Предмет: Алгебра, автор: Аноним

помогите пж дам сверху сто балов очень надо

Приложения:

Ответы

Автор ответа: sangers1959

Объяснение:

$1)\ \left \{ {{x=5-y} \atop {y^2+4xy=33}} \right.\ \ \ \ \left \{ {{x=5-y} \atop {y^2+4y*(5-y)=33}} \right. \ \ \ \ \left \{ {{x=5-y} \atop {y^2+20y-4y^2=33\\}} \right.\ \ \ \ \left \{ {{x=5-y} \atop {3y^2-20y+33}} \right.\ \ \ \ \left \{ {{x=5-y} \atop {D=4\ \ \sqrt{D}=2} }} \right.\ \ \ \ \\\left \{ {{x_1=2\ \ x_2=\frac{4}{3} } \atop {y_1=3\ \ y_2=\frac{11}{3} }} \right. .$

Ответ:(2;3), (4/3;11/3).

$\left \{ {{x+y=8} \atop {xy=-20}} \right.\ \ \ \left \{ {{y=8-x} \atop {x*(8-x)=-20}} \right.\ \ \ \left \{ {{y=8-x} \atop {8x-x^2=-20}} \right.\ \ \ \left \{ {{y=8-x} \atop {x^2-8x-20=0}} \right.\ \ \ \left \{ {{y=8-x} \atop {D=144\ \ \sqrt{D}=12 }} \right. \ \ \ \left \{ {{y_1=-2\ \ y_2=10} \atop {x_1=10\ \ x_2=-2} \right. .$

Ответ: (10;-2), (-2;10).

$3)\ \left \{ {{y-7x=3} \atop {y^2-6xy-x^2=-9}} \right.\ \ \ \left \{ {{y=7x+3} \atop {(7x+3)^2-6x*(7x+3)-x^2=-9}} \right. \ \ \ \left \{ {{y=7x+3} \atop {49x^2+42x+9-42x^2-18x-x^2=-9}} \right. \\\left \{ {{y=7x+3} \atop {6x^2+24x+18=0\ |:6}} \right.\ \ \ \left \{ {{y=7x+3} \atop {x^2+4x+3=0}} \right.\ \ \ \left \{ {{y=7x+3} \atop {D=4\ \ \sqrt{D}=2 }} \right.\ \ \ \left \{ {{y_1=-18\ \ y_2=-4} \atop {x_1=-3\ \ x_2=-1}} \right. .$