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

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