x-4/5-(x-1)=x-3/3-x-2/2一元一次方程

问题描述:

x-4/5-(x-1)=x-3/3-x-2/2一元一次方程

x/(x-1)-(x-1)/(x-2)=[x(x-2)-(x-1)^2]/(x-1)(x-2)=(x^2-2x-x^2+2x-1)/(x-1)(x-2)=-1/(x-1)(x-2)(x-3)/(x-4)-(x-4)/(x-5)=(x^2-8x+15-x^2+8x-16)/(x-4)(x-5)=-1/(x-4)(x-5)原式就可以化简成:(x-4)(x-5)=(x-1)(x-2)...麻烦给讲解一下因为我算到最后连X都算没有了