1／[﹙x﹣1﹚﹙x﹣2﹚]＋1／[﹙x﹣2﹚﹙x﹣3﹚]＝2／﹙x﹣1﹚

1／[﹙x﹣1﹚﹙x﹣2﹚]＋1／[﹙x﹣2﹚﹙x﹣3﹚]＝2／﹙x﹣1﹚
1／[﹙x﹣2﹚﹙x﹣3﹚]＝2／﹙x﹣1﹚-1／[﹙x﹣1﹚﹙x﹣2﹚]
1／[﹙x﹣2﹚﹙x﹣3﹚]＝(2x-5)／[﹙x﹣1﹚﹙x﹣2﹚]
1/(x-3)=(2x-5)/(x-1)
(2x-5)(x-3)=x-1
2x^2-11x+15=x-1
2x^2-12x+16=0
x^2-6x+8=0
(x-2)(x-4)=0
x=2(舍去)或x=4