Question

X+y+z=1 si (x+y)(y+z)(z+x)=14 Calculati xy+yz+zx-xyz
Va rog ajutati ma

Answer 1

desfacem parantezele 
xyz+x^2 y+xz^2+x^2z+zy^2+xy^2+yz^2+xyz=14
xy(x+y)+xz(x+z)+yz(y+z)+2xyz=14 (1)
x+y=1-z
x+z=1-y
y+z=1-x     cu acestea relatia (1) devine:
xy-xyz+xz-xyz+yz-xyz+2xyz=14
xy+xz+yz-xyz=14