八年级分式的运算如果(x^2-yz)/x(1-yz)=(y^2-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.证明：x+y+z=1/x+1/y+1/z

Question

八年级分式的运算
如果(x^2-yz)/x(1-yz)=(y^2-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.
证明：x+y+z=1/x+1/y+1/z

Answer 1

(x^2-yz)/x(1-yz)=(y^2-xz)/y(1-xz)
y(x^2-yz)(1-xz)=x(y^2-xz)(1-yz)
y(x^2-x^3z-yz+xyz^2)=x(y^2-y^3z-xz+xyz^2)
xy(x-y)-xyz(x+y)(x-y)-z(x+y)(y-x)+xyz^2(y-x)=0
两边同除以(x-y)
xy-xyz(x+y)+z(x+y)-xyz^2=0
两边同除以xyz
1/z-x-y+1/y+1/x-z=0
x+y+z=1/x+1/y+1/z
得出结论