Question

计算：(2a^2+3a+2)/(a+1)-(a^2-a-5)/(a+2)-(3a^2-4a-5)/(a-2)+(2a^2-8a+5)/(a-3)

2011届 七年级第一学期数学一课一练上第100页的第四题。
要过程！！！！！

Answer 1

（2a^2+3a+2)/(a+1)-(a^2-a-5)/(a+2)-(3a^2-4a-5)/(a-2)+(2a^2-8a+5)/(a-3)
=(2a+1)+1/(a+1)-(a-3)-1/(a+2)-(3a+2)+1/(a-2)+(2a-2)-1/(a-3)
=1/(a+1)-1/(a+2)+1/(a-2)-1/(a-3)=1/[(a+1)(a+2)]-1/[(a-2)(a-3)]
=[(a^2-5a+6)-(a^2+3a+2)]/[(a+1)(a+2)(a-2)(a-3)]
=(4-8a)/[(a+1)(a+2)(a-2)(a-3)]
=4(1-2a)/[(a+1)(a+2)(a-2)(a-3)].