Question

Очень нужна ваша помощь!!! ЗАРАНЕЕ СПАСИБО ❤️
(Не важно на какой вопрос будет ответ)


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Answer 1

Ответ на файле, прикреплённом ниже.

Задание 1 – графически.

Answer 2

Объяснение:

$\left \{ {{(x+2)*(y+2)*xy=9} \atop {x-xy+y=1}} \right.\ \ \ \ \ \left \{ {{(xy+2x+2y+4)*xy=9} \atop {x+y-xy=1}} \right.\ \ \ \ \ \left \{ {{(xy+2*(x+y)+4)*xy=9} \atop {x+y-xy=1}} \right..$

Пусть x+y=t,  xy=v.       ⇒

$\left \{ {{((v+2t+4)*v=9} \atop {t-v=1}} \right. \ \ \ \ \left \{ {{v^2+2tv+4v=9} \atop {t=v+1}} \right. \ \ \ \ \ \left \{ {{v^2+2v*(v+1)+4v=9} \atop {t=v+1}} \right.\ \ \ \ \left \{ {{v^2+2v^2+2v+4v-9=0} \atop {t=v+1}} \right.$

$\left \{ {{3v^2+6v-9=0 \ |:3} \atop {t=v+1}} \right. \ \ \ \ \ \left \{ {{v^2+2v-3=0} \atop {t=v+1}} \right.\ \ \ \ \left \{ {{D=16\ \ \ \sqrt{D}=4 } \atop {t=v+1}} \right.\ \ \ \ \left \{ {{v_1=-3\ \ \ \ v_2=1} \atop {t_1=-2\ \ \ \ t_2=2}} \right. .$

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

$\left \{ {{y=-x-2} \atop {D=16\ \ \ \ \sqrt{D}=4 }} \right.\ \ \ \ \left \{ {{y_1=1\ \ \ \ y_2=-3} \atop {x_1=-3\ \ \ \ x_2=1}} \right. .$

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

$\left \{ {{y=2-x} \atop {x-1=0}} \right. \ \ \ \ \ \left \{ {{y_3=1} \atop {x_3=1}} \right. .$

Ответ: (-3;1),  (1;-3),  (1;1).