Question

一道简单初一数学题

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

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)].

Answer 2

这个简单吗

Answer 3

2a^2+3a+2/a+1=[(2a²+3a+1)+1]/(a+1)=[(a+1)(2a+1)+1]/(a+1)=2a+1+[1/(a+1)]

a^2-a-5/a+2=[(a²-a-6)+1]/(a+2)=[(a+2)(a-3)+1]/(a+2)=a-3+[1/(a+2)]

3a^2-4a-5/a-2=[(3a²-4a-4)-1]/(a-2)=[(3a+2)(a-2)-1]/(a-2)=3a+2-[1/(a-2)]

2a^2-8a+5/a-3=[(2a²-8a+6)-1]/(a-3)=[(2a-2)(a-3)-1]/(a-3)=2a-2-[1/(a-3)]

所以原式
=2a+1+[1/(a+1)]- 【a-3+[1/(a+2)]】-【3a+2-[1/(a-2)]】+ 2a-2-[1/(a-3)]

=(2a+1-a+3-3a-2+2a-2)+[1/(a+1)-1/(a+2)]+[1/(a-3)-1/(a-2)]

=[2a]+1/(a+1)(a+2)+1/(a-2)(a-3)

=[2a]+[(a-2)(a-3)+(a+1)(a+2)]/(a+1)(a+2)(a-2)(a-3)

=[2a]+(a²-5a+6+a²+3a+2) /(a+1)(a+2)(a-2)(a-3)

=[2a]+(2a²-2a+8)/ (a+1)(a+2)(a-2)(a-3)

=[2a]+2(a²-a+4)/ (a+1)(a+2)(a-2)(a-3)