问题
填空题
计算
|
答案
原式=
+4a (a+1)(a-1)
+-(a+1)2 (a+1)(a-1) (a-1)2 (a+1)(a-1)
=4a-(a+1)2+(a-1)2 (a+1)(a-1)
=0 (a+1)(a-1)
=0．
故答案为0．
搜题请搜索小程序“爱下问搜题”
计算
|
原式=
+4a (a+1)(a-1)
+-(a+1)2 (a+1)(a-1) (a-1)2 (a+1)(a-1)
=4a-(a+1)2+(a-1)2 (a+1)(a-1)
=0 (a+1)(a-1)
=0．
故答案为0．