{ {x^3+y^3=9} \atop {xy=2}} \right. \ \ \ \ \ \left \{ {{(x+y)*(x^2-xy+y^2)=9} \atop {xy=2}} \right. \ \ \ \ \ \left \{ {{(x+y)*(x^2+2xy+y^2-3xy)=9} \atop {x=2}} \right. \ \ \ \\\left \{ {{(x+y)*((x+y)^2-3xy)=9} \atop {xy=2}} \right. \ \ \ \ \ \left \{ {{(x+y)*((x+y)^2-3*2)=9} \atop {xy=2}} \right.\ \ \ \ \left \{ {{(x+y)^3-6*(x+y)-9=0} \atop {xy=2}} \right. .$

Пусть x+y=t.                    ⇒

$t^3-6t-9=0\\t^3-3t^2+3t^2-6t-9=0\\t^2*(t-3)+3*(t^2-2t-3)=0\\t^2*(t-3)+3*(t^2-3t+t-3)=0\\t^2*(t-3)+3*(t*(t-3)+(t-3))=0\\t^2*(t-3)+3*(t-3)*(t+1)=0\\(t-3)*(t^2+3*(t+1)=0\\(t-3)*(t^2+3t+3)=0\\t-3=0\\t_1=x+y=3.\\t^2+3t+3=0\\D=-3\ \ \ \ \Rightarrow$

Уравнение не имеет действительных корней.

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

Ответ: (1;2)   (2;1).