急求(x^2-3x-1)/(x-1)-(x^2+3x+3)/(x+3)+(2x+12)/(x+3) 如何用裂项法来计算这道题。

首先我想问下式子后两项分母都是X+3为什么分开写？
解：1)x^2-3x-1=x^2-x-2x+2-3=x(x-1)-2(x-1)-3
2)-(x^2+3x+3)+2x+12=-x^2-x+9=-[x^2+3x-2x-6-3]=-[x(x+3)-2(x+3)-3]
前面2式分别除仪去分母，化简得x-2-3/x-1-x+2+3/x+3=3/x+3-3/x-1

(x^2-3x-1)/(x-1)-(x^2+3x+3)/(x+3)+(2x+12)/(x+3)
=(x^2-3x-1)/(x-1)-(x^2+x-9)/(x+3)
=[(x-1)(x+1)-3x]/(x-1)-[(x+3)(x-3)+x]/(x+3)
=(x+1)-3x/(x-1)-(x-3)-x/(x+3)
=4-[3x/(x-1)+x/(x+3)]
=4-[3((x-1)+1)/(x-1)+((x+3)-3)/(x+3)]
=4-[3+3/(x-1)+1-3/(x+3)]
=-3[1/(x-1)-1/(x+3)]
=-12/[(x-1)(x+3)]