Puntos de interseccion de la recta x+y-1=0 con la parabola x2-4xy+1 , .
Respuestas a la pregunta
Contestado por
2
x=1-y
(1-y)+4(1-y)y+1=0
1-y+4y-4y^2+1=0
4y^2-3y-2=0
y=(3+(41)^(1/2))/8 x=(5-(41)^(1/2))/8
y:(3-(41)^(1/2))/8 x=(5-(41)^(1/2))/8
(1-y)+4(1-y)y+1=0
1-y+4y-4y^2+1=0
4y^2-3y-2=0
y=(3+(41)^(1/2))/8 x=(5-(41)^(1/2))/8
y:(3-(41)^(1/2))/8 x=(5-(41)^(1/2))/8
Contestado por
1
x=1-y
(1-y)+4(1-y)y+1=0
1-y+4y-4y^2+1=0
4y^2-3y-2=0
y=(3+(41)^(1/2))/8
y:(3-(41)^(1/2))/8
x=(5-(41)^(1/2))/8
x=(5-(41)^(1/2))/8
Otras preguntas