1/[﹙x﹣1﹚﹙x﹣2﹚]+1/[﹙x﹣2﹚﹙x﹣3﹚]=2/﹙x﹣1﹚
问题描述:
1/[﹙x﹣1﹚﹙x﹣2﹚]+1/[﹙x﹣2﹚﹙x﹣3﹚]=2/﹙x﹣1﹚
答
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