解方程:(x-1)(x-2) - (x-2)⼀(x-3) = (x-4)⼀(x-5) - (x-5)⼀(x-6)

2024-10-30 04:41:11
推荐回答(1个)
回答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