Question

a(a+1)(a+2)(a+3)+1因式分解

用完全平方公式哦
要具体过程

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

a.(a+1).(a+2).(a+3)=11880
求a等于多少