x,y,z满足x+y+z=3,x^2+y^2+z^2=29,x^3+y^3+z^3=45,求x,y,z,x^4+y^4+
1个回答

x+y+z=3 (1)

x²+y²+z²=29 (2)

x³+y³+z³=45 (3)

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由方程(1)得

x+y=3-z (4)

(x+y)²=(3-z)²

x²+y²+2xy=9-6z+z²

x²+y²+z²+2xy=9-6z+2z²

29+2xy=9-6z+2z²

xy=z²-3z-10 (5)

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由方程(3)得

(x+y)(x²-xy+y²)+z³=45

代入(4)和(5)得

(3-z)(29-z²-z²+3z+10)+z³=45

(3-z)(39-2z²+3z)+z³=45

z³-3z²-10z+24=0

(z³-6z²+12z-8)+(3z²-12z+12)-10z+20=0

(z-2)³+3(z-2)²-10(z-2)=0

(z-2)[(z-2)²+3(z-2)-10]=0

(z-2)(z-2+5)(z-2-2)=0

(z-2)(z+3)(z-4)=0

z=2,-3,4

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z=2时,xy=-12,x+y=1 解得 x=4,y=-3 或 x=-3,y=4

z=-3时,xy=8,x+y=6 解得 x=4,y=2 或 x=2,y=4

z=4时,xy=-6,x+y=-1 解得 x=2,y=-3 或 x=-3,y=2

原方程组的解是

x = 4,y = 2,z = -3

x = -3,y = 2,z = 4

x = 2,y = 4,z = -3

x = -3,y = 4,z = 2

x = 4,y = -3,z = 2

x = 2,y = -3,z = 4

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x^4+y^4+z^4

=(x²+y²)²-2x²y²+z^4

=(29-z²)²-2(z²-3z-10)²+z^4

=12z³-36z²-120z+641

=12(z-2)(z-4)(z+3)+353

=353