Question

数学题两道求解

1.化简
a³-a²-a+1/a³-3a²+3a-1

2。计算
（1-（a- 1/1-a）²÷a³+1/a²-2a+1）x1/1-a
那是a减去1/1-a的差的平方

麻烦过程

Answer 1

1. 原式=[(a³+1)-(a²+a)]/[(a²-1)-（3a²-3a）]
=[（a+1)(a²-a+1)-a(a+1)]/[(a-1)(a²+a+1)-3a(a-1)]
=(a+1)(a²-2a+1)/(a-1)(a²-2a+1)
=(a+1)/(a-1)
第二个好像表述的不是很清楚

Answer 2

1\ a+1

Answer 3

1.原式=a3-a2-a+1/（a-1）3
=（a2-1）(a-1)/(a-1)3
=a2-1/(a-1)2
=a2-2a+1+2a-2/(a-1)2
=（a-1）（a-1）+2(a-1)/(a-1)2
=(a+1)(a-1)/(a-1)(a-1)
=a+1/a-1