✔ 最佳答案
1. [(x+2)/(x^2-x-6)] - [(x-4)/(x^2-x-12)] + [(x+3)/(x^2+2x-3)]
=[(x+2)/(x-3)(x+2)] - [(x-4)/(x-4)(x+3)] + [(x+3)/(x+3)(x-1)]
=[1/(x-3)] - [1/(x+3)] + [1/(x-1)]
=[(x+3-x+3)/x^2-9] + [1/(x-1)]
=[6/x^2-9] + [1/(x-1)]
=(x^2+6x-15)/(x^3-x^2-9x+9)
2007-04-06 12:02:21 補充:
2. [1/(w-x)(x-y)] - [2/(x-y)(y-w)] + [3/(w-x)(w-y)] =[1/(w-x)(x-y)] - [2/(x-y)(y-w)] - [3/(w-x)(y-w)] ={[(y-w)-2(w-x)-3(x-y)]/(w-x)(x-y)(y-w)} =(y-w-2w+2x-3x+3y)/(w-x)(x-y)(y-w) =(4y-x-3w)/(w-x)(x-y)(y-w)
2007-04-06 12:08:50 補充:
3. (x-3/x+3) - (2/x-3) + (3/x^2-9) =(x-3)^2/(x^2-9) - 2(x+3)/(x^2-9) + (3/x^2-9) =[(x-3)^2-2(x+3)+3]/(x^2-9)
2007-04-06 12:10:42 補充:
3. (x-3/x+3) - (2/x-3) + (3/x^2-9)=(x-3)^2/(x^2-9) - 2(x+3)/(x^2-9) + (3/x^2-9)=[(x-3)^2-2(x+3)+3]/(x^2-9) =(x^2-8x+6)/(x^2-9)