a/x^2-yz=b/y^2-zx=c/z^2-xy xyz=0 求证ax+by+cz=(a+b+c)(x+y+z)

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a/x^2-yz=b/y^2-zx=c/z^2-xy xyz=0 求证ax+by+cz=(a+b+c)(x+y+z)

a/x^2-yz=b/y^2-zx=c/z^2-xy xyz=0 求证ax+by+cz=(a+b+c)(x+y+z)
a/x^2-yz=b/y^2-zx=c/z^2-xy xyz=0 求证ax+by+cz=(a+b+c)(x+y+z)

a/x^2-yz=b/y^2-zx=c/z^2-xy xyz=0 求证ax+by+cz=(a+b+c)(x+y+z)
设a/x^2-yz=b/y^2-zx=c/z^2-xy =k
因为xyz=0
所以x,y,z中至少有一个=0
不妨设x=0
则a/x^2-yz=b/y^2-zx=c/z^2-xy
a/-yz=b/y^2=c/z^2=k
a=-kyz
b=ky^2
c=kz^2
ax+by+cz=by+cz=ky^3+kz^3=k(y^3+z^3)
(a+b+c)(x+y+z)
=k(y^2+z^2-yz)(y+z)
=k(y^3+z^3)
所以ax+by+cz=(a+b+c)(x+y+z)

因为XYZ=0所以知道X Y Z 其中至少有一项为0
可X Y Z 又都可以做分母。。。题错了吧?

设k=a/(x^2-yz)=b/(y^2-xz)=c/(z^2-xy)
则a=k(x^2-yz)
b=k(y^2-xz)
c=k(z^2-xy)
(a+b+c)(x+y+z)
=k(x^2+y^2+z^2-xy-xz-yz)(x+y+z)
=k(x^3+y^3+z^3-3xyz)
ax+by+cz
=k(x^3-xyz+y^3-xyz+z^3-xyz)
=k(x^3+y^3+z^3-3xyz)
得证

xyz=0 x y z中必有一个是0
又原题xyz作除数 不为0 是不是矛盾?

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