假設(shè)xyz=x^2+Y^3+Z^3=4,而且X^2Y+X Y^2+X^2Z+XZ^2+Y^2Z+YZ^2=12 求xy+yz+zx

數(shù)學(xué)人氣：594 ℃時(shí)間：2020-06-25 03:47:07

優(yōu)質(zhì)解答

是不是題目寫錯(cuò)了,第一個(gè)條件里的x^2應(yīng)該是x^3吧
這樣已知
xyz = 4
x3+y3+z3 = 4
x2y+xy2+x2z+xz2+y2z+yz2 = 12
設(shè)xy+yz+zx = t
則(x+y+z)*(xy+yz+zx) = x2y+xy2+x2z+xz2+y2z+yz2 + 3xyz = 12+3*4 = 24
即(x+y+z)*t = 24
x+y+z = 24/t
另有
(x+y+z)^3 = x3+y3+z3+3(x2y+xy2+x2z+xz2+y2z+yz2)+6xyz = 4+3*12+6*4 = 64
即x+y+z = 4
代入得4=24/t
t = 6
即xy+yz+zx = 6
如果您題目沒寫錯(cuò),那真心做不出