Question

解方程(x-1/x-2)+(x-7/x-8)=(x-3/x-4)+(x-5/x-6)

谢谢，要过程

Answer 1

(x-1)/(x-2)
-
(x-2)/(x-3)
=
(x-4)/(x-5)
-
(x-5)/(x-6)
(x-2+1)/(x-2)
-
(x-3+1)/(x-3)
=
(x-5+1)/(x-5)
-
(x-5+1)/(x-6)
[(x-2)/(x-2)+1/(x-2)]-[(x-3)/(x-3)+1/(x-3)]=](x-5)/(x-5)+1/(x-5)]-[(x-6)/(x-6)+1/(x-6)]
1+1/(x-2)-1-1/(x-3)=1+1/(x-5)-1-1/(x-6)
1/(x-2)-1/(x-3)=1+1/(x-5)-1/(x-6)
1/(x-2)+1/(x-6)=1/(x-3)+1/(x-5)
(x-6+x-2)/(x-2)(x-6)-(x-3+x-5)/(x-3)(x-5)=0
(2x-8)/(x^2-8x+12)-(2x-8)/(x^2-8x+15)=0
(2x-8)[1/(x^2-8x+12)-1/(x^2-8x+15)]=0
因为x^2-8x+12不等于x^2-8x+15
所以1/(x^2-8x+12)-1/(x^2-8x+15)不等于0
所以2x-8=0
x=4
分式方程要检验
经检验
x=4是方程的解

Answer 2

(x-1)/(x-2)+(x-7)/(x-8)=(x-3)/(x-4)+(x-5)/(x-6)
(x-2+1)/(x-2)+(x-8+1)/(x-8)=(x-4+1)/(x-4)+(x-6+1)/(x-6)
（此步对分子加1减1，是简便算法）
1+1/(x-2)+1+1/(x-8)=1+1/(x-4)+1+1/(x-6)
1/(x-2)+1/(x-8)=1/(x-4)+1/(x-6)
1/(x-8)-1/(x-4)=1/(x-6)-1/(x-2)
（此步移项成左右相减，是再次简便算法）
(x-4-x+8)/(x-8)(x-4)=(x-2-x+6)/(x-6)(x-2)
4/(x-8)(x-4)=4/(x-6)(x-2)
(x-8)(x-4)=(x-6)(x-2)
x^2-12x+32=x^2-8x+12
4x=20
x=5
验算：
1/3-1/3=1-1（验算也正确）