初二数学【分式方程】，追+哈！@@

x/(x-1)-(x-1)/(x-2)=(x-3)/(x-4)-(x-4)/(x-5)
(x-1+1)/(x-1)-(x-2+1)/(x-2)=( x-4+1)/(x-4) -( x-5+1)/(x-5)
1+1/(x-1) -1-1/(x-2)=1+1/(x-4)-1-1/(x-5)
1/(x-1) -1/(x-2)=1/(x-4)-1/(x-5)
-1/[(x-1)(x-2)]=-1/[(x-4)*(x-5)]
(x-1)(x-2)=(x-4)(x-5)
x^2-3x+2=x^2-9x+20
6x=18
得到x=3

方程化为：
[1+1/(x-1)]-[1+1/(x-2)]=[1+1/(x-4)]-[1+1/(x-5)]
消去1后，移项得
1/(x-1)-1/(x-4)=1/(x-2)-1/(x-5)
通分得
3/(x-1)(x-4)=3/(x-2)(x-5)
(x-1)(x-4)=(x-2)(x-5)
x^2-5x+4=x^2-7x+10
2x=6,x=3

左边分子等于x(x-2)-(x-1)^2 = x^2-2x -x^2+2x-1 = -1
分母为 (x-1)(x-2)
右边分子等于(x-3)(x-5)-(x-4)^2 = x^2-8x+15-x^2+8x-16 = -1
分母为 (x-4)(x-5)
所以只要解 (x-1)(x-2) = (x-4)(-5)->x^2-3x+2 = x^2-9x+20->6x = 18->x = 3

x/（x-1）-（x-1）/（x-2）=（x-3）/（x-4）-（x-4）/（x-5）.....[x(x-2)-(x-1)（x-1）]/(x-1)(x-2)=[(x-3)(x-5)-(x-4)(x-4)]/(x-4)(x-5).....1/(x-1)(x-2)=1/(x-4)(x-5)....x=3