x+y+z=1；x方+y方+z方=2；x立方+y立方+z立方=3 求x的四次方+y的四次方+z的四次方

Question

初中数学题

Answer 1

首先，得知道一个公式：（x+y+z）^2=x^2+y^2+z^2+2xy+2yz+2xz.....①
(x+y+z)(x^2+y^2+z^2)=x^3+y^3+z^3+xy(x+y)+xz(x+z)+yz(y+z)
=3+xy(1-z)+xz(1-y)+yz(1-x)
=3+xy+yz+xz-3xyz
又由①：xy+yz+xz=-1/2
所以 原式=3+（-1/2）-3xyz=2 xyz=1/6
(x+y+z)(x^3+y^3+z^3)=x^4+y^4+z^4+xy(x^2+y^2)+yz(y^2+z^2)+xz(x^2+z^2)
=x^4+y^4+z^4+xy(1-z^2)+yz(1-x^2)+xz(1-y^2)
=x^4+y^4+z^4+xy+yz+xz-xyz(x+y+z)
=x^4+y^4+z^4+(-1/2)-1/6
=3
所以x^4+y^4+z^4=3+1/2+1/6

Answer 2

（x+y+z）^2=x^2+y^2+z^2+2xy+2yz+2xz.....
(x+y+z)(x^2+y^2+z^2)=x^3+y^3+z^3+xy(x+y)+xz(x+z)+yz(y+z)
=3+xy(1-z)+xz(1-y)+yz(1-x)
=3+xy+yz+xz-3xyz
xy+yz+xz=-1/2
原式=3+（-1/2）-3xyz=2 xyz=1/6
(x+y+z)(x^3+y^3+z^3)=x^4+y^4+z^4+xy(x^2+y^2)+yz(y^2+z^2)+xz(x^2+z^2)
=x^4+y^4+z^4+xy(1-z^2)+yz(1-x^2)+xz(1-y^2)
=x^4+y^4+z^4+xy+yz+xz-xyz(x+y+z)
=x^4+y^4+z^4+(-1/2)-1/6
=3
x^4+y^4+z^4=3+1/2+1/6