解分式方程：1/X-2+1/X-6=1/X-7+1/X-11/X-2+1/X-6=1/X-7+1/X-11/(x-2)-1/(x-1)=1/(x-7)-1/(x-6)(x-1-x+2)/(x-2)(x-1)=(x-6-x+7)/(x-7)(x-6)1/(x-2)(x-1)=1/(x-7)(x-6)(x-2)(x-1)=(x-7)(x-6)x^2-3x+2=x^2-13x+4213x-3x=42-210x=40x=4 为什么要移向才能解出来
问题描述：
解分式方程：1/X-2+1/X-6=1/X-7+1/X-1
1/X-2+1/X-6=1/X-7+1/X-1
1/(x-2)-1/(x-1)=1/(x-7)-1/(x-6)
(x-1-x+2)/(x-2)(x-1)=(x-6-x+7)/(x-7)(x-6)
1/(x-2)(x-1)=1/(x-7)(x-6)
(x-2)(x-1)=(x-7)(x-6)
x^2-3x+2=x^2-13x+42
13x-3x=42-2
10x=40
x=4 为什么要移向才能解出来
答
4
答
1/X-2+1/X-6=1/X-7+1/X
x=4
答
x=5
答
1/X-2+1/X-6=1/X-7+1/X-1
1/(x-2)-1/(x-1)=1/(x-7)-1/(x-6)
(x-1-x+2)/(x-2)(x-1)=(x-6-x+7)/(x-7)(x-6)
1/(x-2)(x-1)=1/(x-7)(x-6)
(x-2)(x-1)=(x-7)(x-6)
x^2-3x+2=x^2-13x+42
13x-3x=42-2
10x=40
x=4
答
先看一个式子.1/n - 1/（n+1）=1 / n（n+1）.这个等式容易理解,通分就行了.在原式中,x-2比x-1小1,x-7比x-6小1.所以可以把原等式进行移项,目的是通分后使分子为1,从而达到使分子降次的目的,最终把原方程转化为一元一次方程.这样方程就解出来了.