Question

ПОМОГИТЕЕЕ! 100 БАЛЛ ДАЮ ​

Answer 1

Объяснение:

$9)\ \left \{ {{x^2+y^2=10} \atop {x^4+x^2y^2=90}} \right.\ \ \ \ \left \{ {{x^2+y^2=10} \atop {x^2*(x^2+y^2)=90}} \right. \ \ \ \ \left \{ {{x^2+y^2=10} \atop {x^2*10=90\ |:10}} \right. \ \ \ \ \left \{ {{x^2+y^2=10} \atop {x^2=9}} \right.\\\left \{ {{9+y^2=10} \atop {x_1=-3\ \ x_2=3}} \right. \ \ \ \ \left \{ {{y^2=1} \atop {x=-3\ \ x=3}} \right.\ \ \ \ \left \{ {{y=-1\ \ y=1} \atop {x=-3\ \ x=3}} \right..$

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

$10)\ \left \{ {{x^3-y^3=124} \atop {x^2+xy+y^2=31}} \right. \ \ \ \ \left \{ {{(x-y)*(x^2+xy+y^2)=124} \atop {x^2+xy+y^2=31}} \right. \ \ \ \ \left \{ {{(x-y)*31=124\ |:31} \atop {x^2-2xy+2xy+xy+y^2=31}} \right. \\\left \{ {{x-y=4} \atop {(x-y)^2+3xy=31}} \right.\ \ \ \ \left \{ {{x-y=4} \atop {4^2+3xy=31}} \right. \ \ \ \ \left \{ {{x-y=4} \atop {16+3xy=31}} \right.\ \ \ \ \left \{ {{x-y=4} \atop {3xy=15\ |:3}} \right. \ \ \ \ \left \{ {{x-y=4} \atop {xy=5}} \right.$

$\left \{ {{y=x-4} \atop {x*(x-4)=5}} \right. \ \ \ \ \left \{ {{y=x-4} \atop {x^2-4x-5=0}} \right.\ \ \ \ \left \{ {{y=x-4} \atop {D=36\ \ \sqrt{D}=6 }} \right. \ \ \ \ \left \{ {{y_1=-5\ \ y_2=1} \atop {x_1=-1\ \ x_2=5}} \right..$

Ответ: (-1;-5), (5;1).

$11)\ \left \{ {{x^2y+xy^2=120} \atop {x^2y-xy^2=30}} \right..$

1.

Суммируем эти уравнения:

$2*x^2y=150\ |:2\\x^2y=75.$

2.

$\left \{ {{xy*(x+y)=120} \atop {xy*(x-y)=30}} \right..$

Разделим первое уравнение на второе:

$\frac{x+y}{x-y} =4\\x+y=4*(x-y)\\x+y=4x-4y\\5y=3x\ |:5\\y=0,6x\ \ \ \ \Rightarrow\\\left \{ {{y=0.6x} \atop {x^2y=75}} \right.\ \ \ \ \ \left \{ {{y=0,6x} \atop {x^2*0,6x=75}} \right. \ \ \ \ \ \left \{ {{y=0,6x} \atop {0,6*x^3=75\ |:0,6}} \right. \ \ \ \ \left \{ {{y=0,6x} \atop {x^3=125=5^3}} \right.\ \ \ \ \left \{ {{y=0,6*5=3} \atop {x=5}} \right. .$

Ответ: (5;3).