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的值
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的值
写一下过程，谢谢！！！
细心你一定可以算出25/6
x^3+y^3+z^3+2x^2(y+z)+2z^2(x+y)+2y^2(x+z)+3xyz=1
整理得xyz=0
不妨设x=0
又xy+yz+xz=0.5[(x+y+z)^2-(x^2+y^2+z^2)]=-0.5
则y+z=1 y^3+z^3=1
y^4+z^4=(y+z)(y^3+z^3)-yz(y^2+z^2)=4
故x^4+y^4+z^4=4
参考资料：sweetandbitter - 魔导师 十级
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的值
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已知,实数x、y、z满足:x+y+z=1、x²+y²+z²=2、x³+y³+z...
x4+y4+z4=4
已知,实数x、y、z满足:x+y+z=1、x²+y²+z²=2、x³+y³+z...
答案：25\/6 【不好意思，看到题目时太晚了】因为x+y+z=1 所以 (x+y+z)²=1 x²+y²+z²+2xy+2xz+2yz=1 因为x²+y²+z²=2 所以 xy+xz+yz=-1\/2 所以 (xy+xz+yz)²=1\/4 x²y²+x²z²+y²z&...
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设u=x+y+z=1,v=xy+yz+zx,w=xyz,则 x^2+y^2+z^2=u^2-2v=2,v=-1\/2,x^3+y^3+z^3-3xuz=u(u^2-3v)=5\/2=3-3w,w=1\/6,由恒等式x^4+y^4+z^4=u^4-4u^2*v+2v^2+4uw得 x^4+y^4+z^4=1+2+1\/2+2\/3=25\/6.
已知x+y+z=1 x2+y2+z2=2 x3+y3+z3=3 求x4+y4+z4=?
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已知X+Y+Z=1 X^2+Y^2+Z^2=2 X^3+Y^3+Z^3=3 求1\/xyz
X+Y+Z=1 x^2+y^2+z^2+2xy+2xz+2yz=1 因为x+y^2+z^2=2 xy+xz+yz=-1\/2 X^3+Y^3+Z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-xz)=1*(2+1\/2)所以 3-3xyz=5\/2 xyz=1\/6 1\/xyz=6
