Question

请问这道数学题怎么做：(x+2/x+1)-(x+3/x+2)-(x-4/x-3)+(x-5/x-4)

Answer 1

我给你一种好的解法：

解：分子都比分母大1，所以先整理为：
原式=(x+2)/(x+1)-(x+3)/(x+2)-(x-4)/(x-3)+(x-5)/(x-4)
=[1+1/(x+1)]-[1+1/(x+2)]-[1-1/(x-3)]+[1-1/(x-4)]
=1+1/(x+1)-1-1/(x+2)-1+1/(x-3)+1-1/(x-4)
=[1/(x+1)+1/(x-3)]-[1/(x+2)+1/(x-4)]
=[(x-3)+(x+1)]/[(x+1)(x-3)]-[(x-4)+(x+2)]/[(x+2)(x-4)]
=(2x-2)/(x^2-2x-3)-(2x-2)/(x^2-2x-8)
=(2x-2)*[1/(x^2-2x-3)-1/(x^2-2x-8)]
=(2x-2)*[(x^2-2x-8)-(x^2-2x-3)]/[(x^2-2x-3)(x^2-2x-8)]
=(2x-2)*(-5)/[(x^2-2x-3)(x^2-2x-8)]
=(10-10x)/[(x^2-2x-3)(x^2-2x-8)]

注：为了让你看得更清楚，我一步一步写下来，但在实际书写时，有些步骤可以省略。

Answer 2

将x+2/x+1拆成1+1/（x+1）
后面的也这样拆，再通分计算。

Answer 3

(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=14/x-2