若a/x^2-yz=b/y^2-zx=c/z^2-xy,求证ax+by+cz=(a+b+c)(x+y+z)
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若a/x^2-yz=b/y^2-zx=c/z^2-xy,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
则a=k(x^2-yz)
b=k(y^2-zx)
c=k(z^2-xy)
带入待证式子:
k(x^2-yz)x+k(y^2-zx)y+k(z^2-xy)z
=k(x^3+y^3+z^3-3xyz)
右=k(x^2+y^2+z^2-xy-yz-xz)*(x+y+z)
即证+x^3+y^3+z^3-3xyz=(x^2+y^2+z^2-xy-yz-xz)*(x+y+z)
事实上:
a^3+b^3+c^3-3abc
=(a^3+3a^2b+3ab^2+b^3+c^3)-(3abc+3a^2b+3ab^2)
=[(a+b)^3+c^3]-3ab(a+b+c)
=(a+b+c)(a^2+b^2+2ab-ac-bc+c^2)-3ab(a+b+c)
=(a+b+c)(a^2+b^2+c^2+2ab-3ab-ac-bc)
=(a+b+c)(a^2+b^2+c^2-ab-bc-ac)
故即证
以上
则a=k(x^2-yz)
b=k(y^2-zx)
c=k(z^2-xy)
带入待证式子:
k(x^2-yz)x+k(y^2-zx)y+k(z^2-xy)z
=k(x^3+y^3+z^3-3xyz)
右=k(x^2+y^2+z^2-xy-yz-xz)*(x+y+z)
即证+x^3+y^3+z^3-3xyz=(x^2+y^2+z^2-xy-yz-xz)*(x+y+z)
事实上:
a^3+b^3+c^3-3abc
=(a^3+3a^2b+3ab^2+b^3+c^3)-(3abc+3a^2b+3ab^2)
=[(a+b)^3+c^3]-3ab(a+b+c)
=(a+b+c)(a^2+b^2+2ab-ac-bc+c^2)-3ab(a+b+c)
=(a+b+c)(a^2+b^2+c^2+2ab-3ab-ac-bc)
=(a+b+c)(a^2+b^2+c^2-ab-bc-ac)
故即证
以上
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