Question

СРОЧНО решить систему методом замены
x^2y - xy^2=6, Замена: ху=а, х-у =b
xy+x-y= -5

Ответ (х1,у1).(х2,у2)

Answer 1

Ответ: (-2;1),  (-1;2).

Объяснение:

$\displaystyle\\\left \{ {{x^2y-xy^2=6} \atop {xy+x-y=-5} \right.\ \ \ \ \ \ \ \left \{ {{xy*(x-y)=6} \atop {xy+x-y=-5}} \right.$

Пусть ху=a,   x-y=b.          ⇒

$\displaystyle\\\left \{ {{a*b=6} \atop {a+b=-5}} \right.\ \ \ \ \ \ \left \{ {{(-b-5)*b=6} \atop {a=-b-5}} \right. \ \ \ \ \ \ \left \{ {{-b^2-5b-6=0\ |:0} \atop {a=-b-5}} \right. \\\\\\\left \{ {{b^2+5b+6=0} \atop {a=-b-5}} \right.\ \ \ \ \ \ \left \{ {{b^2+2b+3b+6=0} \atop {a=-b-5}} \right.\ \ \ \ \ \ \left \{ {{b*(b+2)+3*(b+2)=0} \atop {a=-b-5}} \right. \\\\\\\left \{ {{(b+2)*(b+3)=0} \atop {a=-b-5}} \right. \ \ \ \ \ \ \left \{ {b_1=-2\ \ \ \ b_2=-3} \atop {a_1=-3\ \ \ \ a_2=-2}} \right. .$

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

$\displaystyle\\2)\ \left \{ {{xy=-3} \atop {x-y=-2}} \right. \ \ \ \ \ \ \left \{ {{(y-2)*y=-3} \atop {x=y-2}} \right. \ \ \ \ \ \ \left \{ {{y^2-2y+3=0} \atop {x=y-2}} \right.\ \ \ \ \ \ \left \{ {{D=-8} \ \ \ \ y\in \varnothing \atop {x\in\varnothing}} \right..$