Question

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=?

Answer 1

郭敦荣回答：
x+y+z=1 （1）;
x^2+y^2+z^2=2 （2）;
x^3+y^3+z^3=3 （3）
这是个对称性方程组，应有x=y=z的关系，从而由（1）得x=y=z=1/3，
但这与（2）、（3）矛盾，故此题无解。

Answer 2

由题可知:x=1.y与z为0.或y=1.x与z为0.或z=1.x与y为0. 所以x^4+y^4+z^4=4

追答

不知道可以这样理解不

追问

X=1 Y Z 为0 x^2+y^2+z^2怎么会等于2.。。。

Answer 3

(x^3+y^3+z^3)*(x+y+z)=x^4+y^4+z^4+xy*(x^2+y^2)+xz(x^2+z^2)+yz(y^2+z^2)
=x^4+y^4+z^4+xy(1-z^2)+xz(1-y^2)+yz(1-x^2)
=x^4+y^4+z^4+xy+xz+yz-xyz(x+y+z)
=3

x^4+y^4+z^4=3-(xy+xz+yz)+xyz

x²+y²+z² =2=2(x+y+z)
x²-x+y²-y+z²-z=x+y+z=1
x(x-1)+y(y-1)+z(z-1)=1
x(y+z)+y(x+z)+z(x+y)=-1
2(xy+xz+yz)=-1

xy+xz+yz=-1/2 ①
x³+y³+z³=3=3(x+y+z)
x³-x+y³-y+z³-z=2(x+y+z)=2

x²(1-x)+y²(1-y)+z²(1-z ) =x²(y+z)+y²(x+z)+z²(x+y) =xy(x+y)+xz(x+z)+yz(y+z) =xy(1-z)+xz(1-y)+yz(1-x) =xy+xz+yz-3xyz =-1/2-3xyz=2
∴xyz=-5/6 ②
x^4+y^4+z^4=3-2(-1/2)+(-5/6)
=19/4.

x^4+y^4+z^4=3-(-1/2)-5/6=8/3

Answer 4

x+y+z=1
(x+y+z)^2 =1
x^2+y^2+z^2 + 2(xy+yz+yx) = 1
2+2(xy+yz+yx) = 1
xy+yz+yx = -1/2 (1)
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3( xy(x+y) + xz(x+z) + yz(y+z) )
=3xyz ( (x+y)/z + (x+z)/y + (y+z)/x )
=3xyz ( (1-z)/z + (1-y)/y + (1-x)/x )
= 3xyz ( 1/x +1/y +1/z - 3)
=3xyz [ (xy+yz+zx)/(xyz) - 3]
=3(xy+yz+zx) - 9xyz
=-3/2 -9xyz
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(x+y+z)^3 = 1
x^3+y^3+z^3 + 3[ xy(x+y) + yz(y+z) + zx( x+z) ] + 6xyz = 1
3 - 3/2 -9xyz + 6xyz =1
3xyz = 1/2
xyz = 1/6 (2)
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(x^2+y^2+z^2)^2 =x^4+y^4+z^4 + 2( (xy)^2 + (yz)^2 + (zx))^2
x^4+y^4+z^4 + 2[ (xy)^2 + (yz)^2 + (zx) ]^2 = 4
x^4+y^4+z^4 + 2[ (xy+ yz+zx)^2 - 2xyz(x+y+z) ] = 4
x^4+y^4+z^4 + 2[ 1/4 - 2(1/6)(1) ] =4
x^4+y^4+z^4 - 1/6 =4
x^4+y^4+z^4 = 25/4