Question

Помогите пожалуйста, даю 20 баллов
x+y+x^2/y^2=7
((x+y)x^2)/y^2=12

Answer 1

Объяснение:

$\displaystyle\\\left \{ {{x+y+\frac{x^2}{y^2}=7 } \atop {\frac{(x+y)*x^2}{y^2}=12 }} \right. .$

Пусть: x+y=t     x^2/y^2=v       ⇒

$\displaystyle\\\left \{ {{t+v=7} \atop {tv=12}} \right.\ \ \ \ \left \{ {{t=7-v} \atop {(7-v)*v=12}} \right. \ \ \ \ \left \{ {{t=7-v} \atop {7v-v^2=12}} \right. \ \ \ \ \left \{ {{t=7-v} \atop {v^2-7v+12=0\right.\\\\\\\left \{ {{t=7-v} \atop {v^2-3v-4v+12=0}} \right. \ \ \ \ \left \{ {{t=7-v} \atop {v*(v-3)-4*(x-3)=0}} \right. \ \ \ \ \left \{ {{t=7-v} \atop {(v-3)*(v-4)=0}} \right.$

$\displaystyle\\\left \{ {{t_1=4\ \ \ \ t_2=3} \atop {v_1=3\ \ \ \ v_2=4}} \right.$

1)

$\displaystyle\\\left \{ {{x+y=4} \atop {\frac{x^2}{y^2}=3 }} \right. \ \ \ \ \left \{ {{x=4-y} \atop {x^2=3y^2} \right. \ \ \ \ \left \{ {{x=4-y} \atop {(4-y)^2=3y^2 }} \right. \ \ \ \ \left \{ {{x=4-y} \atop {16-8y+y^2=3y^2}} \right. \\\\\\\left \{ {{x=4-y} \atop {2y^2+8y-16=0\ |:2}} \right. \ \ \ \ \left \{ {{x=4-y} \atop {y^2+4y-8=0}} \right. \ \ \ \ \left \{ {{x=4-y} \atop {D=48\ \ \ \ \sqrt{D}=4\sqrt{3} }} \right.$

$\displaystyle\\\left \{ {{x_1=6+2\sqrt{3}\ \ \ \ x_2=6-2\sqrt{3} } \atop {y_1=-2-2\sqrt{3}\ \ \ y_2=-2+2\sqrt{3} }} \right. .$

2)

$\displaystyle\\\left \{ {{x+y=3} \atop {\frac{x^2}{y^2}=4 }} \right. \ \ \ \ \left \{ {{x=3-y} \atop {x^2=4y^&2}} \right. \ \ \ \ \left \{ {{x=3-y} \atop {(3-y)^2=4y^2}} \right. \ \ \ \ \left \{ {{x=3-y} \atop {9-6y+y^2=4y^2}} \right.\\\\\\\left \{ {{x=3-y} \atop {3y^2+6y-9=0\ |:3}} \right. \ \ \ \ \left \{ {{x=3-y} \atop {y^2+2y-3=0}} \right. \ \ \ \ \left \{ {{x=3-y} \atop {y^2+3y-y-3=0}} \right.$

$\displaystyle\\\left \{ {{x=3-y} \atop {y*(y+3)-(y+3)=0}} \right. \ \ \ \ \left \{ {{x=3-y} \atop {(y+3)*((y-1)=0}} \right. \ \ \ \ \left \{ {{x_3=6\ \ \ \ x_4=2} \atop {y_3=-3\ \ \ \ y_4=1}} \right. .$