解分式方程: (2⼀X-3)—(2⼀X-4)=(2⼀X-5)—(2⼀X-6)怎么做?(具体过程)

2024-12-30 09:16:08
推荐回答(2个)
回答1:

先化简下:-2/(x-3)(x-4)=-2/(x-5)(x-6)
两边同乘 (x-3)(x-4)(x-5)(x-6)
-2(x-5)(x-6)=-2(x-3)(x-4)
-2(x^2-11x+30)=-2(x^2-7x+12)
-2x^2+22x-60=-2x^2+14x-24
x=4.5
检验:把x=4.5代入得
(x-3)(x-4)(x-5)(x-6)≠0
∴x=4.5是原方程的根

回答2:

2/(X-3) -2/(X-4) = 2/(X-5) - 2/(X-6)

首先: (X-3) ≠0 , (X-4)≠0 , (X-5)≠0 ,(X-6) ≠0

1/(X-3) -1/(X-4) = 1/(X-5) - 1/(X-6)
{ (x-4)-(x-3) } / { (X-3)(X-4) } = { (x-6)-(x-5) } / { (X-5)(X-6) }
-1 / { (X-3)(X-4) } = -1 / { (X-5)(X-6) }
(X-3)(X-4) = (X-5)(X-6)
x^2-7x+12=x^2-11x+30
4x=18
x=9/2