求解一道初二分式方程

由（x-5)/(x-6)+(x-8)/(x-9)=(x-7)/(x-8）+(x-6)/(x-7)
变形：（x-5)/(x-6)-(x-6)/(x-7)=(x-7)/(x-8）-(x-8)/(x-9)
等式两边通分算一下，很巧，结果是：
1/(x-6)(x-7)=1/(x-8)(x-9)
就有：
(x-6)(x-7)=(x-8)(x-9)
两边乘开：(x平方没了)
42-13x=72-17x
好算了吧！
x=15/2

（x-5)/(x-6)+(x-8)/(x-9)=(x-7)/(x-8）+(x-6)/(x-7)
so 1+1/(x-6)+1+1/(x-9)=1+1/(x-8）+1+1/(x-7)
so 1/(x-6)+1/(x-9)=1/(x-8）+1/(x-7)
so 1/(x-6)-1/(x-7)=1/(x-8）-1/(x-9)
so -1/[(x-6)(x-7)]=-1/[(x-8)(x-9)]
so [(x-6)(x-7)]=[(x-8)(x-9)]
so x^2-13x+42=x^2-17x+72
so 4x=30
so x=7.5

（x-5)/(x-6)+(x-8)/(x-9)=(x-7)/(x-8）+(x-6)/(x-7)
（x-5)/(x-6)-(x-6)/(x-7) =(x-7)/(x-8)-(x-8)/(x-9)
两边同分后展开
-1/(x-6)(x-7)=-1/(x-8)(x-9)
(x-6)(x-7)=(x-8)(x-9)
x^2-13x+42=x^2-17x+72
x=7.5

解：移项

（x-5)/(x-6)-（x-7)/(x-8）=(x-6)/(x-7) )-(x-8)/(x-9）
同分

1/（x-6）(x-8)=1/(x-7)(x-9)

所以（x-6）(x-8)=(x-7)(x-9)

x^2-14x+48=x^2-16x+63