Question

分式计算题: (x+2/x+1) - (x+3/x-2) - (x-4/x-3) + (x-5/x-4) 写一下详细步骤，谢谢（括号内是一个分式）

Answer 1

(x+2/x+1) - (x+3/x-2) - (x-4/x-3) + (x-5/x-4)
=x+2/x+1 - x-3/x+2 - x+4/x+3 + x-5/x-4
=(x-x-x+x)+(2/x-3/x+4/x-5/x)+(1+2+3-4)
=-2/x + 2

Answer 2

通分算。

Answer 3

(x+2)/(x+1) - (x+3)/(x-2) - (x-4)/(x-3) + (x-5)/(x-4)
=[(x+1)+1]/(x+1)-[(x-2)+5]/(x-2)-[(x-3)-1]/(x-3)+[(x-4)-1]/(x-4)
=[1+1/(x+1)]-[1+5/(x-2)]-[1-1/(x-3)]+[1-1/(x-4)]
=1+1/(x+1)-1-5/(x-2)-1+1/(x-3)+1-1/(x-4)
=1/(x+1)-5/(x-2)+1/(x-3)-1/(x-4)
= -(4x+7)/(x+1)(x-2)-1/(x-3)(x-4)
= -[(4x+7)(x-3)(x-4)-(x+1)(x-2)]/(x+1)(x-2)(x-3)(x-4)
= -(4x^3-22x^2+86)/(x+1)(x-2)(x-3)(x-4)
答案如上，欢迎采纳！