Question

помогите пж
алгебра ​

Answer 1

Объяснение:

1)

$\left \{ {{x^2-2y=54} \atop {y=x-3}} \right. \ \ \ \ \left \{ {{x^2-2*(x-3)=54} \atop {y=x-3}} \right. \ \ \ \ \left \{ {{x^2-2x+6-54=0} \atop {y=x-3}} \right. \ \ \ \ \left \{ {{x^2-2x-48=0} \atop {y=x-3}} \right.$

$\left \{ {{D=196\ \ \ \ \sqrt{D}=14} \atop {y=x-3}} \right.\ \ \ \ \left \{ {{x_1=-6\ \ \ \ x_2=8} \atop {y_1=-9\ \ \ \ y_2=5}} \right..$

Ответ: (-6;-9),  (8;5).

2)

$\left \{ {{x+2y=1} \atop {2x+y^2=-1}} \right. \ \ \ \ \left \{ {{x=1-2y} \atop {2*(1-2y)+y^2=-1}} \right. \ \ \ \ \ \left \{ {{x=1-2y} \atop {2-4y+y^2+1=0}} \right. \ \ \ \ \left \{ {{x=1-2y} \atop {y^2-4y+3=0\\}} \right.$

$\left \{ {{x=1-2y} \atop {D=4\ \ \ \ \sqrt{D}=2 }} \right. \ \ \ \ \left \{ {{x_1=-1\ \ \ \ x_2=-5} \atop {y_1=1\ \ \ \ y_2=3}} \right. .$

Ответ: (-1;1),  (-5;3).