已知 (x²-yz)/x(1-yz)=(y²-xz)/y(1-xz)且x不等于y,x不等于0,y不等
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已知 (x²-yz)/x(1-yz)=(y²-xz)/y(1-xz)且x不等于y,x不等于0,y不等于0,z不等于0,求证x+y+z=1/x+1/y+1/z
等式两边同乘以x(1-yz)·y(1-xz)
得:(x²-yz)·y(1-xz)=(y²-xz)·x(1-yz)
→x²y-x³yz-y²z+xy²z²=xy²-xy³z-x²z+x²yz²
移项后:x²y-xy²+x²z-y²z=x²yz²-xy³z+x³yz-xy²z²
→xy(x-y)+z(x²-y²)=xyz(xz-y²+x²-yz)
→(x-y)[xy+z(x+y)]=xyz[z(x-y)+(x+y)(x-y)]
→(x-y)(xy+xz+yz)=xyz(x-y)(x+y+z)
→(xy+xz+yz)=xyz(x+y+z)
两边同除以xyz得:1/x+1/y+1/z=x+y+z
得:(x²-yz)·y(1-xz)=(y²-xz)·x(1-yz)
→x²y-x³yz-y²z+xy²z²=xy²-xy³z-x²z+x²yz²
移项后:x²y-xy²+x²z-y²z=x²yz²-xy³z+x³yz-xy²z²
→xy(x-y)+z(x²-y²)=xyz(xz-y²+x²-yz)
→(x-y)[xy+z(x+y)]=xyz[z(x-y)+(x+y)(x-y)]
→(x-y)(xy+xz+yz)=xyz(x-y)(x+y+z)
→(xy+xz+yz)=xyz(x+y+z)
两边同除以xyz得:1/x+1/y+1/z=x+y+z
已知 (x²-yz)/x(1-yz)=(y²-xz)/y(1-xz)且x不等于y,x不等于0,y不等
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