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证明(x-y)/(1+xy)+(y-z)/(1+yz)+(z-x)/(1+zx)=(x-y)/(1+xy)*(y-z)/(1+yz)*(z-x)/(1+zx)
题目详情
证明(x-y)/(1+xy) +(y-z)/(1+yz )+(z-x)/(1+zx)=(x-y)/(1+xy)*(y-z)/(1+yz)*(z-x)/(1+zx)
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答案和解析
证明(x-y)/(1+xy) +(y-z)/(1+yz )+(z-x)/(1+zx)=(x-y)/(1+xy)*(y-z)/(1+yz)*(z-x)/(1+zx)
证明(x-y)/(1+xy) +(y-z)/(1+yz )+(z-x)/(1+zx)
=[(x-y)(1+yz)(1+zx)+(y-z)(1+xy)(1+zx)+(z-x)(1+xy)(1+yz)]/(1+xy)(1+yz)(1+zx)
=(x+x^2z+xyz+x^2yz^2-y-xyz-y^2z-xy^2z^2+y+xy^2+xyz+x^2y^2z-z-xyz-xz^2-x^2yz^2+z+xyz+yz^2+xy^2z^2-x-x^2y-xyz-x^2y^2z)/(1+xy)(1+yz)(1+zx)
=(x^2z-y^2z+xy^2-xz^2+yz^2-x^2y)/(1+xy)(1+yz)(1+zx)
=[z(x+y)(x-y)-xy(x-y)-z^2(x-y)]/(1+xy)(1+yz)(1+zx)
=(x-y)(zx+zy-xy-z^2)/(1+xy)(1+yz)(1+zx)
=(x-y)(z-x)(y-z)/(1+xy)(1+yz)(1+zx)
=(x-y)/(1+xy)*(y-z)/(1+yz)*(z-x)/(1+zx)
证明(x-y)/(1+xy) +(y-z)/(1+yz )+(z-x)/(1+zx)
=[(x-y)(1+yz)(1+zx)+(y-z)(1+xy)(1+zx)+(z-x)(1+xy)(1+yz)]/(1+xy)(1+yz)(1+zx)
=(x+x^2z+xyz+x^2yz^2-y-xyz-y^2z-xy^2z^2+y+xy^2+xyz+x^2y^2z-z-xyz-xz^2-x^2yz^2+z+xyz+yz^2+xy^2z^2-x-x^2y-xyz-x^2y^2z)/(1+xy)(1+yz)(1+zx)
=(x^2z-y^2z+xy^2-xz^2+yz^2-x^2y)/(1+xy)(1+yz)(1+zx)
=[z(x+y)(x-y)-xy(x-y)-z^2(x-y)]/(1+xy)(1+yz)(1+zx)
=(x-y)(zx+zy-xy-z^2)/(1+xy)(1+yz)(1+zx)
=(x-y)(z-x)(y-z)/(1+xy)(1+yz)(1+zx)
=(x-y)/(1+xy)*(y-z)/(1+yz)*(z-x)/(1+zx)
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