Question

зробіть будь ласка цю систему

Answer 1

Объяснение:

$\left \{ {{x^2+y^2+5x+5y+3xy=15} \atop {x^2+y^2-x-y+xy=1}} \right.\ \ \ \ \left \{ {{x^2+2xy+y^2+5x+5y+xy=15} \atop {x^2+2xy+y^2-x-y-xy=1}} \right.\ \ \ \ \left \{ {{(x+y)^2+5*(x+y)+xy=15} \atop {(x+y)^2-(x+y)-xy=1}} \right. .$

Пусть х+у=t     ху=v.      ⇒

  $\left \{ {{t^2+5t+v=15} \atop {t^2-t-v=1}} \right. .$

Суммируем эти уравнения:

 $2t^2+4t=16\\2t^2+4t-16=0\ |:2\\t^2+2t-8=0\\D=36\ \ \sqrt{D}=6\\t_1=x+y=-4 \\(-4)^2-(-4)-xy=1\\16+4-xy=1\\xy=19\\\left \{ {{x+y=-4} \atop {xy=19}} \right. \ \ \ \ \ \left \{ {{y=-x-4} \atop {x*(-x-4)}=19} \right. \ \ \ \ \left \{ {{y=-x-4} \atop {-x^2-4x=19}} \right. \ \ \ \ \left \{ {{y=-x-4} \atop {x^2+4x+19=0}} \right. \ \ \ \ \left \{ {{y=-x-4} \atop {D=-60\ \ \varnothing}} \right..$

$t_2=x+y=2\\2^2-2-xy=1\\4-2-xy=1\\xy=2-1\\xy=1\\\left \{ {{x+y=2} \atop {xy=1}} \right.\ \ \ \ \left \{ {{y=2-x} \atop {x*(2-x)=1}} \right.\ \ \ \ \left \{ {{y=2-x} \atop {2x-x^2=1}} \right.\ \ \ \ \left \{ {{y=2-x} \atop {x^2-2x+2=0}} \right. \ \ \ \ \left \{ {{y=2-x} \atop {(x-1)^2=0}} \right. \ \ \ \ \left \{ {{y=2-1=1} \atop {x=1}} \right..$

Ответ: x=1    y=1.