已知:x+y+z=1,x^2+y^2+z^2=2,x^3+y^3+z^3=3,求x^4+y^4+z^4=
来源:百度知道 编辑:UC知道 时间:2024/06/06 07:02:45
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解:
X^4+Y^4+Z^4=(X^2+Y^2+Z^2)^2-2(X^2*Y^2+Y^2*Z^2+Z^2*X^2)
(X+Y+Z)^2=X^2+Y^2+Z^2+2(XY+YZ+ZX)
解得XY+YZ+ZX=1/2
X^3+Y^3+Z^3=(X+Y+Z)(X^2+Y^2+Z^2-XY-YZ-ZX)+3XYZ
解得XYZ=1/2
设XY=a,YZ=b,ZX=c.
则X^2*Y^2+Y^2*Z^2+Z^2*X^2=a^2+b^2+c^2=(a+b+c)^2-2(ab+bc+ca)
又因为ab+bc+ca=XYZ(X+Y+Z)=1/2
所以X^2*Y^2+Y^2*Z^2+Z^2*X^2=a^2+b^2+c^2=1/2-2* 1/2=-1/2
X^4+Y^4+Z^4=2^2-2*(-1/2)=5
4
不是4么?
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