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)
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)
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