x+y+z=3,x2+y2+z2=29,x3+y3+z3=45,求xyz和x4+y4+z4的值
发布网友
发布时间：2024-10-04 10:43
共2个回答
热心网友
时间：2024-12-02 21:47
x^4 + y^4 + z^4 = (x^2 + y^2 + z^2)^2 - 2(x^2y^2 + y^2z^2 + z^2x^2)
let (x^2y^2 + y^2z^2 + z^2x^2) = pp
pp = (xy + yz + zx)^2 - 2(xy^2z + yz^2x+zx^2y)
= (xy + yz + zx)^2 - 2xyz(x + y + z)
xy + yz + zx = ((x + y + z)^2 - (x^2 + y^2 + z^2))/2 =(3^2 - 29)/2 = -10
xyz =[ (x^3 + y^3 + z^3) -(x + y + z)(x^2 + y^2 + z^2 - (xy + yz + zx)) ]/3
= [45 - 3 (29-(-10))]/3 = -24
所以，
pp = (-10)^2 - 2 (-24)(3) = 244
原式 = x^4 + y^4 + z^4 = (x^2 + y^2 + z^2)^2 - 2pp = 29^2 - 2*244 = 353
热心网友
时间：2024-12-02 21:47
x=4,y=-3,z=2
xyz=-24,x4+y4+z4=-353
