Ответы и объяснения

2013-07-06T16:55:18+04:00

x+y+z=4

2xy-z^2=16

 

z=4-(x+y)

 

2xy-16+8(x+y)-(x+y)^2=16

2xy-16+8(x+y)-x^2-2xy-y^2=16

-x^2-y^2+8(x+y)=32

(-x^2+8x-16)+(-y^2+8y-16)=0

-(x-4)^2-(y-4)^2=0

-(x-4)^2=(y-4)^2

x-4=0

y-4=0

x=4

y=4

z=-4

 

x+y+2z=0